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{Rings of fractions $S^{-1}A$}
2064
2065\begin{defn}{6.1}[ring of fractions] \label{def.6.1}
2066  let $A$ a ring, $S \subset A$ a multiplicative set ($1 \in S$, and $st \in S
2067  ~\forall s, t \in S$).
2068
2069  Introduce th following relation $\sim$ on $A \times S$:
2070  $$(a, s) \sim (b, t) \Longleftrightarrow \exists y \in S ~\text{such that}~ u(at - bs) = 0$$
2071
2072  (write $a/s$ for the equivalence class of $(a, s)$.
2073
2074  Then, the \emph{ring of fractions of $A$ with respect to $S$} is
2075  $$S^{-1}A = (A \times S)/\sim$$
2076
2077  with ring op'ns defined by the usual arithmetic op'ns on fractions:
2078  $$\frac{a}{s} \pm \frac{b}{t} = \frac{(at \pm bs)}{st} ~~\text{and}~~ \frac{a}{s} \cdot \frac{b}{t} = \frac{ab}{st}$$
2079\end{defn}
2080
2081\begin{prop}{6.1} \label{prop.6.1}
2082  \begin{enumerate}[i.]
2083    \item $\sim$ is an equivalence relation
2084    \item the ring op'ns are well defined, and $S^{-1}A$ is a ring
2085    \item
2086      \begin{align*}
2087        \psi: A &\longrightarrow S^{-1}A\\
2088              a &\longmapsto a/1
2089      \end{align*}
2090      is a ring homomorphism.
2091  \end{enumerate}
2092\end{prop}
2093
2094\begin{eg}{ }
2095  TODO
2096\end{eg}
2097
2098\begin{lemma}{6.2} \label{6.2}
2099  For $f \in A$, write $S = \{ 1, f, f^2, \ldots \}$, and $A_f \cong S^{-1}A$.
2100
2101  Then
2102  $$A_f \cong \frac{A[X]}{(Xf-1)}$$
2103\end{lemma}
2104\begin{proof}
2105  Define the homomorphism
2106  \begin{align*}
2107    \psi: A[X] &\longrightarrow A_f\\
2108    a &\longmapsto a/1\\
2109    X &\longmapsto 1/f
2110  \end{align*}
2111
2112By the 1st isomorphism theorem:
2113
2114  \begin{tikzpicture}[node distance=1.5cm, auto]
2115    \node (G)    {$A[X]$};
2116    \node (H) [right of=G] {$A_f$};
2117    \node (GmodK) [below of=G, xshift=0.75cm] {$A[X]/\ker(\psi)$};
2118
2119    \draw[->] (G) to node {$\psi$} (H);
2120    \draw[->] (G) to node [swap] {$\phi$} (GmodK);
2121    \draw[->] (GmodK) to node [swap] {$\eta$} (H);
2122  \end{tikzpicture}
2123
2124  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.
2125
2126  \vspace{0.3cm}
2127  First, observe that $\psi(Xf-1) = \psi(X)\psi(f)-\psi(1) = \frac{1}{f} \frac{f}{1} - 1 = 1-1=0$,
2128  so $Xf-1 \in ker(\psi)$, ie. $(Xf-1) \subseteq ker(\psi)$.
2129
2130  Now, we want to prove that $ker(\psi) \subseteq (Xf-1)$.
2131
2132  Take $h \in ker(\psi)$, will prove that $h \in (Xf-1)~ \forall~ h \in ker(\psi)$, and thus $ker(\psi) \subseteq(Xf-1)$.
2133
2134  \vspace{0.3cm}
2135  Want to prove that $h(X)$ is a multiple of $(Xf-1)$.
2136
2137  Let
2138  $$h(X)=a_n X^n + a_{n-1} X^{n-1} + \ldots + a_1 X + a_0$$
2139
2140  % Since $h \in ker(\psi) ~~\Longrightarrow~~ \psi(h)=0$, so
2141  % $$\psi(h) = \frac{a_n}{f^n} + \frac{a_{n-1}}{f^{n-1}} + \ldots \frac{a_1}{f} + \frac{a_0}{1} = 0 \in A_f$$
2142
2143  multiply $h(X)$ by $f^n$:
2144  $$f^n \cdot h(X)=a_n (f^n X^n) + a_{n-1} f (f^{n-1} X^{n-1}) + a_{n-2} f^2 (f^{n-2} X^{n-2})  \ldots$$
2145
2146  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 $f^i X^i \equiv 1 \pmod{Xf-1}$.
2147
2148  So,
2149  $$f^n \cdot h(X)=\underbrace{a_n (1) + a_{n-1} f (1) + a_{n-2} f^2 (1) + \ldots + a_0 f^n}_{C~~\text{(constant)}} \pmod{Xf-1}$$
2150
2151  $$\Longrightarrow~~ f^n \cdot h(X)=C \pmod{Xf-1}$$
2152  $$\Longleftrightarrow~~ f^n \cdot h(X)=Q(X) \cdot (Xf-1) + C$$
2153
2154  Want to remove $C$, but it is non-zero. Note that in $A_f$ (ring of
2155  fractions), $\frac{a}{f^n} = 0$ iff $\exists~ k$ such that $f^k \cdot a = 0$ in $A$.
2156
2157  So, multiply both sides by $f^k$:
2158  $$f^k \cdot f^n \cdot h(X)=f^k \cdot (Q(X) \cdot (Xf-1) + C)$$
2159  $$\underbrace{f^k f^n}_{f^{nk}} \cdot h(X)=\underbrace{f^k Q(X)}_{Q'(X)} \cdot (Xf-1) + \underbrace{f^k C}_{0}$$
2160
2161  $$\Longrightarrow~~ f^{n+k} \cdot h(X)=Q'(X) \cdot (Xf-1)$$
2162  $$\Longleftrightarrow~~ f^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2163
2164  multiply it by $X^{n+k}$:
2165  $$X^{n+k} \cdot f^{n+k} \cdot h(X) \equiv X^{n+k} \cdot 0 \pmod{Xf-1}$$
2166  $$(Xf)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2167
2168  Now, since we had $Xf \equiv 1$ in $\frac{A[X]}{(Xf-1)}$,
2169  $$(1)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2170  $$\Longrightarrow~~ h(X) \equiv 0 \pmod{Xf-1}$$
2171  By definition this is saying $h(X) \in (Xf-1) ~\forall~ k \in ker(\psi)$.
2172
2173  Thus $ker(\psi) \subseteq (Xf-1)$.
2174
2175  Initially we saw that $(Xf-1) \subseteq ker(\psi)$. Therefore $ker(\psi)=(Xf-1)$.
2176
2177  Hence,
2178  $$A_f \cong \frac{A[X]}{(Xf-1)}$$
2179\end{proof}
2180
2181
2182\vspace{0.4cm}
2183
2184Given a ring homomorphism $\psi: A \longrightarrow B$, there is a correspondence
2185
2186$$e: \{ \text{ideals of } A \} \longrightarrow \{ \text{ideals of } B \}$$
2187
2188given by $e(I) = \psi(I)B = IB$ (called \emph{extension}),
2189
2190and
2191
2192$$r: \{ \text{ideals of } B \} \longrightarrow \{ \text{ideals of } A \}$$
2193
2194given by $r(J) = \psi^{-1} J$ (called \emph{restriction}, written $A \cap J$).
