math/commutative-algebra-notes.tex
commutative-algebra-notes.texRaw
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}
9\usepackage{amssymb}
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}
36
37\newtheorem{innerdefn}{Definition}
38\newenvironment{defn}[1]
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40{\endinnerdefn}
41
42\newtheorem{innerthm}{Theorem}
43\newenvironment{thm}[1]
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45{\endinnerthm}
46
47\newtheorem{innerlemma}{Lemma}
48\newenvironment{lemma}[1]
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50{\endinnerlemma}
51
52\newtheorem{innerprop}{Proposition}
53\newenvironment{prop}[1]
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55{\endinnerprop}
56
57\newtheorem{innercor}{Corollary}
58\newenvironment{cor}[1]
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60{\endinnercor}
61
62\newtheorem{innerremark}{Remark}
63\newenvironment{remark}[1]
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65{\endinnerremark}
66
67\newtheorem{innereg}{Example}
68\newenvironment{eg}[1]
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70{\endinnereg}
71
72\newtheorem{innerex}{Exercise}
73\newenvironment{ex}[1]
74{\renewcommand\theinnerex{#1}\innerex}
75{\endinnerex}
76
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, and let
1892$$\theta:\frac{A}{m}\stackrel{iso}{\longrightarrow} k$$
1893be the $k$-algebra isomorphism.
1894
1895Define
1896$$\phi=\theta\circ\pi:A\longrightarrow k.$$
1897
1898Thus, $\forall~ i\in[n]$, define
1899$$a_i=\phi(X_i)\in k.$$
1900
1901Since $\theta$ is injective,
1902\begin{align*}
1903 \phi(X_i-a_i)=0
1904 &\Longrightarrow \pi(X_i-a_i)=0\\
1905 &\Longrightarrow X_i-a_i\in m.
1906\end{align*}
1907
1908Since $X_i - a_i \in m ~~\Longrightarrow~~ J=(X_1 - a_1, \ldots, X_n -a_n) \subseteq m$,
1909
1910since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$, then $J$ is a maximal ideal.
1911
1912
1913\begin{adjustwidth}{4em}{0pt}
1914 (Corollary \ref{5.2}) Thus $J=m$, ie. $m=(X_1 - a_1, \ldots, X_n -a_n)$, with $a_i \in k$.
1915
1916 \vspace{0.4cm}
1917 We had $k \stackrel{iso}{\cong} \frac{A}{m} = \frac{k[X_1, \ldots, X_n]}{m}$.
1918
1919 We have the map
1920 \begin{align*}
1921 \phi: k[X_1, \ldots, X_n] &\longrightarrow \frac{k[X_1, \ldots, X_n]}{m} \cong k\\
1922 X_i &\mapsto a_i \in k
1923 \end{align*}
1924 which is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$.
1925
1926 Since $\phi$ is a $k$-algebra homomorphism, then
1927 $$\forall~ f \in A,~ \phi(f(X_1, \ldots, X_n)) = f(\phi(X_1), \ldots, \phi(X_n)) = f(a_1, \ldots, a_n).$$
1928
1929 Thus, $\exists$ a one-to-one correspondence:
1930 \begin{align*}
1931 \text{points in}~ k^n ~~~ &\longleftrightarrow~~~ m-Spec~A, ~\text{maximal ideals in}~ k[X_1, \ldots, X_n]\\
1932 (a_1, \ldots, a_n) ~~~&\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)
1933 \end{align*}
1934\end{adjustwidth}
1935
1936\vspace{0.5cm}
1937Now 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
1938$$J\subseteq m$$
1939
1940We have seen that
1941$$m=(X_1-a_1, \ldots, X_n-a_n)$$
1942for some $P=(a_1,\ldots,a_n)\in k^n$.
1943
1944So $P$ is a zero for every polynomial in $m$, ie.
1945$$\forall~ f \in m ~\Longrightarrow~ f(P)=0.$$
1946
1947Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$; ie. every element of $J$ vanishes at $P$.
1948
1949$\Longrightarrow~ P \in V(J)$, and thus $V(J) \neq \emptyset$. This is what the Nullstellensatz first part states:
1950\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
1951 Let $k$ be an algebraically closed field and let $J\subseteq k[X_1,\ldots,X_n]$ be an ideal.
1952 \begin{enumerate}[a.]
1953 \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
1954 \item $I(V(J)) = rad(J)~~(=\sqrt{J})$, in other words, for $f \in k[X_1, \ldots, X_n]$,
1955 $$f(P)=0 ~\forall~ P \in V(J) ~~\Longleftrightarrow~~ f^N \in J ~\text{for some $N\geq 1$.}$$
1956 \end{enumerate}
1957\end{thm}
1958\begin{proof}
1959 \begin{enumerate}[a.]
1960 \item already proven in the previous steps.
1961
1962 \item $I(V(J)) = \sqrt{J}$\\
1963 ie. vanishing ideal of a variety = radical of the ideal defining the variety
1964
1965 where
1966 $$\sqrt{J} = \{ f \in A ~|~ f^N \in J ~\text{for some}~ N>0 \}.$$
1967
1968 First, want to show that
1969 $$\sqrt{J}\subseteq I(V(J)).$$
1970
1971 Let $f\in\sqrt{J}$. Then $f^N\in J$ for some $N\geq 1$.
1972
1973 Thus, for every $P\in V(J)$,
1974 $$f(P)^N=f^N(P)=0,$$
1975 and therefore $f(P)=0$.
1976
1977 Hence $f\in I(V(J))$, so
1978 $$\sqrt{J}\subseteq I(V(J)).$$
1979
1980 Now, want to show that if a polynomial vanishes at all points where $J$ vanishes, then $f \in \sqrt{J}$.
1981
1982 Since $k[X_1,\ldots,X_n]$ is Noetherian, write
1983 $$J=(g_1,\ldots,g_m).$$
1984
1985 Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$.
1986
1987 Suppose there is a point $(a_1, \ldots, a_n,b)$ that is a zero of $J'$. ie.
1988 $$\exists~ (a_1, \ldots, a_n,b) \in V(J').$$
1989
1990 Since
1991 $$g_i(a_1,\ldots,a_n)=0,$$
1992 our hypothesis says
1993 $$f(a_1,\ldots,a_n)=0.$$
1994
1995 However, the last generator $(1-Yf)$ requires
1996 $$1 - b f(a_1,\ldots,a_n) = 0 ~~\Longrightarrow~ 1 - b \cdot 0 = 0 ~\Longrightarrow~ 1=0,$$
1997 a contradiction.
1998
1999 Therefore, $V(J') = \emptyset$.
2000
2001 \vspace{0.4cm}
2002
2003 By part a., applied to $k[X_1,\ldots,X_n,Y]$,\\
2004 \hspace*{2em} if $V(J')=\emptyset$ then $J'=k[X_1,\ldots,X_n,Y]$, so $1 \in J'$.
