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]
38{\renewcommand\theinnerdefn{#1}\innerdefn}
39{\endinnerdefn}
40
41\newtheorem{innerthm}{Theorem}
42\newenvironment{thm}[1]
43{\renewcommand\theinnerthm{#1}\innerthm}
44{\endinnerthm}
45
46\newtheorem{innerlemma}{Lemma}
47\newenvironment{lemma}[1]
48{\renewcommand\theinnerlemma{#1}\innerlemma}
49{\endinnerlemma}
50
51\newtheorem{innerprop}{Proposition}
52\newenvironment{prop}[1]
53{\renewcommand\theinnerprop{#1}\innerprop}
54{\endinnerprop}
55
56\newtheorem{innercor}{Corollary}
57\newenvironment{cor}[1]
58{\renewcommand\theinnercor{#1}\innercor}
59{\endinnercor}
60
61\newtheorem{innereg}{Example}
62\newenvironment{eg}[1]
63{\renewcommand\theinnereg{#1}\innereg}
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{2026}
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}. For the exercises, I follow the assignments listed at \cite{mit-course}.
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: local 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{Z and K[X], two Principal Ideal Domains}
182
183
184\begin{lemma}{}
185 $\mathbb{Z}$ is a PID.
186\end{lemma}
187\begin{proof}
188 Let $I$ a nonzero ideal of $\mathbb{Z}$.
189
190 Since $I \neq \{0\}$, there is at least one nonzero integer in $I$. Choose the smallest element of $I$, namely $d$.
191
192 Observe that $(d) \subseteq I$, since $d \in I$. Then, every multiple $nd \in I$, since $I$ is an ideal.
193
194 Take $a \in I$. By the Euclidean division algorithm in $\mathbb{Z}$, $a=qd+r$, with $q,r \in \mathbb{Z}$ and $0 \leq r \leq d$.
195
196 Then $r = a - qd \in I$, but $d$ was chosen to be the smallest positive element of $I$, so the only possibility is $r=0$.
197
198 Hence, $a=qd$, so $a \in (d)$, giving $I \subseteq (d)$.
199
200 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).
201\end{proof}
202
203\begin{lemma}{}
204 $K[X]$ is a PID.
205\end{lemma}
206\begin{proof}
207 This proof follows very similarly to the previous proof.\\
208
209 Let $K$ be a field, $K[X]$ a polynomial ring.
210
211 Take $\{0\} \neq I \subseteq K[X]$.
212
213 Since $I \neq \{0\}$, there is at least one non-zero polynomial in $I$.
214
215 Let $p(X) \in I$ be of minimal degree among nonzero elements of $I$.
216
217 Observe that $(p(X)) \subseteq I$, because $p(X) \in I$ and $I$ is an ideal.
218
219 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 eithr $r(X)=0$ or $deg(r) < deg(p)$.
220
221 Since $f,p \in I$, then $r(X) = f(X) - q(X)\cdot p(X) \in I$
222
223 If $r(X) \neq 0$, then $deg(r) < deg(p)$, which contradicts the minimality of $deg(p)$ in $I$.
224
225 Therefore, $r(X)=0$, thus $f(X)=q(X)\cdot p(X)$, hence $f(X) \in (p(X))$.
226 Henceforth, $I \subseteq (p(X))$.
227
228 Then, since $(p(X)) \subseteq I$ and $I \subseteq (p(X))$, we have that $I = (p(X))$.
229
230 So every ideal of $K[X]$ is principal; thus $K[X]$ is a PID.
231
232\end{proof}
233
234
235
236\subsection{Zorn's lemma and Jacobson radicals}
237
238Let $\Sigma$ be a partially orddered set. Given subset $S \subset \Sigma$, an \emph{upper bound} of $S$ is an element $u \in \Sigma$ such that $s<u \forall s \in S$.
239
240A \emph{maximal element} of $\Sigma$, is $m \in \Sigma$ such that $m<s$ does not hold for any $s \in \Sigma$.
241
242A 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 $s_2 \leq s_1$.
