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\newpage
2327
2328\section{Exercises}
2329
2330For the exercises, I follow the assignments listed at \cite{mit-course}.
2331
2332The 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}.
2333
2334\subsection{Exercises Chapter 1}
2335
2336\begin{ex}{R.1.1}
2337  Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$?
2338\end{ex}
2339\begin{proof}
2340  Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$.
2341
2342  $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\
2343  Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$.
2344
2345  Thus $I \cup J$ is not closed under addition; thus is not an ideal.
2346
2347
2348  Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum:
2349
2350  $$I+J = \{ a+b | a \in I, b \in J \}$$
2351
2352  $gcd(2,3)=1$, so $I+J = \mathbb{Z}$.
2353
2354  Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$.
2355\end{proof}
2356
2357\begin{ex}{R.1.5}
2358  let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\
2359  In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$.
2360\end{ex}
2361\begin{proof}
2362  (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$).
2363
2364  Let
2365  $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
2366  The claim is that $\psi^{-1}(P)$ is prime ideal of $A$.
2367
2368  \begin{enumerate}[i.]
2369    \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
2370      $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$).
2371
2372      If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so
2373      $$\psi(a-b)= \psi(a) - \psi(b) \in P$$
2374      hence $a-b \in \psi^{-1}(P)$.
2375
2376      If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\
2377      Thus $ra \in \psi^{-1}(P)$.
2378
2379      $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$.
2380
2381    \item show that $\psi^{-1}(P)$ is prime:\\
2382      $\psi^{-1}(P) \neq A$, 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 $B$.
2383
2384      Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$.
2385
2386      Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\
2387      Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$.
2388
2389      Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$.
2390  \end{enumerate}
2391\end{proof}
2392
2393
2394\begin{ex}{R.1.6}
2395  prove or give a counter example:
2396  \begin{enumerate}[a.]
2397    \item the intersection of two prime ideals is prime
2398    \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
2399    \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
2400    \item the map $\psi^{-1}$ of Proposition 1.2 takes maximal ideals of $A/I$
2401      to maximal ideals of $A$
2402  \end{enumerate}
2403\end{ex}
2404\begin{proof}
2405  \begin{enumerate}[a.]
2406    \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime.
2407
2408      Then $I \cap J = (2) \cap (3) = (6)$.
2409
2410      The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$.
2411
2412      Thus the intersection of two primes can not be prime.
2413
2414    \item $P_1=(2),~ P_2=(3)$, both prime.
2415
2416      Then,
2417      $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$
2418
2419      $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$.
2420
2421      (recall: principal= generated by a single element)
2422
2423      So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$.
2424
2425      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.
2426
2427      Henceforth, the sum of two prime ideals is not necessarily prime.
2428
2429    \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$.
2430
2431      Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$.
2432
2433      Then
2434      \begin{align*}
2435        \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\
2436        \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A
2437      \end{align*}
2438
2439      But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
2440
2441      Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
2442
2443    \item $\psi: A \longrightarrow A/I$ quotient homomorphism, $I \subseteq A$ an ideal.
2444
2445      Let $M$ a maximal ideal of $A/I$, then $\frac{(A/I)}{M}$ is a field (Proposition 1.3).
2446
2447      By the isomorphism theorems,
2448      $$\frac{(A/I)}{M} \cong \frac{A}{\psi^{-1}(M)}$$
2449
2450
2451      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 $A$.
2452
2453      $\Longrightarrow~$ under $\psi$, preimages of maximal ideals are maximal.
2454  \end{enumerate}
2455\end{proof}
2456
2457\begin{ex}{R.1.12.a}
2458  if $I,J$ ideals and $P$ prime ideal, prove that
2459  $$IJ \subset P ~\Longleftrightarrow~ I \cap J \subset P ~\Longleftrightarrow~ I ~\text{or}~ J \subset P$$
2460\end{ex}
2461\begin{proof}
2462  assume $I \subseteq P$ (for $J \subseteq P$ will be the same, symmetric), take $x \in IJ$,
2463
2464  then
2465  $$x = \sum_{k=1}^n a_k b_k$$
2466  with $a_k \in I,~ b_k \in J$.
