1\documentclass{article}
   2\usepackage[utf8]{inputenc}
   3\usepackage{amsfonts}
   4% \usepackage{yfonts} % WIP
   5\usepackage{amsthm}
   6\usepackage{amsmath}
   7\usepackage{enumerate}
   8\usepackage{hyperref}
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  10\usepackage{tikz} % diagram
  11\usepackage{changepage} % for adjustwidth
  12
  13\begin{filecontents}[overwrite]{commutative-algebra-notes.bib}
  14@misc{am,
  15  author = {M. F. Atiyah and I. G. MacDonald},
  16  title = {{Introduction to Commutative Algebra}},
  17  year = {1969}
  18}
  19@misc{reid,
  20  author = {Miles Reid},
  21  title = {{Undergraduate Commutative Algebra}},
  22  year = {1995}
  23}
  24@misc{mit-course,
  25  author = {Steven Kleiman},
  26  title = {{Commutative Algebra - MIT OpenCourseWare}},
  27  year = {2008},
  28  note = {\url{https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/}},
  29  url = {https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/}
  30}
  31\end{filecontents}
  32\nocite{*}
  33
  34
  35\theoremstyle{definition}
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  77\newcommand{\aA}{\mathfrak{a}} % TODO: use goth font
  78\newcommand{\mM}{\mathfrak{m}}
  79
  80\title{Commutative Algebra notes}
  81\author{arnaucube}
  82\date{2026}
  83
  84\begin{document}
  85
  86\maketitle
  87
  88\begin{abstract}
  89    Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}. For the exercises, I follow the assignments listed at \cite{mit-course}.
  90
  91    Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
  92
  93    The proofs may slightly differ from the ones from the books, since I try to extend them for a deeper understanding.
  94\end{abstract}
  95
  96\tableofcontents
  97
  98\section{Ideals}
  99
 100\subsection{Definitions}
 101
 102\begin{defn}{}[ideal]
 103  $I \subset R$ ($R$ ring) such that $0 \in I$ and $\forall x \in I,~ r \in R,~ xr, rx \in I$.\\
 104  \hspace*{2em} ie. $I$ absorbs products in $R$.
 105\end{defn}
 106
 107\begin{defn}{}[prime ideal]
 108  if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a \in P$ or $b \in P$.
 109\end{defn}
 110
 111\begin{defn}{}[principal ideal]
 112  generated by a single element, $(a)$.
 113
 114  $(a)$: principal ideal, the set of all multiples $xa$ with $x \in R$.
 115\end{defn}
 116
 117\begin{defn}{}[maximal ideal]
 118  $\mM \subset A$ ($A$ ring) with $m \neq A$ and there is no ideal $I$ strictly between $\mM$ and $A$. ie. if $\mM$ maximal and $\mM \subseteq I \subseteq A$, either $\mM=I$ or $I=A$.
 119\end{defn}
 120
 121
 122\begin{defn}{}[unit]
 123  $x \in A$ such that $xy=1$ for some $y \in A$. ie. element \emph{which divides 1}.
 124\end{defn}
 125
 126\begin{cor}{1.8} \label{1.8}
 127  $A = A^{\times} \sqcup \bigcup m$ (where $\sqcup$ denotes ``disjoint union"), ie. $f \in A$ is either a unit or it is contained in a maximal ideal, not both.
 128\end{cor}
 129
 130\begin{defn}{}[zerodivisor]
 131  $x \in A$ such that $\exists~ 0 \neq y \in A$ such that $xy=0 \in A$. ie. $x$ \emph{divides 0}..
 132
 133
 134  If a ring does not have zerodivisors is an integral domain.
 135\end{defn}
 136
 137\begin{defn}{}[prime spectrum - $Spec(A)$]
 138  set of prime ideals of $A$. ie.
 139
 140  $$Spec(A) = \{ P ~|~ P \subset A~ \text{is a prime ideal} \}$$
 141\end{defn}
 142
 143\begin{defn}{}[integral domain]
 144  Ring in which the product of any two nonzero elements is nonzero.
 145
 146  ie. no zerodivisors.
 147
 148  ie. $\forall~ 0 \neq a,~ 0 \neq b \in A,~ ab \neq 0 \in A$.
 149
 150  Every field is an integral domain, not the converse.
 151\end{defn}
 152
 153\begin{defn}{}[principal ideal domain - PID]
 154  integral domain in which every ideal is principal. ie.
 155  ie. $\forall I \subset R,~ \exists~ a \in I$ such that $I = (a) = \{ ra ~|~ r \in R \}$.
 156\end{defn}
 157
 158\begin{defn}{}[nilpotent]
 159  $a \in A$ such that $a^n=0$ for some $n>0$.
 160\end{defn}
 161\begin{defn}{}[nilrad A]
 162  set of all nilpotent elements of $A$; is an ideal of $A$.
 163
 164  if $nilrad A = 0 ~\Longrightarrow$ $A$ has no nonzero nilpotents.
 165
 166
 167  $$nilrad A = \bigcap_{P \in Spec(A)} P$$
 168\end{defn}
 169
 170\begin{defn}{}[idempotent]
 171  $e \in A$ such that $e^2=e$.
 172\end{defn}
 173
 174\begin{defn}{}[radical of an ideal]
 175  $$rad I = \{ f \in A | f^n \in I~ \text{for some}~ n \}$$
 176
 177  $rad I$ is an ideal.
 178
 179  $nilrad A = rad 0$
 180
 181  $rad I = \bigcap_{\substack{P \in \operatorname{Spec}(A)\\ P \supset I}} P$
 182\end{defn}
 183
 184\begin{defn}{1.13}[local ring] \label{1.13}
 185  A ring is \emph{local} if it has a unique maximal ideal.
 186
 187  Notation: local ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$:
 188  $$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$
 189
 190  \vspace{0.3cm}
 191  By Corollary \ref{1.8}, $A$ is local\\
 192    \hspace*{4em}$\Longleftrightarrow$ $A$ has only one maximal ideal.\\
 193    \hspace*{4em}$\Longleftrightarrow$ all the nonunits of $A$ form an ideal.
 194\end{defn}
 195
 196\begin{prop}{(R/m=Field)}\label{maximal-quotient-field}
 197  Let $R$ be a commutative ring with identity, and let $m \subsetneq R$ be a maximal ideal. Then the quotient ring $R/m$ is a field.
 198\end{prop}
 199\begin{proof}
 200  Let $x + m$ be a nonzero element of $R/m$. Since
 201  $$x + m \neq m$$
 202  we have $x \not\in m$.
 203
 204  Consider the ideal $m + (x)$, where $(x)$ is the principal ideal generated by $x$.
 205  Since $x \not\in m$, this ideal strictly contains $m$:
 206  $$m \subsetneq m + (x)$$
 207
 208  Because $m$ is maximal, there are no proper ideals strictly between $m$ and $R$. Therefore,
 209  $m + (x) = R$
 210
 211  In particular, $1 \in m + (x)$. Hence, there exist
 212  $a \in m$ and $y \in R$ such that $a + yx = 1$.
 213
 214  Passing to the quotient modulo $m$, we obtain
 215  $$
 216    (y+m)(x+m)
 217    = yx+m
 218    = 1+m,
 219  $$
 220  because $a \in m$ implies $a+m=m$.
 221
 222  Thus, $y+m$ is the multiplicative inverse of
 223  $x+m$. Therefore, every nonzero element of
 224  $R/m$ is invertible, and hence $R/m$ is a field.
 225\end{proof}
 226
 227
 228
 229\subsection{Z and K[X], two Principal Ideal Domains}
 230
 231
 232\begin{lemma}{}
 233  $\mathbb{Z}$ is a PID.
 234\end{lemma}
 235\begin{proof}
 236  Let $I$ a nonzero ideal of $\mathbb{Z}$.
 237
 238  Since $I \neq \{0\}$, there is at least one nonzero integer in $I$. Choose the smallest element of $I$, namely $d$.
 239
 240  Observe that $(d) \subseteq I$, since $d \in I$. Then, every multiple $nd \in I$, since $I$ is an ideal.
 241
 242  Take $a \in I$. By the Euclidean division algorithm in $\mathbb{Z}$, $a=qd+r$, with $q,r \in \mathbb{Z}$ and $0 \leq r \leq d$.
 243
 244  Then $r = a - qd \in I$, but $d$ was chosen to be the smallest positive element of $I$, so the only possibility is $r=0$.
 245
 246  Hence, $a=qd$, so $a \in (d)$, giving $I \subseteq (d)$.
 247
 248  Since we had $(d) \subseteq I$ and now we got $I \subseteq (d)$, we have $I = (d)$, so every ideal of $\mathbb{Z}$ is principal. Thus $\mathbb{Z}$ is a Principal Ideal Domain(PID).
 249\end{proof}
 250
 251\begin{lemma}{}
 252  $K[X]$ is a PID.
 253\end{lemma}
 254\begin{proof}
 255  This proof follows very similarly to the previous proof.\\
 256
 257  Let $K$ be a field, $K[X]$ a polynomial ring.
 258
 259  Take $\{0\} \neq I \subseteq K[X]$.
 260
 261  Since $I \neq \{0\}$, there is at least one non-zero polynomial in $I$.
 262
 263  Let $p(X) \in I$ be of minimal degree among nonzero elements of $I$.
 264
 265  Observe that $(p(X)) \subseteq I$, because $p(X) \in I$ and $I$ is an ideal.
 266
 267  Let $f(X) \in I$. By Euclidean division algorithm in $K[X]$, $\exists q, r \in K[X]$ such that $f(X) = q(X) \cdot p(X) + r(X)$ with either $r(X)=0$ or $deg(r) < deg(p)$.
 268
 269  Since $f,p \in I$, then $r(X) = f(X) - q(X)\cdot p(X) \in I$
 270
 271  If $r(X) \neq 0$, then $deg(r) < deg(p)$, which contradicts the minimality of $deg(p)$ in $I$.
 272
 273  Therefore, $r(X)=0$, thus $f(X)=q(X)\cdot p(X)$, hence $f(X) \in (p(X))$.
 274  Henceforth, $I \subseteq (p(X))$.
 275
 276  Then, since $(p(X)) \subseteq I$ and $I \subseteq (p(X))$, we have that $I = (p(X))$.
 277
 278  So every ideal of $K[X]$ is principal; thus $K[X]$ is a PID.
 279
 280\end{proof}
 281
 282
 283
 284\subsection{Zorn's lemma and Jacobson radicals}
 285
 286Let $\Sigma$ be a partially orddered set. Given subset $S \subset \Sigma$, an \emph{upper bound} of $S$ is an element $u \in \Sigma$ such that $s<u \forall s \in S$.
 287
 288A \emph{maximal element} of $\Sigma$, is $m \in \Sigma$ such that $m<s$ does not hold for any $s \in \Sigma$.
 289
 290A subset $S \subset \Sigma$ is \emph{totally ordered} if for every pair $s_1,s_2 \in S$, either $s_1 \leq s_2$ or $s_2 \leq s_1$.
 291
 292\begin{lemma}{R.1.7}[Zorn's lemma] \label{zorn}
 293  Suppose $\Sigma$ a nonempty partially ordered set (ie. we are given a relation $x \leq y$ on $\Sigma$), and that any totally ordered subset $S \subset \Sigma$ has an upper bound in $\Sigma$.
 294
 295  Then $\Sigma$ has a maximal element.
 296\end{lemma}
 297
 298\begin{thm}{AM.1.3} \label{1.3}
 299  Every ring $A \neq 0$ has at least one maximal ideal.
 300\end{thm}
 301\begin{proof}
 302  By Zorn's lemma \ref{zorn}.
 303\end{proof}
 304
 305\begin{cor}{AM.1.4} \label{1.4}
 306  if $I \neq (1)$ an ideal of $A$, $\exists$ a maximal ideal of $A$ containing $I$.
 307\end{cor}
 308
 309\begin{cor}{AM.1.5} \label{1.5}
 310  Every non-unit of $A$ is contained in a maximal ideal.
 311\end{cor}
 312
 313\begin{defn}{}[Jacobson radical]
 314  The \emph{Jacobson radical} of a ring $A$ is the intersection of all the maximal ideals of $A$.
 315
 316  Denoted $Jac(A)$.
 317  
 318  $Jac(A)$ is an ideal of $A$.
 319\end{defn}
 320
 321\begin{prop}{AM.1.9} \label{1.9}
 322  $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$.
 323\end{prop}
 324\begin{proof}
 325  Suppose $1-xy$ not a unit.
 326
 327  By \ref{1.5}, $1-xy \in \mM$ for $\mM$ some maximal ideal.
 328
 329  But $x \in Jac(A) \subseteq \mM$, since $Jac(A)$ is the intersection of all maximal ideals of $A$.
 330
 331  Hence $xy \in \mM$, and therefore $1 \in \mM$, which is absurd, thus $1-xy$ is a unit.
 332
 333  Conversely:\\
 334  Suppose $x \not\in \mM$ for some maximal ideal $\mM$.
 335
 336  Then $\mM$ and $x$ generate the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$.
 337
 338  Hence $1 -xy \in \mM$, and is therefore not a unit.
 339\end{proof}
 340
 341\section{Modules}
 342
 343\subsection{Modules concepts}
 344
 345Let $A$ be a ring. An $A$-module is an Abelian group $M$ with a multiplication
 346map
 347\begin{align*}
 348  A \times M &\longrightarrow M\\
 349  (f, m) &\longmapsto fm
 350\end{align*}
 351satisfying $\forall~ f,g \in A,~~ m, n \in M$.
 352\begin{enumerate}[i.]
 353  \item $f(m \pm n)=fm \pm fn$
 354  \item $(f \pm g) m = fm \pm gm$
 355  \item $(fg) m = f(gm)$
 356  \item $1_A m = m$
 357\end{enumerate}
 358
 359Let $\psi: M \longrightarrow M$ an $A$-linear endomorphism of $M$.\\
 360$A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi$.
 361\begin{itemize}
 362\item since $\psi$ is $A$-linear, $A[\psi]$ is a commutative ring.
 363\item $M$ is a module over $A[\psi]$, so $\psi$ beomes multiplication by a ring element.
 364\end{itemize}
 365
 366\begin{remark}{}
 367  Vector spaces are defined over fields, a Module is the generalization of a vector space where the scalars come from a ring instead of a field.
 368\end{remark}
 369
 370\subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries}
 371
 372\begin{prop}{AM.2.4}(Cayley-Hamilton Theorem) \label{2.4}
 373  Let $M$ a fingen $A$-module. Let $\aA$ an ideal of $A$, let $\psi$ an
 374  $A$-module endomorphism of $M$ such that $\psi(M) \subseteq \aA M$.
 375
 376  Then $\psi$ satisfies
 377  
 378  $$\psi^n + a_1 \psi^{n-1} + \ldots + a_{n-1} \psi + a_n = 0$$
 379
 380  with $a_i \in \aA$.
 381\end{prop}
 382\begin{proof}
 383  Since $M$ fingen, let $\{ x_1, \ldots, x_n \}$ be generators of $M$.\\
 384  By hypothesis, $\psi(M) \subseteq \aA M$; so for any generator $x_i$, it's image $\psi(x_i) \in \aA M$.
 385
 386  Any element in $\aA M$ is a linear combination of the generators with coefficients in the ideal $\aA$, thus
 387  $$\psi(x_i)= \sum_{j=1}^n a_{ij} x_j$$
 388  with $a_{ij} \in \aA$.
 389  
 390  Thus, for a module with $n$ generators, we have $n$ different $\psi(x_i)$ equations:
 391
 392  $$
 393  \left.
 394  \begin{aligned}
 395    \psi(x_1) &= a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n\\
 396    \psi(x_2) &= a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n\\
 397    \ldots\\
 398    \psi(x_n) &= a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n
 399  \end{aligned}
 400  \right\}
 401  \begin{aligned}
 402      &\text{n elements $\psi(x_i) \in \aA M$ which}\\
 403      &\text{are linear combinations of the}\\
 404      &\text{$n$ generators of $M$}
 405  \end{aligned}
 406  $$
 407  
 408  Next step: rearrange in order to use matrix algebra.
 409
 410  Observe that each row equals $0$, and rearranging the elements at each row we get
 411
 412  $$
 413  \left.
 414  \begin{aligned}
 415    &\psi(x_1) - (a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n) = 0\\
 416    &\psi(x_2) - (a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n) = 0\\
 417    &\ldots\\
 418    &\psi(x_n) - (a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n) = 0
 419  \end{aligned}
 420  \right\}
 421  $$
 422
 423  Then, group the $x_i$ terms together; as example, take the row $i=1$:
 424    $$(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0$$
 425
 426  $$
 427  \left.
 428  \begin{aligned}
 429    &~~~~(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0\\
 430    &-a_{2,1} x_1 + (\psi - a_{2,2}) x_2 - \ldots - a_{2,n} x_n = 0\\
 431    &\ldots\\
 432    &-a_{1,1} x_1 - a_{1,2} x_2 - \ldots + (\psi - a_{1,n}) x_n = 0\\
 433  \end{aligned}
 434  \right\}
 435  $$
 436    
 437  So, $\forall i \in [n]$, as a matrix:
 438  
 439  $$
 440  \begin{pmatrix}
 441    \psi - a_{1,1} & -a_{1,2} & \ldots & -a_{1,n}\\
 442    -a_{2,1} & \psi-a_{2,2} & \ldots & -a_{2,n}\\
 443    \vdots\\
 444    -a_{n,1} & -a_{n,2} & \ldots & \psi-a_{n,n}\\
 445  \end{pmatrix}
 446  \begin{pmatrix}
 447    x_1\\ x_2\\ \vdots\\ x_n
 448  \end{pmatrix}
 449  =
 450  \begin{pmatrix}
 451    0\\ 0\\ \vdots\\ 0
 452  \end{pmatrix}
 453  $$
 454
 455
 456  
 457  Denote the previous matrix by $\Phi$. Let $m$ denote the vector $(x_1, x_2, \ldots, x_n)^T$ (ie. the vector of generators of the $A$-module $M$).\\
 458  \hspace*{4em}Then we can write the previous equality as
 459  \begin{equation}
 460    \Phi \cdot m = 0
 461    \label{eq:2.4.1}
 462  \end{equation}
 463
 464  We know that
 465  \begin{equation}
 466    adj(\Phi) \Phi = det(\Phi) I
 467    \label{eq:2.4.2}
 468  \end{equation}
 469  (aka. \emph{fundamental identity for the adjugate matrix}).
 470
 471  So if at \eqref{eq:2.4.1} we multiply both sides by $adj(\Phi)$,
 472  \begin{align*}
 473    adj(\Phi) \cdot \Phi \cdot &m = 0\\
 474    (\text{recall from \eqref{eq:2.4.2}:}~ &adj(\Phi)\Phi=det(\Phi)\cdot I ~)\\
 475    =det(\Phi) \cdot I \cdot &m = 0
 476  \end{align*}
 477
 478  Thus,
 479  \begin{align*}
 480    det(\Phi) \cdot I \cdot &m = 0:\\
 481  \begin{pmatrix}
 482    det(\Phi) & 0 & \ldots & 0\\
 483    0 & det(\Phi) & \ldots & 0\\
 484    \vdots\\
 485    0 & 0 & \ldots & det(\Phi)
 486  \end{pmatrix}
 487  \cdot
 488  &\begin{pmatrix}
 489    x_1\\ x_2\\ \vdots\\ x_n
 490  \end{pmatrix}
 491                              =
 492  \begin{pmatrix}
 493    0\\ 0\\ \vdots\\ 0
 494  \end{pmatrix}
 495  \end{align*}
 496
 497  $\Longrightarrow$
 498  \begin{equation}
 499    det(\Phi) \cdot x_i = 0 ~~\forall i \in [n]
 500    \label{eq:2.4.3}
 501  \end{equation}
 502
 503  ie. $det(\Phi)$ is an \emph{annihilator} of the generators $x_i$ of $M$, thus
 504  is an annihilator of the entire module $M$.
 505
 506
 507  So, we're interested into calculating the $det(\Phi)$.
 508
 509  By the Leibniz formula,
 510  $$\det(A) = \sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=1}^n a_{i, \sigma(i)}$$
 511
 512  thus,
 513  $$det(\Phi) = \underbrace{(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})}_{\text{diagonal of $\Phi$, leading term of the determinant}} - \ldots$$
 514
 515  The \emph{determinant trick} is that the terms that go after the "leading term of the determinant", will belong to $\aA$ and their combinations with $\psi$ will not be bigger than $\psi^n$. Furthermore, when expanding it
 516  \begin{itemize}
 517    \item highest power is $1 \cdot \psi^n$
 518    \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\
 519      where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$
 520    \item the rest of coefficients of $\psi^k$ are also elements in $\aA$
 521  \end{itemize}
 522
 523  Therefore we have
 524  $$det(\Phi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$
 525  with $a_i \in \aA$.