2195
2196Set $B= S{-1}A$. Then $S^{-1}I = e(I) = \psi(I)B$.
2197
2198\begin{prop}{6.3} \label{prop.6.3}
2199  \begin{enumerate}[a.]
2200    \item $\forall$ ideal $J$ of $S^{-1}A$, $e(r(J))=J$
2201    \item $\forall$ ideal $I$ of $A$,
2202      $$r(e(I)) = \{ a \in A ~|~ as \in I ~\text{for some}~ s \in S \}$$
2203    \item if $P$ prime and $P \cap S = \emptyset$,\\
2204      then $e(P)=S^{-1}P$ is a prime ideal of $S^{-1}A$.
2205  \end{enumerate}
2206\end{prop}
2207\begin{proof}
2208\end{proof}
2209
2210\begin{cor}{6.3} \label{cor.6.3}
2211    for an ideal $I$ of $A$, the necessary and sufficient condition for
2212    $r(e(I))=I$ is
2213      $$as \in I \Longrightarrow a \in I~~ \forall s \in S$$
2214\end{cor}
2215
2216
2217\subsection{Localization}
2218
2219If $P$ prime ideal, then $S=A \setminus P$ is a multiplicative set.
2220Define the set $A_P=S^{-1}A$.
2221
2222\begin{prop}{6.4} \label{6.4}
2223  $a/s \in A_P$ is a unit of $A_P ~\Longleftrightarrow~ a \not\in P$.
2224
2225  Therefore $A_P$ is a \emph{local ring}, with maximal ideal $e(P)=P A_P$.
2226
2227  The local ring $(A_P, PA_P)$ is called the \emph{localization} of $A$ at $P$.
2228\end{prop}
2229\begin{proof}
2230  \begin{enumerate}
2231    \item[$\Longleftarrow$] (if $a \not\in P ~\Longrightarrow~ a/s \in A_P$ is a unit)\\
2232
2233      if $a \not\in P$, then by definition of $S,~~ a \in S$.
2234
2235      In the localization $A_P$, every element of $S$ is invertible.
2236
2237      The inverse of $\frac{a}{s}$ is $\frac{s}{a}$\\
2238      \hspace*{2em}$\longrightarrow~~ \frac{a}{s} \cdot \frac{s}{a} = 1$, thus
2239      $a/s \in A_P$ is a unit.
2240
2241    \item[$\Longrightarrow$] (if $a/s$ unit $~\Longrightarrow~ a \not\in P$)\\
2242
2243      Suppose $a/s$ a unit; then $\exists~ \frac{b}{t} \in A_P$ such that
2244      $\frac{a}{s} \cdot \frac{b}{t} = 1$.
2245
2246      By definition of equality in localization, $\frac{ab}{st}=1$ means $\exists u \in S$ such that
2247      \begin{equation}
2248        u(ab \cdot 1 - st \cdot 1) = 0~~ \Longrightarrow~ uab=ust
2249        \label{eq:6.4}
2250      \end{equation}
2251      with $u,s,t \in S$.
2252
2253      Since $S=A \setminus P$ and $P$ is a prime ideal,\\
2254      the products of elements outside $P$ must also be outside of $P$.
2255      $$u,s,t \not\in P ~\Longrightarrow~ ust \not\in P$$
2256
2257      At \eqref{eq:6.4}, we know that $uab=ust \not\in P$, thus $uab \not\in P$.
2258
2259      If $uab \not\in P$, then $a \not\in P$.
2260
2261
2262      Next we will prove that $A_P$ is a local ring.\\
2263
2264      A ring is local if it has exactly one maimal ideal (\ref{1.13}); so we
2265      show that the set of non-units forms an ideal.
2266
2267      $a/s$ is not a unit iff $a \in P$. Let $m= \{ a/s | a \in P,~ s \not\in P \} = PA_P$.
2268      Want to show that $PA_P$ is the unique maximal ideal:
2269      \begin{itemize}
2270        \item it's an ideal: let $\frac{a}{s}, \frac{b}{t} \in PA_P$ with $a,b \in P$.
2271
2272          Then $\frac{a}{b} + \frac{b}{t} \in PA_P$ with numerator in $P$.
2273
2274          If multiply a fraction in $PA_P$ by any fraction in $A_P$, the numerator stays in $P$, thus $PA_P$ is an ideal.
2275
2276        \item every element outside of $PA_P$ is a unit (proven at the beginning of this proposition's proof ($\Longleftarrow$)).
2277
2278        \item maximality: every ideal strictly larger than $PA_P$ must contain a unit.\\
2279          If an ideal contains a unit, it must be the entire ring $A_P$.\\
2280          \hspace*{2em}$\Longrightarrow$ therefore, $PA_P$ is the unique maximal ideal.
2281      \end{itemize}
2282  \end{enumerate}
2283\end{proof}
2284
2285
2286In other words:
2287
2288\emph{Localization}: formal way to 'force' certain elements to have inverses.
2289
2290It's the smallest ring that contains $A$ and makes all the elements of $S$ invertible.
2291
2292There is a natural map $\psi: A \longrightarrow S^{-1}A$, by $f(a)=\frac{a}{1},~ f(s)=\frac{1}{s}~~ \text{for}~ s \in S$.
2293
2294
2295\subsection{Localization commutes with taking quotients}
2296Let $A$ ring, $S$ multiplicative set, $I$ ideal.
2297
2298Write $T$ for the image of $S$ in $A/I$.
2299
2300Then $S^{-1}I=I \cdot S^{-1}A$ is an ideal of $S^{-1}A$, can take the quotient ring $S^{-1}A / S^{-1} I$.
2301
2302$\longleftrightarrow$ can take the quotient $A/I$ and then localize to get $T^{-1}(A/I)$.
2303
2304
2305\begin{cor}{6.7} \label{6.7}
2306  $$T^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
2307  In particular, for $P$ prime ideal,
2308  $$\frac{A_P}{PA_P} = k(P) = Frac(\underbrace{A/P}_{\substack{\text{integral}\\\text{domain}}})$$
2309
2310  From \ref{6.4}, $A_P$: local ring, $PA_P$: unique maximal ideal.
2311  $k(P)$ and $Frac(A/P)$ are field of fractions.
2312\end{cor}
2313\begin{proof}
2314  the quotient ring $A/I$ can be viewed as an $A$-module and
2315  $$\underbrace{ T^{-1}(A/I) }_{\text{ring of fractions}} \cong
2316  \underbrace{ \frac{S^{-1}A}{S^{-1}I} }_{\text{module of fractions}}$$
2317
2318  The \ref{cor.6.6}.i, gives an isomorphism of modules
2319  $$ T^{-1}(A/I) = S^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
2320  it's easy to see that this is a ring homomorphism.