2467
2468  Each $a_k \in I \subseteq P$. Since $P$ an ideal,
2469  $$\sum_{k=1}^n a_k b_k \in P$$
2470
2471  thus $x \in P$, hence $IJ \subseteq P$.
2472
2473  So $I \subseteq P$ or $J \subseteq P$ $~\Longrightarrow IJ \subseteq P$.
2474
2475  \vspace{0.5cm}
2476  Conversely,\\
2477  assume $P$ prime and $IJ \subseteq P$.
2478
2479  Suppose by contradiction that $I \not\subseteq P$ and $J \not\subseteq P$.
2480  \begin{itemize}
2481    \item[-] since $I \not\subseteq P,~ \exists a \in I$ with $a \not\in P$
2482    \item[-] since $J \not\subseteq P,~ \exists b \in J$ with $b \not\in P$
2483  \end{itemize}
2484  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 $P$.
2485
2486  Thus $I \not\subseteq P$ and $J \not\subseteq P$ are false.
2487
2488  \vspace{0.3cm}
2489  So both directions are proven, hence
2490  $$IJ \subseteq P ~\Longrightarrow~ I \subseteq P ~\text{or}~ J \subseteq P$$
2491
2492\end{proof}
2493
2494\begin{ex}{R.1.18}
2495  Use Zorn's lemma to prove that any prime ideal $P$ contains a minimal prime ideal.
2496\end{ex}
2497\begin{proof}
2498  Let $P$ prime ideal of $R$.
2499
2500  $$S = \{ Q \subseteq R ~|~ Q ~\text{a prime ideal AND}~ Q \subseteq P \}$$
2501
2502  Goal: show that $S$ has a minimal element, the minimal ideal contained in $P$.
2503
2504  $P \subset S$, so $S$ is nonempty.
2505
2506  Let $C \subseteq S$ be a chain (= totally ordered subset) with respect to inclusion.
2507
2508  Define
2509  $$Q_C = \bigcap_{Q \in C} Q$$
2510
2511  Clearly $Q_C \subseteq P$, since each $Q \in C$ is $Q \subseteq P$.
2512
2513  Since $C$ is ordered by inclusion, it is a decreasing chain of prime ideals.
2514
2515  Intersection of a decreasing chain of prime ideals is again a prime ideal:
2516  \begin{itemize}
2517    \item[-] if $ab \in Q_C$, then $ab \in Q ~\forall Q \in C$
2518    \item[-] since $Q$ prime, $\forall Q \in C$ either $a \in Q$ or $b \in Q$
2519  \end{itemize}
2520
2521  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 $Q_2 \subseteq Q_1$.
2522
2523  In either case: contradiction, since the smaller one would have to contain the element that was assumed to be excluded.
2524
2525  Thus $\forall Q \in C$ the same element $a, b$ must lie in all $Q$. $\Longrightarrow~$ lies in the intersection of them, $Q_C$.
2526
2527  Henceforth, $Q_C$ is a prime ideal and lies in $S$, and its a lower bound of $C$ in $S$.
2528
2529  Now, $S$ is nonempty, and every chain in $S$ has a lower bound in $S$ (its intersection).\\
2530  Therefore, $S$ has a minimal element $P_{min}$.
2531
2532  By construction, $P_{min}$ is a prime ideal $P_{min} \subseteq P$, and by minimality there are no strictly smaller prime ideals inside $P$.
2533
2534  So $P_{min}$ is a minimal prime ideal, contained in $P$.
2535\end{proof}
2536
2537\begin{ex}{R.1.10}
2538\end{ex}
2539\begin{proof}
2540\end{proof}
2541
2542\begin{ex}{R.1.11}
2543\end{ex}
2544\begin{proof}
2545\end{proof}
2546
2547\begin{ex}{R.1.4}
2548\end{ex}
2549\begin{proof}
2550\end{proof}
2551
2552\subsection{Exercises Chapter 2}
2553
2554\begin{ex}{R.2.9}
2555$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ is a s.e.s. of $A$-modules. Prove that if $N, L$ are finite over $A$, then $M$ is finite over $A$.