 526
 527  \vspace{0.5cm}
 528
 529  Now, notice that we had $det(\Phi) \cdot x_i = 0 ~\forall~ i\in [n]$.
 530
 531  The matrix $\Phi$ is the \emph{characteristic matrix}, $xI-A$, viewed as an operator. Then,
 532  $$det(\Phi) = det(xI-A) = p(x)$$
 533  where $p(x)$ is the \emph{characteristic polynomial}.
 534
 535  If a linear transformation turns every basis vector ($x_i$) into zero, then that transformation is the zero transformation. So in our case, $det(\Phi)$ is the zero transformation, thus $det(\Phi)=0$.
 536  Therefore,
 537  $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$
 538
 539
 540  %%%%%% OLD START
 541
 542  % \vspace{3cm}
 543  % 
 544  % Kronecker delta:
 545  % $\delta_{ij} =
 546  % \begin{cases}
 547  %   1 & \text{if } i = j,\\
 548  %   0 & \text{otherwise}
 549  % \end{cases}$
 550  % 
 551  % With the Kronecker delta, $\psi(x_i)$ can be expressed as
 552  % $$\psi(x_i) = \sum_{j=1}^n \delta_{ij} \psi(x_j)$$
 553  % so the previous matrix can be characterized as
 554  % $$\sum_{j=1}^n (\delta_{ij} \psi - a_{ij}) x_j = 0$$
 555  % 
 556  % The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$)
 557  % \begin{itemize}
 558  %   \item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$
 559  %   \item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$
 560  % \end{itemize}
 561  % 
 562  % We need a single element $x \in A$ that \emph{annihilates} every $m \in M$ simultaneously, ie. $xM=0$. We're going to use the determinant for getting $x$.
 563  % 
 564  % Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$,
 565  % the ring of operators $A[\psi]$ is a commutative ring.
 566  % 
 567  % $\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant.
 568  % 
 569  % 
 570  % \vspace{0.75cm}
 571  % This is called \emph{"the determinant trick"}.\\
 572  % We're interested in the determinant because it is the only way to turn a system of multiple equations in a single scalar-like equation that describes the endomorphism $\psi$.\\
 573  % $\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\
 574  % $\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$
 575  % 
 576  % $$det(M) \cdot x_i = 0~~ \forall i$$
 577  % where $x_i$ are the generators of $M$.
 578  % 
 579  % Use $\Phi$ to denote the previous matrix. The determinant is the only function that can take that matrix $\Phi$ and produce a single scalar $x=det(\Phi)$ such that the following identity holds: $adj(\Phi)\cdot \Phi=det(\Phi) \cdot \aA$.
 580  % \vspace{0.5cm}
 581  % 
 582  % Since $det(M)$ kills every generator, it must kill every element in $M$\\
 583  % $\Longrightarrow~~ det(M)$ is the zero map.
 584  % 
 585  % Leibniz formula of the determinant of an $n \times n$ matrix:
 586  % $$
 587  % det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)}
 588  % $$
 589  % 
 590  % so,
 591  % $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$
 592  % expanding it,
 593  % \begin{itemize}
 594  %   \item highest power is $1 \cdot \psi^n$
 595  %   \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\
 596  %     where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$
 597  %   \item the rest of coefficients of $\psi^k$ are also elements in $\aA$
 598  % \end{itemize}
 599  % 
 600  % So we have
 601  % $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$
 602  % with $a_i \in \aA$.
 603  % 
 604  % Since this determinant annihilates the generators (ie. $det(M)x_i=0$), the resulting enddomorphism $p(\psi)$ is the zero map on the entire module $M$, so:
 605  % $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$
 606  % with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem.
 607  
 608  %%%%% OLD END
 609\end{proof}
 610
 611
 612\vspace{0.5cm}
 613
 614\begin{cor}{AM.2.5} \label{2.5}
 615  Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$.
 616
 617  Then, $\exists~ x \equiv 1 \pmod \aA$ such that $xM = 0$.
 618\end{cor}
 619\begin{proof}
 620  take $\psi = \text{identity}$. Then in Cayley-Hamilton (\ref{2.4}):
 621  \begin{align*}
 622    &\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0\\
 623    \Longrightarrow~ &id_M + a_1 id_M + a_2 id_M + \ldots + a_{n-1} id_M + a_n = 0\\
 624    \Longrightarrow~ &(1 + a_1 + \ldots + a_n) id_M = 0
 625  \end{align*}
 626  apply it to $m \in M$, where since $id_M(m)=m$ (by definition of the identity), we then have
 627  $$(1 + a_1 + \ldots + a_n) \cdot m = 0$$
 628  with $a_i \in \aA$.
 629
 630  \begin{enumerate}[\text{part} i.]
 631    \item $xM=0$:\\
 632      Thus the scalar $x = (1 + a_1 + \ldots + a_n)$ annihilates every $m \in M$, ie. the entire module $M$.
 633
 634    \item $x \equiv 1 \pmod \aA$:\\
 635      $x \equiv 1 \pmod \aA ~~ \Longleftrightarrow (x-1) \in \aA$\\
 636      then from $x = (1 + \underbrace{a_1 + \ldots + a_n}_b) \in \aA$, set $b=a_1 + \ldots + a_n$,\\
 637      so that $x=(1+b) \in \aA$.\\
 638
 639      Then $x-1 = (1+b)-1 = b \in \aA$\\
 640      so $x-1 \in \aA$, thus $x \equiv 1 \pmod \aA$ as stated.
 641  \end{enumerate}
 642\end{proof}
 643
 644
 645
 646\begin{prop}{AM.2.6}[Nakayama's lemma] \label{2.6}
 647  Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$.
 648
 649  Then $\aA M = M$ implies $M=0$.
 650\end{prop}
 651\begin{proof}
 652  By \ref{2.5}: since $\aA M = M$, we have $x M =0$ for some $x \equiv 1 \pmod {Jac(A)}$. (notice that at \ref{2.5} is $\pmod \aA$ but here we use $\pmod {Jac(A)}$, since we have $\aA \subseteq Jac(A)$).
 653
 654  (recall \ref{1.9}: $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$).\\
 655  By \ref{1.9}, $x$ is a unit in $A$ (thus $x^{-1}\cdot x=1$).
 656
 657  Hence $M = x^{-1} \cdot \underbrace{x~ \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$.
 658
 659  Thus, if $\aA M = M$ then $M=0$.
 660\end{proof}
 661
 662
 663
 664\begin{cor}{AM.2.7} \label{2.7}
 665  Let $M$ a fingen $A$-module, let $N \subseteq M$ a submodule of $M$, let $\aA \subseteq Jac(A)$ an ideal.
 666
 667  Then $M = \aA M + N \stackrel{\text{implies}}{\Longrightarrow} M=N$.
 668\end{cor}
 669\begin{proof}
 670  The idea is to apply Nakayama (\ref{2.6}) to $M/N$.
 671
 672  Since $M$ fingen $\Longrightarrow~~ M/N$ is fingen and an $A$-module.
 673
 674  Since $\aA \subseteq Jac(A) ~\Longrightarrow~$ Nakayama applies to $M/N$ too.
 675
 676  By definition,
 677  $$\aA M = \left\{ \sum a_i \cdot m_i ~~|~~ a_i \in \aA, m_i \in M \right\}$$
 678  where $m_i$ are the generators of $M$.
 679
 680  Then, for $M/N$,
 681  $$\aA (\frac{M}{N}) = \left\{ \sum a_i \cdot (m_i + N) ~~|~~ a_i \in \aA, m_i \in M \right\}$$
 682
 683  observe that $a_i(m_i+N)= a_i m_i +N$, thus
 684  $$\sum_i a_i \cdot (m_i + N) = \underbrace{(\sum_i a_i \cdot m_i)}_{\in \aA M} + N \in \aA M + N$$
 685
 686  Hence,
 687
 688  \begin{equation}
 689    \aA (\frac{M}{N}) = \left\{ x + N ~~|~~ x \in \aA M \right\} = \aA M + N
 690    \label{eq:2.7.1}
 691  \end{equation}
 692
 693  By definition, if we take $\frac{\aA M + N}{N}$, then
 694  $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$
 695
 696  thus every $y \in \aA M +N$ can be written as
 697  $$y=x+n,~~ \text{with}~ x \in \aA M,~ n\in N$$
 698  which comes from \eqref{eq:2.7.1}.
 699
 700  Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient.
 701  
 702  Hence, every element of $\frac{\aA M +N}{N}$ has the form
 703  $$\frac{\aA M + N}{N} = \left\{ x + N ~~|~~ x \in \aA M \right\}$$
 704  as in \eqref{eq:2.7.1}.
 705
 706  Thus
 707  \begin{equation}
 708    \aA (\frac{M}{N}) = \aA M + N = \frac{\aA M +N}{N}
 709    \label{eq:2.7.2}
 710  \end{equation}
 711
 712  By the Collorary assumption, $M = \aA M + N$; quotient it by $N$:
 713  \begin{equation}
 714    \frac{M}{N} = \frac{\aA M +N}{N}
 715  \label{eq:2.7.3}
 716  \end{equation}
 717
 718  So, from \eqref{eq:2.7.2} and \eqref{eq:2.7.3}:
 719  $$\aA (\frac{M}{N}) = \aA M +N = \frac{\aA M +N}{N} = \frac{M}{N}$$
 720  thus, $\aA (\frac{M}{N}) = \frac{M}{N}$.
 721
 722  By Nakayama's lemma \ref{2.6}, if $\aA (\frac{M}{N}) = \frac{M}{N} ~\stackrel{implies}{\Longrightarrow}~ \frac{M}{N}=0$
 723
 724  Note that
 725  $$\frac{M}{N} = \{ m + N ~|~ m \in M \}$$
 726  (the zero element in $\frac{M}{N}$ is the coset $N=0+N$)
 727
 728  Then, $\frac{M}{N}=0$ means that the quotient has exactly one element, the zero coset $N$.
 729
 730  Thus, every coset $m + N$ equals the zero coset $N$, so $m-0 \in N ~\Longrightarrow~ m \in N$.
 731
 732  Hence every $m \in M$ lies in $N$, ie. $\forall m \in M,~ m \in N$.
 733
 734  So $M \subseteq N$. But notice that by the Corollary, we had $N \subseteq M$, therefore $M = N$.
 735
 736  Thus, if $M = \aA M + N \stackrel{implies}{\Longrightarrow} M = N$.
 737\end{proof}
 738
 739
 740
 741\begin{prop}{AM.2.8} \label{2.8}
 742  Let $x_i ~\forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vector space. Then the $x_i$ generate $M$.
 743\end{prop}
 744\begin{proof}
 745  Let $N$ submodule $M$, generated by the $x_i$.
 746
 747  Then the composite map $N \longrightarrow M \longrightarrow \frac{M}{m M}$ maps $N$ onto $\frac{M}{m M}$, hence $N + \aA M = M$, which by \ref{2.7} implies $N = M$.
 748\end{proof}
 749
 750
 751\subsection{Sequences}
 752
 753\begin{defn}{R.2.9.a}[Exact Sequence]
 754  Let a sequence of homomorphisms
 755$$L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N$$
 756It is \emph{exact} at $M$ if $im(\alpha)=ker(\beta)$.
 757
 758ie. $\beta \circ \alpha = 0$ and $\alpha$ maps surjectively to
 759$ker(\beta)$.
 760\end{defn}
 761
 762\begin{defn}{R.2.9.b}[Short Exact Sequence (s.e.s.)] \label{2.9}
 763$$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$$
 764is exact $\Longleftrightarrow~ L \subset M$ and $N=M / L$.
 765
 766Properties:
 767\begin{itemize}
 768  \item $\alpha$ injective
 769  \item $\beta$ surjective
 770  \item $\alpha:~ L \Longrightarrow ker \beta$
 771  \item $\beta$ induces $M/\alpha(L) \longrightarrow N$
 772\end{itemize}
 773\end{defn}
 774
 775\begin{prop}{R.2.10}[Split exact sequence] \label{2.10}
 776  For the previous s.e.s., 3 equivalent conditions:
 777  \begin{enumerate}[i.]
 778    \item $\exists$ isomorphism $M \cong L \oplus N$, with
 779      \begin{align*}
 780        \alpha:~ &m \longmapsto (m,0)\\
 781        \beta:~ &(m, n) \longmapsto n
 782      \end{align*}
 783
 784    \item $\exists$ a \emph{section} of $\beta$, that is, a map $s: N \longrightarrow M$ such that $\beta \circ s = id_N$
 785    \item $\exists$ a \emph{retraction} of $\alpha$, that is, a map $r: M \longrightarrow L$ such that $r \circ \alpha = id_L$
 786  \end{enumerate}
 787  
 788  If all i, ii, iii are satisfied, it is a split exact sequence.
 789\end{prop}
 790\begin{proof}
 791  Intuitively, when a s.e.s. \emph{splits} it means that the middle module $M$ is the direct sum of the other (outer) two modules, ie. $M = L \oplus N$.
 792
 793  \begin{itemize}
 794    \item[(i to ii, iii)]
 795      if $M \cong L \oplus N$ such that $\alpha:~ m \longmapsto (m,0),~~ \beta:~ s(m, n) \longmapsto n$, we can define the maps
 796
 797      for ii:
 798      \begin{align*}
 799        s:~ N &\longrightarrow L \oplus N\\
 800        s(n) &\longmapsto (0, n)
 801      \end{align*}
 802
 803      Then $\beta(s(n))=\beta(0,n)$, so $\beta \circ s = id_N$.
 804
 805      for iii:
 806      \begin{align*}
 807        r:~ L \oplus N &\longrightarrow L\\
 808        r(m,n) &\longmapsto m
 809      \end{align*}
 810
 811      Then $r(\alpha(m))=r(m,0)$, so $r \circ \alpha = id_L$.
 812
 813    \item[(ii to i)]
 814      assume $s: N \longrightarrow M$ such that $\beta \circ s = id_M$
 815
 816      Want to show $M \cong im(\alpha) \oplus im(s)$.
 817
 818      $\forall m \in M$, consider $m - s(\beta(m))$, apply $\beta$ to it:\\
 819      $\beta(m - s(\beta(m))) = \beta(m) - (\beta \circ s)(\beta(m)) = \beta(m) - \beta(m) = 0$
 820
 821      Since $ker(\beta) = im(\alpha),~~ \exists! l \in L ~\text{such that}~ \alpha(l) = m - s(\beta(m))$.
 822
 823      Thus $m = \alpha(l) + s(\beta(m))$.
 824
 825      \vspace{0.3cm}
 826      Now, suppose $x \in im(\alpha) \cap im(s)$, then $x = \alpha(l)=s(n)$, apply $\beta$ to it: $\beta(\alpha(l)) = \beta(s(n)) ~\Longrightarrow~ 0=n$.
 827
 828      If $n=0$, then $s(n)=0$, so the intersection is $\{0\}$.
 829
 830      \vspace{0.3cm}
 831      Define
 832      \begin{align*}
 833        \phi: L \oplus N &\longrightarrow M\\
 834        \phi(l,n) &\longmapsto \alpha(l)+s(n)
 835      \end{align*}
 836      This isomorphism satisfies the required conditions.
 837
 838    \item[(iii to i)] similar to the previous one.
 839  \end{itemize}
 840
 841  \vspace{0.3cm}
 842  Overview:\\
 843
 844$$
 8450 \longrightarrow L
 846  \substack{
 847    \stackrel{\alpha}{\longrightarrow}\\
 848    \stackrel{\longleftarrow}{r}
 849  }
 850  \substack{M \\[0.5ex] \cong L \oplus N}
 851  \substack{
 852    \stackrel{\beta}{\longrightarrow}\\
 853    \stackrel{\longleftarrow}{s}
 854  }
 855  N \longrightarrow 0
 856$$
 857\begin{align*}
 858  \alpha:~ &l \longmapsto (l,0)\\
 859  r:~ &(m,n) \longmapsto m\\
 860  \alpha &\circ r = id_L\\
 861  \beta:~ &(l,n) \longmapsto n\\
 862  s:~ &n \longmapsto (0,n)\\
 863  \beta &\circ s = id_N
 864\end{align*}
 865
 866\end{proof}
 867
 868\section{Noetherian rings (and modules)}
 869
 870\begin{defn}{}[Ascending Chain Condition]
 871  A partially orddered set $\Sigma$ has the \emph{ascending chain condition} (a.c.c.) if every chain
 872  $$s_1 \leq s_2 \leq \ldots \leq s_k \leq \ldots$$
 873  eventually breaks off, that is, $s_k = s_{k+1} = \ldots$ for some $k$.
 874\end{defn}
 875
 876$\Longrightarrow~ \Sigma$ has the a.c.c. iff every non-empty subset $S \subset \Sigma$ has a maximal element.\\
 877\hspace*{2em} if $\empty \neq S \subset \Sigma$ does not have a maximal element, choose $s_1 \in S$, and for each $s_k$, an element $s_{k+1}$ with $s_k < s_{k+1}$, thus contradicting the a.c.c.
 878
 879\subsection{Noetherian rings and modules}
 880\begin{defn}{R.3.2}[Noetherian ring]\\
 881  Let $A$ a ring; 3 equivalent conditions:
 882  \begin{enumerate}[i.]
 883    \item the set $\Sigma$ of ideals of $A$ has the a.c.c.; in other words, every increasing chain of ideals
 884      $$I_1 \subset I_2 \subset \ldots \subset I_k \subset \ldots$$
 885      eventually stops, that is $I_k = I_{k+1}=\ldots$ for some $k$.
 886    \item every nonempty set $S$ of ideals has a maximal element
 887    \item every ideal $I \subset A$ is finitely generated
 888  \end{enumerate}
 889  If these conditions hold, then $A$ is \emph{Noetherian}.
 890\end{defn}
 891\begin{proof}
 892  TODO
 893\end{proof}
 894
 895\begin{defn}{R.3.4.D}[Noetherian modules]\\
 896  An $A$-module $M$ is Noetherian if the submoles of $M$ have the a.c.c.,\\
 897  that is, ay increasing chain
 898  $$M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$$
 899  of submodules eventually stops.
 900\end{defn}
 901
 902As in with rings, it is equivalent to say that
 903\begin{enumerate}[i.]
 904  \item any nonempty set of modulesof $M$ has a maximal element
 905  \item every submodule of $M$ is finite
 906\end{enumerate}
 907
 908
 909
 910\begin{prop}{R.3.4.P}\label{R.3.4.P}
 911  Let $0 \longrightarrow L \xrightarrow{\ \alpha \ } M \xrightarrow{\ \beta \ } N \longrightarrow 0$ be a s.e.s. (split exact sequence, \ref{2.10}).
 912
 913  Then, $M$ is Noetherian $\Longleftrightarrow~ L$ and $N$ are Noetherian.
 914\end{prop}
 915\begin{proof}
 916  $\Longrightarrow$: trivial, since ascending chains of submodules in $L$ and $N$ correspond one-to-one to certain chains in $M$.
 917
 918  $\Longleftarrow$: suppose $M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$ is an increasing chain of submodules of $M$.
 919
 920  Then identifying $\alpha(L)$ with $L$ and taking intersection gives a chain
 921  $$L \cap M_1 \subset L \cap M_2 \subset \ldots \subset L \cap M_k \subset \ldots$$
 922  of submodules of $L$, and applying $\beta$ gives a chain
 923  $$\beta(M_1) \subset \beta(M_2) \subset \ldots \beta(M_k) \subset \ldots$$
 924  of submodules of $N$.
 925
 926  Each of these two chains eventually stop, by the assumption on $L$ and $N$, so that we only need to prove the following lemma which completes the proof.
 927\end{proof}
 928
 929\begin{lemma}{R.3.4.L}
 930  for submodules $M_1 \subset M_2 \subset M$,
 931  $$L \cap M_1 = L \cap M_2 ~\text{and}~ \beta(M_1) = \beta(M_2) ~\Longrightarrow~ M_1 = M_2$$
 932\end{lemma}
 933\begin{proof}
 934  if $m\in M_2$, then $\beta(m) \in \beta(M_1) = \beta(M_2)$, so that there is an $n \in M_1$ such that $\beta(m)=\beta(n)$.
 935
 936  Then $\beta(m-n)=0$, so that
 937  $$m - n \in M_2 \cap ker(\beta)=M_1 \cap ker(\beta)$$
 938
 939  Hence $m \in M_1$, thus $M_1 = M_2$.
 940\end{proof}
 941
 942
 943
 944\begin{cor}{R.3.5}[Properties of Noetherian modules]\label{R.3.5}
 945  \begin{enumerate}[i.]