2321\end{proof}
2322% TODO 6.6 (proposition and corollary)
2323
2324
2325
2326\section{Tensor Product of Modules}
2327
2328\subsection{Constructing the Tensor Product}
2329We use vector spaces but the same applies for modules.
2330
2331Let $Z$ be the vector space with basis $\{ e_{v,w} \} ~\forall~ (v,w) \in V \times W$.
2332
2333Want 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 $a-b$.
2334
2335Consider the quotient space $\frac{Z}{Span\{a-b\}}$.
2336Anything in the span of $a-b$ is equal to zero.
2337
2338$\Longrightarrow$ This is how we use relations to get vector spaces out of $Z$.
2339
2340\paragraph{Direct Sum, $V \oplus W$}.
2341
2342Want
2343\begin{enumerate}[i.]
2344  \item $(v_1, w_1) + (v_2, w_2) = (v_1 + v_2, w_1 + w_2)$
2345  \item $\lambda (v, w) = (\lambda v, \lambda w)$
2346\end{enumerate}
2347
2348Take the quotient
2349$$Z / Span
2350\left\{
2351\substack{
2352(v_1,w_1)+(v_2,w_2)-(v_1+v_2,w_1+w_2)\\
2353\lambda(v,w)-(\lambda v,\lambda w)
2354}
2355\right\}
2356$$
2357
2358\paragraph{Tensor Product, $V \otimes W$}
2359relations:
2360\begin{enumerate}[i.]
2361  \item distribuitivity law:
2362    $$a (b+c) = ab + ac$$
2363    $$(a+b)c =  ac + bc$$
2364  \item scalar multiplication:
2365    $$\lambda(a \cdot b) = (\lambda a) b = a (\lambda b)$$
2366\end{enumerate}
2367
2368$\Longrightarrow$
2369  \begin{align*}
2370    &a (b+c) = ab + ac\\
2371    &(a+b)c =  ac + bc\\
2372    &\lambda (a \cdot b) = (\lambda a) b\\
2373    &\lambda (a \cdot b)= a (\lambda b)
2374  \end{align*}
2375
2376  then we can define the tensor product as
2377$$V \otimes W = Z / Span
2378\left\{
2379\substack{
2380    a \otimes (b+c) - a \otimes b - a \otimes c\\
2381    (a+b) \otimes c -  a \otimes c - b \otimes c\\
2382    \lambda (a \otimes b) - (\lambda a) \otimes b\\
2383    \lambda (a \otimes b) - a \otimes (\lambda b)
2384}
2385\right\}
2386$$
2387
2388
2389
2390\subsection{Universal Property of Tensor Products}
2391\begin{prop}{}[Universal Property of Tensor Products]
2392  For every bilinear map $g: V \times W \longrightarrow U$, $\exists !$ linear map $f: T \longrightarrow U$ such that $f \circ \tau = g$.
2393
2394  \begin{tikzpicture}[node distance=2.5cm, auto]
2395    \node (VW)    {$V \times W$};
2396    \node (T) [right of=VW] {$T = V \otimes W$};
2397    \node (U) [below of=VW] {$U$};
2398
2399    \draw[->] (VW) to node {$\tau$} (T);
2400    \draw[->] (VW) to node [swap] {$g$} (U);
2401    \draw[->, dashed] (T) to node [swap] {$f$} (U);
2402  \end{tikzpicture}
2403\end{prop}
2404
2405\begin{proof}
2406  \begin{enumerate}[i.]
2407    \item uniqueness:\\
2408
2409    suppose we have two different T.P., $(T, \tau)$ and $(T', \tau')$. Want to prove $T \stackrel{iso}{\cong} T'$.
2410
2411
2412
2413    On the right is the same relation as on the left, but switching $T$ by $T'$:
2414
2415  \begin{minipage}[t]{0.48\textwidth}
2416    \centering
2417    \begin{tikzpicture}[node distance=2.0cm, auto]
2418      \node (VW)    {$V \times W$};
2419      \node (T) [right of=VW] {$T$};
2420      \node (T2) [below of=VW] {$T'$};
2421
2422      \draw[->] (VW) to node {$\tau$} (T);
2423      \draw[->] (VW) to node [swap] {$\tau'$} (T2);
2424      \draw[->, dashed] (T) to node [swap] {$f$} (T2);
2425    \end{tikzpicture}
2426
2427    \vspace{0.5em}
2428
2429    by the U.P., $f \circ \tau = \tau'$.
2430  \end{minipage}
2431  \hfill
2432  \begin{minipage}[t]{0.48\textwidth}
2433    \centering
2434    \begin{tikzpicture}[node distance=2.0cm, auto]
2435      \node (VW)    {$V \times W$};
2436      \node (T) [right of=VW] {$T$};
2437      \node (T2) [below of=VW] {$T'$};
2438
2439      \draw[->] (VW) to node {$\tau$} (T);
2440      \draw[->] (VW) to node [swap] {$\tau'$} (T2);
2441      \draw[->, dashed] (T) to node [swap] {$f$} (T2);
2442    \end{tikzpicture}
2443
2444    \vspace{0.5em}
2445
2446    by the U.P., $f \circ \tau = \tau'$.
2447  \end{minipage}
2448
2449  \vspace{0.3cm}
2450  $$\text{thus,}~~ f \circ f' \circ \tau' = \tau'$$
2451
2452  Want to show that $f$ has an inverse, so that it is an isomorphism $T \stackrel{iso}{\cong} T'$.
2453
2454  we just saw that
2455  $$f \circ f' \circ \tau' = \tau'$$
2456  also we can get it the other way
2457  $$f' \circ f \circ \tau = \tau$$
2458
2459  we can see that $f \circ f'$ seems to be the identity, need more info to prove it.
2460
2461  What if we apply the U.P. to the same module on both sides?
2462
2463    \begin{tikzpicture}[node distance=2.0cm, auto]
2464      \node (VW)    {$V \times W$};
2465      \node (T) [right of=VW] {$T$};
2466      \node (T2) [below of=VW] {$T$};
2467
2468      \draw[->] (VW) to node {$\tau$} (T);
2469      \draw[->] (VW) to node [swap] {$\tau$} (T2);
2470      \draw[->, dashed] (T) to node {$? = id_T$} (T2);
2471    \end{tikzpicture}
2472
2473    thus, $id_T \circ \tau = \tau$, hence
2474
2475    \begin{tikzpicture}[node distance=2.0cm, auto]
2476      \node (VW)    {$V \times W$};
2477      \node (T) [right of=VW] {$T$};
2478      \node (T2) [below of=VW] {$T$};
2479
2480      \draw[->] (VW) to node {$\tau$} (T);
2481      \draw[->] (VW) to node [swap] {$\tau$} (T2);
2482      \draw[->, dashed] (T) to node {$id_T,~ f' \circ f$} (T2);
2483    \end{tikzpicture}
2484
2485    and by uniqueness of the U.P., we can not have two different functions ($id_T$ and $f' \circ f$) and must be that $f' \circ f = id_T$.
