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
12\begin{filecontents}[overwrite]{commutative-algebra-notes.bib}
13@misc{am,
14 author = {M. F. Atiyah and I. G. MacDonald},
15 title = {{Introduction to Commutative Algebra}},
16 year = {1969}
17}
18@misc{reid,
19 author = {Miles Reid},
20 title = {{Undergraduate Commutative Algebra}},
21 year = {1995}
22}
23@misc{mit-course,
24 author = {Steven Kleiman},
25 title = {{Commutative Algebra - MIT OpenCourseWare}},
26 year = {2008},
27 note = {\url{https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/}},
28 url = {https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/}
29}
30\end{filecontents}
31\nocite{*}
32
33
34\theoremstyle{definition}
35
36\newtheorem{innerdefn}{Definition}
37\newenvironment{defn}[1]
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39{\endinnerdefn}
40
41\newtheorem{innerthm}{Theorem}
42\newenvironment{thm}[1]
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44{\endinnerthm}
45
46\newtheorem{innerlemma}{Lemma}
47\newenvironment{lemma}[1]
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49{\endinnerlemma}
50
51\newtheorem{innerprop}{Proposition}
52\newenvironment{prop}[1]
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54{\endinnerprop}
55
56\newtheorem{innercor}{Corollary}
57\newenvironment{cor}[1]
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59{\endinnercor}
60
61\newtheorem{innereg}{Example}
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64{\endinnereg}
65
66\newtheorem{innerex}{Exercise}
67\newenvironment{ex}[1]
68{\renewcommand\theinnerex{#1}\innerex}
69{\endinnerex}
70
71\newcommand{\aA}{\mathfrak{a}} % TODO: use goth font
72\newcommand{\mM}{\mathfrak{m}}
73
74\title{Commutative Algebra notes}
75\author{arnaucube}
76\date{}
77
78\begin{document}
79
80\maketitle
81
82\begin{abstract}
83 Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}.
84
85 Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
86
87 The proofs may slightly differ from the ones from the books, since I try to extend them for a deeper understanding.
88\end{abstract}
89
90\tableofcontents
91
92\section{Ideals}
93
94\subsection{Definitions}
95
96\begin{defn}{ideal}
97 $I \subset R$ ($R$ ring) such that $0 \in I$ and $\forall x \in I,~ r \in R,~ xr, rx \in I$.\\
98 \hspace*{2em} ie. $I$ absorbs products in $R$.
99\end{defn}
100
101\begin{defn}{prime ideal}
102 if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a in P$ or $b \in P$.
103\end{defn}
104
105\begin{defn}{principal ideal}
106 generated by a single element, $(a)$.
107
108 $(a)$: principal ideal, the set of all multiples $xa$ with $x \in R$.
109\end{defn}
110
111\begin{defn}{maximal ideal}
112 $\mM \subset A$ ($A$ 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 $I=A$.
113\end{defn}
114
115
116\begin{defn}{unit}
117 $x \in A$ such that $xy=1$ for some $y \in A$. ie. element \emph{which divides 1}.
118\end{defn}
119
120\begin{defn}{zerodivisor}
121 $x \in A$ such that $\exists 0 \neq y \in A$ such that $xy=0 \in A$. ie. $x$ \emph{divides 0}..
122
123
124 If a ring does not have zerodivisors is an integral domain.
125\end{defn}
126
127\begin{defn}{prime spectrum - $Spec(A)$}
128 set of prime ideals of $A$. ie.
129
130 $$Spec(A) = \{ P ~|~ P \subset A~ \text{is a prime ideal} \}$$
131\end{defn}
132
133\begin{defn}{integral domain}
134 Ring in which the product of any two nonzero elements is nonzero.
135
136 ie. no zerodivisors.
137
138 ie. $\forall~ 0 \neq a,~ 0 \neq b \in A,~ ab \neq 0 \in A$.
139
140 Every field is an integral domain, not the converse.
141\end{defn}
142
143\begin{defn}{principal ideal domain - PID}
144 integral domain in which every ideal is principal. ie.
145 ie. $\forall I \subset R,~ \exists~ a \in I$ such that $I = (a) = \{ ra ~|~ r \in R \}$.
146\end{defn}
147
148\begin{defn}{nilpotent}
149 $a \in A$ such that $a^n=0$ for some $n>0$.