 946    \item if $\forall i \in [r],~~M_i$ are Noetherian modules, then
 947      $\bigoplus_{i=1}^r M_i$ is Noetherian.
 948    \item if $A$ a Noetherian ring, then an $A$-module $M$ is Noetherian iff it is finite over $A$.
 949    \item if $A$ a Noetherian ring, $M$ a finite module, then any submodule $N \subset M$ is again finite.
 950    \item if $A$ a Noetherian ring, and $\psi: A \longrightarrow B$ a ring homomorphism such that $B$ is a finite $A$-module, then $B$ is a Noetherian ring.
 951  \end{enumerate}
 952\end{cor}
 953\begin{proof}
 954  \begin{enumerate}[i.]
 955    \item a direct sum $M_1 \oplus M_2$ is a particular case of an exact sequence.
 956
 957      Then, Proposition \ref{R.3.4.P} proves this statement when $r=2$. The case $r>2$ follows by induction.
 958
 959    \item if $M$ finite, then $\exists~$ surjective homomorphism
 960      $$A^r \longrightarrow M \longrightarrow 0$$
 961      for some $r$, so that $M$ is a quotient
 962      $$M \cong A^r / N$$
 963      for some submodule $N \subset A^r$.
 964
 965      $A^r$ is a Noetherian module by i., so $M$ is Noetherian due Proposition \ref{R.3.4.P}.
 966
 967      Conversely, $M$ Noetherian implies $M$ finite.
 968      
 969      item as in previous implications:\\
 970      $M$ finite and $A$ Noetherian $\Longrightarrow$ $M$ is Noetherian,\\
 971      $\Longrightarrow$ since $N \subseteq M$, then $N$ is Noetherian too\\
 972      $\Longrightarrow$ which implies that $N$ is a finite $A$-module.
 973
 974    \item $B$ is Noetherian as an $A$-module; but ideals of $B$ are submodules of $B$ as an $A$-submodule, so that $B$ is a Noetherian ring.
 975  \end{enumerate}
 976\end{proof}
 977
 978
 979\vspace{0.5cm}
 980\subsection{Hilbert basis}
 981\begin{thm}{R.3.6}[Hilbert basis theorem] \label{hilbert-basis}
 982  If $A$ a Noetherian ring, then so is the polynomial ring $A[x]$.
 983
 984  ie. Every ideal in a polynomial ring over a field has a finite set of
 985  generators (is fingen).
 986\end{thm}
 987\begin{proof}
 988  Prove that any ideal $I \subset A[x]$ is fingen.
 989
 990  Define auxiliary sets $J_n \subset A$ by
 991  $$J_n = \{ a \in A ~|~ \exists f \in I ~\text{s.th.}~ f = a x^n + b_{n-1}x^{n-1} + \ldots b_0 \}$$
 992  ie. $J_n$ is the set of leading coefficients of $I$ of degree $n$.
 993
 994  $J_n$ is an ideal, since $I$ is an ideal.
 995
 996  $J_n \subset J_{n+1}$, since for $f \in I$ also $x f \in I$.
 997
 998  Therefore $J_1 \subset J_2 \subset \ldots \subset J_k \subset \ldots$ is an increasing chain of ideals.\\
 999  Using the assumption that $A$ is Noetherian, deduce that $J_n = J_{n+1}$ for some $n$.
1000
1001  For each $m \leq n, ~~ J_m \subset A$ is fingen, ie.
1002  $$J_m = (a_{m,1}, \ldots a_{m, r_m})$$
1003
1004  By definition of $J_m$, for each $a_{m,j}$ with $1 \leq j \leq r_m$,\\
1005  $\exists$ a polynomial $f_{m, j} \in I$ of degree $m$ having the leading coefficient $a_{m, j}$.
1006
1007  $$\Longrightarrow~~ \{ f_{m,j} \}_{m<n; 1 \leq j \leq r_m}$$
1008  the set of elements of $I$.
1009
1010  Claim: this finite set ($\{ f_{m,j} \}$) generates $I$.
1011
1012  $\forall f \in I$, if $\deg f =m$, then its leading coefficient is $a \in J_m$,
1013
1014  hence if $m \geq n$, then $a \in J_m=J_n$, so that
1015  $$a = \sum b_i a_{n,i} ~~ \text{with}~ b_i \in A$$
1016  and
1017  $$f - \sum b_i X^{m-n} \cdot f_{n, i}$$
1018  has degree $<m$.
1019
1020  Similarly, if $m \leq n$, then $a \in J_m$, so that
1021  $$a = \sum b_i a_{m, i} ~~\text{with}~ b_i \in A$$
1022  and
1023  $$f - \sum b_i f_{n, i}$$
1024  has degree $<m$.
1025
1026
1027  \vspace{0.3cm}
1028  By induction on $m$, $f$ can be written as a linear combination of finitely many elements.
1029
1030  Thus, any ideal of $A[x]$ is finitely generated.
1031\end{proof}
1032
1033\vspace{0.5cm}
1034\begin{cor}{R.3.6.C}
1035  if $A$ a Noetherian ring, and $\psi: A \longrightarrow B$ a ring homomorphism such that $B$ is a fingen extension ring of $\psi(A)$, then $B$ is Noetherian.
1036
1037  In particular, any fingen algebra over $\mathbb{Z}$ or over a field $K$ is Noetherian.
1038\end{cor}
1039\begin{proof}
1040  the assumption is that $B$ is a quotient of a polynomial ring,
1041  $$B \cong A[x_1, \ldots, x_n] / I$$
1042  for some ideal $I$.
1043
1044  By the Hilbert basis theorem \ref{hilbert-basis} and induction,\\
1045  $A$ being Noetherian implies that $A[x_1, \ldots, x_n]$ is Noetherian.
1046
1047  And by Corollary \ref{R.3.5}(iv),\\
1048  $A[x_1, \ldots, x_n]$ being Noetherian implies that $A[x_1, \ldots, x_n]/I$ is Noetherian.
1049\end{proof}
1050
1051
1052
1053\vspace{1cm}
1054\section{Finite ring extensions and Noether normalization}
1055
1056\subsection{A-algebras and integral domains}
1057
1058\begin{defn}{}[A-algebra / k-algebra]
1059  An $A$-algebra is a ring $B$ with a ring homomorphism $\psi: A \longrightarrow B$.
1060
1061  $B$ is an $A$-module with multiplication defined by $\psi(a) \cdot b~~~ (a \in A, b \in B)$.
1062
1063  When $A \subset B$, $B$ is an extension ring of $A$; denoted $\psi(A) = A' \subset B$.
1064\end{defn}
1065
1066\begin{defn}{R.4.1}\label{R.4.1}
1067  Let $B$ be an $A$-algebra.
1068
1069  \begin{enumerate}[i.]
1070    \item $B$ is a \emph{finite} $A$-algebra (\emph{finite over $A$}) if it is finite as an $A$-module.
1071    \item $y \in B$ is \emph{integral over} $A$ if $\exists$ a monic polynomial
1072      $$f(Y) = Y^n + a_{n-1} Y^{n-1} + \ldots + a_0 ~\in A'[Y]$$
1073      such that $f(y)=0:$
1074      $$f(y) = y^n + a_{n-1} y^{n-1} + \ldots + a_0 = 0$$
1075
1076      The algebra $B$ is \emph{integral over} $A$ if $\forall~ b \in B$ is integral.
1077  \end{enumerate}
1078\end{defn}
1079
1080
1081\begin{prop}{R.4.2}\label{R.4.2}
1082  Let $\psi: A \longrightarrow B$ be an $A$-algebra, and $y \in B$. Three equivalent conditions:
1083
1084  \begin{enumerate}[i.]
1085    \item $y$ is integral over $A$
1086    \item subring $A'[y] \subset B$ generated by $A' = \psi(A)$ and $y$ is finite over $A$
1087    \item $\exists$ an $A$-subalgebra $C \subset B$ such that $A'[y] \subset C$ and $C$ is finite over $A$
1088  \end{enumerate}
1089
1090  Notes: $A'$ is the image of $A$ in $B$, ie. $A' = \psi(A)$.\\
1091  $A'[y]$ is the smallest subring of $B$ containing both coefficients from $A$ and the element $y$.
1092\end{prop}
1093\begin{proof} .\\
1094  \begin{itemize}
1095    \item[(i to ii):]
1096  since $y$ integral over $A$ $~~\Longrightarrow~$ by \ref{R.4.1} (ii), $y$ satisfies
1097  $$f(y) = y^n + a_{n-1} y^{n-1} + \ldots + a_0 = 0$$
1098
1099  So any power $y^k~~ (k\geq n)$ can be expressed in terms of $\{1, y, y^2, \ldots, y^{n-1} \}$.
1100
1101  Thus the set $\{1, y, y^2, \ldots, y^{n-1} \}$ spans $A'[y]$ as an $A$-module.
1102
1103    \item[(iii to i):]
1104      since $A'[y] \subset C ~~\Longrightarrow~ y \in C$\\
1105      since $C$ finite over $A$, $C$ has finite generators $\{ c_1, \ldots, c_n \}$ such that $C = A \cdot c_1 + A \cdot c_2 + \ldots + A \cdot c_n$
1106
1107      Thus $y \cdot c_i \in C$,
1108      $$y \cdot c_i = \sum_{j=1}^n a_{ij} c_j$$
1109      with $a_{ij} \in A$.
1110
1111      By the Cayley-Hamilton theorem (\ref{2.4}),
1112      $$y^n + a_{n-1} y^{n-1} + \ldots + a_1 y + a_0 = 0$$
1113      Therefore, $y$ is integral (by \ref{R.4.1} (ii)).
1114  \end{itemize}
1115\end{proof}
1116
1117
1118\begin{prop}{R.4.3}[Tower Laws]\label{R.4.3}
1119  \\Let $B$ be an $A$-algebra.
1120  \begin{enumerate}[a.]
1121    \item Transitivity of finiteness: if $A \subset B \subset C$ are extension rings such that $C$ is a finite $B$-algebra and $B$ a finite $A$-algebra,\\
1122      then $C$ is finite over $A$.
1123    \item Finiteness of generated algebras: if $y_1, \ldots, y_m \in B$ are integral over $A$, then $A[y_1, \ldots, y_m]$ is finite over $A$.\\
1124      In particular, every $f \in A[y_1, \ldots, y_m]$ is integral over $A$.
1125    \item Transitivity of integrality: if $A \subset B \subset C$ with $C$ integral over $B$, and $B$ integral over $A$,\\
1126      then $C$ is integral over $A$.
1127    \item Integral closure as a subring: the subset
1128      $$\tilde{A} = \{ y \in B ~|~ y ~\text{is integral over}~ A \} \subset B$$
1129      is a subring of $B$.
1130
1131      Moreover, if $y \in B$ is integral over $\tilde{A}$ then $y \in \tilde{A}$, so that $\tilde{\tilde{A}} = \tilde{A}$.
1132  \end{enumerate}
1133\end{prop}
1134\begin{proof}.\\
1135  \begin{enumerate}[a.]
1136    \item if $\{ \beta_1, \ldots, \beta_n \}$ generate $B$ as an $A$-module and $\{ \gamma_1, \ldots, \gamma_n \}$ generate $C$ as an $B$-module,\\
1137      then the set of products $\{ \beta_i \gamma_j \}$ generates $C$ as an $A$-module.
1138
1139      Since there are $n \times m$ generators (ie. finite), $C$ is finite over $A$.
1140
1141    \item proof by induction:
1142
1143      base case: if $y_1$ integral over $A ~\Longrightarrow~$ it satisfies a monic polynomial.
1144
1145      Thus $A[y_1]$ is generated as an $A$-module by $\{1, y_1, y_1^2, \ldots, y_1^{n-1} \}$, making it a finite $A$-algebra.
1146
1147      inductive step: let $R_k = A[y_1, \ldots, y_k]$. Assume $R_k$ is finite over $A$.
1148
1149      Since $y_{k+1}$ is integral over $A~~ \Longrightarrow~$ it is also integral over $R_k$.
1150
1151      Thus $R_{k+1} = R_k[y_{k+1}]$ is finite over $R_k$.
1152
1153      Applying part (a) (transitivity of finiteness), if $R_{k+1}$ is finite over $R_k$ and $R_k$ finite over $A$, then $R_{k+1}$ is finite over $A$.
1154
1155      Consequence: since any $f \in A[y_1, \ldots, y_m]$ belongs to a finite $A$-algebra, $f$ must be integral over $A$ (since an element is integral iff it is contained in a finite extension).
1156
1157    \item let $x \in C$, since $x$ integral over $B$, it satisfies:
1158      $$x^n + b_{n-1} x^{n-1} + \ldots + b_1 x + b_0 = 0,~~~~~b_i \in B$$
1159
1160      Let $B''=A[b_0, b_1, \ldots, b_{n-1}]$. Since each $b_i \in B$ and $B$ is integral over $A$\\
1161      \hspace*{2em} $\Longrightarrow~$ each $b_i$ is integral over $A$.
1162
1163      Since all $b_i$ are integral over $B'$ $~~\Longrightarrow~ B'[x]$ is a finite $B'$-algebra.
1164
1165      By part (a) (transitivity of finiteness), $B'[x]$ is a finite $A$-algebra.
1166
1167      Therefore, $x$ is integral over $A$.
1168
1169    \item
1170      \begin{enumerate}[I.]
1171        \item subring:\\
1172          let $x,y \in \tilde{A}$. Want to show $x+y, xy \in \tilde{A}$:
1173
1174          by part (b), the algebra $A[x,y]$ is finite over $A$.
1175
1176          Since $x+y,xy \in A[x,y]$, they are integral over $A$.
1177
1178          Thus $x+y, xy \in \tilde{A}$, since $\tilde{A} = \{ b \in B ~|~ b ~\text{integral over}~ A \}$.
1179
1180        \item idempotence\\
1181          let $z \in B$ be integral over $\tilde{A}$
1182          
1183          we have a chain $A \subseteq \tilde{A} \subseteq \tilde{A}[x]$.
1184
1185          By definition, $\tilde{A}$ is integral over $A$, and $z$ is integral over $\tilde{A}$\\
1186          thus by part (c), $z$ is integral over $A$.
1187
1188          Therefore, $z \in \tilde{A}$.
1189      \end{enumerate}
1190  \end{enumerate}
1191\end{proof}
1192
1193\begin{lemma}{4.3.Aux}[Integrality implies finiteness] \label{integral-implies-finite}
1194  If $y$ integral over $A$ then $A[y]$ is finite over $A$.
1195
1196  This extends on point (b) from the previous proposition \ref{R.4.3}.
1197\end{lemma}
1198\begin{proof}
1199  Suppose $y$ is integral over $A$. By definition $\exists~ f \in A[T]$, with
1200  $f$ monic, such that $f(y)=0$. (here $T$ is a formal variable).
1201
1202  Let $deg(f)=d$, so that for $f(y)=0$ we have
1203  $$y^d + a_{d-1} y^{d-1} + \ldots + a_1 y + a_0 = 0 ~~~~ a_i \in A$$
1204
1205  Since it is monic (leading coefficient is $1$), we can rearrange it to isolate the highest power:
1206  \begin{equation}
1207    y^d = -(a_{d-1} y^{d-1} + \ldots + a_1 y + a_0)
1208    \label{eq:yn}
1209  \end{equation}
1210  Thus $y^d$ can be written using lower powers of $y$ with coefficients in $A$.
1211
1212  \vspace{0.5cm}
1213  Consider any element $p \in A[y]$, $p = c_m y^m + c_{m-1} y^{m-1} + \ldots + c_0$.
1214
1215  if $m<d$, leave it as it is.\\
1216  if $m \geq d$, use the monic equation \eqref{eq:yn} to replace $y^d$ with lower powers.
1217
1218  Repeating this process, can reduce any power of $y$ down to a linear
1219  combination of $\{1, y, y^2, \ldots, y^{d-1} \}$.
1220
1221  Thus every element in $A[y]$ can be expressed as
1222  $$\lambda_{d-1} y^{d-1} + \ldots + \lambda_2 y^2 + \lambda_1 y + \lambda_0 \cdot 1~~~~ \lambda_i \in A$$
1223
1224  Henceforth, the set $\{1, y, y^2, \ldots, y^{d-1} \}$ generates $A[y]$ as a finite $A$-module.
1225\end{proof}
1226
1227
1228
1229\vspace{0.5cm}
1230\begin{defn}{4.4}[Integral closure]
1231  Given the ring $\tilde{A}$ from \ref{R.4.3}.(d), ie. $\tilde{A} = \{ y \in B ~|~ y ~\text{integral over}~ A \} \subset B$,
1232  $\tilde{A}$ is the \emph{integral closure} of $A$ in $B$.
1233
1234  If $A=\tilde{A}$, then $A$ is \emph{integrally closed} in $B$.
1235
1236  An integral domain $A$ is \emph{normal} if it is \emph{integrally closed in its field of fractions}, that is if
1237  $$A = \tilde{A} \subset K = Frac(A)$$
1238
1239  For any integral domain $A$, the integral closure of $A$ in its field of fractions $K=Frac(A)$ is also called the \emph{normalization} of $A$.
1240\end{defn}
1241
1242
1243\vspace{0.5cm}
1244
1245\subsection{Noether normalization}
1246
1247\begin{defn}{4.6}[Algebraically independent] \label{R.4.6.D}
1248  $y_1, \ldots, y_n \in A$ are \emph{algebraically independent} over $K$ if the natural surjection $K[Y_1, \ldots, Y_n] \longrightarrow K[y_1, \ldots, y_n]$ is an isomorphism.
1249
1250  $\Longrightarrow~~ \nexists~ F(y_1, \ldots, y_n)=0$ ($F$ nonzero) with coefficients in $K$.
1251\end{defn}
1252
1253Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some finite set $y_1, \ldots, y_n$.
1254
1255
1256\vspace{0.5cm}
1257\begin{lemma}{R.4.6.L} \label{R.4.6.L}
1258  Let $A = K[y_1, \ldots, y_n]$ and $0 \neq F \in K[Y_1, \ldots, Y_n]$ such that $F(y_1, \ldots, y_n)=0$.
1259
1260  Then $\exists~ y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over
1261  $$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~~\text{and}~~ A=A^*[y_n]$$
1262\end{lemma}
1263\begin{proof}
1264  Set $y_i^* = y_i - y_n^{r_i}$ for $i \in [n-1]$ and $r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.\\
1265  \hspace*{3em}(ie. $y_i=y_i^* + y_n^{r_i}$)
1266
1267  Define $G \in A$ by
1268  $$G(y_1^*, \ldots, y_{n-1}^*, y_n) = F(y_i^* + y_n^{r_i}, y_n)=0$$
1269  viewed as a relation for $y_n$ over $K[y_1^*, \ldots, y_{n-1}^*]$.
1270
1271  Since $F$ is a polynomial in $y_1, \ldots, y_{n-1}^*$, can write it as a sum of monomials
1272  $$F= \sum_m a_m y^m = \sum_m a_m \prod y_i^{m_i}$$
1273  where $m=(m_1, \ldots, m_n)$ and each $a_m \neq 0$.
1274
1275
1276  Therefore,
1277  $$G= \sum a_m \prod (y_i^* + y_n^{r_i})^{m_i}$$
1278  which when expanding out, each summand $a_m \prod (y_i^* + y_n^{r_i})^{m_i}$ has a unique term of highest order in $y_n$, namely $a_m y_n^{(\sum r_i m_i)}$.
1279
1280  Suppose we can arrange so that
1281  $$m \neq m' ~\Longrightarrow~ \sum r_i m_i \neq \sum r_i m_i'$$
1282
1283  Then $max \{ \sum r_i m_i ~|~ m ~\text{s.th.}~ a_m \neq 0 \}$ is achieved in only one summand, so that here is no cancellation; thus the highest order term in $G$ is $a_m y_n^{(\sum r_i m_i)}$ (ie. $a_m$ times a pure power of $y_n$).
1284\end{proof}
1285
1286
1287\vspace{0.5cm}
1288\begin{thm}{R.4.6}[Noether normalization lemma] \label{noether-normalization}
1289  Let $K$ a field, $A$ a fingen $K$-algebra.
1290
1291  Then $\exists~ z_1, \ldots, z_m \in A$ such that
1292  \begin{enumerate}[i.]
1293    \item $z_1, \ldots, z_m$ are algebraically independent over $K$
1294    \item $A$ is finite over $B=K[z_1, \ldots, z_m]$
1295  \end{enumerate}
1296
1297  That is, a fingen extension $K \subset A$ can be written as a composite
1298  $$K \subset B = K[z_1, \ldots, z_m] \subset A$$
1299  where $K \subset B$ is a polynomial extension, and $B \subset A$ is finite.