2486
2487    Similarly for $f \circ f' = id_{T'}$.
2488
2489    Therefore, $f'$ is the inverse of $f$, and $T \stackrel{iso}{\cong} T'$, which means that the tensor prodduct is unique.
2490
2491    % TODO (ii.) existence
2492  \end{enumerate}
2493
2494
2495  \vspace{1.0cm}
2496  This is stated in the Atiyah and MacDonald book (\cite{am}) for multiple nodes and multilinear maps:
2497
2498  \begin{prop}{2.12}
2499    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:
2500
2501    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 $f' \circ g = f$.
2502
2503    Moreover, if $(T,g)$ and $(T',g')$ are two pairs with this property, then $\exists !~ j: T \stackrel{iso}{\longrightarrow} T'$ such that $j \circ g = g'$.
2504  \end{prop}
2505
2506  \subsection{Usefulness of Tensor Product's Universal Property}
2507  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..
2508
2509  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.
2510
2511  \paragraph{Using the U.P. to show that a function on the T.P. is well defined and linear:}
2512  \begin{enumerate}[i.]
2513    \item define a map $\phi$ on the Cartesian product
2514      $$(x,y) \in M \times N$$
2515      
2516    \item prove bilinearity of $\phi$:\\
2517      show that $\phi$ is $A$-linear in each variable separately.
2518
2519      ie. check
2520      \begin{align*}
2521        \phi(x + x', y) &= \phi(x, y) + \phi(x', y)\\
2522        \phi(x, y + y') &= \phi(x, y) + \phi(x, y')\\
2523        \phi(ax, y) &= a \phi(x, y) = \phi(x, ay)
2524      \end{align*}
2525
2526    \item invoke the Universal Property.\\\
2527
2528      Once bilinearity is proven on $\phi$, the U.P. guarantees the existence of a unique map $f$ on the tensor product $M \otimes N$
2529
2530    \item automatic well-definedness and linearity: $f$ is linear by definition.
2531
2532    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).
2533  \end{enumerate}
2534
2535
2536\end{proof}
2537
2538
2539
2540%%%%%%%%%%%%%%%%%%%%%%%%
2541%%%%%%
2542%%%%%% end of notes,
2543%%%%%% beginning of exercises
2544%%%%%%
2545%%%%%%%%%%%%%%%%%%%%%%%%
2546
2547\newpage
2548
2549\section{Exercises}
2550
2551For the exercises, I follow the assignments listed at \cite{mit-course}.
2552
2553The 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}.
2554
2555\subsection{Exercises Chapter 1}
2556
2557\begin{ex}{R.1.1}
2558  Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$?
2559\end{ex}
2560\begin{proof}
2561  Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$.
2562
2563  $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\
2564  Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$.
2565
2566  Thus $I \cup J$ is not closed under addition; thus is not an ideal.
2567
2568
2569  Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum:
2570
2571  $$I+J = \{ a+b | a \in I, b \in J \}$$
2572
2573  $gcd(2,3)=1$, so $I+J = \mathbb{Z}$.
2574
2575  Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$.
2576\end{proof}
2577
2578\begin{ex}{R.1.5}
2579  let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\
2580  In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$.
2581\end{ex}
2582\begin{proof}
2583  (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$).
2584
2585  Let
2586  $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
2587  The claim is that $\psi^{-1}(P)$ is prime ideal of $A$.
2588
2589  \begin{enumerate}[i.]
2590    \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
2591      $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$).
2592
2593      If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so
2594      $$\psi(a-b)= \psi(a) - \psi(b) \in P$$
2595      hence $a-b \in \psi^{-1}(P)$.
2596
2597      If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\
2598      Thus $ra \in \psi^{-1}(P)$.
2599
2600      $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$.
2601
2602    \item show that $\psi^{-1}(P)$ is prime:\\
2603      $\psi^{-1}(P) \neq A$, 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 $B$.
2604
2605      Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$.
2606
2607      Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\
2608      Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$.
2609
2610      Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$.
2611  \end{enumerate}
2612\end{proof}
2613
2614
2615\begin{ex}{R.1.6}
2616  prove or give a counter example:
2617  \begin{enumerate}[a.]
2618    \item the intersection of two prime ideals is prime
2619    \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
2620    \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
2621    \item the map $\psi^{-1}$ of Proposition 1.2 takes maximal ideals of $A/I$
2622      to maximal ideals of $A$
2623  \end{enumerate}
2624\end{ex}
2625\begin{proof}
2626  \begin{enumerate}[a.]
2627    \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime.
2628
2629      Then $I \cap J = (2) \cap (3) = (6)$.
2630
2631      The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$.
2632
2633      Thus the intersection of two primes can not be prime.
2634
2635    \item $P_1=(2),~ P_2=(3)$, both prime.
2636
2637      Then,
2638      $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$
2639
2640      $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$.
2641
2642      (recall: principal= generated by a single element)
2643
2644      So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$.
2645
2646      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.
2647
2648      Henceforth, the sum of two prime ideals is not necessarily prime.
2649
2650    \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$.
2651
2652      Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$.
2653
2654      Then
2655      \begin{align*}
2656        \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\
2657        \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A
2658      \end{align*}
2659
2660      But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
2661
2662      Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
2663
2664    \item $\psi: A \longrightarrow A/I$ quotient homomorphism, $I \subseteq A$ an ideal.
2665
2666      Let $M$ a maximal ideal of $A/I$, then $\frac{(A/I)}{M}$ is a field (Proposition 1.3).
2667
2668      By the isomorphism theorems,
2669      $$\frac{(A/I)}{M} \cong \frac{A}{\psi^{-1}(M)}$$
2670
2671
2672      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 $A$.
2673
2674      $\Longrightarrow~$ under $\psi$, preimages of maximal ideals are maximal.
2675  \end{enumerate}
2676\end{proof}
2677
2678\begin{ex}{R.1.12.a}
2679  if $I,J$ ideals and $P$ prime ideal, prove that
2680  $$IJ \subset P ~\Longleftrightarrow~ I \cap J \subset P ~\Longleftrightarrow~ I ~\text{or}~ J \subset P$$
2681\end{ex}
2682\begin{proof}
2683  assume $I \subseteq P$ (for $J \subseteq P$ will be the same, symmetric), take $x \in IJ$,
2684
2685  then
2686  $$x = \sum_{k=1}^n a_k b_k$$
2687  with $a_k \in I,~ b_k \in J$.
2688
2689  Each $a_k \in I \subseteq P$. Since $P$ an ideal,
2690  $$\sum_{k=1}^n a_k b_k \in P$$
2691
2692  thus $x \in P$, hence $IJ \subseteq P$.