2556\end{ex}
2557\begin{proof}
2558  Denote the generators of $L$ and $N$ respectively as
2559  \begin{align*}
2560    \{l_1, \ldots, l_k \} &\subseteq L\\
2561    \{n_1, \ldots, n_p \} &\subseteq N
2562  \end{align*}
2563
2564  By s.e.s. definition, 
2565  \begin{itemize}
2566    \item[-] $\alpha$ is injective (one-to-one), so
2567    $$\forall l_i \in L,~ \exists~ x_i \in M ~\text{s.th.}~ \alpha(l_i)=x_i$$
2568
2569  \item[-] $\beta$ is surjective (onto), so
2570    $$\forall n_j \in N,~ \exists~ y_j \in M ~\text{s.th.}~ \beta(y_j)=n_j$$
2571\end{itemize}
2572
2573  We will show that $\{x_1, \ldots, x_k, y_1, \ldots, y_p \}$ generate $M$, and thus $M$ is finite:
2574
2575  Let $m \in M$, then $\beta(m) \in N$, and
2576  $$\beta(m) = \sum_{j=1}^p a_j n_j ~~~\text{with}~ a_j \in A$$
2577
2578  Take $m' \in M$, with $m' = \sum a_j y_j$, then
2579  $$\beta(m') = \sum a_j \beta(y_j) = \sum a_j n_j = \beta(m)$$
2580
2581  Then, since $\beta(m)=\beta(m')~~ \Longrightarrow~~ \beta(m-m')=0$, thus
2582  $$(m-m') \in ker(\beta)$$
2583
2584  By \emph{exactness} property, since $\alpha: L \longrightarrow ker(\beta)$, we have $ker(\beta)=im(\alpha)$.
2585
2586  Therefore, $\exists~ l \in L$ such that $\alpha(l)= m-m'$.
2587
2588  Since $\{l_i\}_k$ generate $L$,
2589  $$l = \sum^k b_i l_i$$
2590  thus
2591  $$m-m' = \alpha(l) = \alpha(\underbrace{\sum b_i l_i}_{l}) = \sum b_i \underbrace{\alpha(l_i)}_{x_i} = \sum b_i x_i$$
2592
2593  Rearrange,
2594  $$m = m' + \sum b_i x_i = \sum_{j=1}^p a_j y_j + \sum_{i=1}^k b_i x_i ~~~~~ \forall m \in M$$
2595
2596  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.\\
2597  Thus $M$ is finitely generated over $A$.
2598\end{proof}
2599
2600
2601\subsection{Exercises Chapter 3}
2602
2603\begin{ex}{R.3.2}
2604  $K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space.
2605   Prove that $A$ is Noetherian and Artinian.
2606\end{ex}
2607\begin{proof}
2608  $dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in \aA$ and $c \in K$, then $c \cdot x \in \aA$.
2609
2610  \begin{enumerate}
2611    \item Noetherian:\\
2612      let $I_1 \subseteq I_2 \subseteq \ldots$ be an ascending chain of ideals in $A$.
2613
2614      Since each $I_i$ is a subspace, we have
2615      $$dim_K(I_1) \leq dim_K(I_2) \leq \ldots \leq n$$
2616      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.
2617      
2618    \item Artinian:\\
2619      Similarly, if $I_1 \supseteq I_2 \supseteq \ldots$ a descending chain of ideals in $A$.
2620
2621      then
2622      $$n \geq dim_K(I_1) \geq dim_K(I_2) \geq \ldots \geq 0$$
2623      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.
2624  \end{enumerate}
2625\end{proof}
2626
2627\begin{ex}{R.3.5}
2628  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 $M$.