1300\end{thm}
1301\begin{proof}
1302  % (by induction on the number of generators ($n$) of $A$).
1303  Let $y_1, \ldots, y_n$ be generators of $A = K[y_1, \ldots, y_n]$.
1304
1305  if $n=0$, nothing to prove since $A$ is generated by $0$ elements $~\Longrightarrow~ A=K$, and $K$ is finite.
1306
1307  if $n>0$ we have two cases:
1308  \begin{itemize}
1309    \item[-] $y_1, \ldots, y_n$ are algebraically independent over $K$, then by
1310      definition \ref{R.4.6.D} $A \cong K[y_1, \ldots, y_n]$, so that $A$ is a finite module over itself, with $m=n$.
1311    \item[-] $y_1, \ldots, y_n$ are algebraically dependent over $K$,
1312      $$\exists~ 0 \neq f \in K[y_1, \ldots, y_n] ~\text{s.th}~ f(y_1, \ldots, y_n)=0$$
1313  \end{itemize}
1314
1315  Want $f$ to be \emph{monic}, so that $y_n$ is integral over new defined variables $y_1^*, \ldots, y_{n-1}^*$. In other words, want some polynomial like
1316  $$y_n^d+ a_{d-1} y_n^{d-1} + \ldots + a_1 y_n + a_0 = 0~~~~~~~a_i \in K[y_1^*,
1317  \ldots, y_{n-1}^*]$$
1318  \hspace*{2em} ie. monic, so that by definition (\ref{R.4.1}), $y_n$ is
1319  integral over $K[y_1^*, \ldots, y_{n-1}^*]$.
1320
1321  $~\longrightarrow~$ Change variables so that $f$ becomes monic in one of the variables ($y_n$); this allows to express one generator ($y_n$) as an integral element over the others.
1322
1323
1324  \vspace{0.3cm}
1325  Following from Lemma \ref{R.4.6.L}, define the new variables $y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over
1326  $$A^* = K[y^*_1, \ldots, y^*_{n-1}], ~~\text{and}~ A=A^*[y_n]$$
1327
1328  Setting $y_i^* = y_i - y_n^{r_i}$, so that $y_i = y_i^* + y_n^{r_i}$ $\forall i \in [n-1],~~ r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.
1329
1330  Use those new variables at $f(y_1, \ldots, y_n)=0$:
1331  $$f(y_1^* + y_n^{r_1}, y_2^* + y_n^{r_2}, \ldots, y_{n-1}^* + y_n^{r_{n-1}}, y_n) = 0$$
1332
1333  Then the highest power of $y_n$ in each term of $f$ will look like $y_n^{(\sum a_i r_i)}$, and with $r_i$ growing fast enough we ensure that each monomial in $f$ produces a unique power of $y_n$.
1334
1335  Then we have $c \cdot y_n^D + \text{(terms with lower powers of $y_n$)} = 0$ with $c \in K \setminus \{0\}$. So that dividing by $c$ we get the shape $y_n^D + \ldots =0$, thus $y_n$ is integral over $A^* = K[y_1^*, \ldots, y_{n-1}^*]$.
1336
1337  \vspace{0.3cm}
1338  Induction:\\
1339  Since $y_n$ integral over $A^* ~~\Longrightarrow~ A=A^*[y_n]$ is finite over $A^*=K[y_1^*, \ldots, y_{n-1}^*]$ (by \ref{integral-implies-finite}).
1340
1341  By inductive hypothesis on $A^*,~~ \exists~ z_1, \ldots, z_m \in A^*$ algebraically independent over $K$ and with $A^*$ finite over $B=K[z_1, \ldots, z_m]$.
1342
1343  Therefore, each step of $B \subset A^* \subset A^*[y_n]=A$ is finite, and $A$ is finite over $B$ as required.
1344\end{proof}
1345
1346
1347\vspace{0.5cm}
1348\begin{eg}{ }
1349  $A = K[X,Y]/(XY-1)$. $Y$ is algebraic over $K[X]$, but not integral over $K[Y]$.
1350
1351  This corresponds to the fact that the hyperbola $XY=1$ has the line $X=0$ as an asymptotic line (so that its projection to the $X$-axis misses a root over $X=0$).
1352
1353  Take $X' = X- \epsilon Y$ as the element of $A$ instead of $X$; then the relation becomes $(X' + \epsilon Y) Y=1$, monic in $Y$ if $\epsilon \neq 0$.
1354
1355  This corresponds geometrically to tilting the hyperbola a little before projecting, so that no longer has a vertical asymptotic line.
1356\end{eg}
1357
1358\vspace{1cm}
1359
1360\subsection{Weak Nullstellensatz}
1361
1362\begin{prop}{R.4.9}\label{R.4.9}
1363  let $A \subset B$ be an integral extension of integral domain,\\
1364  then $A$ is a field $\Longleftrightarrow~ B$ is a field.
1365\end{prop}
1366\begin{proof}
1367  $\Longrightarrow$:\\
1368  let $0 \neq x \in B$, then $\exists~~ x^n + a_{n-1} x^{n-1} + \ldots + a_0 = 0 ~~~~ a_i \in A$, monic.
1369
1370  Since $A$ is a field, $\exists$ inverse, observe that:
1371  \begin{align*}
1372    x^n &+ a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0\\
1373    x(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = - a_0\\
1374    -a_0^{-1}(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = x^{-1} \in B
1375  \end{align*}
1376  thus there exists inverse in $B$, so $B$ is a field too.
1377
1378  $\Longleftarrow$:\\
1379  if $B$ is a field and $0 \neq x \in A$, then $x^{-1} \in B$, so $x^{-1}$ is integral over $A$.
1380
1381  So there is a relation of the form
1382  $$(x^{-1})^n + a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 =0$$
1383
1384  Therefore
1385  \begin{align*}
1386    (x^{-1})^n &+ a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 = 0\\
1387    (x^{-1})^n &= -a_{n-1} (x^{-1})^{n-1} - \ldots - a_0\\
1388    x^{-n} &= -a_{n-1} x^{-n+1} - \ldots - a_0 ~~\text{(mult by $x^{n-1}$)}\\
1389    x^{-n+(n-1)} &= -a_{n-1} x^{-n+1+(n-1)} - \ldots - a_0 x^{n-1}\\
1390    x^{-1} &= -a_{n-1} - \ldots - a_0 x^{n-1} \in A
1391  \end{align*}
1392
1393  thus there exists inverse in $A$, so $A$ is a field too.
1394\end{proof}
1395
1396\begin{thm}{4.10.prev}[Zariski's lemma] \label{zariski}
1397  $k$ a field, $L$ fingen $k$-algebra and a field. Then $L$ is a finite algebraic extension of $k$.
1398
1399  If $k$ is algebraically closed, then $L=k$.
1400\end{thm}
1401
1402\vspace{0.4cm}
1403From Zariski's lemma we can see that:\\
1404let maximal ideal $m \subset k[X_1, \ldots, X_n]$, then $k[X_1, \ldots, X_n]/m$ is a (quotient) field.
1405
1406It's a fingen $k$-algebra, so by Zariski's lemma, $k[X_1, \ldots, X_n]/m$ is algebraic over $k$.
1407
1408Since $k[X_1, \ldots, X_n]/m$ is algebraically closed, it must equal $k$; ie.
1409$$\frac{k[X_1, \ldots, X_n]}{m}=k$$
1410
1411
1412Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, we have the natural map
1413$$\psi: k \longrightarrow \frac{k[X_1, \ldots, X_n]}{m}$$
1414so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $k[X_1, \ldots, X_n]/m$ must be an element of $k$).\\
1415Then, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$.
1416Thus,
1417$$a_i = X_i +m$$
1418$$\Longrightarrow X_i - a_i \in m$$
1419
1420  % So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$.
1421  % $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$
1422  % $$\Longrightarrow x_i - a_i \in m$$
1423
1424The ideal generated by these terms is a subset of $m$:
1425$$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
1426
1427Thus $m=(X_1 - a_1, \ldots, X_n - a_n)$ (shown also at \ref{5.2}).
1428
1429\vspace{0.4cm}
1430Then:
1431For $k$ algebraically closed, every maximal ideal of $k[X_1, \ldots, X_n]$ has the form 
1432$(X_1 - a_1, \ldots, X_n - a_n)$ for some $a_i \in k$.
1433
1434Equivalently, $V(I) = \emptyset ~~\Longleftrightarrow  1 \in I$.
1435
1436\vspace{0.4cm}
1437
1438\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski2}
1439  let $k$ a field, $K$ a $k$-algebra which
1440  \begin{enumerate}
1441    \item is finitely generated as a $k$-algebra
1442    \item is a field
1443  \end{enumerate}
1444  Then $K$ is algebraic over $k$, so that $k \subset K$ is a finite field
1445  extension. That is, $[K:k] < \infty$.
1446\end{thm}
1447\begin{proof}
1448  $K=k[z_1, \ldots, z_m]$ a field; want to show that $K$ is algebraic over $k$.
1449  \vspace{0.3cm}
1450
1451  Since $K$ is a fingen $k$-algebra, by Noether normalization lemma (\ref{noether-normalization}),\\
1452  $\exists~ z_1, \ldots, z_m \in K$ such that
1453  \begin{itemize}
1454    \item are algebraically independent
1455    \item $K$ is integral over the polynomial ring $A=k[z_1, \ldots, z_m]$ (which by \ref{integral-implies-finite} is finite)
1456  \end{itemize}
1457\\
1458  Now we're at the situation of \ref{R.4.9}:
1459
1460  $A \subset K$ is integral, $K$ is a field $~~\Longrightarrow~$ therefore $A$ is a field.
1461
1462  \vspace{0.2cm}
1463  Since $z_1, \ldots, z_m \in K$ are algebraically independent,\\
1464  \hspace*{2em}$\Longrightarrow~ A=k[z_1, \ldots, z_m]$ is a polynomial ring in $m$ indeterminates, and this is a field only if $m=0$
1465
1466  (since in $k[z_1]$ the element $z_1$ is not invertible, since $1/z_1$ is a rational function, not a polynomial).
1467
1468  So $A=k$; which by Noether normalization we saw that $K$ is integral over $A=k$, and by \ref{integral-implies-finite} that it is finite, thus $K$ is finite over $k$, ie. $[K:k],\infty$, and $K$ is algebraic over $k$.
1469\end{proof}
1470
1471
1472\vspace{1cm}
1473
1474\section{Nullstellensatz}
1475
1476\subsection{Corollary from Weak Nullstellensatz}
1477Note: for $k$ a field, $k[X_1, \ldots, X_n]$, $m$ maximal ideal; the residue field $K=k[X_1, \ldots, X_n]/m$ satisfies the Zariski's lemma (\ref{zariski}), thus $K$ is a finite algebraic extension of $k$.
1478
1479\vspace{0.3cm}
1480\begin{cor}{5.2} \label{5.2}
1481  $k$ algebraically closed. Then every maximal ideal of $A = k[X_1, \ldots,
1482  X_n]$ is of the form
1483  $$m = (X_1 - a_1, \ldots, X_n -a_n),~~ a_i \in k$$
1484
1485  The map $k[X_1, \ldots, X_n] \longrightarrow k[X_1, \ldots, X_n]/m=k$ is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$.
1486
1487  Thus
1488  \begin{align*}
1489    k^n &\longleftrightarrow m-Spec~ A\\
1490    (a_1, \ldots, a_n) &\longleftrightarrow f(a_1, \ldots, a_n)
1491    \end{align*}
1492\end{cor}
1493\begin{proof}
1494  let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.
1495
1496  By fundamental property of maximal ideals, $K=A/m$ is a field.
1497
1498  Since $A$ is a fingen $k$-algebra (generated by $X_1, \ldots, X_n$), then $K=A/m$ is also a fingen $k$-algebra, generated by residues $\psi(x_i) = x_i' = x_i +m$ (where $\psi: k \longrightarrow A/m$).
1499
1500  By Zariski's lemma (\ref{zariski}), $K=A/m$ is algeraic over $k$.
1501
1502  Since by hypothesis $k$ is algebraically closed, it has no proper algebraic extensions\\
1503  \hspace*{2em} $\Longrightarrow~~ K=k~~ \Longrightarrow~~ k \cong A/m$.
1504  
1505  \vspace{0.3cm}
1506  Recall that since $A$ is a $k$-algebra, we have the natural map $\psi: k \longrightarrow A/m$, so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $A/m$ must be an element of $k$).\\
1507  Also, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$.
1508  Thus,
1509  $$a_i = X_i +m$$
1510  $$\Longrightarrow X_i - a_i \in m$$
1511
1512  The ideal generated by these terms is a subset of $m$:
1513  $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
1514
1515  Since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$,
1516  then $J$ is a maximal ideal. Together with $J \subseteq m$, then we have
1517  $J=m$, ie. $$m = (X_1 -a_1, \ldots, X_n -a_n)$$
1518
1519  \vspace{0.3cm}
1520  Let
1521  \begin{align*}
1522    \phi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\
1523    \phi: x_i &\longmapsto a_i
1524  \end{align*}
1525
1526  Which is the natural evaluation map $f(X_1, \ldots, X_n) \longrightarrow f(a_1, \ldots, a_n)$.
1527  \vspace{0.2cm}
1528
1529  Since $\phi$ is a $k$-algebra homomorphism, then $\forall f \in A$:
1530  $$\phi(f(X_1, \ldots, X_n)) = f(\phi(x_1), \ldots, \phi(x_n))= f(a_1, \ldots, a_n)$$
1531
1532  Thus there is a one-to-one correspondence:
1533
1534  points in $k^n ~~~ \longleftrightarrow~~~ m-Spec~ A$ (maximal ideals in $k[X_1, \ldots, X_n]$
1535
1536  $(a_1, \ldots, a_n) ~~~\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)$
1537\end{proof}
1538
1539
1540\vspace{0.5cm}
1541\subsection{Variety}
1542
1543\begin{defn}{5.3}[Variety]
1544  A \emph{variety} $V \subset k^n$:
1545  $$ V = V(J) = \{ P=(a_1, \ldots, a_n) \in k^n | f(P)=0 ~\forall~ f \in J \}$$
1546
1547  \begin{itemize}
1548    \item[$\rightarrow$] $V$ is defined by $f_1(P)= \ldots = f_m(P) = 0$
1549    \item[$\rightarrow$] $V$ is defined as the simultaneous solutions of a number of polynomial equations.
1550  \end{itemize}
1551\end{defn}
1552
1553\begin{prop}{5.3} \label{5.3}
1554  $k$ an algebraically closed field, and $A=k[X_1, \ldots, X_n]$ a fingen $k$-algebra of the form $A=k[X_1, \ldots, X_n]/J$, where $J$ is an ideal of $k[X_1, \ldots, X_n]$.
1555  (notation: $x_i = X_i \pmod J$)
1556
1557  Then every maximal ideal of $A$ is of the form
1558  $$(x_1 -a_1, \ldots, x_n - a_n)$$
1559  for some point $(a_1, \ldots, a_n) \in V(J)$.
1560
1561  Therefore, $\exists$ a one-to-one correspondence
1562  \begin{align*}
1563    V(X) &\longleftrightarrow m-Spec~ A\\
1564    \text{given by}~~~(a_1, \ldots, a_n) &\longleftrightarrow (x_1 -a_1, \ldots, x_n - a_n)
1565  \end{align*}
1566\end{prop}
1567\begin{proof}
1568  the ideals of $A$ are given by ideals of $k[X_1, \ldots, X_n]$ containing $J$, since for $Q=R/I$, $\exists$ one-to-one correspondence between ideals of $Q$ and ideals of $R$ that contain $I$, ie.
1569  \begin{align*}
1570    m \subset A \longleftrightarrow &m' \subseteq k[X_1, \ldots, X_n]\\
1571                &\text{s.th.}~ J \subseteq m'
1572  \end{align*}
1573
1574  Thus, every maximal ideal of $A$ is of the form
1575  $$\underbrace{(x_1 -a_1, \ldots, x_n - a_n)}_{i.} ~\text{such that}~ \underbrace{J \subset (X_1 - a_1, \ldots, X_n - a_n)}_{ii.}$$
1576
1577  \begin{enumerate}[i.]
1578    \item Since $k$ is algebraically closed, the maximal ideals of $k[X_1, \ldots, X_n]$ look like $m=(X_1 - a_1, \ldots, X_n - a_n)$ for some point $(a_1, \ldots, a_n) \in k^n$.
1579
1580      Which when projected to the quotient ring $A$, $X_i \longmapsto x_i$ (residue class), giving $(x_1 - a_1, \ldots, x_n - a_n)$.
1581      
1582    \item for $m$ to exist in $A$, the corresponding $m'$ must contain $J$; since if it didn't contain $J$ it wouldn't "survive" the quotient process.
1583  \end{enumerate}
1584
1585  \vspace{0.3cm}
1586  However, since $(X_1 - a_1, \ldots, X_n - a_n)$ is the kernel of the evaluation map $f \longmapsto f(a_1, \ldots, a_n)$
1587
1588  $\Longrightarrow~$ means that $m'=(X_1 - a_1, \ldots, X_n - a_n)$ consists of all polynomials that vanish at point $P=(a_1, \ldots, a_n)$.
1589
1590  If $J \subseteq m'$, then $\forall~ f \in J$ must vanish at $P$.
1591
1592  By definition, the set of points where all polynomials in $J$ vanish is the \emph{variety}, $V(J)$.
1593
1594  \vspace{0.4cm}
1595  Thus,\\
1596  every maximal ideal in $A$ corresponds to a point $(a_1, \ldots, a_n) \in k^n$, ie.
1597  $$m-Spec~ A \longleftrightarrow k^n$$
1598
1599  The condition that the ideal belongs to the quotient ring $A=k[X_1, \ldots, X_n]/J$ forces that point to lie in $V(J)$, so
1600  \begin{align*}
1601    m-Spec~ A &\longleftrightarrow V(J)\\
1602    \text{maximal spectrum} &\longleftrightarrow \text{variety}
1603  \end{align*}
1604\end{proof}
1605
1606\begin{prop}{5.5}[Correspondeces $V$ and $I$] \label{5.5}
1607  A variety $X \subset k^n$ is by definition $X=V(J)$ (J an ideal of $k[X_1, \ldots, X_n]$).
1608
1609  So $V$ gives a map:\\
1610  $$\{ \text{ideals of}~ k[X_1, \ldots, X_n] \} \stackrel{V}{\longrightarrow} \{ \text{subsets}~ X ~\text{of}~ k^n \}$$
1611
1612  correspondence going the other way:
1613  $$\{ \text{subsets}~ X ~\text{of}~ k^n \} \stackrel{I}{\longrightarrow} \{ \text{ideals of} ~k[X_1, \ldots, X_n] \}$$
1614
1615  defined by taking a subset $X \subset k^n$ into the ideal
1616  $$I(X) = \{ f \in k[X_1, \ldots, X_n] | f(P)=0 ~\forall~ P \in X \}$$
1617
1618  \vspace{0.4cm}
1619
1620  $V,~I$ satisfy reverse inclusions:
1621  $$J \subset J' ~\Longrightarrow~ V(J) \supset V(J') ~~~~\text{and}~~~~ X \subset Y ~\Longrightarrow~ I(X) \supset I(Y)$$
1622\end{prop}
1623
1624
1625\vspace{0.4cm}
1626\subsection{Nullstellensatz}
1627
1628\vspace{0.5cm}
1629\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
1630  Let $k$ algebraically closed field.
1631
1632  \begin{enumerate}[a.]
1633    \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
1634    \item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$,
1635      $$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$
1636  \end{enumerate}
1637\end{thm}
1638\begin{proof}
1639  \begin{enumerate}[a.]
1640    \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$:\\
1641  Let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.\\
1642
1643  Now,
1644  \begin{itemize}
1645    \item[-] since $m$ maximal, $L= k[X_1, \ldots, X_n]/m$ is a field.
1646    \item[-] since $k[X_1, \ldots, X_n]$ is a fingen $k$-algebra, $L$ is a fingen $k$-algebra.
1647  \end{itemize}
1648  $\Longrightarrow~$ thus $L$ is a finite field extension of $k$.
1649
1650  (Recall: if $k$ algebraically closed and $L$ a fingen field extension of $k$,
1651  then $L \cong k$.)