150\end{defn}
151\begin{defn}{nilrad A}
152 set of all nilpotent elements of $A$; is an ideal of $A$.
153
154 if $nilrad A = 0 ~\Longrightarrow$ $A$ has no nonzero nilpotents.
155
156
157 $$nilrad A = \bigcap_{P \in Spec(A)} P$$
158\end{defn}
159
160\begin{defn}{idempotent}
161 $e \in A$ such that $e^2=e$.
162\end{defn}
163
164\begin{defn}{radical of an ideal}
165 $$rad I = \{ f \in A | f^n \in I~ \text{for some} n \}$$
166
167 $rad I$ is an ideal.
168
169 $nilrad A = rad 0$
170
171 $rad I = \bigcap_{\substack{P \in \operatorname{Spec}(A)\\ P \supset I}} P$
172\end{defn}
173
174\begin{defn}{local ring}
175 A \emph{local ring} has a unique maximal ideal.
176
177 Notation: locall ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$:
178 $$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$
179\end{defn}
180
181\subsection{Lemmas, propositions and corollaries}
182\begin{thm}{AM.1.X}{Zorn's lemma} \label{zorn}
183 TODO
184\end{thm}
185
186\begin{thm}{AM.1.3} \label{1.3}
187 Every ring $A \neq 0$ has at lleast one maximal ideal.
188\end{thm}
189\begin{proof}
190 By Zorn's lemma \ref{zorn}.
191\end{proof}
192
193\begin{cor}{AM.1.4} \label{1.4}
194 if $I \neq (1)$ an ideal of $A$, $\exists$ a maximal ideal of $A$ containing $I$.
195\end{cor}
196
197\begin{cor}{AM.1.5} \label{1.5}
198 Every non-unit of $A$ is contained in a maximal ideal.
199\end{cor}
200
201\begin{defn}{Jacobson radical}
202 The \emph{Jacobson radical} of a ring $A$ is the intersection of all the maximal ideals of $A$.
203
204 Denoted $Jac(A)$.
205
206 $Jac(A)$ is an ideal of $A$.
207\end{defn}
208
209\begin{prop}{AM.1.9} \label{1.9}
210 $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$.
211\end{prop}
212\begin{proof}
213 Suppose $1-xy$ not a unit.
214
215 By \ref{1.5}, $1-xy \in \mM$ for $\mM$ some maximal ideal.
216
217 But $x \in Jac(A) \subseteq \mM$, since $Jac(A)$ is the intersection of all maximal ideals of $A$.
218
219 Hence $xy \in \mM$, and therefore $1 \in \mM$, which is absurd, thus $1-xy$ is a unit.
220
221 Conversely:\\
222 Suppose $x \not\in \mM$ for some maximal ideal $\mM$.
223
224 Then $\mM$ and $x$ generte the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$.
225
226 Hence $1 -xy \in \mM$, and is therefore not a unit.
227\end{proof}
228
229\section{Modules}
230
231\subsection{Modules}
232
233Let $A$ be a ring. An $A$-module is an Abelian group $M$ with a multiplication
234map
235\begin{align*}
236 A \times M &\longrightarrow M\\
237 (f, m) &\longmapsto fm
238\end{align*}
239satisfying $\forall~ f,g \in A,~~ m, n \in M$.
240\begin{enumerate}[i.]
241 \item $f(m \pm n)=fm \pm fn$
242 \item $(f \pm g) m = fm \pm gm$
243 \item $(fg) m = f(gm)$
244 \item $1_A m = m$
245\end{enumerate}
246
247Let $\psi: M \longrightarrow M$ an $A$-linear endomorphism of $M$.\\
248$A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi$.
249\begin{itemize}
250\item since $\psi$ is $A$-linear, $A[\psi]$ is a commutative ring.
251\item $M$ is a module over $A[\psi]$, so $\psi$ beomes multiplication by a ring element.
252\end{itemize}
253
254\subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries}
255
256\begin{prop}{AM.2.4}(Cayley-Hamilton Theorem) \label{2.4}
257 Let $M$ a fingen $A$-module. Let $\aA$ an ideal of $A$, let $\psi$ an
258 $A$-module endomorphism of $M$ such that $\psi(M) \subseteq \aA M$.
259
260 Then $\psi$ satisfies
261
262 $$\psi^n + a_1 \psi^{n-1} + \ldots + a_{n-1} \psi + a_n = 0$$
263
264 with $a_i \in \aA$.