2693
2694  So $I \subseteq P$ or $J \subseteq P$ $~\Longrightarrow IJ \subseteq P$.
2695
2696  \vspace{0.5cm}
2697  Conversely,\\
2698  assume $P$ prime and $IJ \subseteq P$.
2699
2700  Suppose by contradiction that $I \not\subseteq P$ and $J \not\subseteq P$.
2701  \begin{itemize}
2702    \item[-] since $I \not\subseteq P,~ \exists a \in I$ with $a \not\in P$
2703    \item[-] since $J \not\subseteq P,~ \exists b \in J$ with $b \not\in P$
2704  \end{itemize}
2705  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 $P$.
2706
2707  Thus $I \not\subseteq P$ and $J \not\subseteq P$ are false.
2708
2709  \vspace{0.3cm}
2710  So both directions are proven, hence
2711  $$IJ \subseteq P ~\Longrightarrow~ I \subseteq P ~\text{or}~ J \subseteq P$$
2712
2713\end{proof}
2714
2715\begin{ex}{R.1.18}
2716  Use Zorn's lemma to prove that any prime ideal $P$ contains a minimal prime ideal.
2717\end{ex}
2718\begin{proof}
2719  Let $P$ prime ideal of $R$.
2720
2721  $$S = \{ Q \subseteq R ~|~ Q ~\text{a prime ideal AND}~ Q \subseteq P \}$$
2722
2723  Goal: show that $S$ has a minimal element, the minimal ideal contained in $P$.
2724
2725  $P \subset S$, so $S$ is nonempty.
2726
2727  Let $C \subseteq S$ be a chain (= totally ordered subset) with respect to inclusion.
2728
2729  Define
2730  $$Q_C = \bigcap_{Q \in C} Q$$
2731
2732  Clearly $Q_C \subseteq P$, since each $Q \in C$ is $Q \subseteq P$.
2733
2734  Since $C$ is ordered by inclusion, it is a decreasing chain of prime ideals.
2735
2736  Intersection of a decreasing chain of prime ideals is again a prime ideal:
2737  \begin{itemize}
2738    \item[-] if $ab \in Q_C$, then $ab \in Q ~\forall Q \in C$
2739    \item[-] since $Q$ prime, $\forall Q \in C$ either $a \in Q$ or $b \in Q$
2740  \end{itemize}
2741
2742  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 $Q_2 \subseteq Q_1$.
2743
2744  In either case: contradiction, since the smaller one would have to contain the element that was assumed to be excluded.
2745
2746  Thus $\forall Q \in C$ the same element $a, b$ must lie in all $Q$. $\Longrightarrow~$ lies in the intersection of them, $Q_C$.
2747
2748  Henceforth, $Q_C$ is a prime ideal and lies in $S$, and its a lower bound of $C$ in $S$.
2749
2750  Now, $S$ is nonempty, and every chain in $S$ has a lower bound in $S$ (its intersection).\\
2751  Therefore, $S$ has a minimal element $P_{min}$.
2752
2753  By construction, $P_{min}$ is a prime ideal $P_{min} \subseteq P$, and by minimality there are no strictly smaller prime ideals inside $P$.
2754
2755  So $P_{min}$ is a minimal prime ideal, contained in $P$.
2756\end{proof}
2757
2758\begin{ex}{R.1.10}
2759\end{ex}
2760\begin{proof}
2761\end{proof}
2762
2763\begin{ex}{R.1.11}
2764\end{ex}
2765\begin{proof}
2766\end{proof}
2767
2768\begin{ex}{R.1.4}
2769\end{ex}
2770\begin{proof}
2771\end{proof}
2772
2773\subsection{Exercises Chapter 2}
2774
2775\begin{ex}{R.2.9}
2776$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ is a s.e.s. of $A$-modules. Prove that if $N, L$ are finite over $A$, then $M$ is finite over $A$.
2777\end{ex}
2778\begin{proof}
2779  Denote the generators of $L$ and $N$ respectively as
2780  \begin{align*}
2781    \{l_1, \ldots, l_k \} &\subseteq L\\
2782    \{n_1, \ldots, n_p \} &\subseteq N
2783  \end{align*}
2784
2785  By s.e.s. definition, 
2786  \begin{itemize}
2787    \item[-] $\alpha$ is injective (one-to-one), so
2788    $$\forall l_i \in L,~ \exists~ x_i \in M ~\text{s.th.}~ \alpha(l_i)=x_i$$
2789
2790  \item[-] $\beta$ is surjective (onto), so
2791    $$\forall n_j \in N,~ \exists~ y_j \in M ~\text{s.th.}~ \beta(y_j)=n_j$$
2792\end{itemize}
2793
2794  We will show that $\{x_1, \ldots, x_k, y_1, \ldots, y_p \}$ generate $M$, and thus $M$ is finite:
2795
2796  Let $m \in M$, then $\beta(m) \in N$, and
2797  $$\beta(m) = \sum_{j=1}^p a_j n_j ~~~\text{with}~ a_j \in A$$
2798
2799  Take $m' \in M$, with $m' = \sum a_j y_j$, then
2800  $$\beta(m') = \sum a_j \beta(y_j) = \sum a_j n_j = \beta(m)$$
2801
2802  Then, since $\beta(m)=\beta(m')~~ \Longrightarrow~~ \beta(m-m')=0$, thus
2803  $$(m-m') \in ker(\beta)$$
2804
2805  By \emph{exactness} property, since $\alpha: L \longrightarrow ker(\beta)$, we have $ker(\beta)=im(\alpha)$.
2806
2807  Therefore, $\exists~ l \in L$ such that $\alpha(l)= m-m'$.
2808
2809  Since $\{l_i\}_k$ generate $L$,
2810  $$l = \sum^k b_i l_i$$
2811  thus
2812  $$m-m' = \alpha(l) = \alpha(\underbrace{\sum b_i l_i}_{l}) = \sum b_i \underbrace{\alpha(l_i)}_{x_i} = \sum b_i x_i$$
2813
2814  Rearrange,
2815  $$m = m' + \sum b_i x_i = \sum_{j=1}^p a_j y_j + \sum_{i=1}^k b_i x_i ~~~~~ \forall m \in M$$
2816
2817  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.\\
2818  Thus $M$ is finitely generated over $A$.
2819\end{proof}
2820
2821
2822\subsection{Exercises Chapter 3}
2823
2824\begin{ex}{R.3.2}
2825  $K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space.
2826   Prove that $A$ is Noetherian and Artinian.
2827\end{ex}
2828\begin{proof}
2829  $dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in \aA$ and $c \in K$, then $c \cdot x \in \aA$.
2830
2831  \begin{enumerate}
2832    \item Noetherian:\\
2833      let $I_1 \subseteq I_2 \subseteq \ldots$ be an ascending chain of ideals in $A$.
2834
2835      Since each $I_i$ is a subspace, we have
2836      $$dim_K(I_1) \leq dim_K(I_2) \leq \ldots \leq n$$
2837      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.