2005
2006 Every element in an ideal is a linear combination of its generators:\\
2007 $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
2008
2009 $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf).$$
2010
2011 which, since $1 \in J'$,
2012 $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X)).$$
2013
2014 Now work in the localization
2015 $$A_f=A[f^{-1}],$$
2016 where $f$ is invertible, and apply the homomorphism
2017 $$A[Y]\longrightarrow A_f,\qquad Y\longmapsto f^{-1}.$$
2018
2019 Then
2020 $$1= \left( \sum_{i=1}^m p_i(X,f^{-1}) g_i(X) \right)
2021 +q(X,f^{-1})\cdot \underbrace{(1-f^{-1}f(X))}_{0},$$
2022 thus
2023 $$1= \sum_{i=1}^m p_i(X,f^{-1}) g_i(X).$$
2024
2025 Since there are finitely many polynomials $p_i$, $\exists~ N\geq 0$ such that
2026 $$f^N p_i(X,f^{-1})\in A ~~\forall~ i.$$
2027
2028 Multiplying by $f^N$,
2029 $$f^N= \sum_{i=1}^m A_i(X) g_i(X),$$
2030 where
2031 $$A_i(X)=f^N p_i(X,f^{-1})\in A.$$
2032
2033 Thus $f^N$ is a linear combination of $g_i$.\\
2034 Therefore $f^N \in J$, so $f \in \sqrt{J}$.
2035
2036 Hence
2037 $$I(V(J))\subseteq\sqrt{J}.$$
2038
2039 Combining both inclusions,
2040 $$I(V(J))=\sqrt{J}.$$
2041 \end{enumerate}
2042\end{proof}
2043
2044
2045
2046
2047\section{Rings of fractions $S^{-1}A$ and localization}
2048
2049\subsection{Rings of fractions $S^{-1}A$}
2050
2051\begin{defn}{6.1}[ring of fractions] \label{def.6.1}
2052 let $A$ a ring, $S \subset A$ a multiplicative set ($1 \in S$, and $st \in S
2053 ~\forall s, t \in S$).
2054
2055 Introduce th following relation $\sim$ on $A \times S$:
2056 $$(a, s) \sim (b, t) \Longleftrightarrow \exists y \in S ~\text{such that}~ u(at - bs) = 0$$
2057
2058 (write $a/s$ for the equivalence class of $(a, s)$.
2059
2060 Then, the \emph{ring of fractions of $A$ with respect to $S$} is
2061 $$S^{-1}A = (A \times S)/\sim$$
2062
2063 with ring op'ns defined by the usual arithmetic op'ns on fractions:
2064 $$\frac{a}{s} \pm \frac{b}{t} = \frac{(at \pm bs)}{st} ~~\text{and}~~ \frac{a}{s} \cdot \frac{b}{t} = \frac{ab}{st}$$
2065\end{defn}
2066
2067\begin{prop}{6.1} \label{prop.6.1}
2068 \begin{enumerate}[i.]
2069 \item $\sim$ is an equivalence relation
2070 \item the ring op'ns are well defined, and $S^{-1}A$ is a ring
2071 \item
2072 \begin{align*}
2073 \psi: A &\longrightarrow S^{-1}A\\
2074 a &\longmapsto a/1
2075 \end{align*}
2076 is a ring homomorphism.
2077 \end{enumerate}
2078\end{prop}
2079
2080\begin{eg}{ }
2081 TODO
2082\end{eg}
2083
2084\begin{lemma}{6.2} \label{6.2}
2085 For $f \in A$, write $S = \{ 1, f, f^2, \ldots \}$, and $A_f \cong S^{-1}A$.
2086
2087 Then
2088 $$A_f \cong \frac{A[X]}{(Xf-1)}$$
2089\end{lemma}
2090\begin{proof}
2091 Define the homomorphism
2092 \begin{align*}
2093 \psi: A[X] &\longrightarrow A_f\\
2094 a &\longmapsto a/1\\
2095 X &\longmapsto 1/f
2096 \end{align*}
2097
2098By the 1st isomorphism theorem:
2099
2100 \begin{tikzpicture}[node distance=1.5cm, auto]
2101 \node (G) {$A[X]$};
2102 \node (H) [right of=G] {$A_f$};
2103 \node (GmodK) [below of=G, xshift=0.75cm] {$A[X]/\ker(\psi)$};
2104
2105 \draw[->] (G) to node {$\psi$} (H);
2106 \draw[->] (G) to node [swap] {$\phi$} (GmodK);
2107 \draw[->] (GmodK) to node [swap] {$\eta$} (H);
2108 \end{tikzpicture}
2109
2110 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.
2111
2112 \vspace{0.3cm}
2113 First, observe that $\psi(Xf-1) = \psi(X)\psi(f)-\psi(1) = \frac{1}{f} \frac{f}{1} - 1 = 1-1=0$,
2114 so $Xf-1 \in ker(\psi)$, ie. $(Xf-1) \subseteq ker(\psi)$.
2115
2116 Now, we want to prove that $ker(\psi) \subseteq (Xf-1)$.
2117
2118 Take $h \in ker(\psi)$, will prove that $h \in (Xf-1)~ \forall~ h \in ker(\psi)$, and thus $ker(\psi) \subseteq(Xf-1)$.
2119
2120 \vspace{0.3cm}
2121 Want to prove that $h(X)$ is a multiple of $(Xf-1)$.
2122
2123 Let
2124 $$h(X)=a_n X^n + a_{n-1} X^{n-1} + \ldots + a_1 X + a_0$$
2125
2126 % Since $h \in ker(\psi) ~~\Longrightarrow~~ \psi(h)=0$, so
2127 % $$\psi(h) = \frac{a_n}{f^n} + \frac{a_{n-1}}{f^{n-1}} + \ldots \frac{a_1}{f} + \frac{a_0}{1} = 0 \in A_f$$
2128
2129 multiply $h(X)$ by $f^n$:
2130 $$f^n \cdot h(X)=a_n (f^n X^n) + a_{n-1} f (f^{n-1} X^{n-1}) + a_{n-2} f^2 (f^{n-2} X^{n-2}) \ldots$$
2131
2132 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 $f^i X^i \equiv 1 \pmod{Xf-1}$.
2133
2134 So,
2135 $$f^n \cdot h(X)=\underbrace{a_n (1) + a_{n-1} f (1) + a_{n-2} f^2 (1) + \ldots + a_0 f^n}_{C~~\text{(constant)}} \pmod{Xf-1}$$
2136
2137 $$\Longrightarrow~~ f^n \cdot h(X)=C \pmod{Xf-1}$$
2138 $$\Longleftrightarrow~~ f^n \cdot h(X)=Q(X) \cdot (Xf-1) + C$$
2139
2140 Want to remove $C$, but it is non-zero. Note that in $A_f$ (ring of
2141 fractions), $\frac{a}{f^n} = 0$ iff $\exists~ k$ such that $f^k \cdot a = 0$ in $A$.