265\end{prop}
266\begin{proof}
267 Since $M$ fingen, let $\{ x_1, \ldots, x_n \}$ be generators of $M$.\\
268 By hypothesis, $\psi(M) \subseteq \aA M$; so for any generator $x_i$, it's image $\psi(x_i) \in \aA M$.
269
270 Any element in $\aA M$ is a linear combination of the generators with coefficients in the ideal $\aA$, thus
271 $$\psi(x_i)= \sum_{j=1}^n a_{ij} x_j$$
272 with $a_{ij} \in \aA$.
273
274 Thus, for a module with $n$ generators, we have $n$ different $\psi(x_i)$ equations:
275
276 $$
277 \left.
278 \begin{aligned}
279 \psi(x_1) &= a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n\\
280 \psi(x_2) &= a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n\\
281 \ldots\\
282 \psi(x_n) &= a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n
283 \end{aligned}
284 \right\}
285 \begin{aligned}
286 &\text{n elements $\psi(x_i) \in \aA M$ which}\\
287 &\text{are linear combinations of the}\\
288 &\text{$n$ generators of $M$}
289 \end{aligned}
290 $$
291
292 Next step: rearrange in order to use matrix algebra.
293
294 Observe that each row equals $0$, and rearranging the elements at each row we get
295
296 \begin{align*}
297 &\psi(x_1) - (a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n) = 0\\
298 &\psi(x_2) - (a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n) = 0\\
299 &\ldots\\
300 &\psi(x_n) - (a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n) = 0
301 \end{align*}
302
303 Then, group the $x_i$ terms together; as example, take the row $i=1$:
304 $$(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0$$
305
306 for $i=2$:
307 $$-a_{2,1} x_1 + (\psi - a_{2,2}) x_2 - \ldots - a_{2,n} x_n = 0$$
308
309 So, $\forall i \in [n]$, as a matrix:
310
311 $$
312 \begin{pmatrix}
313 \psi - a_{1,1} & -a_{1,2} & \ldots & -a_{1,n}\\
314 -a_{2,1} & \psi-a_{2,2} & \ldots & -a_{2,n}\\
315 \vdots\\
316 -a_{n,1} & -a_{n,2} & \ldots & \psi-a_{n,n}\\
317 \end{pmatrix}
318 \begin{pmatrix}
319 x_1\\ x_2\\ \vdots\\ x_n
320 \end{pmatrix}
321 =
322 \begin{pmatrix}
323 0\\ 0\\ \vdots\\ 0
324 \end{pmatrix}
325 $$
326
327 Kronecker delta:
328 $\delta_{ij} =
329 \begin{cases}
330 1 & \text{if } i = j,\\
331 0 & \text{otherwise}
332 \end{cases}$
333
334 With the Kronecker delta, $\psi(x_i)$ can be expressed as
335 $$\psi(x_i) = \sum_{j=1}^n \delta_{ij} \psi(x_j)$$
336 so the previous matrix can be characterized as
337 $$\sum_{j=1}^n (\delta_{ij} \psi - a_{ij}) x_j = 0$$
338
339 The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$)
340 \begin{itemize}
341 \item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$
342 \item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$
343 \end{itemize}
344
345 Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$,
346 the ring of operators $A[\psi]$ is a commutative ring.
347
348 $\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant.
349
350 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 $\psi$.\\
351 $\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\
352 $\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$
353
354 $$det(M) \cdot x_i = 0~~ \forall i$$
355 where $x_i$ are the generators of $M$.
356
357 Since $det(M)$ kills every generator, it must kill every element in $M$\\
358 $\Longrightarrow~~ det(M)$ is the zero map.
359
360 Leibniz formula of the determinant of an $n \times n$ matrix:
361 $$
362 det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)}
363 $$
364
365 so,
366 $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$
367 expanding it,
368 \begin{itemize}
369 \item highest power is $1 \cdot \psi^n$
370 \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\
371 where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$
372 \item the rest of coefficients of $\psi^k$ are also elements in $\aA$
373 \end{itemize}
374
375 So we have
376 $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$
377 with $a_i \in \aA$.
378
379 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:
380 $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$
381 with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem.
382\end{proof}
383
384
385
386\begin{cor}{AM.2.5} \label{2.5}
387 Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$.