1652
1653  Therefore,
1654  $$L=\frac{k[X_1, \ldots, X_n]}{m} \cong k$$
1655
1656  Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, $\exists$ a surjective homomorphism
1657  $$\psi: k[X_1, \ldots, X_n] \longrightarrow k$$
1658
1659  Let $a_i = \psi(x_i)$. Then $x_i - a_i \in ker(\psi) = m ~\forall~ i \in [n]$.
1660
1661  Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$. (as in \ref{5.2})
1662
1663  Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$,
1664  $$f \in m ~\Longleftrightarrow~ f(P)=0$$
1665
1666  Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$, ie. every element of $J$ vanishes at $P$.\\
1667  $\Longrightarrow~$ therefore $P \in V(J)$, and thus $V(J) \neq \emptyset$.
1668
1669\item $I(V(J)) = rad J$:\\
1670  \begin{align*}
1671    I(V(J)) &= rad J\\
1672    \text{vanishing ideal of a variety} &= \text{radical of the ideal defining the variety}
1673  \end{align*}
1674  where $rad~J = \{ f \in R ~|~ f^n \in J ~\text{for some}~ n>0 \}$.
1675
1676  Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$.
1677
1678  Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$
1679
1680  Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie.
1681  $$\exists~ (a_1, \ldots, a_n, a_{n+1}) \in V(J')$$
1682
1683  Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires
1684  $$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$
1685  a contradiction.
1686
1687  Therefore, $V(J') = \emptyset$.
1688
1689  \vspace{0.4cm}
1690
1691  Since $V(J')=\emptyset$, by the Weak Nullstellensatz/Zariski (\ref{zariski}),\\
1692  \hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$.
1693
1694  Every element in an ideal is a linear combination of its generators: $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
1695
1696  $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf)$$
1697
1698  which, since $1 \in J'$,
1699  $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$
1700
1701  substitute $Y=1/f$,
1702  $$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$
1703  thus
1704  $$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$
1705  multiply by $f^n$,
1706  $$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$
1707
1708  thus $f^n$ is a linear combination of $g_i$.\\
1709  Thus $f^n \in J$, so $f \in rad~J$.
1710\end{enumerate}
1711\end{proof}
1712
1713
1714\vspace{0.4cm}
1715\subsection{Irreducible varieties}
1716
1717\begin{defn}{5.7}[Irreducible variety]
1718  a variety $X \subset k^n$ is \emph{irreducible} if it is nonempty and not the union of two proper subvarieties; that is, if
1719  $$X = X_1 \cup X_2 ~~\text{for varieties}~ X_1, X_2 ~\Longrightarrow~ X= X_1 ~\text{or}~ X_2$$
1720\end{defn}
1721
1722\begin{prop}{5.7} \label{5.7}
1723  a variety $X$ is irreducible iff $I(X)$ is prime.
1724\end{prop}
1725\begin{proof}
1726  set $I=I(X)$.
1727
1728  if $I$ not prime, then $f,g \in A \setminus I$ be such that $fg \in I$.
1729
1730  Define new ideals
1731  $$J_1=(I, f) ~~\text{and}~~ J_2=(I, g)$$
1732
1733  Then, since $f \not\in I(X)$, it follows that $V(J_1) \subsetneq X$
1734
1735  \hspace*{2em}$\Longrightarrow$ so $X=V(J_1) \cup V(J_2)$ is reducible.
1736
1737  The converse is similar.
1738\end{proof}
1739
1740\begin{cor}{5.8} \label{cor.5.8}
1741  let $k$ algebraically closed field. Then $V$ and $I$ induce one-to-one correspondences
1742
1743  \begin{align*}
1744    \{ \text{radical ideals}~J~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{varieties}~ X \subset k^n \}\\
1745    \{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}
1746  \end{align*}
1747
1748  Therefore,
1749  $$Spec~ k[X_1, \ldots, X_n] = \{ \text{irreducible varieties}~ X \subset k^n \}$$
1750\end{cor}
1751
1752\begin{prop}{5.8} \label{5.8}
1753  let $A= k[x_1, \ldots, x_n]$ a fingen $k$-algebra ($k$ an algebraically closed field).
1754
1755  Write $J$ for the ideal of relations holding between $x_1, \ldots, x_n$, so that $A=k[X_1, \ldots, X_n]/J$.
1756
1757  Then there is a one-to-one correspondence
1758  $$Spec~ A \longleftrightarrow \{ \text{irreducible subvarieties}~ X \subset V(J) \}$$
1759\end{prop}
1760\begin{proof}
1761  By definition, $Spec~ A = \{ P ~|~ P \subset A ~\text{is prime ideal} \}$.
1762
1763  By Corollary \ref{cor.5.8}:
1764    $$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} \longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$
1765
1766  About varieties:
1767  \begin{itemize}
1768    \item $I(X)$ in $R=k[X_1, \ldots, X_n]$ is the ideal of the variety $X$\\
1769      \hspace*{2em}ie. the set of all polynomials that vanish on every point of $X$.
1770    \item $I(X)$ in $A=k[X_1, \ldots, X_n]/J$, we're not looking at all possible polynomials but at the residue classes.
1771  \end{itemize}
1772
1773  If $P$ is prime in $A~~\Longrightarrow~$ it must correspond to some prime ideal $\mathfrak{P}$ in $R$ (also $J \subset R$).
1774
1775  Then from Nullstellensatz (\ref{nullstellensatz}), every prime ideal $\mathfrak{P}$ in $R$ is the ideal of some irreducible variety $X$.\\
1776  \hspace*{2em}$\Longrightarrow~~ \mathfrak{P}=I(X)$.
1777
1778
1779  Since we're restricted to the ring $A=k[X_1, \ldots, X_n]/J$, the ideal $P$ are the elements of $I(X)$ viewed through the lens of the quotient\\
1780  $$\Longrightarrow~~ P = \{ f+J ~|~ f \in I(X) \}~~ = I(X) ~\text{mod}~J$$
1781
1782  Now, $J$ is the set of equations defining our "universe" $V(J)$.
1783
1784  Since $A=R/J ~~ \Longrightarrow~$ we thus have $J \subseteq R$.
1785
1786  Also we have a correspondence between $A=R/J$ and $R$.
1787
1788  Let $\mathfrak{P}$ be the preimage of $P$ in $R=k[X_1, \ldots, X_n]$, ie.
1789  \begin{align*}
1790    R &\longrightarrow A=R/J\\
1791    \mathfrak{P} &\longmapsto P\\
1792    I(X) &\longmapsto I(X) ~\text{mod}~J
1793  \end{align*}
1794
1795  $\Longrightarrow~ \text{thus}~ J \subseteq \mathfrak{P}$.
1796
1797
1798  Now, $J \subset \mathfrak{P} = I(X)$;\\
1799  by \ref{5.5}, the set of points where $\mathfrak{P}$ vanishes must be inside the set of points where $J$ vanishes, so
1800  $$V(J) \supseteq V(\mathfrak{P}) = V(I(X)) = X$$
1801  $\Longrightarrow~~ X \subseteq V(J)$, ie. the irreducible variety $X$ must be a subvariety of $V(J)$.
1802
1803  Therefore,
1804  $$\mathfrak{P} \in Spec~ A \longleftrightarrow X \subseteq V(J)$$
1805  where $X = V(\mathfrak{P})$.
1806\end{proof}
1807
1808\subsection{Again: Zariski, Weak Nullstellensatz, Nullstllensatz}
1809This subsection proves again Zariski's lemma, Weak Nullstellensatz and Nullstellensatz, in a more interconnected way than the proofs described previously.
1810
1811\begin{lemma}{}[Zariski's lemma]
1812  $k$ a field, $K$ a fingen $k$-algebra and a field.
1813
1814  Then $K$ is a finite algebraic extension of $k$.
1815
1816  If $k$ algebraically closed, then $K=k$.
1817\end{lemma}
1818\begin{proof}
1819  $K$ a fingen $k$-algebra $\Longleftrightarrow$ $\exists~ x_1,\ldots,x_n\in K$ such that
1820  $$K=k[x_1,\ldots,x_n].$$
1821  Want to show that $K$ is algebraic over $k$.
1822
1823  Since $K$ is a fingen $k$-algebra, by Noether normalization \ref{noether-normalization},\\
1824  $\exists~ \{y_1, \ldots, y_r\} \subset K$ such that
1825  \begin{itemize}
1826    \item the set $\{y_1, \ldots, y_r\}$ is algebraically independent over $k$
1827    \item $K$ is integral over the polynomial ring $R=k[y_1, \ldots, y_r] \subset K$.
1828  \end{itemize}
1829
1830  By \ref{R.4.9}:\\
1831  if $A \subset B$ integral extension of integral domains,\\
1832  then $A$ is a field $\Longleftrightarrow~ B$ is a field.
1833
1834  which applied to our case:
1835  $$B:=K ~\text{(a field)},~~ A:= R=k[y_1, \ldots, y_r]$$
1836  since $K$ is integral over $R$ and $K$ a field $\Longrightarrow$ by \ref{R.4.9} $R$ must be a field.
1837
1838  \vspace{0.3cm}
1839  Since $y_i \in K$ are algebraically independent\\
1840  $\Longrightarrow~ R=k[y_1, \ldots, y_r]$ is a polynomial ring in $r$ indeterminates.
1841
1842  Then, if $r \geq 1$, we have at least one variable $y_1$ (with $k[y_1]$);\\
1843  but the only invertible elements in a polynomial ring over a field are the non-zero constants of the field $k$, so $y_1$ is not invertible in $R$.
1844  (alternatively: the element $y_1$ is not invertible in the polynomial ring, since $1/y_1$ is a rational function, not a polynomial).
1845
1846  Therefore, $R=k[y_1, \ldots, y_r]$ can be a field iff $r=0$, and we know that it is a field, therefore $r=0$ and $R=k$.
1847
1848  By Noether normalization, at the previous step we knew that $K$ is integral over $R$ ($=k$).
1849
1850  \vspace{0.3cm}
1851  By \ref{integral-implies-finite}, a fingen algebra that is integral over its base field is a finite vector space over that field.
1852
1853  Thus $[K:k]<\infty$, and $K$ is algebraic over $k$.
1854
1855  If $k$ is algebraically closed, every element $\alpha\in K$ is algebraic over $k$, and its minimal polynomial over $k$ must have degree one. Therefore $\alpha\in k$, and thus $K=k$.
1856\end{proof}
1857
1858\vspace{0.5cm}
1859This brings us to the \emph{Weak Nullstellensatz}:
1860
1861\begin{thm}{4.10}[Weak Nullstellensatz]\label{weak-nullstellensatz-again}
1862  $k$ a field, $K$ a $k$-algebra with
1863  \begin{itemize}
1864    \item $K$ fingen as a $k$-algebra
1865    \item $K$ a field
1866  \end{itemize}
1867  Then $K$ is finite-dimensional over $k$, and hence algebraic over $k$; that is $[K:k] < \infty$.
1868
1869  If $k$ is algebraically closed, then $K=k$.
1870\end{thm}
1871\begin{proof}
1872  proven at previous proof.
1873\end{proof}
1874
1875
1876
1877\vspace{0.8cm}
1878Now, assume that $k$ is algebraically closed, and let $m\subset k[X_1, \ldots, X_n]$ be a maximal ideal. Let $A=k[X_1, \ldots, X_n]$.
1879
1880Then $\frac{k[X_1, \ldots, X_n]}{m} = \frac{A}{m}$ is a field (by basic ideal quotient theory \ref{maximal-quotient-field}).
1881
1882Moreover, $\frac{A}{m}$ is a fingen $k$-algebra, generated by the classes
1883$$X_1+m,\ldots,X_n+m.$$
1884
1885Therefore, by Zariski's lemma, $\frac{A}{m}$ is algebraic over $k$.
1886
1887And since $k$ is algebraically closed $\Longrightarrow~ \frac{A}{m}\cong k$.
1888
1889Let
1890$$\pi:A\longrightarrow \frac{A}{m}$$
1891be the quotient map. By definition, $\pi$ is the surjective homomorphism that sends each polynomial to its coset modulo $m$, and the kernel of this map is exactly $ker(\pi)=m$.
1892
1893By Zariski's lemma, $\frac{A}{m}$ is a finite field extension of $k$, and since $k$ is algebraically closed, by Zariski's lemma, this extension must be $k$ itself; denote this relation by
1894$$\theta:\frac{A}{m}\stackrel{iso}{\longrightarrow} k$$
1895the $k$-algebra isomorphism.
1896
1897Define the composite map
1898$$\phi=\theta\circ\pi:A\longrightarrow k.$$
1899which is a $k$-algebra homomorphism.
1900
1901Thus, $\forall~ i\in[n]$, define
1902$$a_i=\phi(X_i)\in k.$$
1903
1904Since $\phi$ is a $k$-algebra homomorphism, it must map any scalar $c \in k$ to itself, ie. $\phi(c) = c$. Therefore
1905$$\phi(X_i - a_i) = \phi(X_i) - \phi(a_i) = a_i - a_i = 0$$
1906
1907Since $\phi = \theta \circ \pi$, and $\theta$ is an isomorphism (by definition injective), the only way for $\phi(f)$ to be $0$ is if $\pi(f)=0$:
1908\begin{align*}
1909  &\phi(X_i-a_i)=\theta(\pi(X_i - a_i)) = 0\\
1910  &\Longrightarrow \pi(X_i-a_i)=0\\
1911\end{align*}
1912
1913Since $ker(\pi)=m$, any element that maps to $0$ under $\pi$ must belong to $m$, thus
1914$$\pi(X_i - a_i) =0 ~~\Longrightarrow~ X_i-a_i\in m$$
1915
1916Alternative way: by definition of a quotient ring, and elements of $\frac{A}{m}$ are cosets of the form $f+m$. $\pi(x_i)$ sends $X_i$ to the coset $X_i+m$, ie. $\pi(x_i)=X_i+m$. Thus, if $a_i$ is the image of $X_i$, then
1917$$X_i \equiv a_i \pmod m,$$
1918which by the definition of congruence modulo ideal, this means $X_i -a_i \in m$.
1919
1920\vspace{0.5cm}
1921
1922Since $X_i - a_i \in m ~~\Longrightarrow~~ J=(X_1 - a_1, \ldots, X_n -a_n) \subseteq m$,
1923
1924since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$, then $J$ is a maximal ideal.
1925
1926
1927\begin{adjustwidth}{4em}{0pt}
1928  (Corollary \ref{5.2}) Thus $J=m$, ie. $m=(X_1 - a_1, \ldots, X_n -a_n)$, with $a_i \in k$.
1929
1930  \vspace{0.4cm}
1931  We had $k \stackrel{iso}{\cong} \frac{A}{m} = \frac{k[X_1, \ldots, X_n]}{m}$.
1932
1933  We have the map
1934  \begin{align*}
1935    \phi: k[X_1, \ldots, X_n] &\longrightarrow \frac{k[X_1, \ldots, X_n]}{m} \cong k\\
1936    X_i &\mapsto a_i \in k
1937  \end{align*}
1938  which is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$.
1939
1940  Since $\phi$ is a $k$-algebra homomorphism, then
1941  $$\forall~ f \in A,~ \phi(f(X_1, \ldots, X_n)) = f(\phi(X_1), \ldots, \phi(X_n)) = f(a_1, \ldots, a_n).$$
1942
1943  Thus, $\exists$ a one-to-one correspondence:
1944  \begin{align*}
1945    \text{points in}~ k^n ~~~ &\longleftrightarrow~~~ m-Spec~A, ~\text{maximal ideals in}~ k[X_1, \ldots, X_n]\\
1946    (a_1, \ldots, a_n) ~~~&\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)
1947  \end{align*}
1948\end{adjustwidth}
1949
1950\vspace{0.5cm}
1951Now let $J\subsetneq A$ be an arbitrary proper ideal. Since every proper ideal is contained in a maximal ideal, $\exists$ a maximal ideal $m$ such that
1952$$J\subseteq m$$
1953
1954We have seen that
1955$$m=(X_1-a_1, \ldots, X_n-a_n)$$
1956for some $P=(a_1,\ldots,a_n)\in k^n$.
1957
1958So $P$ is a zero for every polynomial in $m$, ie.
1959$$\forall~ f \in m ~\Longrightarrow~ f(P)=0.$$
1960
1961Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$; ie. every element of $J$ vanishes at $P$.
1962
1963$\Longrightarrow~ P \in V(J)$, and thus $V(J) \neq \emptyset$. This is what the Nullstellensatz first part states:
1964\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz-again}
1965  Let $k$ be an algebraically closed field and let $J\subseteq k[X_1,\ldots,X_n]$ be an ideal.
1966  \begin{enumerate}[a.]
1967    \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
1968    \item $I(V(J)) = rad(J)~~(=\sqrt{J})$, in other words, for $f \in k[X_1, \ldots, X_n]$,
1969      $$f(P)=0 ~\forall~ P \in V(J) ~~\Longleftrightarrow~~ f^N \in J ~\text{for some $N\geq 1$.}$$
1970  \end{enumerate}
1971\end{thm}
1972\begin{proof}
1973  \begin{enumerate}[a.]
1974  \item already proven in the previous steps.
1975    
1976  \item $I(V(J)) = \sqrt{J}$\\
1977  ie. vanishing ideal of a variety = radical of the ideal defining the variety
1978
1979  where
1980  $$\sqrt{J} = \{ f \in A ~|~ f^N \in J ~\text{for some}~ N>0 \}.$$
1981
1982  First, want to show that
1983  $$\sqrt{J}\subseteq I(V(J)).$$
1984
1985  Let $f\in\sqrt{J}$. Then $f^N\in J$ for some $N\geq 1$.
1986
1987  Thus, for every $P\in V(J)$,
1988  $$f(P)^N=f^N(P)=0,$$
1989  and therefore $f(P)=0$.
1990
1991  Hence $f\in I(V(J))$, so
1992  $$\sqrt{J}\subseteq I(V(J)).$$
1993
1994  Now, want to show that if a polynomial vanishes at all points where $J$ vanishes, then $f \in \sqrt{J}$.
1995
1996  Since $k[X_1,\ldots,X_n]$ is Noetherian, write
1997  $$J=(g_1,\ldots,g_m).$$
1998
1999  Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$.
2000
2001  Suppose there is a point $(a_1, \ldots, a_n,b)$ that is a zero of $J'$. ie.
2002  $$\exists~ (a_1, \ldots, a_n,b) \in V(J').$$
2003
2004  Since
2005  $$g_i(a_1,\ldots,a_n)=0,$$
2006  our hypothesis says
2007  $$f(a_1,\ldots,a_n)=0.$$
2008
2009  However, the last generator $(1-Yf)$ requires
2010  $$1 - b f(a_1,\ldots,a_n) = 0 ~~\Longrightarrow~ 1 - b \cdot 0 = 0 ~\Longrightarrow~ 1=0,$$
2011  a contradiction.
2012
2013  Therefore, $V(J') = \emptyset$.
2014
2015  \vspace{0.4cm}
2016
2017  By part a., applied to $k[X_1,\ldots,X_n,Y]$,\\
2018  \hspace*{2em} if $V(J')=\emptyset$ then $J'=k[X_1,\ldots,X_n,Y]$, so $1 \in J'$.
2019
2020  Every element in an ideal is a linear combination of its generators:\\
2021  $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
2022
2023  $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf).$$
2024
2025  which, since $1 \in J'$,
2026  $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X)).$$
2027
2028  Now work in the localization
2029  $$A_f=A[f^{-1}],$$
2030  where $f$ is invertible, and apply the homomorphism
2031  $$A[Y]\longrightarrow A_f,\qquad Y\longmapsto f^{-1}.$$
2032
2033  Then
2034  $$1= \left( \sum_{i=1}^m p_i(X,f^{-1}) g_i(X) \right)
2035  +q(X,f^{-1})\cdot \underbrace{(1-f^{-1}f(X))}_{0},$$
2036  thus
2037  $$1= \sum_{i=1}^m p_i(X,f^{-1}) g_i(X).$$
2038
2039  Since there are finitely many polynomials $p_i$, $\exists~ N\geq 0$ such that
2040  $$f^N p_i(X,f^{-1})\in A ~~\forall~ i.$$
2041
2042  Multiplying by $f^N$,
2043  $$f^N= \sum_{i=1}^m A_i(X) g_i(X),$$
2044  where
2045  $$A_i(X)=f^N p_i(X,f^{-1})\in A.$$
2046
2047  Thus $f^N$ is a linear combination of $g_i$.\\
2048  Therefore $f^N \in J$, so $f \in \sqrt{J}$.
2049
2050  Hence
2051  $$I(V(J))\subseteq\sqrt{J}.$$
2052
2053  Combining both inclusions,
2054  $$I(V(J))=\sqrt{J}.$$
2055  \end{enumerate}
2056\end{proof}
2057
2058
2059
2060
2061\section{Rings of fractions $S^{-1}A$ and localization}
2062
2063\subsection{Small recap}
2064\paragraph{principal ideal:} generated by a single element, ie. for $a \in A$, $(a)=aA=\{ra ~|~ r\in A\}$.