2142
2143 So, multiply both sides by $f^k$:
2144 $$f^k \cdot f^n \cdot h(X)=f^k \cdot (Q(X) \cdot (Xf-1) + C)$$
2145 $$\underbrace{f^k f^n}_{f^{nk}} \cdot h(X)=\underbrace{f^k Q(X)}_{Q'(X)} \cdot (Xf-1) + \underbrace{f^k C}_{0}$$
2146
2147 $$\Longrightarrow~~ f^{n+k} \cdot h(X)=Q'(X) \cdot (Xf-1)$$
2148 $$\Longleftrightarrow~~ f^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2149
2150 multiply it by $X^{n+k}$:
2151 $$X^{n+k} \cdot f^{n+k} \cdot h(X) \equiv X^{n+k} \cdot 0 \pmod{Xf-1}$$
2152 $$(Xf)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2153
2154 Now, since we had $Xf \equiv 1$ in $\frac{A[X]}{(Xf-1)}$,
2155 $$(1)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2156 $$\Longrightarrow~~ h(X) \equiv 0 \pmod{Xf-1}$$
2157 By definition this is saying $h(X) \in (Xf-1) ~\forall~ k \in ker(\psi)$.
2158
2159 Thus $ker(\psi) \subseteq (Xf-1)$.
2160
2161 Initially we saw that $(Xf-1) \subseteq ker(\psi)$. Therefore $ker(\psi)=(Xf-1)$.
2162
2163 Hence,
2164 $$A_f \cong \frac{A[X]}{(Xf-1)}$$
2165\end{proof}
2166
2167
2168\vspace{0.4cm}
2169
2170Given a ring homomorphism $\psi: A \longrightarrow B$, there is a correspondence
2171
2172$$e: \{ \text{ideals of } A \} \longrightarrow \{ \text{ideals of } B \}$$
2173
2174given by $e(I) = \psi(I)B = IB$ (called \emph{extension}),
2175
2176and
2177
2178$$r: \{ \text{ideals of } B \} \longrightarrow \{ \text{ideals of } A \}$$
2179
2180given by $r(J) = \psi^{-1} J$ (called \emph{restriction}, written $A \cap J$).
2181
2182Set $B= S{-1}A$. Then $S^{-1}I = e(I) = \psi(I)B$.
2183
2184\begin{prop}{6.3} \label{prop.6.3}
2185 \begin{enumerate}[a.]
2186 \item $\forall$ ideal $J$ of $S^{-1}A$, $e(r(J))=J$
2187 \item $\forall$ ideal $I$ of $A$,
2188 $$r(e(I)) = \{ a \in A ~|~ as \in I ~\text{for some}~ s \in S \}$$
2189 \item if $P$ prime and $P \cap S = \emptyset$,\\
2190 then $e(P)=S^{-1}P$ is a prime ideal of $S^{-1}A$.
2191 \end{enumerate}
2192\end{prop}
2193\begin{proof}
2194\end{proof}
2195
2196\begin{cor}{6.3} \label{cor.6.3}
2197 for an ideal $I$ of $A$, the necessary and sufficient condition for
2198 $r(e(I))=I$ is
2199 $$as \in I \Longrightarrow a \in I~~ \forall s \in S$$
2200\end{cor}
2201
2202
2203\subsection{Localization}
2204
2205If $P$ prime ideal, then $S=A \setminus P$ is a multiplicative set.
2206Define the set $A_P=S^{-1}A$.
2207
2208\begin{prop}{6.4} \label{6.4}
2209 $a/s \in A_P$ is a unit of $A_P ~\Longleftrightarrow~ a \not\in P$.
2210
2211 Therefore $A_P$ is a \emph{local ring}, with maximal ideal $e(P)=P A_P$.
2212
2213 The local ring $(A_P, PA_P)$ is called the \emph{localization} of $A$ at $P$.
2214\end{prop}
2215\begin{proof}
2216 \begin{enumerate}
2217 \item[$\Longleftarrow$] (if $a \not\in P ~\Longrightarrow~ a/s \in A_P$ is a unit)\\
2218
2219 if $a \not\in P$, then by definition of $S,~~ a \in S$.
2220
2221 In the localization $A_P$, every element of $S$ is invertible.
2222
2223 The inverse of $\frac{a}{s}$ is $\frac{s}{a}$\\
2224 \hspace*{2em}$\longrightarrow~~ \frac{a}{s} \cdot \frac{s}{a} = 1$, thus
2225 $a/s \in A_P$ is a unit.
2226
2227 \item[$\Longrightarrow$] (if $a/s$ unit $~\Longrightarrow~ a \not\in P$)\\
2228
2229 Suppose $a/s$ a unit; then $\exists~ \frac{b}{t} \in A_P$ such that
2230 $\frac{a}{s} \cdot \frac{b}{t} = 1$.
2231
2232 By definition of equality in localization, $\frac{ab}{st}=1$ means $\exists u \in S$ such that
2233 \begin{equation}
2234 u(ab \cdot 1 - st \cdot 1) = 0~~ \Longrightarrow~ uab=ust
2235 \label{eq:6.4}
2236 \end{equation}
2237 with $u,s,t \in S$.
2238
2239 Since $S=A \setminus P$ and $P$ is a prime ideal,\\
2240 the products of elements outside $P$ must also be outside of $P$.
2241 $$u,s,t \not\in P ~\Longrightarrow~ ust \not\in P$$
2242
2243 At \eqref{eq:6.4}, we know that $uab=ust \not\in P$, thus $uab \not\in P$.
2244
2245 If $uab \not\in P$, then $a \not\in P$.
2246
2247
2248 Next we will prove that $A_P$ is a local ring.\\
2249
2250 A ring is local if it has exactly one maimal ideal (\ref{1.13}); so we
2251 show that the set of non-units forms an ideal.
2252
2253 $a/s$ is not a unit iff $a \in P$. Let $m= \{ a/s | a \in P,~ s \not\in P \} = PA_P$.
2254 Want to show that $PA_P$ is the unique maximal ideal:
2255 \begin{itemize}
2256 \item it's an ideal: let $\frac{a}{s}, \frac{b}{t} \in PA_P$ with $a,b \in P$.
2257
2258 Then $\frac{a}{b} + \frac{b}{t} \in PA_P$ with numerator in $P$.
2259
2260 If multiply a fraction in $PA_P$ by any fraction in $A_P$, the numerator stays in $P$, thus $PA_P$ is an ideal.
2261
2262 \item every element outside of $PA_P$ is a unit (proven at the beginning of this proposition's proof ($\Longleftarrow$)).
2263
2264 \item maximality: every ideal strictly larger than $PA_P$ must contain a unit.\\
2265 If an ideal contains a unit, it must be the entire ring $A_P$.\\
2266 \hspace*{2em}$\Longrightarrow$ therefore, $PA_P$ is the unique maximal ideal.