2629
2630  Prove if the following holds or not:
2631  $$\beta(M_1)=\beta(M_2) ~\text{and}~ \alpha^{-1}(M_1)=\alpha^{-1}(M_2)~~\Longrightarrow~ M_1=M_2$$
2632\end{ex}
2633\begin{proof}
2634  Counterexample showing that it does not hold:
2635
2636  Let $K$ a field, $M = K \oplus K~, L=K,~N=K$.
2637
2638  Set, for $l \in L,~ (m_1, m_2) \in M$,
2639  \begin{align*}
2640    \alpha:~ &l \longmapsto (l, 0)\\
2641    \beta:~ &(m_1, m_2) \longmapsto m_2
2642  \end{align*}
2643
2644  So we have
2645  $$0 \longrightarrow K \stackrel{\alpha}{\longrightarrow} K^2 \stackrel{\beta}{\longrightarrow} K \longrightarrow 0$$
2646
2647  Then,
2648  \begin{align*}
2649    M_1 &= \{ (x, x) ~|~ x \in K\} ~~~~\sim\text{(diagonal line)} \\
2650    M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)}
2651  \end{align*}
2652
2653  (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 $M$).
2654
2655  Then,
2656  \begin{align*}
2657    \beta(M_1) &= \{ x ~|~ x \in K\} = K \\
2658    \beta(M_2) &= \{ x ~|~ x \in K\} = K
2659  \end{align*}
2660  thus, $\beta(M_1) = \beta(M_2)$.
2661
2662  For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1) = \{0\}$,\\
2663  for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2) = \{0\}$,\\
2664  thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$.
2665
2666
2667  So we've seen that
2668  \begin{align*}
2669    \beta(M_1) = \beta(M_2)\\
2670    \alpha^{-1}(M_1)=\alpha^{-1}(M_2)
2671  \end{align*}
2672  while having $M_1 \neq M_2$.
2673\end{proof}
2674
2675
2676
2677\begin{ex}{R.3.3}
2678Let $A$ a ring, $I_1, \ldots, I_k$ ideals such that each $A/I_i$ is a Noetherian ring.
2679Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bigcap I_i = 0$ then $A$ is also Noetherian.
2680\end{ex}
2681\begin{proof}
2682\begin{enumerate}[i.]
2683  \item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian.
2684    $\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian.
2685  \item Take the canonical homomorphism
2686    $$\phi: A \longrightarrow \bigoplus_{i=1}^n A/ I_i$$
2687    by $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$.
2688
2689    $\phi$ is injective: $ker(\phi)= \{ a \in A | a \in I_i \forall i \}$.
2690
2691    Since we're given $\cap I_i = 0$, then $ker(\phi)=\cap I_i$, and $\phi$ is injective.
2692
2693    Thus, $\phi$ is the isomorphism $A \cong im(\phi)$, where $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$.
2694
2695    We know that any submodule of a Noetherian module is Noetherian, thus, since
2696    \begin{itemize}
2697      \item $A/I_i$ is Noetherian by hypothesis of the exercise
2698      \item $A \cong im(\phi)$
2699      \item $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$
2700  \end{itemize}
2701  then, $A$ is Noetherian.
2702\end{enumerate}
2703\end{proof}
2704
2705\begin{ex}{R.3.4}
2706  Prove that if A is a Noetherian ring and M a finite A-module, then there
2707  exists an exact sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$.
2708  That is, M has a presentation as an A-module in terms of finitely many generators and relations.
2709\end{ex}
2710\begin{proof}
2711  since $M$ fingen $~\Longrightarrow~$ generators $\{m_1, \ldots, m_2 \} \subseteq M$ span $M$.
2712
2713  Let $\beta$ be a surjective $A$-linear map, which forms a free $A$-module of rank $p$ onto $M$:
2714  \begin{align*}
2715    \beta: A^p &\longrightarrow M\\
2716    (a_1, \ldots, a_p) &\longmapsto \sum_{i=1}^p a_i m_i
2717  \end{align*}
2718
2719  Let $K=ker(\beta)$. By the 1st Isomorphism Theorem,
2720  $$M \cong A^p / K$$
2721
2722  Since $A$ is a Noetherian ring, then every free $A$-module of finite rank (eg. $A^p$) is a Noetherian module.
2723
2724  Every submodule of a Noetherian module is fingen.
2725
2726  $\Longrightarrow~$ since $K \subseteq A^p, ~\Longrightarrow~ K ~~(=ker(\beta))$ is fingen.