2838      
2839    \item Artinian:\\
2840      Similarly, if $I_1 \supseteq I_2 \supseteq \ldots$ a descending chain of ideals in $A$.
2841
2842      then
2843      $$n \geq dim_K(I_1) \geq dim_K(I_2) \geq \ldots \geq 0$$
2844      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.
2845  \end{enumerate}
2846\end{proof}
2847
2848\begin{ex}{R.3.5}
2849  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 $M$.
2850
2851  Prove if the following holds or not:
2852  $$\beta(M_1)=\beta(M_2) ~\text{and}~ \alpha^{-1}(M_1)=\alpha^{-1}(M_2)~~\Longrightarrow~ M_1=M_2$$
2853\end{ex}
2854\begin{proof}
2855  Counterexample showing that it does not hold:
2856
2857  Let $K$ a field, $M = K \oplus K~, L=K,~N=K$.
2858
2859  Set, for $l \in L,~ (m_1, m_2) \in M$,
2860  \begin{align*}
2861    \alpha:~ &l \longmapsto (l, 0)\\
2862    \beta:~ &(m_1, m_2) \longmapsto m_2
2863  \end{align*}
2864
2865  So we have
2866  $$0 \longrightarrow K \stackrel{\alpha}{\longrightarrow} K^2 \stackrel{\beta}{\longrightarrow} K \longrightarrow 0$$
2867
2868  Then,
2869  \begin{align*}
2870    M_1 &= \{ (x, x) ~|~ x \in K\} ~~~~\sim\text{(diagonal line)} \\
2871    M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)}
2872  \end{align*}
2873
2874  (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 $M$).
2875
2876  Then,
2877  \begin{align*}
2878    \beta(M_1) &= \{ x ~|~ x \in K\} = K \\
2879    \beta(M_2) &= \{ x ~|~ x \in K\} = K
2880  \end{align*}
2881  thus, $\beta(M_1) = \beta(M_2)$.
2882
2883  For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1) = \{0\}$,\\
2884  for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2) = \{0\}$,\\
2885  thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$.
2886
2887
2888  So we've seen that
2889  \begin{align*}
2890    \beta(M_1) = \beta(M_2)\\
2891    \alpha^{-1}(M_1)=\alpha^{-1}(M_2)
2892  \end{align*}
2893  while having $M_1 \neq M_2$.
2894\end{proof}
2895
2896
2897
2898\begin{ex}{R.3.3}
2899Let $A$ a ring, $I_1, \ldots, I_k$ ideals such that each $A/I_i$ is a Noetherian ring.
2900Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bigcap I_i = 0$ then $A$ is also Noetherian.
2901\end{ex}
2902\begin{proof}
2903\begin{enumerate}[i.]
2904  \item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian.
2905    $\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian.
2906  \item Take the canonical homomorphism
2907    $$\phi: A \longrightarrow \bigoplus_{i=1}^n A/ I_i$$
2908    by $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$.
2909
2910    $\phi$ is injective: $ker(\phi)= \{ a \in A | a \in I_i \forall i \}$.
2911
2912    Since we're given $\cap I_i = 0$, then $ker(\phi)=\cap I_i$, and $\phi$ is injective.
2913
2914    Thus, $\phi$ is the isomorphism $A \cong im(\phi)$, where $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$.
2915
2916    We know that any submodule of a Noetherian module is Noetherian, thus, since
2917    \begin{itemize}
2918      \item $A/I_i$ is Noetherian by hypothesis of the exercise
2919      \item $A \cong im(\phi)$
2920      \item $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$
2921  \end{itemize}
2922  then, $A$ is Noetherian.
2923\end{enumerate}
2924\end{proof}
2925
2926\begin{ex}{R.3.4}
2927  Prove that if A is a Noetherian ring and M a finite A-module, then there
2928  exists an exact sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$.
2929  That is, M has a presentation as an A-module in terms of finitely many generators and relations.
2930\end{ex}
2931\begin{proof}
2932  since $M$ fingen $~\Longrightarrow~$ generators $\{m_1, \ldots, m_2 \} \subseteq M$ span $M$.
2933
2934  Let $\beta$ be a surjective $A$-linear map, which forms a free $A$-module of rank $p$ onto $M$:
2935  \begin{align*}
2936    \beta: A^p &\longrightarrow M\\
2937    (a_1, \ldots, a_p) &\longmapsto \sum_{i=1}^p a_i m_i
2938  \end{align*}
2939
2940  Let $K=ker(\beta)$. By the 1st Isomorphism Theorem,
2941  $$M \cong A^p / K$$
2942
2943  Since $A$ is a Noetherian ring, then every free $A$-module of finite rank (eg. $A^p$) is a Noetherian module.
2944
2945  Every submodule of a Noetherian module is fingen.
2946
2947  $\Longrightarrow~$ since $K \subseteq A^p, ~\Longrightarrow~ K ~~(=ker(\beta))$ is fingen.
2948
2949  Since $K$ fingen, let $\{k_1, \ldots, l_q\}$ be generators of $K$.
2950
2951  Define $\psi: A^q \longrightarrow K$.
2952
2953  Compose it with the inclusion map $i: K \longrightarrow A^p$,
2954  $$\alpha = i \circ \psi:~ A^q \longrightarrow A^p$$
2955
2956  So we have the whole sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$, where
2957  \begin{itemize}
2958    \item $\beta$ is surjective
2959    \item $im(\alpha)=ker(\beta)$
2960  \end{itemize}
2961  so that it is a exact sequence, thus, $M$ has a finite presentation.
2962\end{proof}
2963
2964\subsection{Exercises Chapter 4}
2965
2966\begin{ex}{R.4.1.a}
2967  $k[X^2] \subset k[X]$ is a finite extension, hence integral. Find the integral dependence relation for any $f \in k[X]$.
2968\end{ex}
2969\begin{proof}
2970  $\forall f(X) \in k[X]$ can be uniquely decomposed into its even and odd parts:
2971  $$f(X) = p(X^2) + X \cdot q(X^2)$$
2972
2973  with $p(X^2),~ q(X^2) \in k[X^2]$, and\\
2974  $p(X^2)$: sum of all terms with even exponents\\
2975  $q(X^2)$: sum of all terms with odd exponents, and then factoring out $X$.
2976
2977  \vspace{0.3cm}
2978  (Observation: this is used in FRI cryptographic protocol\\
2979  \href{https://github.com/arnaucube/math/blob/master/notes_fri_stir.pdf}{https://github.com/arnaucube/math/blob/master/notes\_fri\_stir.pdf})
2980
2981
2982  Rearrange it
2983  \begin{align*}
2984    f(X) &- p(X^2) = X \cdot q(X^2), ~~\text{square:}\\
2985    (f(X) &- p(X^2))^2 = X^2 \cdot q(X^2)^2\\
2986    f(X)^2-2 p(X^2) f(X) &+ p(X^2)^2 = X^2 \cdot q(X^2)^2\\
2987    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
2988  \end{align*}
2989
2990   Denote the last polynomial as $P(T) \in k[X^2]$, where $f(X)$ is a root of $P(T)$.
2991
2992   \vspace{0.2cm}
2993   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 $a_i \in k[X^2]$.