2267 \end{itemize}
2268 \end{enumerate}
2269\end{proof}
2270
2271
2272In other words:
2273
2274\emph{Localization}: formal way to 'force' certain elements to have inverses.
2275
2276It's the smallest ring that contains $A$ and makes all the elements of $S$ invertible.
2277
2278There is a natural map $\psi: A \longrightarrow S^{-1}A$, by $f(a)=\frac{a}{1},~ f(s)=\frac{1}{s}~~ \text{for}~ s \in S$.
2279
2280
2281\subsection{Localization commutes with taking quotients}
2282Let $A$ ring, $S$ multiplicative set, $I$ ideal.
2283
2284Write $T$ for the image of $S$ in $A/I$.
2285
2286Then $S^{-1}I=I \cdot S^{-1}A$ is an ideal of $S^{-1}A$, can take the quotient ring $S^{-1}A / S^{-1} I$.
2287
2288$\longleftrightarrow$ can take the quotient $A/I$ and then localize to get $T^{-1}(A/I)$.
2289
2290
2291\begin{cor}{6.7} \label{6.7}
2292 $$T^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
2293 In particular, for $P$ prime ideal,
2294 $$\frac{A_P}{PA_P} = k(P) = Frac(\underbrace{A/P}_{\substack{\text{integral}\\\text{domain}}})$$
2295
2296 From \ref{6.4}, $A_P$: local ring, $PA_P$: unique maximal ideal.
2297 $k(P)$ and $Frac(A/P)$ are field of fractions.
2298\end{cor}
2299\begin{proof}
2300 the quotient ring $A/I$ can be viewed as an $A$-module and
2301 $$\underbrace{ T^{-1}(A/I) }_{\text{ring of fractions}} \cong
2302 \underbrace{ \frac{S^{-1}A}{S^{-1}I} }_{\text{module of fractions}}$$
2303
2304 The \ref{cor.6.6}.i, gives an isomorphism of modules
2305 $$ T^{-1}(A/I) = S^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
2306 it's easy to see that this is a ring homomorphism.
2307\end{proof}
2308% TODO 6.6 (proposition and corollary)
2309
2310
2311
2312\newpage
2313
2314\section{Exercises}
2315
2316For the exercises, I follow the assignments listed at \cite{mit-course}.
2317
2318The 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}.
2319
2320\subsection{Exercises Chapter 1}
2321
2322\begin{ex}{R.1.1}
2323 Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$?
2324\end{ex}
2325\begin{proof}
2326 Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$.
2327
2328 $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\
2329 Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$.
2330
2331 Thus $I \cup J$ is not closed under addition; thus is not an ideal.
2332
2333
2334 Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum:
2335
2336 $$I+J = \{ a+b | a \in I, b \in J \}$$
2337
2338 $gcd(2,3)=1$, so $I+J = \mathbb{Z}$.
2339
2340 Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$.
2341\end{proof}
2342
2343\begin{ex}{R.1.5}
2344 let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\
2345 In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$.
2346\end{ex}
2347\begin{proof}
2348 (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$).
2349
2350 Let
2351 $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
2352 The claim is that $\psi^{-1}(P)$ is prime ideal of $A$.
2353
2354 \begin{enumerate}[i.]
2355 \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
2356 $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$).
2357
2358 If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so
2359 $$\psi(a-b)= \psi(a) - \psi(b) \in P$$
2360 hence $a-b \in \psi^{-1}(P)$.
2361
2362 If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\
2363 Thus $ra \in \psi^{-1}(P)$.
2364
2365 $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$.
2366
2367 \item show that $\psi^{-1}(P)$ is prime:\\
2368 $\psi^{-1}(P) \neq A$, 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 $B$.
2369
2370 Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$.
2371
2372 Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\
2373 Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$.
2374
2375 Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$.
2376 \end{enumerate}
2377\end{proof}
2378
2379
2380\begin{ex}{R.1.6}
2381 prove or give a counter example:
2382 \begin{enumerate}[a.]
2383 \item the intersection of two prime ideals is prime
2384 \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
2385 \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
2386 \item the map $\psi^{-1}$ of Proposition 1.2 takes maximal ideals of $A/I$
2387 to maximal ideals of $A$
2388 \end{enumerate}
2389\end{ex}
2390\begin{proof}
2391 \begin{enumerate}[a.]
2392 \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime.
2393
2394 Then $I \cap J = (2) \cap (3) = (6)$.
2395
2396 The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$.
2397
2398 Thus the intersection of two primes can not be prime.
2399
2400 \item $P_1=(2),~ P_2=(3)$, both prime.
2401
2402 Then,
2403 $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$
2404
2405 $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$.
2406
2407 (recall: principal= generated by a single element)
2408
2409 So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$.
2410
2411 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.
2412
2413 Henceforth, the sum of two prime ideals is not necessarily prime.
2414
2415 \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$.
2416
2417 Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$.
2418
2419 Then
2420 \begin{align*}
2421 \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\
2422 \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A
2423 \end{align*}
2424
2425 But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
2426
2427 Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
2428
2429 \item $\psi: A \longrightarrow A/I$ quotient homomorphism, $I \subseteq A$ an ideal.
2430
2431 Let $M$ a maximal ideal of $A/I$, then $\frac{(A/I)}{M}$ is a field (Proposition 1.3).
2432
2433 By the isomorphism theorems,
2434 $$\frac{(A/I)}{M} \cong \frac{A}{\psi^{-1}(M)}$$
2435
2436
2437 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 $A$.
2438
2439 $\Longrightarrow~$ under $\psi$, preimages of maximal ideals are maximal.
2440 \end{enumerate}
2441\end{proof}
2442
2443\begin{ex}{R.1.12.a}
2444 if $I,J$ ideals and $P$ prime ideal, prove that
2445 $$IJ \subset P ~\Longleftrightarrow~ I \cap J \subset P ~\Longleftrightarrow~ I ~\text{or}~ J \subset P$$
2446\end{ex}
2447\begin{proof}
2448 assume $I \subseteq P$ (for $J \subseteq P$ will be the same, symmetric), take $x \in IJ$,
2449
2450 then
2451 $$x = \sum_{k=1}^n a_k b_k$$
2452 with $a_k \in I,~ b_k \in J$.
2453
2454 Each $a_k \in I \subseteq P$. Since $P$ an ideal,
2455 $$\sum_{k=1}^n a_k b_k \in P$$
2456
2457 thus $x \in P$, hence $IJ \subseteq P$.
2458
2459 So $I \subseteq P$ or $J \subseteq P$ $~\Longrightarrow IJ \subseteq P$.
2460
2461 \vspace{0.5cm}
2462 Conversely,\\
2463 assume $P$ prime and $IJ \subseteq P$.