243
244\begin{lemma}{R.1.7}[Zorn's lemma] \label{zorn}
245 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 $\Sigma$.
246
247 Then $\Sigma$ has a maximal element.
248\end{lemma}
249
250\begin{thm}{AM.1.3} \label{1.3}
251 Every ring $A \neq 0$ has at least one maximal ideal.
252\end{thm}
253\begin{proof}
254 By Zorn's lemma \ref{zorn}.
255\end{proof}
256
257\begin{cor}{AM.1.4} \label{1.4}
258 if $I \neq (1)$ an ideal of $A$, $\exists$ a maximal ideal of $A$ containing $I$.
259\end{cor}
260
261\begin{cor}{AM.1.5} \label{1.5}
262 Every non-unit of $A$ is contained in a maximal ideal.
263\end{cor}
264
265\begin{defn}{}[Jacobson radical]
266 The \emph{Jacobson radical} of a ring $A$ is the intersection of all the maximal ideals of $A$.
267
268 Denoted $Jac(A)$.
269
270 $Jac(A)$ is an ideal of $A$.
271\end{defn}
272
273\begin{prop}{AM.1.9} \label{1.9}
274 $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$.
275\end{prop}
276\begin{proof}
277 Suppose $1-xy$ not a unit.
278
279 By \ref{1.5}, $1-xy \in \mM$ for $\mM$ some maximal ideal.
280
281 But $x \in Jac(A) \subseteq \mM$, since $Jac(A)$ is the intersection of all maximal ideals of $A$.
282
283 Hence $xy \in \mM$, and therefore $1 \in \mM$, which is absurd, thus $1-xy$ is a unit.
284
285 Conversely:\\
286 Suppose $x \not\in \mM$ for some maximal ideal $\mM$.
287
288 Then $\mM$ and $x$ generate the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$.
289
290 Hence $1 -xy \in \mM$, and is therefore not a unit.
291\end{proof}
292
293\section{Modules}
294
295\subsection{Modules concepts}
296
297Let $A$ be a ring. An $A$-module is an Abelian group $M$ with a multiplication
298map
299\begin{align*}
300 A \times M &\longrightarrow M\\
301 (f, m) &\longmapsto fm
302\end{align*}
303satisfying $\forall~ f,g \in A,~~ m, n \in M$.
304\begin{enumerate}[i.]
305 \item $f(m \pm n)=fm \pm fn$
306 \item $(f \pm g) m = fm \pm gm$
307 \item $(fg) m = f(gm)$
308 \item $1_A m = m$
309\end{enumerate}
310
311Let $\psi: M \longrightarrow M$ an $A$-linear endomorphism of $M$.\\
312$A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi$.
313\begin{itemize}
314\item since $\psi$ is $A$-linear, $A[\psi]$ is a commutative ring.
315\item $M$ is a module over $A[\psi]$, so $\psi$ beomes multiplication by a ring element.
316\end{itemize}
317
318\subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries}
319
320\begin{prop}{AM.2.4}(Cayley-Hamilton Theorem) \label{2.4}
321 Let $M$ a fingen $A$-module. Let $\aA$ an ideal of $A$, let $\psi$ an
322 $A$-module endomorphism of $M$ such that $\psi(M) \subseteq \aA M$.
323
324 Then $\psi$ satisfies
325
326 $$\psi^n + a_1 \psi^{n-1} + \ldots + a_{n-1} \psi + a_n = 0$$
327
328 with $a_i \in \aA$.
329\end{prop}
330\begin{proof}
331 Since $M$ fingen, let $\{ x_1, \ldots, x_n \}$ be generators of $M$.\\
332 By hypothesis, $\psi(M) \subseteq \aA M$; so for any generator $x_i$, it's image $\psi(x_i) \in \aA M$.
333
334 Any element in $\aA M$ is a linear combination of the generators with coefficients in the ideal $\aA$, thus
335 $$\psi(x_i)= \sum_{j=1}^n a_{ij} x_j$$
336 with $a_{ij} \in \aA$.