2727
2728  Since $K$ fingen, let $\{k_1, \ldots, l_q\}$ be generators of $K$.
2729
2730  Define $\psi: A^q \longrightarrow K$.
2731
2732  Compose it with the inclusion map $i: K \longrightarrow A^p$,
2733  $$\alpha = i \circ \psi:~ A^q \longrightarrow A^p$$
2734
2735  So we have the whole sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$, where
2736  \begin{itemize}
2737    \item $\beta$ is surjective
2738    \item $im(\alpha)=ker(\beta)$
2739  \end{itemize}
2740  so that it is a exact sequence, thus, $M$ has a finite presentation.
2741\end{proof}
2742
2743\subsection{Exercises Chapter 4}
2744
2745\begin{ex}{R.4.1.a}
2746  $k[X^2] \subset k[X]$ is a finite extension, hence integral. Find the integral dependence relation for any $f \in k[X]$.
2747\end{ex}
2748\begin{proof}
2749  $\forall f(X) \in k[X]$ can be uniquely decomposed into its even and odd parts:
2750  $$f(X) = p(X^2) + X \cdot q(X^2)$$
2751
2752  with $p(X^2),~ q(X^2) \in k[X^2]$, and\\
2753  $p(X^2)$: sum of all terms with even exponents\\
2754  $q(X^2)$: sum of all terms with odd exponents, and then factoring out $X$.
2755
2756  \vspace{0.3cm}
2757  (Observation: this is used in FRI cryptographic protocol\\
2758  \href{https://github.com/arnaucube/math/blob/master/notes_fri_stir.pdf}{https://github.com/arnaucube/math/blob/master/notes\_fri\_stir.pdf})
2759
2760
2761  Rearrange it
2762  \begin{align*}
2763    f(X) &- p(X^2) = X \cdot q(X^2), ~~\text{square:}\\
2764    (f(X) &- p(X^2))^2 = X^2 \cdot q(X^2)^2\\
2765    f(X)^2-2 p(X^2) f(X) &+ p(X^2)^2 = X^2 \cdot q(X^2)^2\\
2766    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
2767  \end{align*}
2768
2769   Denote the last polynomial as $P(T) \in k[X^2]$, where $f(X)$ is a root of $P(T)$.
2770
2771   \vspace{0.2cm}
2772   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 $a_i \in k[X^2]$.
2773
2774   We have that
2775   \begin{align*}
2776     a_1 &= -2 p(X^2)\\
2777     a_0 &= p(X^2)^2 - X^2 q(X^2)^2
2778   \end{align*}
2779
2780
2781   So for example, for $f(X)= X^3 + X^2 + X + 1$:
2782   \begin{align*}
2783     f(X) = (X^2 + 1) &+ X (X^2 + 1)\\
2784     (f(X) - (X^2+1))^2 &= X^2 (X^2 + 1)^2\\
2785     (f(X) - p(X))^2 &= X^2 (q(X))^2
2786   \end{align*}
2787\end{proof}
2788
2789
2790\begin{ex}{R.4.5}
2791  Let $A=k[X,Y]/(Y^2 - X^2 - X^3)$. Prove that the normalization of $A$ is $k[t]$ where $t=Y/X$.
2792\end{ex}
2793\begin{proof}
2794$A=k[X,Y]/(Y^2 - X^2 - X^3)$, express $X$ and $Y$ in terms of $t$:
2795
2796Since $t=Y/X$ then $Y=tX$, and combined with $Y^2 = X^2 + X^3$, then
2797\begin{align*}
2798  (tX)^2 &= X^2 + X^3\\
2799  t^2 X^2 &= X^2 + X^3, ~~\text{assuming}~X \neq 0:\\
2800  t^2 &= 1+X, ~\text{thus}\\
2801  X&=t^2-1 \in k[X]
2802\end{align*}
2803
2804Then, $Y=tX=t(t^2-1)=t^3-t \in k[X]$.
2805
2806Hence $X, Y \in k[X]$.