2464
2465 Suppose by contradiction that $I \not\subseteq P$ and $J \not\subseteq P$.
2466 \begin{itemize}
2467 \item[-] since $I \not\subseteq P,~ \exists a \in I$ with $a \not\in P$
2468 \item[-] since $J \not\subseteq P,~ \exists b \in J$ with $b \not\in P$
2469 \end{itemize}
2470 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 $P$.
2471
2472 Thus $I \not\subseteq P$ and $J \not\subseteq P$ are false.
2473
2474 \vspace{0.3cm}
2475 So both directions are proven, hence
2476 $$IJ \subseteq P ~\Longrightarrow~ I \subseteq P ~\text{or}~ J \subseteq P$$
2477
2478\end{proof}
2479
2480\begin{ex}{R.1.18}
2481 Use Zorn's lemma to prove that any prime ideal $P$ contains a minimal prime ideal.
2482\end{ex}
2483\begin{proof}
2484 Let $P$ prime ideal of $R$.
2485
2486 $$S = \{ Q \subseteq R ~|~ Q ~\text{a prime ideal AND}~ Q \subseteq P \}$$
2487
2488 Goal: show that $S$ has a minimal element, the minimal ideal contained in $P$.
2489
2490 $P \subset S$, so $S$ is nonempty.
2491
2492 Let $C \subseteq S$ be a chain (= totally ordered subset) with respect to inclusion.
2493
2494 Define
2495 $$Q_C = \bigcap_{Q \in C} Q$$
2496
2497 Clearly $Q_C \subseteq P$, since each $Q \in C$ is $Q \subseteq P$.
2498
2499 Since $C$ is ordered by inclusion, it is a decreasing chain of prime ideals.
2500
2501 Intersection of a decreasing chain of prime ideals is again a prime ideal:
2502 \begin{itemize}
2503 \item[-] if $ab \in Q_C$, then $ab \in Q ~\forall Q \in C$
2504 \item[-] since $Q$ prime, $\forall Q \in C$ either $a \in Q$ or $b \in Q$
2505 \end{itemize}
2506
2507 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 $Q_2 \subseteq Q_1$.
2508
2509 In either case: contradiction, since the smaller one would have to contain the element that was assumed to be excluded.
2510
2511 Thus $\forall Q \in C$ the same element $a, b$ must lie in all $Q$. $\Longrightarrow~$ lies in the intersection of them, $Q_C$.
2512
2513 Henceforth, $Q_C$ is a prime ideal and lies in $S$, and its a lower bound of $C$ in $S$.
2514
2515 Now, $S$ is nonempty, and every chain in $S$ has a lower bound in $S$ (its intersection).\\
2516 Therefore, $S$ has a minimal element $P_{min}$.
2517
2518 By construction, $P_{min}$ is a prime ideal $P_{min} \subseteq P$, and by minimality there are no strictly smaller prime ideals inside $P$.
2519
2520 So $P_{min}$ is a minimal prime ideal, contained in $P$.
2521\end{proof}
2522
2523\begin{ex}{R.1.10}
2524\end{ex}
2525\begin{proof}
2526\end{proof}
2527
2528\begin{ex}{R.1.11}
2529\end{ex}
2530\begin{proof}
2531\end{proof}
2532
2533\begin{ex}{R.1.4}
2534\end{ex}
2535\begin{proof}
2536\end{proof}
2537
2538\subsection{Exercises Chapter 2}
2539
2540\begin{ex}{R.2.9}
2541$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ is a s.e.s. of $A$-modules. Prove that if $N, L$ are finite over $A$, then $M$ is finite over $A$.
2542\end{ex}
2543\begin{proof}
2544 Denote the generators of $L$ and $N$ respectively as
2545 \begin{align*}
2546 \{l_1, \ldots, l_k \} &\subseteq L\\
2547 \{n_1, \ldots, n_p \} &\subseteq N
2548 \end{align*}
2549
2550 By s.e.s. definition,
2551 \begin{itemize}
2552 \item[-] $\alpha$ is injective (one-to-one), so
2553 $$\forall l_i \in L,~ \exists~ x_i \in M ~\text{s.th.}~ \alpha(l_i)=x_i$$
2554
2555 \item[-] $\beta$ is surjective (onto), so
2556 $$\forall n_j \in N,~ \exists~ y_j \in M ~\text{s.th.}~ \beta(y_j)=n_j$$
2557\end{itemize}
2558
2559 We will show that $\{x_1, \ldots, x_k, y_1, \ldots, y_p \}$ generate $M$, and thus $M$ is finite:
2560
2561 Let $m \in M$, then $\beta(m) \in N$, and
2562 $$\beta(m) = \sum_{j=1}^p a_j n_j ~~~\text{with}~ a_j \in A$$
2563
2564 Take $m' \in M$, with $m' = \sum a_j y_j$, then
2565 $$\beta(m') = \sum a_j \beta(y_j) = \sum a_j n_j = \beta(m)$$
2566
2567 Then, since $\beta(m)=\beta(m')~~ \Longrightarrow~~ \beta(m-m')=0$, thus
2568 $$(m-m') \in ker(\beta)$$
2569
2570 By \emph{exactness} property, since $\alpha: L \longrightarrow ker(\beta)$, we have $ker(\beta)=im(\alpha)$.
2571
2572 Therefore, $\exists~ l \in L$ such that $\alpha(l)= m-m'$.
2573
2574 Since $\{l_i\}_k$ generate $L$,
2575 $$l = \sum^k b_i l_i$$
2576 thus
2577 $$m-m' = \alpha(l) = \alpha(\underbrace{\sum b_i l_i}_{l}) = \sum b_i \underbrace{\alpha(l_i)}_{x_i} = \sum b_i x_i$$
2578
2579 Rearrange,
2580 $$m = m' + \sum b_i x_i = \sum_{j=1}^p a_j y_j + \sum_{i=1}^k b_i x_i ~~~~~ \forall m \in M$$
2581
2582 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.\\
2583 Thus $M$ is finitely generated over $A$.
2584\end{proof}
2585
2586
2587\subsection{Exercises Chapter 3}
2588
2589\begin{ex}{R.3.2}
2590 $K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space.
2591 Prove that $A$ is Noetherian and Artinian.
2592\end{ex}
2593\begin{proof}
2594 $dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in \aA$ and $c \in K$, then $c \cdot x \in \aA$.
2595
2596 \begin{enumerate}
2597 \item Noetherian:\\
2598 let $I_1 \subseteq I_2 \subseteq \ldots$ be an ascending chain of ideals in $A$.
2599
2600 Since each $I_i$ is a subspace, we have
2601 $$dim_K(I_1) \leq dim_K(I_2) \leq \ldots \leq n$$
2602 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.
2603
2604 \item Artinian:\\
2605 Similarly, if $I_1 \supseteq I_2 \supseteq \ldots$ a descending chain of ideals in $A$.
2606
2607 then
2608 $$n \geq dim_K(I_1) \geq dim_K(I_2) \geq \ldots \geq 0$$
2609 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.