337
338 Thus, for a module with $n$ generators, we have $n$ different $\psi(x_i)$ equations:
339
340 $$
341 \left.
342 \begin{aligned}
343 \psi(x_1) &= a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n\\
344 \psi(x_2) &= a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n\\
345 \ldots\\
346 \psi(x_n) &= a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n
347 \end{aligned}
348 \right\}
349 \begin{aligned}
350 &\text{n elements $\psi(x_i) \in \aA M$ which}\\
351 &\text{are linear combinations of the}\\
352 &\text{$n$ generators of $M$}
353 \end{aligned}
354 $$
355
356 Next step: rearrange in order to use matrix algebra.
357
358 Observe that each row equals $0$, and rearranging the elements at each row we get
359
360 $$
361 \left.
362 \begin{aligned}
363 &\psi(x_1) - (a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n) = 0\\
364 &\psi(x_2) - (a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n) = 0\\
365 &\ldots\\
366 &\psi(x_n) - (a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n) = 0
367 \end{aligned}
368 \right\}
369 $$
370
371 Then, group the $x_i$ terms together; as example, take the row $i=1$:
372 $$(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0$$
373
374 $$
375 \left.
376 \begin{aligned}
377 &~~~~(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0\\
378 &-a_{2,1} x_1 + (\psi - a_{2,2}) x_2 - \ldots - a_{2,n} x_n = 0\\
379 &\ldots\\
380 &-a_{1,1} x_1 - a_{1,2} x_2 - \ldots + (\psi - a_{1,n}) x_n = 0\\
381 \end{aligned}
382 \right\}
383 $$
384
385 So, $\forall i \in [n]$, as a matrix:
386
387 $$
388 \begin{pmatrix}
389 \psi - a_{1,1} & -a_{1,2} & \ldots & -a_{1,n}\\
390 -a_{2,1} & \psi-a_{2,2} & \ldots & -a_{2,n}\\
391 \vdots\\
392 -a_{n,1} & -a_{n,2} & \ldots & \psi-a_{n,n}\\
393 \end{pmatrix}
394 \begin{pmatrix}
395 x_1\\ x_2\\ \vdots\\ x_n
396 \end{pmatrix}
397 =
398 \begin{pmatrix}
399 0\\ 0\\ \vdots\\ 0
400 \end{pmatrix}
401 $$
402
403
404
405 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 $M$).\\
406 \hspace*{4em}Then we can write the previous equality as
407 \begin{equation}
408 \Phi \cdot m = 0
409 \label{eq:2.4.1}
410 \end{equation}
411
412 We know that
413 \begin{equation}
414 adj(\Phi) \Phi = det(\Phi) I
415 \label{eq:2.4.2}
416 \end{equation}
417 (aka. \emph{fundamental identity for the adjugate matrix}).
418
419 So if at \eqref{eq:2.4.1} we multiply both sides by $adj(\Phi)$,
420 \begin{align*}
421 adj(\Phi) \cdot \Phi \cdot &m = 0\\
422 (\text{recall from \eqref{eq:2.4.2}:}~ &adj(\Phi)\Phi=det(\Phi)\cdot I ~)\\
423 =det(\Phi) \cdot I \cdot &m = 0
424 \end{align*}
425
426 Thus,
427 \begin{align*}
428 det(\Phi) \cdot I \cdot &m = 0:\\
429 \begin{pmatrix}
430 det(\Phi) & 0 & \ldots & 0\\
431 0 & det(\Phi) & \ldots & 0\\
432 \vdots\\
433 0 & 0 & \ldots & det(\Phi)
434 \end{pmatrix}
435 \cdot
436 &\begin{pmatrix}
437 x_1\\ x_2\\ \vdots\\ x_n
438 \end{pmatrix}
439 =
440 \begin{pmatrix}
441 0\\ 0\\ \vdots\\ 0
442 \end{pmatrix}
443 \end{align*}
444
445 $\Longrightarrow$
446 \begin{equation}
447 det(\Phi) \cdot x_i = 0 ~~\forall i \in [n]
448 \label{eq:2.4.3}
449 \end{equation}
450
451 ie. $det(\Phi)$ is an \emph{annihilator} of the generators $x_i$ of $M$, thus
452 is an annihilator of the entire module $M$.