2065
2066\paragraph{unit:} $u \in A$ is a unit if $\exists v \in A$ with $uv=1$.
2067
2068\paragraph{proper ideal:} $I \subsetneq A$, equivalently $1 \not\in I$. Therefore, a proper ideal does not have units.
2069
2070\paragraph{maximal ideal:}
2071$\mM \subset A$ ($A$ ring) with $m \neq A$ and there is no ideal $I$ strictly between $\mM$ and $A$. ie. if $\mM$ maximal and $\mM \subseteq I \subseteq A$, either $\mM=I$ or $I=A$.
2072\begin{itemize}
2073  \item is proper
2074  \item for $m$ maximal ideal, $\frac{A}{m}$ is a field.
2075  \item every maximal ideal is prime
2076  \item elements of a maximal ideal are always non-units\\
2077    \hspace*{2em} since if $x \in m$ were a unit, then $1=x^{-1}x \in m$, a contradiction since if $1 \in m$ then $m=A$, which would mean that it is not proper; and it is maximal, hence proper.
2078\end{itemize}
2079
2080\paragraph{prime ideal:} proper ideal $P$ such that
2081$$ab \in P ~~\Longrightarrow~ a \in P ~\text{or}~ b \in P$$
2082for $P$ prime, $\frac{A}{P}$ is an integral domain.
2083
2084\paragraph{Spec(A):} prime spectrum, $Spec(A)=\{ P \subset A ~|~ P ~\text{is prime}\}$
2085
2086
2087\vspace{0.6cm}
2088
2089
2090
2091\subsection{Rings of fractions $S^{-1}A$}
2092
2093\begin{defn}{6.1}[ring of fractions] \label{def.6.1}
2094  let $A$ a ring, $S \subset A$ a multiplicative set ($1 \in S$, and $st \in S
2095  ~\forall s, t \in S$).
2096
2097  Introduce th following relation $\sim$ on $A \times S$:
2098  $$(a, s) \sim (b, t) \Longleftrightarrow \exists y \in S ~\text{such that}~ u(at - bs) = 0$$
2099
2100  (write $a/s$ for the equivalence class of $(a, s)$.
2101
2102  Then, the \emph{ring of fractions of $A$ with respect to $S$} is
2103  $$S^{-1}A = (A \times S)/\sim$$
2104
2105  with ring op'ns defined by the usual arithmetic op'ns on fractions:
2106  $$\frac{a}{s} \pm \frac{b}{t} = \frac{(at \pm bs)}{st} ~~\text{and}~~ \frac{a}{s} \cdot \frac{b}{t} = \frac{ab}{st}$$
2107\end{defn}
2108
2109\begin{prop}{6.1} \label{prop.6.1}
2110  \begin{enumerate}[i.]
2111    \item $\sim$ is an equivalence relation
2112    \item the ring op'ns are well defined, and $S^{-1}A$ is a ring
2113    \item
2114      \begin{align*}
2115        \psi: A &\longrightarrow S^{-1}A\\
2116              a &\longmapsto a/1
2117      \end{align*}
2118      is a ring homomorphism.
2119  \end{enumerate}
2120\end{prop}
2121
2122\begin{lemma}{6.2} \label{6.2}
2123  For $f \in A$, write $S = \{ 1, f, f^2, \ldots \}$, and $A_f \cong S^{-1}A$.
2124
2125  Then
2126  $$A_f \cong \frac{A[X]}{(Xf-1)}$$
2127\end{lemma}
2128\begin{proof}
2129  (adjoining a variable $X$ and imposing $Xf=1$ is the algebraic way of saying "add $f^{-1}$".)\\
2130  Define the homomorphism
2131  \begin{align*}
2132    \psi: A[X] &\longrightarrow A_f\\
2133    a &\longmapsto a/1\\
2134    X &\longmapsto 1/f
2135  \end{align*}
2136
2137By the 1st isomorphism theorem:
2138
2139  \begin{tikzpicture}[node distance=1.5cm, auto]
2140    \node (G)    {$A[X]$};
2141    \node (H) [right of=G] {$A_f$};
2142    \node (GmodK) [below of=G, xshift=0.75cm] {$A[X]/\ker(\psi)$};
2143
2144    \draw[->] (G) to node {$\psi$} (H);
2145    \draw[->] (G) to node [swap] {$\phi$} (GmodK);
2146    \draw[->] (GmodK) to node [swap] {$\eta$} (H);
2147  \end{tikzpicture}
2148
2149  Thus we want to prove that $ker(\psi) = (Xf-1)$, so that $\frac{A[X]}{ker(\psi)} = \frac{A[X]}{(Xf-1)}$, and the lemma is proven.
2150
2151  \vspace{0.3cm}
2152  First, observe that $\psi(Xf-1) = \psi(X)\psi(f)-\psi(1) = \frac{1}{f} \frac{f}{1} - 1 = 1-1=0$,
2153  so $Xf-1 \in ker(\psi)$, ie. $(Xf-1) \subseteq ker(\psi)$.
2154
2155  Now, we want to prove that $ker(\psi) \subseteq (Xf-1)$.
2156
2157  Take $h \in ker(\psi)$, will prove that $h \in (Xf-1)~ \forall~ h \in ker(\psi)$, and thus $ker(\psi) \subseteq(Xf-1)$.
2158
2159  \vspace{0.3cm}
2160  Want to prove that $h(X)$ is a multiple of $(Xf-1)$.
2161
2162  Let
2163  $$h(X)=a_n X^n + a_{n-1} X^{n-1} + \ldots + a_1 X + a_0$$
2164
2165  % Since $h \in ker(\psi) ~~\Longrightarrow~~ \psi(h)=0$, so
2166  % $$\psi(h) = \frac{a_n}{f^n} + \frac{a_{n-1}}{f^{n-1}} + \ldots \frac{a_1}{f} + \frac{a_0}{1} = 0 \in A_f$$
2167
2168  multiply $h(X)$ by $f^n$:
2169  $$f^n \cdot h(X)=a_n (f^n X^n) + a_{n-1} f (f^{n-1} X^{n-1}) + a_{n-2} f^2 (f^{n-2} X^{n-2})  \ldots$$
2170
2171  Note that since $\forall~ i \geq 1,~~ f^i X^i = (Xf -1)\cdot (f^{i-1} X^{i-1} + f^{i-2} X^{i-2} + \ldots + 1)$, then $f^i X^i \equiv 1 \pmod{Xf-1}$.
2172
2173  So,
2174  $$f^n \cdot h(X)=\underbrace{a_n (1) + a_{n-1} f (1) + a_{n-2} f^2 (1) + \ldots + a_0 f^n}_{C~~\text{(constant)}} \pmod{Xf-1}$$
2175
2176  $$\Longrightarrow~~ f^n \cdot h(X)=C \pmod{Xf-1}$$
2177  $$\Longleftrightarrow~~ f^n \cdot h(X)=Q(X) \cdot (Xf-1) + C$$
2178
2179  Want to remove $C$, but it is non-zero. Note that in $A_f$ (ring of
2180  fractions), $\frac{a}{f^n} = 0$ iff $\exists~ k$ such that $f^k \cdot a = 0$ in $A$.
2181
2182  So, multiply both sides by $f^k$:
2183  $$f^k \cdot f^n \cdot h(X)=f^k \cdot (Q(X) \cdot (Xf-1) + C)$$
2184  $$\underbrace{f^k f^n}_{f^{nk}} \cdot h(X)=\underbrace{f^k Q(X)}_{Q'(X)} \cdot (Xf-1) + \underbrace{f^k C}_{0}$$
2185
2186  $$\Longrightarrow~~ f^{n+k} \cdot h(X)=Q'(X) \cdot (Xf-1)$$
2187  $$\Longleftrightarrow~~ f^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2188
2189  multiply it by $X^{n+k}$:
2190  $$X^{n+k} \cdot f^{n+k} \cdot h(X) \equiv X^{n+k} \cdot 0 \pmod{Xf-1}$$
2191  $$(Xf)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2192
2193  Now, since we had $Xf \equiv 1$ in $\frac{A[X]}{(Xf-1)}$,
2194  $$(1)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2195  $$\Longrightarrow~~ h(X) \equiv 0 \pmod{Xf-1}$$
2196  By definition this is saying $h(X) \in (Xf-1) ~\forall~ k \in ker(\psi)$.
2197
2198  Thus $ker(\psi) \subseteq (Xf-1)$.
2199
2200  Initially we saw that $(Xf-1) \subseteq ker(\psi)$. Therefore $ker(\psi)=(Xf-1)$.
2201
2202  Hence,
2203  $$A_f \cong \frac{A[X]}{(Xf-1)}$$
2204\end{proof}
2205
2206
2207\vspace{0.4cm}
2208
2209Given a ring homomorphism $\psi: A \longrightarrow B$, there is a correspondence
2210
2211$$e: \{ \text{ideals of } A \} \longrightarrow \{ \text{ideals of } B \}$$
2212
2213given by $e(I) = \psi(I)B = IB$ (called \emph{extension}),
2214
2215and
2216
2217$$r: \{ \text{ideals of } B \} \longrightarrow \{ \text{ideals of } A \}$$
2218
2219given by $r(J) = \psi^{-1} J$ (called \emph{restriction}, written $A \cap J$).
2220
2221Set $B= S{-1}A$. Then $S^{-1}I = e(I) = \psi(I)B$.
2222
2223\begin{prop}{6.3} \label{prop.6.3}
2224  \begin{enumerate}[a.]
2225    \item $\forall$ ideal $J$ of $S^{-1}A$, $e(r(J))=J$
2226    \item $\forall$ ideal $I$ of $A$,
2227      $$r(e(I)) = \{ a \in A ~|~ as \in I ~\text{for some}~ s \in S \}$$
2228    \item if $P$ prime and $P \cap S = \emptyset$,\\
2229      then $e(P)=S^{-1}P$ is a prime ideal of $S^{-1}A$.
2230  \end{enumerate}
2231\end{prop}
2232\begin{proof}
2233\end{proof}
2234
2235\begin{cor}{6.3} \label{cor.6.3}
2236    for an ideal $I$ of $A$, the necessary and sufficient condition for
2237    $r(e(I))=I$ is
2238      $$as \in I \Longrightarrow a \in I~~ \forall s \in S$$
2239\end{cor}
2240
2241
2242\subsection{Localization}
2243
2244If $P$ prime ideal, then $S=A \setminus P$ is a multiplicative set.
2245Define the set $A_P=S^{-1}A$.
2246
2247\begin{prop}{6.4} \label{6.4}
2248  $a/s \in A_P$ is a unit of $A_P ~\Longleftrightarrow~ a \not\in P$.
2249
2250  Therefore $A_P$ is a \emph{local ring}, with maximal ideal $e(P)=P A_P$.
2251
2252  The local ring $(A_P, PA_P)$ is called the \emph{localization} of $A$ at $P$.
2253\end{prop}
2254\begin{proof}
2255  \begin{enumerate}
2256    \item[$\Longleftarrow$] (if $a \not\in P ~\Longrightarrow~ a/s \in A_P$ is a unit)\\
2257
2258      if $a \not\in P$, then by definition of $S,~~ a \in S$.
2259
2260      In the localization $A_P$, every element of $S$ is invertible.
2261
2262      The inverse of $\frac{a}{s}$ is $\frac{s}{a}$\\
2263      \hspace*{2em}$\longrightarrow~~ \frac{a}{s} \cdot \frac{s}{a} = 1$, thus
2264      $a/s \in A_P$ is a unit.
2265
2266    \item[$\Longrightarrow$] (if $a/s$ unit $~\Longrightarrow~ a \not\in P$)\\
2267
2268      Suppose $a/s$ a unit; then $\exists~ \frac{b}{t} \in A_P$ such that
2269      $\frac{a}{s} \cdot \frac{b}{t} = 1$.
2270
2271      By definition of equality in localization, $\frac{ab}{st}=1$ means $\exists u \in S$ such that
2272      \begin{equation}
2273        u(ab \cdot 1 - st \cdot 1) = 0~~ \Longrightarrow~ uab=ust
2274        \label{eq:6.4}
2275      \end{equation}
2276      with $u,s,t \in S$.
2277
2278      Since $S=A \setminus P$ and $P$ is a prime ideal,\\
2279      the products of elements outside $P$ must also be outside of $P$.
2280      $$u,s,t \not\in P ~\Longrightarrow~ ust \not\in P$$
2281
2282      At \eqref{eq:6.4}, we know that $uab=ust \not\in P$, thus $uab \not\in P$.
2283
2284      If $uab \not\in P$, then $a \not\in P$.
2285
2286
2287      Next we will prove that $A_P$ is a local ring.\\
2288
2289      A ring is local if it has exactly one maimal ideal (\ref{1.13}); so we
2290      show that the set of non-units forms an ideal.
2291
2292      $a/s$ is not a unit iff $a \in P$. Let $m= \{ a/s | a \in P,~ s \not\in P \} = PA_P$.
2293      Want to show that $PA_P$ is the unique maximal ideal:
2294      \begin{itemize}
2295        \item it's an ideal: let $\frac{a}{s}, \frac{b}{t} \in PA_P$ with $a,b \in P$.
2296
2297          Then $\frac{a}{b} + \frac{b}{t} \in PA_P$ with numerator in $P$.
2298
2299          If multiply a fraction in $PA_P$ by any fraction in $A_P$, the numerator stays in $P$, thus $PA_P$ is an ideal.
2300
2301        \item every element outside of $PA_P$ is a unit (proven at the beginning of this proposition's proof ($\Longleftarrow$)).
2302
2303        \item maximality: every ideal strictly larger than $PA_P$ must contain a unit.\\
2304          If an ideal contains a unit, it must be the entire ring $A_P$.\\
2305          \hspace*{2em}$\Longrightarrow$ therefore, $PA_P$ is the unique maximal ideal.
2306      \end{itemize}
2307  \end{enumerate}
2308\end{proof}
2309
2310
2311In other words:
2312
2313\emph{Localization}: formal way to 'force' certain elements to have inverses.
2314
2315It's the smallest ring that contains $A$ and makes all the elements of $S$ invertible.
2316
2317There is a natural map $\psi: A \longrightarrow S^{-1}A$, by $f(a)=\frac{a}{1},~ f(s)=\frac{1}{s}~~ \text{for}~ s \in S$.
2318
2319
2320\subsection{Localization commutes with taking quotients}
2321Let $A$ ring, $S$ multiplicative set, $I$ ideal.
2322
2323Write $T$ for the image of $S$ in $A/I$.
2324
2325Then $S^{-1}I=I \cdot S^{-1}A$ is an ideal of $S^{-1}A$, can take the quotient ring $S^{-1}A / S^{-1} I$.
2326
2327$\longleftrightarrow$ can take the quotient $A/I$ and then localize to get $T^{-1}(A/I)$.
2328
2329
2330\begin{cor}{6.7} \label{6.7}
2331  $$T^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
2332  In particular, for $P$ prime ideal,
2333  $$\frac{A_P}{PA_P} = k(P) = Frac(\underbrace{A/P}_{\substack{\text{integral}\\\text{domain}}})$$
2334
2335  From \ref{6.4}, $A_P$: local ring, $PA_P$: unique maximal ideal.
2336  $k(P)$ and $Frac(A/P)$ are field of fractions.
2337\end{cor}
2338\begin{proof}
2339  the quotient ring $A/I$ can be viewed as an $A$-module and
2340  $$\underbrace{ T^{-1}(A/I) }_{\text{ring of fractions}} \cong
2341  \underbrace{ \frac{S^{-1}A}{S^{-1}I} }_{\text{module of fractions}}$$
2342
2343  The \ref{cor.6.6}.i, gives an isomorphism of modules
2344  $$ T^{-1}(A/I) = S^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
2345  it's easy to see that this is a ring homomorphism.
2346\end{proof}
2347% TODO 6.6 (proposition and corollary)
2348
2349
2350\vspace{1.0cm}
2351\subsection{Key ideas}
2352Main idea: make some choosen elements invertible.
2353
2354For localization, it is important to not get confused between "localization at an element" and "localization at a prime ideal".
2355
2356\paragraph{Localization at an element $f$.}
2357
2358Let $S=\{1, f, f^2, f^3, \ldots \}$ a multiplicative set. Then $S^{-1}A$ is the ring of fractions, and
2359$$A_f= S^{-1}A = A \big[ \frac{1}{f} \big] = \{\frac{a}{f^n} ~|~ a \in A, n \geq 0\}$$
2360
2361ie. "make $f$ invertible", then $\frac{1}{f} \in A_f$.\\
2362$\Longrightarrow~ A_f$ is the ring obtained from $A$ by making every $f \in S$ invertible.
2363
2364Lemma \ref{6.2} will show that $A_f \cong \frac{A[X]}{(Xf-1)}$.
2365
2366\paragraph{Localization at a prime ideal P.}
2367
2368For $P \in Spec(A)$ (ie. $P$ is a prime ideal),
2369
2370$$A_P = ( \underbrace{A \setminus P}_{S} )^{-1} A = \{ \frac{a}{s} ~|~ a \in A, s \not\in P \}$$
2371
2372ie. "make every element of $A$ not in $P$ invertible". Since $p \in P$, then $\frac{1}{p} \not\in A_P$.\\
2373
2374Given the ideal $I \subseteq A$, its extension to the localized ring is
2375$$I A_P = \{ \frac{a}{s} ~|~ a \in I, s \not\in P \}$$
2376
2377\vspace{0.6cm}
2378
2379The only nonunits are fractions whose numerator lies in $P$: $\{ \frac{a}{s} ~|~ a \in P, s \not\in P \} = P A_P$
2380
2381Hence, $A_P$ has one maximal ideal: $P A_P$.
2382
2383$(A_P, P A_P)$ denotes the local ring $A_P$ together with $P A_P$, often called the localization of $A$ at $P$ as a local ring.
2384
2385\vspace{1.0cm}
2386
2387
2388
2389\section{Tensor Product of Modules}
2390
2391\subsection{Constructing the Tensor Product}
2392We use vector spaces but the same applies for modules.
2393
2394Let $Z$ be the vector space with basis $\{ e_{v,w} \} ~\forall~ (v,w) \in V \times W$.
2395
2396Want to introduce the relation $a = b \in Z$. Consider the element $a-b \in Z$; take the span $Span\{a-b\}$, that is, all of multiples of $a-b$.
2397
2398Consider the quotient space $\frac{Z}{Span\{a-b\}}$.
2399Anything in the span of $a-b$ is equal to zero.
2400
2401$\Longrightarrow$ This is how we use relations to get vector spaces out of $Z$.
2402
2403\paragraph{Direct Sum, $V \oplus W$}.
2404
2405Want
2406\begin{enumerate}[i.]
2407  \item $(v_1, w_1) + (v_2, w_2) = (v_1 + v_2, w_1 + w_2)$
2408  \item $\lambda (v, w) = (\lambda v, \lambda w)$
2409\end{enumerate}
2410
2411Take the quotient
2412$$Z / Span
2413\left\{
2414\substack{
2415(v_1,w_1)+(v_2,w_2)-(v_1+v_2,w_1+w_2)\\
2416\lambda(v,w)-(\lambda v,\lambda w)
2417}
2418\right\}
2419$$
2420
2421\paragraph{Tensor Product, $V \otimes W$}
2422relations:
2423\begin{enumerate}[i.]
2424  \item distribuitivity law:
2425    $$a (b+c) = ab + ac$$
2426    $$(a+b)c =  ac + bc$$
2427  \item scalar multiplication:
2428    $$\lambda(a \cdot b) = (\lambda a) b = a (\lambda b)$$
2429\end{enumerate}
2430
2431$\Longrightarrow$
2432  \begin{align*}
2433    &a (b+c) = ab + ac\\
2434    &(a+b)c =  ac + bc\\
2435    &\lambda (a \cdot b) = (\lambda a) b\\
2436    &\lambda (a \cdot b)= a (\lambda b)
2437  \end{align*}
2438
2439  then we can define the tensor product as
2440$$V \otimes W = Z / Span
2441\left\{
2442\substack{
2443    a \otimes (b+c) - a \otimes b - a \otimes c\\
2444    (a+b) \otimes c -  a \otimes c - b \otimes c\\
2445    \lambda (a \otimes b) - (\lambda a) \otimes b\\
2446    \lambda (a \otimes b) - a \otimes (\lambda b)
2447}
2448\right\}
2449$$
2450
2451
2452
2453\subsection{Universal Property of Tensor Products}
2454\begin{prop}{}[Universal Property of Tensor Products]
2455  For every bilinear map $g: V \times W \longrightarrow U$, $\exists !$ linear map $f: T \longrightarrow U$ such that $f \circ \tau = g$.