388
389 Then, $\exists~ x \equiv 1 \pmod \aA$ such that $xM = 0$.
390\end{cor}
391\begin{proof}
392 take $\psi = \text{identity}$. Then in Cayley-Hamilton (\ref{2.4}):
393 \begin{align*}
394 &\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0\\
395 \Longrightarrow~ &id_M + a_1 id_M + a_2 id_M + \ldots + a_{n-1} id_M + a_n = 0\\
396 \Longrightarrow~ &(1 + a_1 + \ldots + a_n) id_M = 0
397 \end{align*}
398 apply it to $m \in M$, where since $id_M(m)=m$ (by definition of the identity), we then have
399 $$(1 + a_1 + \ldots + a_n) \cdot m = 0$$
400 with $a_i \in \aA$.
401
402 \begin{enumerate}[\text{part} i.]
403 \item $xM=0$:\\
404 Thus the scalar $x = (1 + a_1 + \ldots + a_n)$ annihilates every $m \in M$, ie. the entire module $M$.
405
406 \item $x \equiv 1 \pmod \aA$:\\
407 $x \equiv 1 \pmod \aA ~~ \Longleftrightarrow (x-1) \in \aA$\\
408 then from $x = (1 + \underbrace{a_1 + \ldots + a_n}_b) \in \aA$, set $b=a_1 + \ldots + a_n$,\\
409 so that $x=(1+b) \in \aA$.\\
410
411 Then $x-1 = (1+b)-1 = b \in \aA$\\
412 so $x-1 \in \aA$, thus $x \equiv 1 \pmod \aA$ as stated.
413 \end{enumerate}
414\end{proof}
415
416
417
418\begin{prop}{AM.2.6}{Nakayama's lemma} \label{2.6}
419 Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$.
420
421 Then $\aA M = M$ implies $M=0$.
422\end{prop}
423\begin{proof}
424 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 $\aA \subseteq Jac(A)$).
425
426 (recall \ref{1.9}: $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$).\\
427 By \ref{1.9}, $x$ is a unit in $A$ (thus $x^{-1}\cdot x=1$).
428
429 Hence $M = x^{-1} \cdot \underbrace{x~ \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$.
430
431 Thus, if $\aA M = M$ then $M=0$.
432\end{proof}
433
434
435
436\begin{cor}{AM.2.7} \label{2.7}
437 Let $M$ a fingen $A$-module, let $N \subseteq M$ a submodule of $M$, let $\aA \subseteq Jac(A)$ an ideal.
438
439 Then $M = \aA M + N \stackrel{\text{implies}}{\Longrightarrow} M=N$.
440\end{cor}
441\begin{proof}
442 The idea is to apply Nakayama (\ref{2.6}) to $M/N$.
443
444 Since $M$ fingen $\Longrightarrow~~ M/N$ is fingen and an $A$-module.
445
446 Since $\aA \subseteq Jac(A) ~\Longrightarrow~$ Nakayama applies to $M/N$ too.
447
448 By definition,
449 $$\aA M = \left\{ \sum a_i \cdot m_i ~~|~~ a_i \in \aA, m_i \in M \right\}$$
450 where $m_i$ are the generators of $M$.
451
452 Then, for $M/N$,
453 $$\aA (\frac{M}{N}) = \left\{ \sum a_i \cdot (m_i + N) ~~|~~ a_i \in \aA, m_i \in M \right\}$$
454
455 observe that $a_i(m_i+N)= a_i m_i +N$, thus
456 $$\sum_i a_i \cdot (m_i + N) = \underbrace{(\sum_i a_i \cdot m_i)}_{\in \aA M} + N \in \aA M + N$$
457
458 Hence,
459
460 \begin{equation}
461 \aA (\frac{M}{N}) = \left\{ x + N ~~|~~ x \in \aA M \right\} = \aA M + N
462 \label{eq:2.7.1}
463 \end{equation}
464
465 By definition, if we take $\frac{\aA M + N}{N}$, then
466 $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$
467
468 thus every $y \in \aA M +N$ can be written as
469 $$y=x+n,~~ \text{with} x \in \aA M,~ n\in N$$
470 which comes from \eqref{eq:2.7.1}.
471
472 Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient.
473
474 Hence, every element of $\frac{\aA M +N}{N}$ has the form
475 $$\frac{\aA M + N}{N} = \left\{ x + N ~~|~~ x \in \aA M \right\}$$
476 as in \eqref{eq:2.7.1}.