453
454
455 So, we're interested into calculating the $det(\Phi)$.
456
457 By the Leibniz formula,
458 $$\det(A) = \sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=1}^n a_{i, \sigma(i)}$$
459
460 thus,
461 $$det(\Phi) = \underbrace{(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})}_{\text{diagonal of $\Phi$, leading term of the determinant}} - \ldots$$
462
463 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
464 \begin{itemize}
465 \item highest power is $1 \cdot \psi^n$
466 \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\
467 where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$
468 \item the rest of coefficients of $\psi^k$ are also elements in $\aA$
469 \end{itemize}
470
471 Therefore we have
472 $$det(\Phi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$
473 with $a_i \in \aA$.
474
475 \vspace{0.5cm}
476
477 Now, notice that we had $det(\Phi) \cdot x_i = 0 ~\forall~ i\in [n]$.
478
479 The matrix $\Phi$ is the \emph{characteristic matrix}, $xI-A$, viewed as an operator. Then,
480 $$det(\Phi) = det(xI-A) = p(x)$$
481 where $p(x)$ is the \emph{characteristic polynomial}.
482
483 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 $det(\Phi)=0$.
484 Therefore,
485 $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$
486
487
488 %%%%%% OLD START
489
490 % \vspace{3cm}
491 %
492 % Kronecker delta:
493 % $\delta_{ij} =
494 % \begin{cases}
495 % 1 & \text{if } i = j,\\
496 % 0 & \text{otherwise}
497 % \end{cases}$
498 %
499 % With the Kronecker delta, $\psi(x_i)$ can be expressed as
500 % $$\psi(x_i) = \sum_{j=1}^n \delta_{ij} \psi(x_j)$$
501 % so the previous matrix can be characterized as
502 % $$\sum_{j=1}^n (\delta_{ij} \psi - a_{ij}) x_j = 0$$
503 %
504 % The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$)
505 % \begin{itemize}
506 % \item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$
507 % \item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$
508 % \end{itemize}
509 %
510 % 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 $x$.
511 %
512 % Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$,
513 % the ring of operators $A[\psi]$ is a commutative ring.
514 %
515 % $\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant.
516 %
517 %
518 % \vspace{0.75cm}
519 % This is called \emph{"the determinant trick"}.\\
520 % 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 $\psi$.\\
521 % $\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\
522 % $\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$
523 %
524 % $$det(M) \cdot x_i = 0~~ \forall i$$
525 % where $x_i$ are the generators of $M$.
526 %
527 % 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: $adj(\Phi)\cdot \Phi=det(\Phi) \cdot \aA$.
528 % \vspace{0.5cm}
529 %
530 % Since $det(M)$ kills every generator, it must kill every element in $M$\\
531 % $\Longrightarrow~~ det(M)$ is the zero map.
532 %
533 % Leibniz formula of the determinant of an $n \times n$ matrix:
534 % $$
535 % det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)}
536 % $$
537 %
538 % so,
539 % $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$
540 % expanding it,
541 % \begin{itemize}
542 % \item highest power is $1 \cdot \psi^n$
543 % \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\
544 % where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$
545 % \item the rest of coefficients of $\psi^k$ are also elements in $\aA$
546 % \end{itemize}
547 %
548 % So we have
549 % $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$
550 % with $a_i \in \aA$.
551 %
552 % 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:
553 % $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$
554 % with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem.
555
556 %%%%% OLD END
557\end{proof}
558
559
560\vspace{0.5cm}
561
562\begin{cor}{AM.2.5} \label{2.5}
563 Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$.