2610 \end{enumerate}
2611\end{proof}
2612
2613\begin{ex}{R.3.5}
2614 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 $M$.
2615
2616 Prove if the following holds or not:
2617 $$\beta(M_1)=\beta(M_2) ~\text{and}~ \alpha^{-1}(M_1)=\alpha^{-1}(M_2)~~\Longrightarrow~ M_1=M_2$$
2618\end{ex}
2619\begin{proof}
2620 Counterexample showing that it does not hold:
2621
2622 Let $K$ a field, $M = K \oplus K~, L=K,~N=K$.
2623
2624 Set, for $l \in L,~ (m_1, m_2) \in M$,
2625 \begin{align*}
2626 \alpha:~ &l \longmapsto (l, 0)\\
2627 \beta:~ &(m_1, m_2) \longmapsto m_2
2628 \end{align*}
2629
2630 So we have
2631 $$0 \longrightarrow K \stackrel{\alpha}{\longrightarrow} K^2 \stackrel{\beta}{\longrightarrow} K \longrightarrow 0$$
2632
2633 Then,
2634 \begin{align*}
2635 M_1 &= \{ (x, x) ~|~ x \in K\} ~~~~\sim\text{(diagonal line)} \\
2636 M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)}
2637 \end{align*}
2638
2639 (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 $M$).
2640
2641 Then,
2642 \begin{align*}
2643 \beta(M_1) &= \{ x ~|~ x \in K\} = K \\
2644 \beta(M_2) &= \{ x ~|~ x \in K\} = K
2645 \end{align*}
2646 thus, $\beta(M_1) = \beta(M_2)$.
2647
2648 For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1) = \{0\}$,\\
2649 for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2) = \{0\}$,\\
2650 thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$.
2651
2652
2653 So we've seen that
2654 \begin{align*}
2655 \beta(M_1) = \beta(M_2)\\
2656 \alpha^{-1}(M_1)=\alpha^{-1}(M_2)
2657 \end{align*}
2658 while having $M_1 \neq M_2$.
2659\end{proof}
2660
2661
2662
2663\begin{ex}{R.3.3}
2664Let $A$ a ring, $I_1, \ldots, I_k$ ideals such that each $A/I_i$ is a Noetherian ring.
2665Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bigcap I_i = 0$ then $A$ is also Noetherian.
2666\end{ex}
2667\begin{proof}
2668\begin{enumerate}[i.]
2669 \item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian.
2670 $\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian.
2671 \item Take the canonical homomorphism
2672 $$\phi: A \longrightarrow \bigoplus_{i=1}^n A/ I_i$$
2673 by $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$.
2674
2675 $\phi$ is injective: $ker(\phi)= \{ a \in A | a \in I_i \forall i \}$.
2676
2677 Since we're given $\cap I_i = 0$, then $ker(\phi)=\cap I_i$, and $\phi$ is injective.
2678
2679 Thus, $\phi$ is the isomorphism $A \cong im(\phi)$, where $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$.
2680
2681 We know that any submodule of a Noetherian module is Noetherian, thus, since
2682 \begin{itemize}
2683 \item $A/I_i$ is Noetherian by hypothesis of the exercise
2684 \item $A \cong im(\phi)$
2685 \item $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$
2686 \end{itemize}
2687 then, $A$ is Noetherian.
2688\end{enumerate}
2689\end{proof}
2690
2691\begin{ex}{R.3.4}
2692 Prove that if A is a Noetherian ring and M a finite A-module, then there
2693 exists an exact sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$.
2694 That is, M has a presentation as an A-module in terms of finitely many generators and relations.
2695\end{ex}
2696\begin{proof}
2697 since $M$ fingen $~\Longrightarrow~$ generators $\{m_1, \ldots, m_2 \} \subseteq M$ span $M$.
2698
2699 Let $\beta$ be a surjective $A$-linear map, which forms a free $A$-module of rank $p$ onto $M$:
2700 \begin{align*}
2701 \beta: A^p &\longrightarrow M\\
2702 (a_1, \ldots, a_p) &\longmapsto \sum_{i=1}^p a_i m_i
2703 \end{align*}
2704
2705 Let $K=ker(\beta)$. By the 1st Isomorphism Theorem,
2706 $$M \cong A^p / K$$
2707
2708 Since $A$ is a Noetherian ring, then every free $A$-module of finite rank (eg. $A^p$) is a Noetherian module.
2709
2710 Every submodule of a Noetherian module is fingen.
2711
2712 $\Longrightarrow~$ since $K \subseteq A^p, ~\Longrightarrow~ K ~~(=ker(\beta))$ is fingen.
2713
2714 Since $K$ fingen, let $\{k_1, \ldots, l_q\}$ be generators of $K$.
2715
2716 Define $\psi: A^q \longrightarrow K$.
2717
2718 Compose it with the inclusion map $i: K \longrightarrow A^p$,
2719 $$\alpha = i \circ \psi:~ A^q \longrightarrow A^p$$
2720
2721 So we have the whole sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$, where
2722 \begin{itemize}
2723 \item $\beta$ is surjective
2724 \item $im(\alpha)=ker(\beta)$
2725 \end{itemize}
2726 so that it is a exact sequence, thus, $M$ has a finite presentation.
2727\end{proof}
2728
2729\subsection{Exercises Chapter 4}
2730
2731\begin{ex}{R.4.1.a}
2732 $k[X^2] \subset k[X]$ is a finite extension, hence integral. Find the integral dependence relation for any $f \in k[X]$.
2733\end{ex}
2734\begin{proof}
2735 $\forall f(X) \in k[X]$ can be uniquely decomposed into its even and odd parts:
2736 $$f(X) = p(X^2) + X \cdot q(X^2)$$
2737
2738 with $p(X^2),~ q(X^2) \in k[X^2]$, and\\
2739 $p(X^2)$: sum of all terms with even exponents\\
2740 $q(X^2)$: sum of all terms with odd exponents, and then factoring out $X$.
2741
2742 \vspace{0.3cm}
2743 (Observation: this is used in FRI cryptographic protocol\\
2744 \href{https://github.com/arnaucube/math/blob/master/notes_fri_stir.pdf}{https://github.com/arnaucube/math/blob/master/notes\_fri\_stir.pdf})
2745
2746
2747 Rearrange it
2748 \begin{align*}
2749 f(X) &- p(X^2) = X \cdot q(X^2), ~~\text{square:}\\
2750 (f(X) &- p(X^2))^2 = X^2 \cdot q(X^2)^2\\
2751 f(X)^2-2 p(X^2) f(X) &+ p(X^2)^2 = X^2 \cdot q(X^2)^2\\
2752 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
2753 \end{align*}
2754
2755 Denote the last polynomial as $P(T) \in k[X^2]$, where $f(X)$ is a root of $P(T)$.
2756
2757 \vspace{0.2cm}
2758 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 $a_i \in k[X^2]$.