2807
2808Therefore, $k[X,Y]/(Y^2 - X^2 - X^3) \subseteq k[t]$.
2809
2810\vspace{0.4cm}
2811By \ref{noether-normalization} (Noether normalization lemma), to show that $k[t]$ is the \emph{normalization}, must show that $k[t]$ is \emph{integral} over $A$.
2812
2813From $X=t^2-1 ~~\Longrightarrow~~ t^2-1-X=0 ~~\Longrightarrow~ t^2 - (1+X) = 0$.
2814
2815$(1+X) \in A$, so $t$ satisfies the monic polynomial
2816$$P(T) = T^2 - (1+X) \in A[T]$$
2817Thus $t$ is integral over $A$.
2818
2819Since $k[t]$ is generated by $t$ over $k$, and $k \subset A$, then the entire ring $k[t]$ is integral over $A$.
2820
2821Since $k[t]$ is a polynomial ring over a field, which is a UFD, it is integrally closed (since all UFD are integrally closed).
2822
2823$Frac A = k(X,Y)$, since $X=t^2-1, ~Y=t^3-t ~~\Longrightarrow~ k(X,Y) \subseteq k(t)$
2824
2825and $t=Y/X \in k(X,Y)$, thus $k(X,Y)=k(t)$.
2826
2827Since $k[t]$ is integrally closed and is the integral closure of $A$ in its fraction field $k(t)$, we conclude that the normalization of $A$ is $k[t]$.
2828\end{proof}
2829
2830
2831
2832\begin{ex}{R.4.9}
2833  $k$ 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.
2834\end{ex}
2835\begin{proof}
2836  (want to find a linear combination of the coordinates such that the original variables satisfy monic polynomials over the new ring $B$)
2837
2838  The relations defining $A$ are
2839  \begin{align*}
2840    X^2-Y^3-1=0 ~~&\Longrightarrow~ Y^3 = X^2 -1 ~~~(*)\\
2841    XZ -1=0 ~~&\Longrightarrow~ Z= 1/X = X^{-1}
2842  \end{align*}
2843
2844Thus $A$ can be denoted as $A = k[X,Y,X^{-1}]/(Y^3 - X^2 - 1)$.
2845
2846
2847Now, $Y$ is inegral over $k[X]$, since $Y^3 - (X^2 - 1) = 0$ is monic in $Y$ with coefficients in $k[X]$.
2848
2849$Z$ is not integral over $k[X]$, since $Z=1/X$ and $X$ is not a unit in $k[X]$.
2850
2851Choose $\alpha, \beta \in k$ such that $X$ (and thus $Z$) becomes integral over $B$:\\
2852set $\alpha=0,~\beta=1 ~~\Longrightarrow~~ B=k[X+\alpha Y + \beta Z]= k[X+Z]$.
2853
2854Let $w=X+Z$; since $XZ=1$, we have
2855$$w=X+\frac{1}{X} ~~\Longrightarrow~~ Xw=X^2+1 ~~\Longrightarrow~~ X^2 -w X +1 = 0 ~~(**)$$
2856
2857which is monic with coefficients in $k[w]$, thus $X$ is integral over $B$.
2858
2859Since $Z=w-X$, $Z$ is also integral.
2860
2861\vspace{0.4cm}
2862Generators of $A$ as a $B$-module:
2863
2864we hadd $B=k[w]$ with $w=X+Z$.
2865
2866From $(**)$ we have $X^2 - wX + 1=0$, so $X^2 = wX -1$.
2867
2868Thus 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 $\{1, X\}$.
2869
2870Fitting $X^2$ into $(*)$,
2871\begin{align*}
2872  X^2 &- Y^3 -1 =0\\
2873  Y^3 &= X^2-1\\
2874  Y^3 &= wX -2
2875\end{align*}
2876
2877thus any power of $Y$ higher than $2$ can be reduced (eg. $Y^4 = Y(wX-2) = w(XY) - 2Y$).
2878
2879So its partial basis is $\{1, Y, Y^2\}$.
2880
2881For $Z$, since $XZ=1$ and $w=X+Z \Longrightarrow Z=w-X$, thus $Z$ is a $B$-linear combination of $\{1,X\}$.