564
565 Then, $\exists~ x \equiv 1 \pmod \aA$ such that $xM = 0$.
566\end{cor}
567\begin{proof}
568 take $\psi = \text{identity}$. Then in Cayley-Hamilton (\ref{2.4}):
569 \begin{align*}
570 &\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0\\
571 \Longrightarrow~ &id_M + a_1 id_M + a_2 id_M + \ldots + a_{n-1} id_M + a_n = 0\\
572 \Longrightarrow~ &(1 + a_1 + \ldots + a_n) id_M = 0
573 \end{align*}
574 apply it to $m \in M$, where since $id_M(m)=m$ (by definition of the identity), we then have
575 $$(1 + a_1 + \ldots + a_n) \cdot m = 0$$
576 with $a_i \in \aA$.
577
578 \begin{enumerate}[\text{part} i.]
579 \item $xM=0$:\\
580 Thus the scalar $x = (1 + a_1 + \ldots + a_n)$ annihilates every $m \in M$, ie. the entire module $M$.
581
582 \item $x \equiv 1 \pmod \aA$:\\
583 $x \equiv 1 \pmod \aA ~~ \Longleftrightarrow (x-1) \in \aA$\\
584 then from $x = (1 + \underbrace{a_1 + \ldots + a_n}_b) \in \aA$, set $b=a_1 + \ldots + a_n$,\\
585 so that $x=(1+b) \in \aA$.\\
586
587 Then $x-1 = (1+b)-1 = b \in \aA$\\
588 so $x-1 \in \aA$, thus $x \equiv 1 \pmod \aA$ as stated.
589 \end{enumerate}
590\end{proof}
591
592
593
594\begin{prop}{AM.2.6}[Nakayama's lemma] \label{2.6}
595 Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$.
596
597 Then $\aA M = M$ implies $M=0$.
598\end{prop}
599\begin{proof}
600 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 $\aA \subseteq Jac(A)$).
601
602 (recall \ref{1.9}: $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$).\\
603 By \ref{1.9}, $x$ is a unit in $A$ (thus $x^{-1}\cdot x=1$).
604
605 Hence $M = x^{-1} \cdot \underbrace{x~ \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$.
606
607 Thus, if $\aA M = M$ then $M=0$.
608\end{proof}
609
610
611
612\begin{cor}{AM.2.7} \label{2.7}
613 Let $M$ a fingen $A$-module, let $N \subseteq M$ a submodule of $M$, let $\aA \subseteq Jac(A)$ an ideal.
614
615 Then $M = \aA M + N \stackrel{\text{implies}}{\Longrightarrow} M=N$.
616\end{cor}
617\begin{proof}
618 The idea is to apply Nakayama (\ref{2.6}) to $M/N$.
619
620 Since $M$ fingen $\Longrightarrow~~ M/N$ is fingen and an $A$-module.
621
622 Since $\aA \subseteq Jac(A) ~\Longrightarrow~$ Nakayama applies to $M/N$ too.
623
624 By definition,
625 $$\aA M = \left\{ \sum a_i \cdot m_i ~~|~~ a_i \in \aA, m_i \in M \right\}$$
626 where $m_i$ are the generators of $M$.
627
628 Then, for $M/N$,
629 $$\aA (\frac{M}{N}) = \left\{ \sum a_i \cdot (m_i + N) ~~|~~ a_i \in \aA, m_i \in M \right\}$$
630
631 observe that $a_i(m_i+N)= a_i m_i +N$, thus
632 $$\sum_i a_i \cdot (m_i + N) = \underbrace{(\sum_i a_i \cdot m_i)}_{\in \aA M} + N \in \aA M + N$$
633
634 Hence,
635
636 \begin{equation}
637 \aA (\frac{M}{N}) = \left\{ x + N ~~|~~ x \in \aA M \right\} = \aA M + N
638 \label{eq:2.7.1}
639 \end{equation}
640
641 By definition, if we take $\frac{\aA M + N}{N}$, then
642 $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$
643
644 thus every $y \in \aA M +N$ can be written as
645 $$y=x+n,~~ \text{with}~ x \in \aA M,~ n\in N$$
646 which comes from \eqref{eq:2.7.1}.
647
648 Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient.
649
650 Hence, every element of $\frac{\aA M +N}{N}$ has the form
651 $$\frac{\aA M + N}{N} = \left\{ x + N ~~|~~ x \in \aA M \right\}$$
652 as in \eqref{eq:2.7.1}.