477
478 Thus
479 \begin{equation}
480 \aA (\frac{M}{N}) = \aA M + N = \frac{\aA M +N}{N}
481 \label{eq:2.7.2}
482 \end{equation}
483
484 By the Collorary assumption, $M = \aA M + N$; quotient it by $N$:
485 \begin{equation}
486 \frac{M}{N} = \frac{\aA M +N}{N}
487 \label{eq:2.7.3}
488 \end{equation}
489
490 So, from \eqref{eq:2.7.2} and \eqref{eq:2.7.3}:
491 $$\aA (\frac{M}{N}) = \aA M +N = \frac{\aA M +N}{N} = \frac{M}{N}$$
492 thus, $\aA (\frac{M}{N}) = \frac{M}{N}$.
493
494 By Nakayama's lemma \ref{2.6}, if $\aA (\frac{M}{N}) = \frac{M}{N} ~\stackrel{implies}{\Longrightarrow}~ \frac{M}{N}=0$
495
496 Note that
497 $$\frac{M}{N} = \{ m + N ~|~ m \in M \}$$
498 (the zero element in $\frac{M}{N}$ is the coset $N=0+N$)
499
500 Then, $\frac{M}{N}=0$ means that the quotient has exactly one element, the zero coset $N$.
501
502 Thus, every coset $m + N$ equals the zero coset $N$, so $m-0 \in N ~\Longrightarrow~ m \in N$.
503
504 Hence every $m \in M$ lies in $N$, ie. $\forall m \in M,~ m \in N$.
505
506 So $M \subseteq N$. But notice that by the Corollary, we had $N \subseteq M$, therefore $M = N$.
507
508 Thus, if $M = \aA M + N \stackrel{implies}{\Longrightarrow} M = N$.
509\end{proof}
510
511
512
513\begin{prop}{AM.2.8} \label{2.8}
514 Let $x_i \forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vecctor space. Then the $x_i$ generate $M$.
515\end{prop}
516\begin{proof}
517 Let $N$ submodule $M$, generated by the $x_i$.
518
519 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 $N = M$.
520\end{proof}
521
522
523\begin{prop}{AM.2.10} \label{2.10}
524 Split exact sequence. TODO
525\end{prop}
526
527\section{Noetherean rings}
528
529\begin{defn}{}{Ascending Chain Condition}
530 A partially orddered set $\Sigma$ has the \emph{ascending chain condition} (a.c.c.) if every chain
531 $$s_1 \leq s_2 \leq \ldots \leq s_k \leq \ldots$$
532 eventually breaks off, that is, $s_k = s_{k+1} = \ldots$ for some $k$.
533\end{defn}
534
535$\Longrightarrow~ \Sigma$ has the a.c.c. iff every non-empty subset $S \subset \Sigma$ has a maximal element.\\
536\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.
537
538\begin{defn}{R.3.2}{Noetherian ring}
539 Let $A$ a ring; 3 equivalent conditions:
540 \begin{enumerate}[i.]
541 \item the set $\Sigma$ of ideals of $A$ has the a.c.c.; in other words, every increasing chain of ideals
542 $$I_1 \subset I_2 \subset \ldots \subset I_k \subset \ldots$$
543 eventually stops, that is $I_k = I_{k+1}=\ldots$ for some $k$.
544 \item every nonempty set $S$ of iddeals has a maximal element
545 \item every iddeal $I \subset A$ is finitely generated
546 \end{enumerate}
547 If these conditions hold, then $A$ is \emph{Noetherian}.
548\end{defn}
549\begin{proof}
550 TODO
551\end{proof}
552
553\begin{defn}{R.3.4.D}{Noetherian modules}
554 An $A$-module $M$ is Noetherian if the submoles of $M$ have the a.c.c.,\\
555 that is, ay increasing chain
556 $$M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$$
557 of submodules eventually stops.
558\end{defn}
559
560As in with rings, it is equivalent to say that
561\begin{enumerate}[i.]
562 \item any nonempty set of modulesof $M$ has a maximal element
563 \item every submodule of $M$ is finite
564\end{enumerate}
565
566
567
568\begin{prop}{R.3.4.P}
569 Let $0 \longrightarrow L \xrightarrow{\ \alpha \ } M \xrightarrow{\ \beta \ } N \longrightarrow 0$ be a s.e.s. (split exact sequence, \ref{2.10}).