2994
2995   We have that
2996   \begin{align*}
2997     a_1 &= -2 p(X^2)\\
2998     a_0 &= p(X^2)^2 - X^2 q(X^2)^2
2999   \end{align*}
3000
3001
3002   So for example, for $f(X)= X^3 + X^2 + X + 1$:
3003   \begin{align*}
3004     f(X) = (X^2 + 1) &+ X (X^2 + 1)\\
3005     (f(X) - (X^2+1))^2 &= X^2 (X^2 + 1)^2\\
3006     (f(X) - p(X))^2 &= X^2 (q(X))^2
3007   \end{align*}
3008\end{proof}
3009
3010
3011\begin{ex}{R.4.5}
3012  Let $A=k[X,Y]/(Y^2 - X^2 - X^3)$. Prove that the normalization of $A$ is $k[t]$ where $t=Y/X$.
3013\end{ex}
3014\begin{proof}
3015$A=k[X,Y]/(Y^2 - X^2 - X^3)$, express $X$ and $Y$ in terms of $t$:
3016
3017Since $t=Y/X$ then $Y=tX$, and combined with $Y^2 = X^2 + X^3$, then
3018\begin{align*}
3019  (tX)^2 &= X^2 + X^3\\
3020  t^2 X^2 &= X^2 + X^3, ~~\text{assuming}~X \neq 0:\\
3021  t^2 &= 1+X, ~\text{thus}\\
3022  X&=t^2-1 \in k[X]
3023\end{align*}
3024
3025Then, $Y=tX=t(t^2-1)=t^3-t \in k[X]$.
3026
3027Hence $X, Y \in k[X]$.
3028
3029Therefore, $k[X,Y]/(Y^2 - X^2 - X^3) \subseteq k[t]$.
3030
3031\vspace{0.4cm}
3032By \ref{noether-normalization} (Noether normalization lemma), to show that $k[t]$ is the \emph{normalization}, must show that $k[t]$ is \emph{integral} over $A$.
3033
3034From $X=t^2-1 ~~\Longrightarrow~~ t^2-1-X=0 ~~\Longrightarrow~ t^2 - (1+X) = 0$.
3035
3036$(1+X) \in A$, so $t$ satisfies the monic polynomial
3037$$P(T) = T^2 - (1+X) \in A[T]$$
3038Thus $t$ is integral over $A$.
3039
3040Since $k[t]$ is generated by $t$ over $k$, and $k \subset A$, then the entire ring $k[t]$ is integral over $A$.
3041
3042Since $k[t]$ is a polynomial ring over a field, which is a UFD, it is integrally closed (since all UFD are integrally closed).
3043
3044$Frac A = k(X,Y)$, since $X=t^2-1, ~Y=t^3-t ~~\Longrightarrow~ k(X,Y) \subseteq k(t)$
3045
3046and $t=Y/X \in k(X,Y)$, thus $k(X,Y)=k(t)$.
3047
3048Since $k[t]$ is integrally closed and is the integral closure of $A$ in its fraction field $k(t)$, we conclude that the normalization of $A$ is $k[t]$.
3049\end{proof}
3050
3051
3052
3053\begin{ex}{R.4.9}
3054  $k$ 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.
3055\end{ex}
3056\begin{proof}
3057  (want to find a linear combination of the coordinates such that the original variables satisfy monic polynomials over the new ring $B$)
3058
3059  The relations defining $A$ are
3060  \begin{align*}
3061    X^2-Y^3-1=0 ~~&\Longrightarrow~ Y^3 = X^2 -1 ~~~(*)\\
3062    XZ -1=0 ~~&\Longrightarrow~ Z= 1/X = X^{-1}
3063  \end{align*}
3064
3065Thus $A$ can be denoted as $A = k[X,Y,X^{-1}]/(Y^3 - X^2 - 1)$.
3066
3067
3068Now, $Y$ is inegral over $k[X]$, since $Y^3 - (X^2 - 1) = 0$ is monic in $Y$ with coefficients in $k[X]$.
3069
3070$Z$ is not integral over $k[X]$, since $Z=1/X$ and $X$ is not a unit in $k[X]$.
3071
3072Choose $\alpha, \beta \in k$ such that $X$ (and thus $Z$) becomes integral over $B$:\\
3073set $\alpha=0,~\beta=1 ~~\Longrightarrow~~ B=k[X+\alpha Y + \beta Z]= k[X+Z]$.
3074
3075Let $w=X+Z$; since $XZ=1$, we have
3076$$w=X+\frac{1}{X} ~~\Longrightarrow~~ Xw=X^2+1 ~~\Longrightarrow~~ X^2 -w X +1 = 0 ~~(**)$$
3077
3078which is monic with coefficients in $k[w]$, thus $X$ is integral over $B$.
3079
3080Since $Z=w-X$, $Z$ is also integral.
3081
3082\vspace{0.4cm}
3083Generators of $A$ as a $B$-module:
3084
3085we hadd $B=k[w]$ with $w=X+Z$.
3086
3087From $(**)$ we have $X^2 - wX + 1=0$, so $X^2 = wX -1$.
3088
3089Thus 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 $\{1, X\}$.
3090
3091Fitting $X^2$ into $(*)$,
3092\begin{align*}
3093  X^2 &- Y^3 -1 =0\\
3094  Y^3 &= X^2-1\\
3095  Y^3 &= wX -2
3096\end{align*}
3097
3098thus any power of $Y$ higher than $2$ can be reduced (eg. $Y^4 = Y(wX-2) = w(XY) - 2Y$).
3099
3100So its partial basis is $\{1, Y, Y^2\}$.
3101
3102For $Z$, since $XZ=1$ and $w=X+Z \Longrightarrow Z=w-X$, thus $Z$ is a $B$-linear combination of $\{1,X\}$.
3103
3104\vspace{0.2cm}
3105Combining the previous partial basis, the generators are
3106$$\{ 1, X \} \times \{ 1,Y, Y^2 \} = \{ 1, Y, Y^2, X, XY, XY^2 \}$$
3107
3108\end{proof}
3109
3110
3111\subsection{Exercises Chapter 6}
3112
3113\begin{ex}{R.6.3.a}
3114  Let $A = A' \times A''$; prove that $A'$ and $A''$ are rings of fractions of $A$.
3115\end{ex}
3116\begin{proof}
3117  ring of fractions of $A$ = $S^{-1}A$, so want to prove that $S^{-1}A \cong A'$.
3118
3119  Localization map
3120  \begin{align*}
3121    \psi: A &\longrightarrow A'\\
3122    (a', a'') &\longmapsto a'
3123  \end{align*}
3124
3125  note that $\psi$ is surjective, since $\forall~ a' \in A',~ \psi(a', 0)=a'$.