653
654 Thus
655 \begin{equation}
656 \aA (\frac{M}{N}) = \aA M + N = \frac{\aA M +N}{N}
657 \label{eq:2.7.2}
658 \end{equation}
659
660 By the Collorary assumption, $M = \aA M + N$; quotient it by $N$:
661 \begin{equation}
662 \frac{M}{N} = \frac{\aA M +N}{N}
663 \label{eq:2.7.3}
664 \end{equation}
665
666 So, from \eqref{eq:2.7.2} and \eqref{eq:2.7.3}:
667 $$\aA (\frac{M}{N}) = \aA M +N = \frac{\aA M +N}{N} = \frac{M}{N}$$
668 thus, $\aA (\frac{M}{N}) = \frac{M}{N}$.
669
670 By Nakayama's lemma \ref{2.6}, if $\aA (\frac{M}{N}) = \frac{M}{N} ~\stackrel{implies}{\Longrightarrow}~ \frac{M}{N}=0$
671
672 Note that
673 $$\frac{M}{N} = \{ m + N ~|~ m \in M \}$$
674 (the zero element in $\frac{M}{N}$ is the coset $N=0+N$)
675
676 Then, $\frac{M}{N}=0$ means that the quotient has exactly one element, the zero coset $N$.
677
678 Thus, every coset $m + N$ equals the zero coset $N$, so $m-0 \in N ~\Longrightarrow~ m \in N$.
679
680 Hence every $m \in M$ lies in $N$, ie. $\forall m \in M,~ m \in N$.
681
682 So $M \subseteq N$. But notice that by the Corollary, we had $N \subseteq M$, therefore $M = N$.
683
684 Thus, if $M = \aA M + N \stackrel{implies}{\Longrightarrow} M = N$.
685\end{proof}
686
687
688
689\begin{prop}{AM.2.8} \label{2.8}
690 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 $M$.
691\end{prop}
692\begin{proof}
693 Let $N$ submodule $M$, generated by the $x_i$.
694
695 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 $N = M$.
696\end{proof}
697
698
699\subsection{Sequences}
700
701\begin{defn}{R.2.9.a}[Exact Sequence]
702 Let a sequence of homomorphisms
703$$L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N$$
704It is \emph{exact} at $M$ if $im(\alpha)=ker(\beta)$.
705
706ie. $\beta \circ \alpha = 0$ and $\alpha$ maps surjectively to
707$ker(\beta)$.
708\end{defn}
709
710\begin{defn}{R.2.9.b}[Short Exact Sequence (s.e.s.)] \label{2.9}
711$$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$$
712is exact $\Longleftrightarrow~ L \subset M$ and $N=M / L$.
713
714Properties:
715\begin{itemize}
716 \item $\alpha$ injective
717 \item $\beta$ surjective
718 \item $\alpha:~ L \Longrightarrow ker \beta$
719 \item $\beta$ induces $M/\alpha(L) \longrightarrow N$
720\end{itemize}
721\end{defn}
722
723\begin{prop}{R.2.10}[Split exact sequence] \label{2.10}
724 For the previous s.e.s., 3 equivalent conditions:
725 \begin{enumerate}[i.]
726 \item $\exists$ isomorphism $M \cong L \oplus N$, with
727 \begin{align*}
728 \alpha:~ &m \longmapsto (m,0)\\
729 \beta:~ &(m, n) \longmapsto n
730 \end{align*}
731
732 \item $\exists$ a \emph{section} of $\beta$, that is, a map $s: N \longrightarrow M$ such that $\beta \circ s = id_N$
733 \item $\exists$ a \emph{retraction} of $\alpha$, that is, a map $r: M \longrightarrow L$ such that $r \circ \alpha = id_L$
734 \end{enumerate}
735
736 If all i, ii, iii are satisfied, it is a split exact sequence.
737\end{prop}
738\begin{proof}
739 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. $M = L \oplus N$.