2882
2883\vspace{0.2cm}
2884Combining the previous partial basis, the generators are
2885$$\{ 1, X \} \times \{ 1,Y, Y^2 \} = \{ 1, Y, Y^2, X, XY, XY^2 \}$$
2886
2887\end{proof}
2888
2889
2890\subsection{Exercises Chapter 6}
2891
2892\begin{ex}{R.6.3.a}
2893  Let $A = A' \times A''$; prove that $A'$ and $A''$ are rings of fractions of $A$.
2894\end{ex}
2895\begin{proof}
2896  ring of fractions of $A$ = $S^{-1}A$, so want to prove that $S^{-1}A \cong A'$.
2897
2898  Localization map
2899  \begin{align*}
2900    \psi: A &\longrightarrow A'\\
2901    (a', a'') &\longmapsto a'
2902  \end{align*}
2903
2904  note that $\psi$ is surjective, since $\forall~ a' \in A',~ \psi(a', 0)=a'$.
2905
2906  Let the multiplicative set $S$ be $S=\{ e_1 \}$ with $e_1 = (1,0)$.
2907
2908  Want $\underbrace{S^{-1}A}_{\text{localization}} \cong A'$.
2909
2910  In $S^{-1}A$, $x \in A$ maps to $0$ iff $s x = 0$ for some $s \in S$.
2911  
2912  Let $x=(a', a'')\in A$; then $sx=0 ~\Longrightarrow~ s \cdot (a', a'') =0$,
2913
2914  with $s \in S$, so $s=(1,0)$, hence
2915  $$(1,0)\cdot(a',a'')=(0,0)$$
2916  $$\Longrightarrow~ (a', 0) = (0, 0)$$
2917  \hspace*{4em} which implies $a' = 0$, so that $(a', a'')=(0, a')$.
2918
2919  $\Longrightarrow$ the elements that become zero in the localization are of the form $(0, a'')$
2920
2921  $$ker(\psi) = \{ (a',a'')\in A ~|~ \psi(a', a'')=0 \} = \{(0, a'') ~|~ a'' \in A'' \}$$
2922
2923  By the 1st isomorphism theorem, 
2924
2925  \begin{tikzpicture}[node distance=1.5cm, auto]
2926    \node (G)    {$A$};
2927    \node (H) [right of=G] {$A'$};
2928    \node (GmodK) [below of=G, xshift=0.75cm] {$\frac{A}{ker(\psi)}$};
2929
2930    \draw[->] (G) to node {$\psi$} (H);
2931    \draw[->] (G) to node [swap] {$\phi$} (GmodK);
2932    \draw[<->] (GmodK) to node [swap] {$\eta$} (H);
2933  \end{tikzpicture}
2934  
2935  $\Longrightarrow~ \frac{A}{ker(\psi)} \cong A'$
2936
2937
2938  \vspace{0.5cm}
2939  Take the localization map
2940  \begin{align*}
2941    \phi: A &\longrightarrow S^{-1}A\\
2942    a &\longmapsto a/1
2943  \end{align*}
2944
2945  $$ker(\phi) = \{ x \in A ~|~ sx=0 ~\text{for some}~ s \in S \}$$
2946
2947  and we've seen that $a'=0$, so $x \in A ~\Rightarrow~ x=(a', a'') = (0, a'')$
2948
2949  $$ker(\phi) = \{ (0, a'') ~|~ a'' \in A'' \}$$
2950
2951  which is the same as $ker(\psi)$; $ker(\psi) = ker(\phi)$.
2952
2953  \vspace{0.2cm}
2954  $\Longrightarrow~~ A'$ and $S^{-1}A$ are both surjective images of $A$ with exact same kernel,
2955  thus $A' \cong S^{-1}A$.
2956\end{proof}
2957
2958\begin{ex}{R.6.4}
2959  \begin{enumerate}[a.]
2960  \item Give an example of a ring $A$ and distinct multiplicative sets $S,~ T$ such that $S^{-1}A = T^{-1}A$.