2759
2760 We have that
2761 \begin{align*}
2762 a_1 &= -2 p(X^2)\\
2763 a_0 &= p(X^2)^2 - X^2 q(X^2)^2
2764 \end{align*}
2765
2766
2767 So for example, for $f(X)= X^3 + X^2 + X + 1$:
2768 \begin{align*}
2769 f(X) = (X^2 + 1) &+ X (X^2 + 1)\\
2770 (f(X) - (X^2+1))^2 &= X^2 (X^2 + 1)^2\\
2771 (f(X) - p(X))^2 &= X^2 (q(X))^2
2772 \end{align*}
2773\end{proof}
2774
2775
2776\begin{ex}{R.4.5}
2777 Let $A=k[X,Y]/(Y^2 - X^2 - X^3)$. Prove that the normalization of $A$ is $k[t]$ where $t=Y/X$.
2778\end{ex}
2779\begin{proof}
2780$A=k[X,Y]/(Y^2 - X^2 - X^3)$, express $X$ and $Y$ in terms of $t$:
2781
2782Since $t=Y/X$ then $Y=tX$, and combined with $Y^2 = X^2 + X^3$, then
2783\begin{align*}
2784 (tX)^2 &= X^2 + X^3\\
2785 t^2 X^2 &= X^2 + X^3, ~~\text{assuming}~X \neq 0:\\
2786 t^2 &= 1+X, ~\text{thus}\\
2787 X&=t^2-1 \in k[X]
2788\end{align*}
2789
2790Then, $Y=tX=t(t^2-1)=t^3-t \in k[X]$.
2791
2792Hence $X, Y \in k[X]$.
2793
2794Therefore, $k[X,Y]/(Y^2 - X^2 - X^3) \subseteq k[t]$.
2795
2796\vspace{0.4cm}
2797By \ref{noether-normalization} (Noether normalization lemma), to show that $k[t]$ is the \emph{normalization}, must show that $k[t]$ is \emph{integral} over $A$.
2798
2799From $X=t^2-1 ~~\Longrightarrow~~ t^2-1-X=0 ~~\Longrightarrow~ t^2 - (1+X) = 0$.
2800
2801$(1+X) \in A$, so $t$ satisfies the monic polynomial
2802$$P(T) = T^2 - (1+X) \in A[T]$$
2803Thus $t$ is integral over $A$.
2804
2805Since $k[t]$ is generated by $t$ over $k$, and $k \subset A$, then the entire ring $k[t]$ is integral over $A$.
2806
2807Since $k[t]$ is a polynomial ring over a field, which is a UFD, it is integrally closed (since all UFD are integrally closed).
2808
2809$Frac A = k(X,Y)$, since $X=t^2-1, ~Y=t^3-t ~~\Longrightarrow~ k(X,Y) \subseteq k(t)$
2810
2811and $t=Y/X \in k(X,Y)$, thus $k(X,Y)=k(t)$.
2812
2813Since $k[t]$ is integrally closed and is the integral closure of $A$ in its fraction field $k(t)$, we conclude that the normalization of $A$ is $k[t]$.
2814\end{proof}
2815
2816
2817
2818\begin{ex}{R.4.9}
2819 $k$ 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.
2820\end{ex}
2821\begin{proof}
2822 (want to find a linear combination of the coordinates such that the original variables satisfy monic polynomials over the new ring $B$)
2823
2824 The relations defining $A$ are
2825 \begin{align*}
2826 X^2-Y^3-1=0 ~~&\Longrightarrow~ Y^3 = X^2 -1 ~~~(*)\\
2827 XZ -1=0 ~~&\Longrightarrow~ Z= 1/X = X^{-1}
2828 \end{align*}
2829
2830Thus $A$ can be denoted as $A = k[X,Y,X^{-1}]/(Y^3 - X^2 - 1)$.
2831
2832
2833Now, $Y$ is inegral over $k[X]$, since $Y^3 - (X^2 - 1) = 0$ is monic in $Y$ with coefficients in $k[X]$.
2834
2835$Z$ is not integral over $k[X]$, since $Z=1/X$ and $X$ is not a unit in $k[X]$.
2836
2837Choose $\alpha, \beta \in k$ such that $X$ (and thus $Z$) becomes integral over $B$:\\
2838set $\alpha=0,~\beta=1 ~~\Longrightarrow~~ B=k[X+\alpha Y + \beta Z]= k[X+Z]$.
2839
2840Let $w=X+Z$; since $XZ=1$, we have
2841$$w=X+\frac{1}{X} ~~\Longrightarrow~~ Xw=X^2+1 ~~\Longrightarrow~~ X^2 -w X +1 = 0 ~~(**)$$
2842
2843which is monic with coefficients in $k[w]$, thus $X$ is integral over $B$.
2844
2845Since $Z=w-X$, $Z$ is also integral.
2846
2847\vspace{0.4cm}
2848Generators of $A$ as a $B$-module:
2849
2850we hadd $B=k[w]$ with $w=X+Z$.
2851
2852From $(**)$ we have $X^2 - wX + 1=0$, so $X^2 = wX -1$.
2853
2854Thus 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 $\{1, X\}$.
2855
2856Fitting $X^2$ into $(*)$,
2857\begin{align*}
2858 X^2 &- Y^3 -1 =0\\
2859 Y^3 &= X^2-1\\
2860 Y^3 &= wX -2
2861\end{align*}
2862
2863thus any power of $Y$ higher than $2$ can be reduced (eg. $Y^4 = Y(wX-2) = w(XY) - 2Y$).
2864
2865So its partial basis is $\{1, Y, Y^2\}$.
2866
2867For $Z$, since $XZ=1$ and $w=X+Z \Longrightarrow Z=w-X$, thus $Z$ is a $B$-linear combination of $\{1,X\}$.
2868
2869\vspace{0.2cm}
2870Combining the previous partial basis, the generators are
2871$$\{ 1, X \} \times \{ 1,Y, Y^2 \} = \{ 1, Y, Y^2, X, XY, XY^2 \}$$
2872
2873\end{proof}
2874
2875
2876\subsection{Exercises Chapter 6}
2877
2878\begin{ex}{R.6.3.a}
2879 Let $A = A' \times A''$; prove that $A'$ and $A''$ are rings of fractions of $A$.
2880\end{ex}
2881\begin{proof}
2882 ring of fractions of $A$ = $S^{-1}A$, so want to prove that $S^{-1}A \cong A'$.
2883
2884 Localization map
2885 \begin{align*}
2886 \psi: A &\longrightarrow A'\\
2887 (a', a'') &\longmapsto a'
2888 \end{align*}
2889
2890 note that $\psi$ is surjective, since $\forall~ a' \in A',~ \psi(a', 0)=a'$.