740
741 \begin{itemize}
742 \item[(i to ii, iii)]
743 if $M \cong L \oplus N$ such that $\alpha:~ m \longmapsto (m,0),~~ \beta:~ s(m, n) \longmapsto n$, we can define the maps
744
745 for ii:
746 \begin{align*}
747 s:~ N &\longrightarrow L \oplus N\\
748 s(n) &\longmapsto (0, n)
749 \end{align*}
750
751 Then $\beta(s(n))=\beta(0,n)$, so $\beta \circ s = id_N$.
752
753 for iii:
754 \begin{align*}
755 r:~ L \oplus N &\longrightarrow L\\
756 r(m,n) &\longmapsto m
757 \end{align*}
758
759 Then $r(\alpha(m))=r(m,0)$, so $r \circ \alpha = id_L$.
760
761 \item[(ii to i)]
762 assume $s: N \longrightarrow M$ such that $\beta \circ s = id_M$
763
764 Want to show $M \cong im(\alpha) \oplus im(s)$.
765
766 $\forall m \in M$, consider $m - s(\beta(m))$, apply $\beta$ to it:\\
767 $\beta(m - s(\beta(m))) = \beta(m) - (\beta \circ s)(\beta(m)) = \beta(m) - \beta(m) = 0$
768
769 Since $ker(\beta) = im(\alpha),~~ \exists! l \in L ~\text{such that}~ \alpha(l) = m - s(\beta(m))$.
770
771 Thus $m = \alpha(l) + s(\beta(m))$.
772
773 \vspace{0.3cm}
774 Now, suppose $x \in im(\alpha) \cap im(s)$, then $x = \alpha(l)=s(n)$, apply $\beta$ to it: $\beta(\alpha(l)) = \beta(s(n)) ~\Longrightarrow~ 0=n$.
775
776 If $n=0$, then $s(n)=0$, so the intersection is $\{0\}$.
777
778 \vspace{0.3cm}
779 Define
780 \begin{align*}
781 \phi: L \oplus N &\longrightarrow M\\
782 \phi(l,n) &\longmapsto \alpha(l)+s(n)
783 \end{align*}
784 This isomorphism satisfies the required conditions.
785
786 \item[(iii to i)] similar to the previous one.
787 \end{itemize}
788
789 \vspace{0.3cm}
790 Overview:\\
791
792$$
7930 \longrightarrow L
794 \substack{
795 \stackrel{\alpha}{\longrightarrow}\\
796 \stackrel{\longleftarrow}{r}
797 }
798 \substack{M \\[0.5ex] \cong L \oplus N}
799 \substack{
800 \stackrel{\beta}{\longrightarrow}\\
801 \stackrel{\longleftarrow}{s}
802 }
803 N \longrightarrow 0
804$$
805\begin{align*}
806 \alpha:~ &l \longmapsto (l,0)\\
807 r:~ &(m,n) \longmapsto m\\
808 \alpha &\circ r = id_L\\
809 \beta:~ &(l,n) \longmapsto n\\
810 s:~ &n \longmapsto (0,n)\\
811 \beta &\circ s = id_N
812\end{align*}
813
814\end{proof}
815
816\section{Noetherian rings (and modules)}
817
818\begin{defn}{}[Ascending Chain Condition]
819 A partially orddered set $\Sigma$ has the \emph{ascending chain condition} (a.c.c.) if every chain
820 $$s_1 \leq s_2 \leq \ldots \leq s_k \leq \ldots$$
821 eventually breaks off, that is, $s_k = s_{k+1} = \ldots$ for some $k$.
822\end{defn}
823
824$\Longrightarrow~ \Sigma$ has the a.c.c. iff every non-empty subset $S \subset \Sigma$ has a maximal element.\\
825\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.
826
827\subsection{Noetherian rings and modules}
828\begin{defn}{R.3.2}[Noetherian ring]\\
829 Let $A$ a ring; 3 equivalent conditions:
830 \begin{enumerate}[i.]