2961  \item Prove that for fixed $S$, there is a maximal multiplicative set $T$ with this property defined by
2962    $$T = \{ t \in A ~|~ at \in S ~\text{for some}~ a \in A \}$$
2963  \end{enumerate}
2964\end{ex}
2965\begin{proof}
2966  \begin{enumerate}[a.]
2967    \item
2968  (\emph{saturated sets})\\
2969
2970  General rule of saturation:\\
2971  two multiplicative sets $S,~T$ yield the same localization, $S^{-1}A = T^{-1}A$, iff the have the same saturation.
2972
2973  Saturation of $S$: $\hat{S}$, set of all elements in $A$ that divide some element of $S$:
2974  $$\hat{S} = \{ a \in A ~|~ \exists~ b \in A ~\text{s.th.}~ ab \in S \}$$
2975
2976  \vspace{0.4cm}
2977  Example 1: $A=\mathbb{Z}$,\\
2978
2979  $$S = \{ 2^n ~|~ n \geq 0 \} = \{ 1, 2, 4, 8, 16, \ldots \}$$
2980  localization:
2981  $$S^{-1} \mathbb{Z} = \{ \frac{a}{2^n} ~|~ a \in \mathbb{Z},~ n \in \mathbb{N} \}$$
2982
2983  $$T = \{ \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \ldots \}$$
2984
2985  Notice that despite $S \neq T$, we have $S^{-1}\mathbb{Z} = T^{-1} \mathbb{Z}$,\\
2986  since once $n \in \mathbb{Z}$ is invertible, $-n$ is automatically invertible: $(-n)^{-1} = - (n^{-1})$.
2987
2988  \vspace{0.4cm}
2989  Example 2: $A= k[x]$,
2990  $$S = \{ x^n ~|~ n \geq 0 \}$$
2991  $$T = \{ (2x)^n ~|~ n \geq 0 \}$$
2992  $$\Longrightarrow~ S^{-1}A = T^{-1} A = k[x, x^{-1}]$$
2993
2994  \item
2995    Show that $T$ is a multiplicative set:
2996
2997    (must contain $1$ and be closed under multiplication)\\
2998    Since $1 \in A$ and $1 \cdot 1 = 1 \in S$, then $1 \in T$.
2999
3000    Let $t_1,~ t_2 \in T$; by definition of $T$,
3001    $$\exists~ a_1, a_2 \in A ~\text{such that}~ a_1 t_1,~ a_2 t_2 \in S$$
3002
3003    Since $S$ a multiplicative set, $(a_1 t_1)(a_2 t_2) \in S$\\
3004    rearrange it: $(a_1 a_2)(t_1 t_2) \in S$.\\
3005    Since $a_1 a_2 \in A,~~ t_1 t_2 \in T$.
3006
3007    Thus $T$ is a multiplicative set.
3008
3009    Next, we show that $T$ is maximal:
3010
3011    Suppose $U$ is the maximal instead of $T$, such that $U^{-1}A \cong S^{-1}A$.
3012
3013    Let $u \in U$; the image of $u$ in $S^{-1}A$ must be a unit.
3014
3015    Units in $S^{-1}A$ are the fractions $\frac{s}{a}$ such that their inverse exists.\\
3016    \hspace*{2em} $\exists~ x=\frac{a}{s}$ such that $u \frac{a}{s} = \frac{1}{1}$\\
3017    \hspace*{2em} $\Longrightarrow~ \frac{ua}{s} = \frac{1}{1} ~\Longrightarrow~ \exists s' \in S$ such that $s'(ua-s)=0$\\
3018    $\Longrightarrow~ s'ua=s's \in S$
3019
3020    since $s,s' \in S$, their product is also in $S$.
3021
3022    Let $b=s'a$, then $ub \in S$.
3023
3024    By definition of $T$, $u \in T$.
3025
3026    Therefore, $U \subseteq T$; thus $T$ is maximal.
3027\end{enumerate}
3028\end{proof}
3029
3030
3031\bibliographystyle{unsrt}
3032\bibliography{commutative-algebra-notes.bib}
3033
3034\end{document}