2456
2457  \begin{tikzpicture}[node distance=2.5cm, auto]
2458    \node (VW)    {$V \times W$};
2459    \node (T) [right of=VW] {$T = V \otimes W$};
2460    \node (U) [below of=VW] {$U$};
2461
2462    \draw[->] (VW) to node {$\tau$} (T);
2463    \draw[->] (VW) to node [swap] {$g$} (U);
2464    \draw[->, dashed] (T) to node [swap] {$f$} (U);
2465  \end{tikzpicture}
2466\end{prop}
2467
2468\begin{proof}
2469  \begin{enumerate}[i.]
2470    \item uniqueness:\\
2471
2472    suppose we have two different T.P., $(T, \tau)$ and $(T', \tau')$. Want to prove $T \stackrel{iso}{\cong} T'$.
2473
2474
2475
2476    On the right is the same relation as on the left, but switching $T$ by $T'$:
2477
2478  \begin{minipage}[t]{0.48\textwidth}
2479    \centering
2480    \begin{tikzpicture}[node distance=2.0cm, auto]
2481      \node (VW)    {$V \times W$};
2482      \node (T) [right of=VW] {$T$};
2483      \node (T2) [below of=VW] {$T'$};
2484
2485      \draw[->] (VW) to node {$\tau$} (T);
2486      \draw[->] (VW) to node [swap] {$\tau'$} (T2);
2487      \draw[->, dashed] (T) to node [swap] {$f$} (T2);
2488    \end{tikzpicture}
2489
2490    \vspace{0.5em}
2491
2492    by the U.P., $f \circ \tau = \tau'$.
2493  \end{minipage}
2494  \hfill
2495  \begin{minipage}[t]{0.48\textwidth}
2496    \centering
2497    \begin{tikzpicture}[node distance=2.0cm, auto]
2498      \node (VW)    {$V \times W$};
2499      \node (T) [right of=VW] {$T$};
2500      \node (T2) [below of=VW] {$T'$};
2501
2502      \draw[->] (VW) to node {$\tau$} (T);
2503      \draw[->] (VW) to node [swap] {$\tau'$} (T2);
2504      \draw[->, dashed] (T) to node [swap] {$f$} (T2);
2505    \end{tikzpicture}
2506
2507    \vspace{0.5em}
2508
2509    by the U.P., $f \circ \tau = \tau'$.
2510  \end{minipage}
2511
2512  \vspace{0.3cm}
2513  $$\text{thus,}~~ f \circ f' \circ \tau' = \tau'$$
2514
2515  Want to show that $f$ has an inverse, so that it is an isomorphism $T \stackrel{iso}{\cong} T'$.
2516
2517  we just saw that
2518  $$f \circ f' \circ \tau' = \tau'$$
2519  also we can get it the other way
2520  $$f' \circ f \circ \tau = \tau$$
2521
2522  we can see that $f \circ f'$ seems to be the identity, need more info to prove it.
2523
2524  What if we apply the U.P. to the same module on both sides?
2525
2526    \begin{tikzpicture}[node distance=2.0cm, auto]
2527      \node (VW)    {$V \times W$};
2528      \node (T) [right of=VW] {$T$};
2529      \node (T2) [below of=VW] {$T$};
2530
2531      \draw[->] (VW) to node {$\tau$} (T);
2532      \draw[->] (VW) to node [swap] {$\tau$} (T2);
2533      \draw[->, dashed] (T) to node {$? = id_T$} (T2);
2534    \end{tikzpicture}
2535
2536    thus, $id_T \circ \tau = \tau$, hence
2537
2538    \begin{tikzpicture}[node distance=2.0cm, auto]
2539      \node (VW)    {$V \times W$};
2540      \node (T) [right of=VW] {$T$};
2541      \node (T2) [below of=VW] {$T$};
2542
2543      \draw[->] (VW) to node {$\tau$} (T);
2544      \draw[->] (VW) to node [swap] {$\tau$} (T2);
2545      \draw[->, dashed] (T) to node {$id_T,~ f' \circ f$} (T2);
2546    \end{tikzpicture}
2547
2548    and by uniqueness of the U.P., we can not have two different functions ($id_T$ and $f' \circ f$) and must be that $f' \circ f = id_T$.
2549
2550    Similarly for $f \circ f' = id_{T'}$.
2551
2552    Therefore, $f'$ is the inverse of $f$, and $T \stackrel{iso}{\cong} T'$, which means that the tensor prodduct is unique.
2553
2554    % TODO (ii.) existence
2555  \end{enumerate}
2556
2557
2558  \vspace{1.0cm}
2559  This is stated in the Atiyah and MacDonald book (\cite{am}) for multiple nodes and multilinear maps:
2560
2561  \begin{prop}{2.12}
2562    Let $M_1, \ldots, M_r$ be $A$-modules. Then $\exists ~ (T, g)$ with $T$ an $A$-module, and $g$ a multilinear map $g: M_1 \times \ldots \times M_r \longrightarrow T$, with the following property:
2563
2564    Given any $A$-module $P$ and any $A$-multilinear map $f: M_1 \times \ldots \times M_r \longrightarrow T$, $\exists !$ an $A$-homomorphism $f': T \longrightarrow P$ such that $f' \circ g = f$.
2565
2566    Moreover, if $(T,g)$ and $(T',g')$ are two pairs with this property, then $\exists !~ j: T \stackrel{iso}{\longrightarrow} T'$ such that $j \circ g = g'$.
2567  \end{prop}
2568
2569  \subsection{Usefulness of Tensor Product's Universal Property}
2570  Anytime we define a function on the T.P. we need to know that it is well defined and linear. This is hard in general because of all the relations of the elements of the T.P..
2571
2572  But using the U.P. of T.P. we can remove all of the extra work and just look at a bilinear map on the cartesian product, which is a much easier object to work with.
2573
2574  \paragraph{Using the U.P. to show that a function on the T.P. is well defined and linear:}
2575  \begin{enumerate}[i.]
2576    \item define a map $\phi$ on the Cartesian product
2577      $$(x,y) \in M \times N$$
2578      
2579    \item prove bilinearity of $\phi$:\\
2580      show that $\phi$ is $A$-linear in each variable separately.
2581
2582      ie. check
2583      \begin{align*}
2584        \phi(x + x', y) &= \phi(x, y) + \phi(x', y)\\
2585        \phi(x, y + y') &= \phi(x, y) + \phi(x, y')\\
2586        \phi(ax, y) &= a \phi(x, y) = \phi(x, ay)
2587      \end{align*}
2588
2589    \item invoke the Universal Property.\\\
2590
2591      Once bilinearity is proven on $\phi$, the U.P. guarantees the existence of a unique map $f$ on the tensor product $M \otimes N$
2592
2593    \item automatic well-definedness and linearity: $f$ is linear by definition.
2594
2595    And since it is a single map originating from the bilinear map $\phi$ on the product, it is automatically well-defined on the tensor product (ie. automatically "respects" the relations that define tensors).
2596  \end{enumerate}
2597
2598
2599\end{proof}
2600
2601
2602
2603%%%%%%%%%%%%%%%%%%%%%%%%
2604%%%%%%
2605%%%%%% end of notes,
2606%%%%%% beginning of exercises
2607%%%%%%
2608%%%%%%%%%%%%%%%%%%%%%%%%
2609
2610\newpage
2611
2612\section{Exercises}
2613
2614For the exercises, I follow the assignments listed at \cite{mit-course}.
2615
2616The exercises that start with \textbf{R} are the ones from the book \cite{reid}, and the ones starting with \textbf{AM} are the ones from the book \cite{am}.
2617
2618\subsection{Exercises Chapter 1}
2619
2620\begin{ex}{R.1.1}
2621  Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$?
2622\end{ex}
2623\begin{proof}
2624  Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$.
2625
2626  $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\
2627  Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$.
2628
2629  Thus $I \cup J$ is not closed under addition; thus is not an ideal.
2630
2631
2632  Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum:
2633
2634  $$I+J = \{ a+b | a \in I, b \in J \}$$
2635
2636  $gcd(2,3)=1$, so $I+J = \mathbb{Z}$.
2637
2638  Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$.
2639\end{proof}
2640
2641\begin{ex}{R.1.5}
2642  let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\
2643  In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$.
2644\end{ex}
2645\begin{proof}
2646  (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$).
2647
2648  Let
2649  $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
2650  The claim is that $\psi^{-1}(P)$ is prime ideal of $A$.
2651
2652  \begin{enumerate}[i.]
2653    \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
2654      $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$).
2655
2656      If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so
2657      $$\psi(a-b)= \psi(a) - \psi(b) \in P$$
2658      hence $a-b \in \psi^{-1}(P)$.
2659
2660      If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\
2661      Thus $ra \in \psi^{-1}(P)$.
2662
2663      $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$.
2664
2665    \item show that $\psi^{-1}(P)$ is prime:\\
2666      $\psi^{-1}(P) \neq A$, since if $\psi^{-1}(P)=A$, then $1_A \in \psi^{-1}(P)$, so $\psi(1_A)=1_B \in P$, which would mean that $P=B$, a contradiction since $P$ is prime ideal of $B$.
2667
2668      Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$.
2669
2670      Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\
2671      Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$.
2672
2673      Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$.
2674  \end{enumerate}
2675\end{proof}
2676
2677
2678\begin{ex}{R.1.6}
2679  prove or give a counter example:
2680  \begin{enumerate}[a.]
2681    \item the intersection of two prime ideals is prime
2682    \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
2683    \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
2684    \item the map $\psi^{-1}$ of Proposition 1.2 takes maximal ideals of $A/I$
2685      to maximal ideals of $A$
2686  \end{enumerate}
2687\end{ex}
2688\begin{proof}
2689  \begin{enumerate}[a.]
2690    \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime.
2691
2692      Then $I \cap J = (2) \cap (3) = (6)$.
2693
2694      The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$.
2695
2696      Thus the intersection of two primes can not be prime.
2697
2698    \item $P_1=(2),~ P_2=(3)$, both prime.
2699
2700      Then,
2701      $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$
2702
2703      $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$.
2704
2705      (recall: principal= generated by a single element)
2706
2707      So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$.
2708
2709      The whole ring is not a prime ideal (by the definition of the prime ideal), so $P_1+P_2$ is not a prime ideal.
2710
2711      Henceforth, the sum of two prime ideals is not necessarily prime.
2712
2713    \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$.
2714
2715      Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$.
2716
2717      Then
2718      \begin{align*}
2719        \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\
2720        \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A
2721      \end{align*}
2722
2723      But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
2724
2725      Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
2726
2727    \item $\psi: A \longrightarrow A/I$ quotient homomorphism, $I \subseteq A$ an ideal.
2728
2729      Let $M$ a maximal ideal of $A/I$, then $\frac{(A/I)}{M}$ is a field (Proposition 1.3).
2730
2731      By the isomorphism theorems,
2732      $$\frac{(A/I)}{M} \cong \frac{A}{\psi^{-1}(M)}$$
2733
2734
2735      Since $\frac{(A/I)}{M}$ is a field, the quotient $\frac{A}{\psi^{-1}(M)}$ is a field, so $\psi^{-1}(M)$ is a maximal ideal of $A$.
2736
2737      $\Longrightarrow~$ under $\psi$, preimages of maximal ideals are maximal.
2738  \end{enumerate}
2739\end{proof}
2740
2741\begin{ex}{R.1.12.a}
2742  if $I,J$ ideals and $P$ prime ideal, prove that
2743  $$IJ \subset P ~\Longleftrightarrow~ I \cap J \subset P ~\Longleftrightarrow~ I ~\text{or}~ J \subset P$$
2744\end{ex}
2745\begin{proof}
2746  assume $I \subseteq P$ (for $J \subseteq P$ will be the same, symmetric), take $x \in IJ$,
2747
2748  then
2749  $$x = \sum_{k=1}^n a_k b_k$$
2750  with $a_k \in I,~ b_k \in J$.
2751
2752  Each $a_k \in I \subseteq P$. Since $P$ an ideal,
2753  $$\sum_{k=1}^n a_k b_k \in P$$
2754
2755  thus $x \in P$, hence $IJ \subseteq P$.
2756
2757  So $I \subseteq P$ or $J \subseteq P$ $~\Longrightarrow IJ \subseteq P$.
2758
2759  \vspace{0.5cm}
2760  Conversely,\\
2761  assume $P$ prime and $IJ \subseteq P$.
2762
2763  Suppose by contradiction that $I \not\subseteq P$ and $J \not\subseteq P$.
2764  \begin{itemize}
2765    \item[-] since $I \not\subseteq P,~ \exists a \in I$ with $a \not\in P$
2766    \item[-] since $J \not\subseteq P,~ \exists b \in J$ with $b \not\in P$
2767  \end{itemize}
2768  Since $a \in I,~ b \in J,~~ ab \in IJ \subseteq P$, but $P$ is prime, so $ab \in P$ implies that $a \in P$ or $b \in P$. This contradicts $a,b$ being taken outside of $P$.
2769
2770  Thus $I \not\subseteq P$ and $J \not\subseteq P$ are false.
2771
2772  \vspace{0.3cm}
2773  So both directions are proven, hence
2774  $$IJ \subseteq P ~\Longrightarrow~ I \subseteq P ~\text{or}~ J \subseteq P$$
2775
2776\end{proof}
2777
2778\begin{ex}{R.1.18}
2779  Use Zorn's lemma to prove that any prime ideal $P$ contains a minimal prime ideal.
2780\end{ex}
2781\begin{proof}
2782  Let $P$ prime ideal of $R$.
2783
2784  $$S = \{ Q \subseteq R ~|~ Q ~\text{a prime ideal AND}~ Q \subseteq P \}$$
2785
2786  Goal: show that $S$ has a minimal element, the minimal ideal contained in $P$.
2787
2788  $P \subset S$, so $S$ is nonempty.
2789
2790  Let $C \subseteq S$ be a chain (= totally ordered subset) with respect to inclusion.
2791
2792  Define
2793  $$Q_C = \bigcap_{Q \in C} Q$$
2794
2795  Clearly $Q_C \subseteq P$, since each $Q \in C$ is $Q \subseteq P$.
2796
2797  Since $C$ is ordered by inclusion, it is a decreasing chain of prime ideals.
2798
2799  Intersection of a decreasing chain of prime ideals is again a prime ideal:
2800  \begin{itemize}
2801    \item[-] if $ab \in Q_C$, then $ab \in Q ~\forall Q \in C$
2802    \item[-] since $Q$ prime, $\forall Q \in C$ either $a \in Q$ or $b \in Q$
2803  \end{itemize}
2804
2805  If there were some $Q_1,~ Q_2 \in C$ with $a \in Q_1$ and $b \not\in Q_2$, then by total ordering, either $Q_1 \subseteq Q_2$ or $Q_2 \subseteq Q_1$.
2806
2807  In either case: contradiction, since the smaller one would have to contain the element that was assumed to be excluded.
2808
2809  Thus $\forall Q \in C$ the same element $a, b$ must lie in all $Q$. $\Longrightarrow~$ lies in the intersection of them, $Q_C$.
2810
2811  Henceforth, $Q_C$ is a prime ideal and lies in $S$, and its a lower bound of $C$ in $S$.
2812
2813  Now, $S$ is nonempty, and every chain in $S$ has a lower bound in $S$ (its intersection).\\
2814  Therefore, $S$ has a minimal element $P_{min}$.
2815
2816  By construction, $P_{min}$ is a prime ideal $P_{min} \subseteq P$, and by minimality there are no strictly smaller prime ideals inside $P$.
2817
2818  So $P_{min}$ is a minimal prime ideal, contained in $P$.
2819\end{proof}
2820
2821\begin{ex}{R.1.10}
2822\end{ex}
2823\begin{proof}
2824\end{proof}
2825
2826\begin{ex}{R.1.11}
2827\end{ex}
2828\begin{proof}
2829\end{proof}
2830
2831\begin{ex}{R.1.4}
2832\end{ex}
2833\begin{proof}
2834\end{proof}
2835
2836\subsection{Exercises Chapter 2}
2837
2838\begin{ex}{R.2.9}
2839$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ is a s.e.s. of $A$-modules. Prove that if $N, L$ are finite over $A$, then $M$ is finite over $A$.
2840\end{ex}
2841\begin{proof}
2842  Denote the generators of $L$ and $N$ respectively as
2843  \begin{align*}
2844    \{l_1, \ldots, l_k \} &\subseteq L\\
2845    \{n_1, \ldots, n_p \} &\subseteq N
2846  \end{align*}
2847
2848  By s.e.s. definition, 
2849  \begin{itemize}
2850    \item[-] $\alpha$ is injective (one-to-one), so
2851    $$\forall l_i \in L,~ \exists~ x_i \in M ~\text{s.th.}~ \alpha(l_i)=x_i$$
2852
2853  \item[-] $\beta$ is surjective (onto), so
2854    $$\forall n_j \in N,~ \exists~ y_j \in M ~\text{s.th.}~ \beta(y_j)=n_j$$
2855\end{itemize}
2856
2857  We will show that $\{x_1, \ldots, x_k, y_1, \ldots, y_p \}$ generate $M$, and thus $M$ is finite:
2858
2859  Let $m \in M$, then $\beta(m) \in N$, and
2860  $$\beta(m) = \sum_{j=1}^p a_j n_j ~~~\text{with}~ a_j \in A$$
2861
2862  Take $m' \in M$, with $m' = \sum a_j y_j$, then
2863  $$\beta(m') = \sum a_j \beta(y_j) = \sum a_j n_j = \beta(m)$$
2864
2865  Then, since $\beta(m)=\beta(m')~~ \Longrightarrow~~ \beta(m-m')=0$, thus
2866  $$(m-m') \in ker(\beta)$$
2867
2868  By \emph{exactness} property, since $\alpha: L \longrightarrow ker(\beta)$, we have $ker(\beta)=im(\alpha)$.
2869
2870  Therefore, $\exists~ l \in L$ such that $\alpha(l)= m-m'$.
2871
2872  Since $\{l_i\}_k$ generate $L$,
2873  $$l = \sum^k b_i l_i$$
2874  thus
2875  $$m-m' = \alpha(l) = \alpha(\underbrace{\sum b_i l_i}_{l}) = \sum b_i \underbrace{\alpha(l_i)}_{x_i} = \sum b_i x_i$$
2876
2877  Rearrange,
2878  $$m = m' + \sum b_i x_i = \sum_{j=1}^p a_j y_j + \sum_{i=1}^k b_i x_i ~~~~~ \forall m \in M$$
2879
2880  So, $L$ provides $k$ generators for the kernel part of $M$, $N$ provides $p$ "lifts" for the quotient part of $M$; thus $M$ is generated by $k+p$ elements.\\
2881  Thus $M$ is finitely generated over $A$.
2882\end{proof}
2883
2884
2885\subsection{Exercises Chapter 3}
2886
2887\begin{ex}{R.3.2}
2888  $K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space.
2889   Prove that $A$ is Noetherian and Artinian.
2890\end{ex}
2891\begin{proof}
2892  $dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in \aA$ and $c \in K$, then $c \cdot x \in \aA$.
2893
2894  \begin{enumerate}
2895    \item Noetherian:\\
2896      let $I_1 \subseteq I_2 \subseteq \ldots$ be an ascending chain of ideals in $A$.
2897
2898      Since each $I_i$ is a subspace, we have
2899      $$dim_K(I_1) \leq dim_K(I_2) \leq \ldots \leq n$$
2900      where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Noetherian.
2901      
2902    \item Artinian:\\
2903      Similarly, if $I_1 \supseteq I_2 \supseteq \ldots$ a descending chain of ideals in $A$.
2904
2905      then
2906      $$n \geq dim_K(I_1) \geq dim_K(I_2) \geq \ldots \geq 0$$
2907      where at some $i=m$ we have $dim_K(I_m)=dim_K(I_{m+1})$; then since $I_m \subseteq I_{m+1}$, we have $I_m = I_{m+1}$. So $A$ is Artinian.
2908  \end{enumerate}
2909\end{proof}
2910
2911\begin{ex}{R.3.5}
2912  Let $0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ an exact sequence. Let $M_1, M_2 \subseteq M$ be submodules of $M$.