570
571 Then, $M$ is Noetherian $\Longleftrightarrow~ L$ and $N$ are Noetherian.
572\end{prop}
573\begin{proof}
574 $\Longrightarrow$: trivial, since ascending chains of submodules in $L$ and $N$ correspond one-to-one to certain chains in $M$.
575
576 $\Longleftarrow$: suppose $M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$ is an increasing chain of submodules of $M$.
577
578 Then identifying $\alpha(L)$ with $L$ and taking intersection gives a chain
579 $$L \cap M_1 \subset L \cap M_2 \subset \ldots \subset L \cap M_k \subset \ldots$$
580 of submodules of $L$, and applying $\beta$ gives a chain
581 $$\beta(M_1) \subset \beta(M_2) \subset \ldots \beta(M_k) \subset \ldots$$
582 of submodules of $N$.
583
584 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.
585\end{proof}
586
587\begin{lemma}{R.3.4.L}
588 for submodules $M_1 \subset M_2 \subset M$,
589 $$L \cap M_1 = L \cap M_2 ~\text{and}~ \beta(M_1) = \beta(M_2) ~\Longrightarrow~ M_1 = M_2$$
590\end{lemma}
591\begin{proof}
592 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 $\beta(m)=\beta(m)$.
593
594 Then $\beta(m-n)=0$, so that
595 $$m - n \in M_2 \cap ker(\beta)=M_1 \cap ker(\beta)$$
596
597 Hence $m \in M_1$, thus $M_1 = M_2$.
598\end{proof}
599
600\newpage
601
602\section{Exercises}
603
604For the exercises, I follow the assignements listed at \cite{mit-course}.
605
606The 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}.
607
608\subsection{Exercises Chapter 1}
609
610\begin{ex}{R.1.1}
611 Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$?
612\end{ex}
613\begin{proof}
614 Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$.
615
616 $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\
617 Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$.
618
619 Thus $I \cup J$ is not closed under addition; thus is not an ideal.
620
621
622 Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum:
623
624 $$I+J = \{ a+b | a \in I, b \in J \}$$
625
626 $gcd(2,3)=1$, so $I+J = \mathbb{Z}$.
627
628 Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$.
629\end{proof}
630
631\begin{ex}{R.1.5}
632 let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\
633 In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$.
634\end{ex}
635\begin{proof}
636 (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$).
637
638 Let
639 $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
640 The claim is that $\psi^{-1}(P)$ is prime iddeal of $A$.
641
642 \begin{enumerate}[i.]
643 \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
644 $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$).
645
646 If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so
647 $$\psi(a-b)= \psi(a) - \psi(b) \in P$$
648 hence $a-b \in \psi^{-1}(P)$.
649
650 If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\
651 Thus $ra \in \psi^{-1}(P)$.
652
653 $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$.
654
655 \item show that $\psi^{-1}(P)$ is prime:\\
656 $\psi^{-1}(P) \neq A$, 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 $B$.
657
658 Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$.
659
660 Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\
661 Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$.
662
663 Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$.
664 \end{enumerate}
665\end{proof}
666
667
668\begin{ex}{R.1.6}
669 prove or give a counter example:
670 \begin{enumerate}[a.]
671 \item the intersection of two prime ideals is prime
672 \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
673 \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
674 \end{enumerate}
675\end{ex}
676\begin{proof}
677 \begin{enumerate}[a.]
678 \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime.
679
680 Then $I \cap J = (2) \cap (3) = (6)$.
681
682 The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$.
683
684 Thus the intersection of two primes can not be prime.
685
686 \item $P_1=(2),~ P_2=(3)$, both prime.
687
688 Then,
689 $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$
690
691 $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$.
692
693 (recall: principal= generated by a single element)
694
695 So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$.
696
697 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.
698
699 Henceforth, the sum of two prime ideals is not necessarily prime.
700
701 \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$.
702
703 Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$.
704
705 Then
706 \begin{align*}
707 \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\
708 \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A
709 \end{align*}
710
711 But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
712
713 Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
714 \end{enumerate}
715\end{proof}
716
717\subsection{Exercises Chapter 2}
718
719\bibliographystyle{unsrt}
720\bibliography{commutative-algebra-notes.bib}
721
722\end{document}