2891
2892 Let the multiplicative set $S$ be $S=\{ e_1 \}$ with $e_1 = (1,0)$.
2893
2894 Want $\underbrace{S^{-1}A}_{\text{localization}} \cong A'$.
2895
2896 In $S^{-1}A$, $x \in A$ maps to $0$ iff $s x = 0$ for some $s \in S$.
2897
2898 Let $x=(a', a'')\in A$; then $sx=0 ~\Longrightarrow~ s \cdot (a', a'') =0$,
2899
2900 with $s \in S$, so $s=(1,0)$, hence
2901 $$(1,0)\cdot(a',a'')=(0,0)$$
2902 $$\Longrightarrow~ (a', 0) = (0, 0)$$
2903 \hspace*{4em} which implies $a' = 0$, so that $(a', a'')=(0, a')$.
2904
2905 $\Longrightarrow$ the elements that become zero in the localization are of the form $(0, a'')$
2906
2907 $$ker(\psi) = \{ (a',a'')\in A ~|~ \psi(a', a'')=0 \} = \{(0, a'') ~|~ a'' \in A'' \}$$
2908
2909 By the 1st isomorphism theorem,
2910
2911 \begin{tikzpicture}[node distance=1.5cm, auto]
2912 \node (G) {$A$};
2913 \node (H) [right of=G] {$A'$};
2914 \node (GmodK) [below of=G, xshift=0.75cm] {$\frac{A}{ker(\psi)}$};
2915
2916 \draw[->] (G) to node {$\psi$} (H);
2917 \draw[->] (G) to node [swap] {$\phi$} (GmodK);
2918 \draw[<->] (GmodK) to node [swap] {$\eta$} (H);
2919 \end{tikzpicture}
2920
2921 $\Longrightarrow~ \frac{A}{ker(\psi)} \cong A'$
2922
2923
2924 \vspace{0.5cm}
2925 Take the localization map
2926 \begin{align*}
2927 \phi: A &\longrightarrow S^{-1}A\\
2928 a &\longmapsto a/1
2929 \end{align*}
2930
2931 $$ker(\phi) = \{ x \in A ~|~ sx=0 ~\text{for some}~ s \in S \}$$
2932
2933 and we've seen that $a'=0$, so $x \in A ~\Rightarrow~ x=(a', a'') = (0, a'')$
2934
2935 $$ker(\phi) = \{ (0, a'') ~|~ a'' \in A'' \}$$
2936
2937 which is the same as $ker(\psi)$; $ker(\psi) = ker(\phi)$.
2938
2939 \vspace{0.2cm}
2940 $\Longrightarrow~~ A'$ and $S^{-1}A$ are both surjective images of $A$ with exact same kernel,
2941 thus $A' \cong S^{-1}A$.
2942\end{proof}
2943
2944\begin{ex}{R.6.4}
2945 \begin{enumerate}[a.]
2946 \item Give an example of a ring $A$ and distinct multiplicative sets $S,~ T$ such that $S^{-1}A = T^{-1}A$.
2947 \item Prove that for fixed $S$, there is a maximal multiplicative set $T$ with this property defined by
2948 $$T = \{ t \in A ~|~ at \in S ~\text{for some}~ a \in A \}$$
2949 \end{enumerate}
2950\end{ex}
2951\begin{proof}
2952 \begin{enumerate}[a.]
2953 \item
2954 (\emph{saturated sets})\\
2955
2956 General rule of saturation:\\
2957 two multiplicative sets $S,~T$ yield the same localization, $S^{-1}A = T^{-1}A$, iff the have the same saturation.
2958
2959 Saturation of $S$: $\hat{S}$, set of all elements in $A$ that divide some element of $S$:
2960 $$\hat{S} = \{ a \in A ~|~ \exists~ b \in A ~\text{s.th.}~ ab \in S \}$$
2961
2962 \vspace{0.4cm}
2963 Example 1: $A=\mathbb{Z}$,\\
2964
2965 $$S = \{ 2^n ~|~ n \geq 0 \} = \{ 1, 2, 4, 8, 16, \ldots \}$$
2966 localization:
2967 $$S^{-1} \mathbb{Z} = \{ \frac{a}{2^n} ~|~ a \in \mathbb{Z},~ n \in \mathbb{N} \}$$
2968
2969 $$T = \{ \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \ldots \}$$
2970
2971 Notice that despite $S \neq T$, we have $S^{-1}\mathbb{Z} = T^{-1} \mathbb{Z}$,\\
2972 since once $n \in \mathbb{Z}$ is invertible, $-n$ is automatically invertible: $(-n)^{-1} = - (n^{-1})$.
2973
2974 \vspace{0.4cm}
2975 Example 2: $A= k[x]$,
2976 $$S = \{ x^n ~|~ n \geq 0 \}$$
2977 $$T = \{ (2x)^n ~|~ n \geq 0 \}$$
2978 $$\Longrightarrow~ S^{-1}A = T^{-1} A = k[x, x^{-1}]$$
2979
2980 \item
2981 Show that $T$ is a multiplicative set:
2982
2983 (must contain $1$ and be closed under multiplication)\\
2984 Since $1 \in A$ and $1 \cdot 1 = 1 \in S$, then $1 \in T$.
2985
2986 Let $t_1,~ t_2 \in T$; by definition of $T$,
2987 $$\exists~ a_1, a_2 \in A ~\text{such that}~ a_1 t_1,~ a_2 t_2 \in S$$
2988
2989 Since $S$ a multiplicative set, $(a_1 t_1)(a_2 t_2) \in S$\\
2990 rearrange it: $(a_1 a_2)(t_1 t_2) \in S$.\\
2991 Since $a_1 a_2 \in A,~~ t_1 t_2 \in T$.
2992
2993 Thus $T$ is a multiplicative set.
2994
2995 Next, we show that $T$ is maximal:
2996
2997 Suppose $U$ is the maximal instead of $T$, such that $U^{-1}A \cong S^{-1}A$.
2998
2999 Let $u \in U$; the image of $u$ in $S^{-1}A$ must be a unit.
3000
3001 Units in $S^{-1}A$ are the fractions $\frac{s}{a}$ such that their inverse exists.\\
3002 \hspace*{2em} $\exists~ x=\frac{a}{s}$ such that $u \frac{a}{s} = \frac{1}{1}$\\
3003 \hspace*{2em} $\Longrightarrow~ \frac{ua}{s} = \frac{1}{1} ~\Longrightarrow~ \exists s' \in S$ such that $s'(ua-s)=0$\\
3004 $\Longrightarrow~ s'ua=s's \in S$
3005
3006 since $s,s' \in S$, their product is also in $S$.
3007
3008 Let $b=s'a$, then $ub \in S$.
3009
3010 By definition of $T$, $u \in T$.
3011
3012 Therefore, $U \subseteq T$; thus $T$ is maximal.
3013\end{enumerate}
3014\end{proof}
3015
3016
3017\bibliographystyle{unsrt}
3018\bibliography{commutative-algebra-notes.bib}
3019
3020\end{document}