2913
2914  Prove if the following holds or not:
2915  $$\beta(M_1)=\beta(M_2) ~\text{and}~ \alpha^{-1}(M_1)=\alpha^{-1}(M_2)~~\Longrightarrow~ M_1=M_2$$
2916\end{ex}
2917\begin{proof}
2918  Counterexample showing that it does not hold:
2919
2920  Let $K$ a field, $M = K \oplus K~, L=K,~N=K$.
2921
2922  Set, for $l \in L,~ (m_1, m_2) \in M$,
2923  \begin{align*}
2924    \alpha:~ &l \longmapsto (l, 0)\\
2925    \beta:~ &(m_1, m_2) \longmapsto m_2
2926  \end{align*}
2927
2928  So we have
2929  $$0 \longrightarrow K \stackrel{\alpha}{\longrightarrow} K^2 \stackrel{\beta}{\longrightarrow} K \longrightarrow 0$$
2930
2931  Then,
2932  \begin{align*}
2933    M_1 &= \{ (x, x) ~|~ x \in K\} ~~~~\sim\text{(diagonal line)} \\
2934    M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)}
2935  \end{align*}
2936
2937  (Geometric interpretation: $M_1,~ M_2$ are the \emph{diagonal line} and \emph{y-axis} respectively; and $\alpha,~\beta$ capture information about the \emph{vertical} components (x-axis, y-axis respectively), but not about the \emph{diagonal} way a submodule is embedded in $M$).
2938
2939  Then,
2940  \begin{align*}
2941    \beta(M_1) &= \{ x ~|~ x \in K\} = K \\
2942    \beta(M_2) &= \{ x ~|~ x \in K\} = K
2943  \end{align*}
2944  thus, $\beta(M_1) = \beta(M_2)$.
2945
2946  For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1) = \{0\}$,\\
2947  for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2) = \{0\}$,\\
2948  thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$.
2949
2950
2951  So we've seen that
2952  \begin{align*}
2953    \beta(M_1) = \beta(M_2)\\
2954    \alpha^{-1}(M_1)=\alpha^{-1}(M_2)
2955  \end{align*}
2956  while having $M_1 \neq M_2$.
2957\end{proof}
2958
2959
2960
2961\begin{ex}{R.3.3}
2962Let $A$ a ring, $I_1, \ldots, I_k$ ideals such that each $A/I_i$ is a Noetherian ring.
2963Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bigcap I_i = 0$ then $A$ is also Noetherian.
2964\end{ex}
2965\begin{proof}
2966\begin{enumerate}[i.]
2967  \item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian.
2968    $\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian.
2969  \item Take the canonical homomorphism
2970    $$\phi: A \longrightarrow \bigoplus_{i=1}^n A/ I_i$$
2971    by $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$.
2972
2973    $\phi$ is injective: $ker(\phi)= \{ a \in A | a \in I_i \forall i \}$.
2974
2975    Since we're given $\cap I_i = 0$, then $ker(\phi)=\cap I_i$, and $\phi$ is injective.
2976
2977    Thus, $\phi$ is the isomorphism $A \cong im(\phi)$, where $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$.
2978
2979    We know that any submodule of a Noetherian module is Noetherian, thus, since
2980    \begin{itemize}
2981      \item $A/I_i$ is Noetherian by hypothesis of the exercise
2982      \item $A \cong im(\phi)$
2983      \item $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$
2984  \end{itemize}
2985  then, $A$ is Noetherian.
2986\end{enumerate}
2987\end{proof}
2988
2989\begin{ex}{R.3.4}
2990  Prove that if A is a Noetherian ring and M a finite A-module, then there
2991  exists an exact sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$.
2992  That is, M has a presentation as an A-module in terms of finitely many generators and relations.
2993\end{ex}
2994\begin{proof}
2995  since $M$ fingen $~\Longrightarrow~$ generators $\{m_1, \ldots, m_2 \} \subseteq M$ span $M$.
2996
2997  Let $\beta$ be a surjective $A$-linear map, which forms a free $A$-module of rank $p$ onto $M$:
2998  \begin{align*}
2999    \beta: A^p &\longrightarrow M\\
3000    (a_1, \ldots, a_p) &\longmapsto \sum_{i=1}^p a_i m_i
3001  \end{align*}
3002
3003  Let $K=ker(\beta)$. By the 1st Isomorphism Theorem,
3004  $$M \cong A^p / K$$
3005
3006  Since $A$ is a Noetherian ring, then every free $A$-module of finite rank (eg. $A^p$) is a Noetherian module.
3007
3008  Every submodule of a Noetherian module is fingen.
3009
3010  $\Longrightarrow~$ since $K \subseteq A^p, ~\Longrightarrow~ K ~~(=ker(\beta))$ is fingen.
3011
3012  Since $K$ fingen, let $\{k_1, \ldots, l_q\}$ be generators of $K$.
3013
3014  Define $\psi: A^q \longrightarrow K$.
3015
3016  Compose it with the inclusion map $i: K \longrightarrow A^p$,
3017  $$\alpha = i \circ \psi:~ A^q \longrightarrow A^p$$
3018
3019  So we have the whole sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$, where
3020  \begin{itemize}
3021    \item $\beta$ is surjective
3022    \item $im(\alpha)=ker(\beta)$
3023  \end{itemize}
3024  so that it is a exact sequence, thus, $M$ has a finite presentation.
3025\end{proof}
3026
3027\subsection{Exercises Chapter 4}
3028
3029\begin{ex}{R.4.1.a}
3030  $k[X^2] \subset k[X]$ is a finite extension, hence integral. Find the integral dependence relation for any $f \in k[X]$.
3031\end{ex}
3032\begin{proof}
3033  $\forall f(X) \in k[X]$ can be uniquely decomposed into its even and odd parts:
3034  $$f(X) = p(X^2) + X \cdot q(X^2)$$
3035
3036  with $p(X^2),~ q(X^2) \in k[X^2]$, and\\
3037  $p(X^2)$: sum of all terms with even exponents\\
3038  $q(X^2)$: sum of all terms with odd exponents, and then factoring out $X$.
3039
3040  \vspace{0.3cm}
3041  (Observation: this is used in FRI cryptographic protocol\\
3042  \href{https://github.com/arnaucube/math/blob/master/notes_fri_stir.pdf}{https://github.com/arnaucube/math/blob/master/notes\_fri\_stir.pdf})
3043
3044
3045  Rearrange it
3046  \begin{align*}
3047    f(X) &- p(X^2) = X \cdot q(X^2), ~~\text{square:}\\
3048    (f(X) &- p(X^2))^2 = X^2 \cdot q(X^2)^2\\
3049    f(X)^2-2 p(X^2) f(X) &+ p(X^2)^2 = X^2 \cdot q(X^2)^2\\
3050    f(X)^2 \underbrace{-[2 p(X^2)]}_{a_1} f(X) &+ \underbrace{[p(X^2)^2 - X^2 \cdot q(X^2)^2]}_{a_0} = 0
3051  \end{align*}
3052
3053   Denote the last polynomial as $P(T) \in k[X^2]$, where $f(X)$ is a root of $P(T)$.
3054
3055   \vspace{0.2cm}
3056   The integral dependence relation for any $f \in k[X]$ is given by the monic polynomial from \ref{R.4.1}.ii, in this case $T^2 + a_1 T + a_0 = 0$ with $a_i \in k[X^2]$.
3057
3058   We have that
3059   \begin{align*}
3060     a_1 &= -2 p(X^2)\\
3061     a_0 &= p(X^2)^2 - X^2 q(X^2)^2
3062   \end{align*}
3063
3064
3065   So for example, for $f(X)= X^3 + X^2 + X + 1$:
3066   \begin{align*}
3067     f(X) = (X^2 + 1) &+ X (X^2 + 1)\\
3068     (f(X) - (X^2+1))^2 &= X^2 (X^2 + 1)^2\\
3069     (f(X) - p(X))^2 &= X^2 (q(X))^2
3070   \end{align*}
3071\end{proof}
3072
3073
3074\begin{ex}{R.4.5}
3075  Let $A=k[X,Y]/(Y^2 - X^2 - X^3)$. Prove that the normalization of $A$ is $k[t]$ where $t=Y/X$.
3076\end{ex}
3077\begin{proof}
3078$A=k[X,Y]/(Y^2 - X^2 - X^3)$, express $X$ and $Y$ in terms of $t$:
3079
3080Since $t=Y/X$ then $Y=tX$, and combined with $Y^2 = X^2 + X^3$, then
3081\begin{align*}
3082  (tX)^2 &= X^2 + X^3\\
3083  t^2 X^2 &= X^2 + X^3, ~~\text{assuming}~X \neq 0:\\
3084  t^2 &= 1+X, ~\text{thus}\\
3085  X&=t^2-1 \in k[X]
3086\end{align*}
3087
3088Then, $Y=tX=t(t^2-1)=t^3-t \in k[X]$.
3089
3090Hence $X, Y \in k[X]$.
3091
3092Therefore, $k[X,Y]/(Y^2 - X^2 - X^3) \subseteq k[t]$.
3093
3094\vspace{0.4cm}
3095By \ref{noether-normalization} (Noether normalization lemma), to show that $k[t]$ is the \emph{normalization}, must show that $k[t]$ is \emph{integral} over $A$.
3096
3097From $X=t^2-1 ~~\Longrightarrow~~ t^2-1-X=0 ~~\Longrightarrow~ t^2 - (1+X) = 0$.
3098
3099$(1+X) \in A$, so $t$ satisfies the monic polynomial
3100$$P(T) = T^2 - (1+X) \in A[T]$$
3101Thus $t$ is integral over $A$.
3102
3103Since $k[t]$ is generated by $t$ over $k$, and $k \subset A$, then the entire ring $k[t]$ is integral over $A$.
3104
3105Since $k[t]$ is a polynomial ring over a field, which is a UFD, it is integrally closed (since all UFD are integrally closed).
3106
3107$Frac A = k(X,Y)$, since $X=t^2-1, ~Y=t^3-t ~~\Longrightarrow~ k(X,Y) \subseteq k(t)$
3108
3109and $t=Y/X \in k(X,Y)$, thus $k(X,Y)=k(t)$.
3110
3111Since $k[t]$ is integrally closed and is the integral closure of $A$ in its fraction field $k(t)$, we conclude that the normalization of $A$ is $k[t]$.
3112\end{proof}
3113
3114
3115
3116\begin{ex}{R.4.9}
3117  $k$ a field, $A= k[X,Y,Z]/(X^2- Y^3 -1, XZ - 1)$, find $\alpha, \beta \in k$ such that $A$ is integral over $B = k[X + \alpha Y + \beta z]$, and write a set of generators of $A$ as a $B$-module.
3118\end{ex}
3119\begin{proof}
3120  (want to find a linear combination of the coordinates such that the original variables satisfy monic polynomials over the new ring $B$)
3121
3122  The relations defining $A$ are
3123  \begin{align*}
3124    X^2-Y^3-1=0 ~~&\Longrightarrow~ Y^3 = X^2 -1 ~~~(*)\\
3125    XZ -1=0 ~~&\Longrightarrow~ Z= 1/X = X^{-1}
3126  \end{align*}
3127
3128Thus $A$ can be denoted as $A = k[X,Y,X^{-1}]/(Y^3 - X^2 - 1)$.
3129
3130
3131Now, $Y$ is inegral over $k[X]$, since $Y^3 - (X^2 - 1) = 0$ is monic in $Y$ with coefficients in $k[X]$.
3132
3133$Z$ is not integral over $k[X]$, since $Z=1/X$ and $X$ is not a unit in $k[X]$.
3134
3135Choose $\alpha, \beta \in k$ such that $X$ (and thus $Z$) becomes integral over $B$:\\
3136set $\alpha=0,~\beta=1 ~~\Longrightarrow~~ B=k[X+\alpha Y + \beta Z]= k[X+Z]$.
3137
3138Let $w=X+Z$; since $XZ=1$, we have
3139$$w=X+\frac{1}{X} ~~\Longrightarrow~~ Xw=X^2+1 ~~\Longrightarrow~~ X^2 -w X +1 = 0 ~~(**)$$
3140
3141which is monic with coefficients in $k[w]$, thus $X$ is integral over $B$.
3142
3143Since $Z=w-X$, $Z$ is also integral.
3144
3145\vspace{0.4cm}
3146Generators of $A$ as a $B$-module:
3147
3148we hadd $B=k[w]$ with $w=X+Z$.
3149
3150From $(**)$ we have $X^2 - wX + 1=0$, so $X^2 = wX -1$.
3151
3152Thus any polynomial in $X$ can be reduced to a linear form $b_1 X + b_0$ with $b_i \in k[w]$. Hence it's partial basis is $\{1, X\}$.
3153
3154Fitting $X^2$ into $(*)$,
3155\begin{align*}
3156  X^2 &- Y^3 -1 =0\\
3157  Y^3 &= X^2-1\\
3158  Y^3 &= wX -2
3159\end{align*}
3160
3161thus any power of $Y$ higher than $2$ can be reduced (eg. $Y^4 = Y(wX-2) = w(XY) - 2Y$).
3162
3163So its partial basis is $\{1, Y, Y^2\}$.
3164
3165For $Z$, since $XZ=1$ and $w=X+Z \Longrightarrow Z=w-X$, thus $Z$ is a $B$-linear combination of $\{1,X\}$.
3166
3167\vspace{0.2cm}
3168Combining the previous partial basis, the generators are
3169$$\{ 1, X \} \times \{ 1,Y, Y^2 \} = \{ 1, Y, Y^2, X, XY, XY^2 \}$$
3170
3171\end{proof}
3172
3173
3174\subsection{Exercises Chapter 6}
3175
3176\begin{ex}{R.6.3.a}
3177  Let $A = A' \times A''$; prove that $A'$ and $A''$ are rings of fractions of $A$.
3178\end{ex}
3179\begin{proof}
3180  ring of fractions of $A$ = $S^{-1}A$, so want to prove that $S^{-1}A \cong A'$.
3181
3182  Localization map
3183  \begin{align*}
3184    \psi: A &\longrightarrow A'\\
3185    (a', a'') &\longmapsto a'
3186  \end{align*}
3187
3188  note that $\psi$ is surjective, since $\forall~ a' \in A',~ \psi(a', 0)=a'$.
3189
3190  Let the multiplicative set $S$ be $S=\{ e_1 \}$ with $e_1 = (1,0)$.
3191
3192  Want $\underbrace{S^{-1}A}_{\text{localization}} \cong A'$.
3193
3194  In $S^{-1}A$, $x \in A$ maps to $0$ iff $s x = 0$ for some $s \in S$.
3195  
3196  Let $x=(a', a'')\in A$; then $sx=0 ~\Longrightarrow~ s \cdot (a', a'') =0$,
3197
3198  with $s \in S$, so $s=(1,0)$, hence
3199  $$(1,0)\cdot(a',a'')=(0,0)$$
3200  $$\Longrightarrow~ (a', 0) = (0, 0)$$
3201  \hspace*{4em} which implies $a' = 0$, so that $(a', a'')=(0, a')$.
3202
3203  $\Longrightarrow$ the elements that become zero in the localization are of the form $(0, a'')$
3204
3205  $$ker(\psi) = \{ (a',a'')\in A ~|~ \psi(a', a'')=0 \} = \{(0, a'') ~|~ a'' \in A'' \}$$
3206
3207  By the 1st isomorphism theorem, 
3208
3209  \begin{tikzpicture}[node distance=1.5cm, auto]
3210    \node (G)    {$A$};
3211    \node (H) [right of=G] {$A'$};
3212    \node (GmodK) [below of=G, xshift=0.75cm] {$\frac{A}{ker(\psi)}$};
3213
3214    \draw[->] (G) to node {$\psi$} (H);
3215    \draw[->] (G) to node [swap] {$\phi$} (GmodK);
3216    \draw[<->] (GmodK) to node [swap] {$\eta$} (H);
3217  \end{tikzpicture}
3218  
3219  $\Longrightarrow~ \frac{A}{ker(\psi)} \cong A'$
3220
3221
3222  \vspace{0.5cm}
3223  Take the localization map
3224  \begin{align*}
3225    \phi: A &\longrightarrow S^{-1}A\\
3226    a &\longmapsto a/1
3227  \end{align*}
3228
3229  $$ker(\phi) = \{ x \in A ~|~ sx=0 ~\text{for some}~ s \in S \}$$
3230
3231  and we've seen that $a'=0$, so $x \in A ~\Rightarrow~ x=(a', a'') = (0, a'')$
3232
3233  $$ker(\phi) = \{ (0, a'') ~|~ a'' \in A'' \}$$
3234
3235  which is the same as $ker(\psi)$; $ker(\psi) = ker(\phi)$.
3236
3237  \vspace{0.2cm}
3238  $\Longrightarrow~~ A'$ and $S^{-1}A$ are both surjective images of $A$ with exact same kernel,
3239  thus $A' \cong S^{-1}A$.
3240\end{proof}
3241
3242\begin{ex}{R.6.4}
3243  \begin{enumerate}[a.]
3244  \item Give an example of a ring $A$ and distinct multiplicative sets $S,~ T$ such that $S^{-1}A = T^{-1}A$.
3245  \item Prove that for fixed $S$, there is a maximal multiplicative set $T$ with this property defined by
3246    $$T = \{ t \in A ~|~ at \in S ~\text{for some}~ a \in A \}$$
3247  \end{enumerate}
3248\end{ex}
3249\begin{proof}
3250  \begin{enumerate}[a.]
3251    \item
3252  (\emph{saturated sets})\\
3253
3254  General rule of saturation:\\
3255  two multiplicative sets $S,~T$ yield the same localization, $S^{-1}A = T^{-1}A$, iff the have the same saturation.
3256
3257  Saturation of $S$: $\hat{S}$, set of all elements in $A$ that divide some element of $S$:
3258  $$\hat{S} = \{ a \in A ~|~ \exists~ b \in A ~\text{s.th.}~ ab \in S \}$$
3259
3260  \vspace{0.4cm}
3261  Example 1: $A=\mathbb{Z}$,\\
3262
3263  $$S = \{ 2^n ~|~ n \geq 0 \} = \{ 1, 2, 4, 8, 16, \ldots \}$$
3264  localization:
3265  $$S^{-1} \mathbb{Z} = \{ \frac{a}{2^n} ~|~ a \in \mathbb{Z},~ n \in \mathbb{N} \}$$
3266
3267  $$T = \{ \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \ldots \}$$
3268
3269  Notice that despite $S \neq T$, we have $S^{-1}\mathbb{Z} = T^{-1} \mathbb{Z}$,\\
3270  since once $n \in \mathbb{Z}$ is invertible, $-n$ is automatically invertible: $(-n)^{-1} = - (n^{-1})$.
3271
3272  \vspace{0.4cm}
3273  Example 2: $A= k[x]$,
3274  $$S = \{ x^n ~|~ n \geq 0 \}$$
3275  $$T = \{ (2x)^n ~|~ n \geq 0 \}$$
3276  $$\Longrightarrow~ S^{-1}A = T^{-1} A = k[x, x^{-1}]$$
3277
3278  \item
3279    Show that $T$ is a multiplicative set:
3280
3281    (must contain $1$ and be closed under multiplication)\\
3282    Since $1 \in A$ and $1 \cdot 1 = 1 \in S$, then $1 \in T$.
3283
3284    Let $t_1,~ t_2 \in T$; by definition of $T$,
3285    $$\exists~ a_1, a_2 \in A ~\text{such that}~ a_1 t_1,~ a_2 t_2 \in S$$
3286
3287    Since $S$ a multiplicative set, $(a_1 t_1)(a_2 t_2) \in S$\\
3288    rearrange it: $(a_1 a_2)(t_1 t_2) \in S$.\\
3289    Since $a_1 a_2 \in A,~~ t_1 t_2 \in T$.
3290
3291    Thus $T$ is a multiplicative set.
3292
3293    Next, we show that $T$ is maximal:
3294
3295    Suppose $U$ is the maximal instead of $T$, such that $U^{-1}A \cong S^{-1}A$.
3296
3297    Let $u \in U$; the image of $u$ in $S^{-1}A$ must be a unit.
3298
3299    Units in $S^{-1}A$ are the fractions $\frac{s}{a}$ such that their inverse exists.\\
3300    \hspace*{2em} $\exists~ x=\frac{a}{s}$ such that $u \frac{a}{s} = \frac{1}{1}$\\
3301    \hspace*{2em} $\Longrightarrow~ \frac{ua}{s} = \frac{1}{1} ~\Longrightarrow~ \exists s' \in S$ such that $s'(ua-s)=0$\\
3302    $\Longrightarrow~ s'ua=s's \in S$
3303
3304    since $s,s' \in S$, their product is also in $S$.
3305
3306    Let $b=s'a$, then $ub \in S$.
3307
3308    By definition of $T$, $u \in T$.
3309
3310    Therefore, $U \subseteq T$; thus $T$ is maximal.
3311\end{enumerate}
3312\end{proof}
3313
3314
3315\bibliographystyle{unsrt}
3316\bibliography{commutative-algebra-notes.bib}
3317
3318\end{document}