3126
3127  Let the multiplicative set $S$ be $S=\{ e_1 \}$ with $e_1 = (1,0)$.
3128
3129  Want $\underbrace{S^{-1}A}_{\text{localization}} \cong A'$.
3130
3131  In $S^{-1}A$, $x \in A$ maps to $0$ iff $s x = 0$ for some $s \in S$.
3132  
3133  Let $x=(a', a'')\in A$; then $sx=0 ~\Longrightarrow~ s \cdot (a', a'') =0$,
3134
3135  with $s \in S$, so $s=(1,0)$, hence
3136  $$(1,0)\cdot(a',a'')=(0,0)$$
3137  $$\Longrightarrow~ (a', 0) = (0, 0)$$
3138  \hspace*{4em} which implies $a' = 0$, so that $(a', a'')=(0, a')$.
3139
3140  $\Longrightarrow$ the elements that become zero in the localization are of the form $(0, a'')$
3141
3142  $$ker(\psi) = \{ (a',a'')\in A ~|~ \psi(a', a'')=0 \} = \{(0, a'') ~|~ a'' \in A'' \}$$
3143
3144  By the 1st isomorphism theorem, 
3145
3146  \begin{tikzpicture}[node distance=1.5cm, auto]
3147    \node (G)    {$A$};
3148    \node (H) [right of=G] {$A'$};
3149    \node (GmodK) [below of=G, xshift=0.75cm] {$\frac{A}{ker(\psi)}$};
3150
3151    \draw[->] (G) to node {$\psi$} (H);
3152    \draw[->] (G) to node [swap] {$\phi$} (GmodK);
3153    \draw[<->] (GmodK) to node [swap] {$\eta$} (H);
3154  \end{tikzpicture}
3155  
3156  $\Longrightarrow~ \frac{A}{ker(\psi)} \cong A'$
3157
3158
3159  \vspace{0.5cm}
3160  Take the localization map
3161  \begin{align*}
3162    \phi: A &\longrightarrow S^{-1}A\\
3163    a &\longmapsto a/1
3164  \end{align*}
3165
3166  $$ker(\phi) = \{ x \in A ~|~ sx=0 ~\text{for some}~ s \in S \}$$
3167
3168  and we've seen that $a'=0$, so $x \in A ~\Rightarrow~ x=(a', a'') = (0, a'')$
3169
3170  $$ker(\phi) = \{ (0, a'') ~|~ a'' \in A'' \}$$
3171
3172  which is the same as $ker(\psi)$; $ker(\psi) = ker(\phi)$.
3173
3174  \vspace{0.2cm}
3175  $\Longrightarrow~~ A'$ and $S^{-1}A$ are both surjective images of $A$ with exact same kernel,
3176  thus $A' \cong S^{-1}A$.
3177\end{proof}
3178
3179\begin{ex}{R.6.4}
3180  \begin{enumerate}[a.]
3181  \item Give an example of a ring $A$ and distinct multiplicative sets $S,~ T$ such that $S^{-1}A = T^{-1}A$.
3182  \item Prove that for fixed $S$, there is a maximal multiplicative set $T$ with this property defined by
3183    $$T = \{ t \in A ~|~ at \in S ~\text{for some}~ a \in A \}$$
3184  \end{enumerate}
3185\end{ex}
3186\begin{proof}
3187  \begin{enumerate}[a.]
3188    \item
3189  (\emph{saturated sets})\\
3190
3191  General rule of saturation:\\
3192  two multiplicative sets $S,~T$ yield the same localization, $S^{-1}A = T^{-1}A$, iff the have the same saturation.
3193
3194  Saturation of $S$: $\hat{S}$, set of all elements in $A$ that divide some element of $S$:
3195  $$\hat{S} = \{ a \in A ~|~ \exists~ b \in A ~\text{s.th.}~ ab \in S \}$$
3196
3197  \vspace{0.4cm}
3198  Example 1: $A=\mathbb{Z}$,\\
3199
3200  $$S = \{ 2^n ~|~ n \geq 0 \} = \{ 1, 2, 4, 8, 16, \ldots \}$$
3201  localization:
3202  $$S^{-1} \mathbb{Z} = \{ \frac{a}{2^n} ~|~ a \in \mathbb{Z},~ n \in \mathbb{N} \}$$
3203
3204  $$T = \{ \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \ldots \}$$
3205
3206  Notice that despite $S \neq T$, we have $S^{-1}\mathbb{Z} = T^{-1} \mathbb{Z}$,\\
3207  since once $n \in \mathbb{Z}$ is invertible, $-n$ is automatically invertible: $(-n)^{-1} = - (n^{-1})$.
3208
3209  \vspace{0.4cm}
3210  Example 2: $A= k[x]$,
3211  $$S = \{ x^n ~|~ n \geq 0 \}$$
3212  $$T = \{ (2x)^n ~|~ n \geq 0 \}$$
3213  $$\Longrightarrow~ S^{-1}A = T^{-1} A = k[x, x^{-1}]$$
3214
3215  \item
3216    Show that $T$ is a multiplicative set:
3217
3218    (must contain $1$ and be closed under multiplication)\\
3219    Since $1 \in A$ and $1 \cdot 1 = 1 \in S$, then $1 \in T$.
3220
3221    Let $t_1,~ t_2 \in T$; by definition of $T$,
3222    $$\exists~ a_1, a_2 \in A ~\text{such that}~ a_1 t_1,~ a_2 t_2 \in S$$
3223
3224    Since $S$ a multiplicative set, $(a_1 t_1)(a_2 t_2) \in S$\\
3225    rearrange it: $(a_1 a_2)(t_1 t_2) \in S$.\\
3226    Since $a_1 a_2 \in A,~~ t_1 t_2 \in T$.
3227
3228    Thus $T$ is a multiplicative set.
3229
3230    Next, we show that $T$ is maximal:
3231
3232    Suppose $U$ is the maximal instead of $T$, such that $U^{-1}A \cong S^{-1}A$.
3233
3234    Let $u \in U$; the image of $u$ in $S^{-1}A$ must be a unit.
3235
3236    Units in $S^{-1}A$ are the fractions $\frac{s}{a}$ such that their inverse exists.\\
3237    \hspace*{2em} $\exists~ x=\frac{a}{s}$ such that $u \frac{a}{s} = \frac{1}{1}$\\
3238    \hspace*{2em} $\Longrightarrow~ \frac{ua}{s} = \frac{1}{1} ~\Longrightarrow~ \exists s' \in S$ such that $s'(ua-s)=0$\\
3239    $\Longrightarrow~ s'ua=s's \in S$
3240
3241    since $s,s' \in S$, their product is also in $S$.
3242
3243    Let $b=s'a$, then $ub \in S$.
3244
3245    By definition of $T$, $u \in T$.
3246
3247    Therefore, $U \subseteq T$; thus $T$ is maximal.
3248\end{enumerate}
3249\end{proof}
3250
3251
3252\bibliographystyle{unsrt}
3253\bibliography{commutative-algebra-notes.bib}
3254
3255\end{document}