831 \item the set $\Sigma$ of ideals of $A$ has the a.c.c.; in other words, every increasing chain of ideals
832 $$I_1 \subset I_2 \subset \ldots \subset I_k \subset \ldots$$
833 eventually stops, that is $I_k = I_{k+1}=\ldots$ for some $k$.
834 \item every nonempty set $S$ of ideals has a maximal element
835 \item every ideal $I \subset A$ is finitely generated
836 \end{enumerate}
837 If these conditions hold, then $A$ is \emph{Noetherian}.
838\end{defn}
839\begin{proof}
840 TODO
841\end{proof}
842
843\begin{defn}{R.3.4.D}[Noetherian modules]\\
844 An $A$-module $M$ is Noetherian if the submoles of $M$ have the a.c.c.,\\
845 that is, ay increasing chain
846 $$M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$$
847 of submodules eventually stops.
848\end{defn}
849
850As in with rings, it is equivalent to say that
851\begin{enumerate}[i.]
852 \item any nonempty set of modulesof $M$ has a maximal element
853 \item every submodule of $M$ is finite
854\end{enumerate}
855
856
857
858\begin{prop}{R.3.4.P}\label{R.3.4.P}
859 Let $0 \longrightarrow L \xrightarrow{\ \alpha \ } M \xrightarrow{\ \beta \ } N \longrightarrow 0$ be a s.e.s. (split exact sequence, \ref{2.10}).
860
861 Then, $M$ is Noetherian $\Longleftrightarrow~ L$ and $N$ are Noetherian.
862\end{prop}
863\begin{proof}
864 $\Longrightarrow$: trivial, since ascending chains of submodules in $L$ and $N$ correspond one-to-one to certain chains in $M$.
865
866 $\Longleftarrow$: suppose $M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$ is an increasing chain of submodules of $M$.
867
868 Then identifying $\alpha(L)$ with $L$ and taking intersection gives a chain
869 $$L \cap M_1 \subset L \cap M_2 \subset \ldots \subset L \cap M_k \subset \ldots$$
870 of submodules of $L$, and applying $\beta$ gives a chain
871 $$\beta(M_1) \subset \beta(M_2) \subset \ldots \beta(M_k) \subset \ldots$$
872 of submodules of $N$.
873
874 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.
875\end{proof}
876
877\begin{lemma}{R.3.4.L}
878 for submodules $M_1 \subset M_2 \subset M$,
879 $$L \cap M_1 = L \cap M_2 ~\text{and}~ \beta(M_1) = \beta(M_2) ~\Longrightarrow~ M_1 = M_2$$
880\end{lemma}
881\begin{proof}
882 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 $\beta(m)=\beta(n)$.
883
884 Then $\beta(m-n)=0$, so that
885 $$m - n \in M_2 \cap ker(\beta)=M_1 \cap ker(\beta)$$
886
887 Hence $m \in M_1$, thus $M_1 = M_2$.
888\end{proof}
889
890
891
892\begin{cor}{R.3.5}[Properties of Noetherian modules]\label{R.3.5}
893 \begin{enumerate}[i.]
894 \item if $\forall i \in [r],~~M_i$ are Noetherian modules, then
895 $\bigoplus_{i=1}^r M_i$ is Noetherian.
896 \item if $A$ a Noetherian ring, then an $A$-module $M$ is Noetherian iff it is finite over $A$.
897 \item if $A$ a Noetherian ring, $M$ a finite module, then any submodule $N \subset M$ is again finite.
898 \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.
899 \end{enumerate}
900\end{cor}
901\begin{proof}
902 \begin{enumerate}[i.]
903 \item a direct sum $M_1 \oplus M_2$ is a particular case of an exact sequence.
904
905 Then, Proposition \ref{R.3.4.P} proves this statement when $r=2$. The case $r>2$ follows by induction.
906
907 \item if $M$ finite, then $\exists~$ surjective homomorphism
908 $$A^r \longrightarrow M \longrightarrow 0$$
909 for some $r$, so that $M$ is a quotient
910 $$M \cong A^r / N$$
911 for some submodule $N \subset A^r$.
912
913 $A^r$ is a Noetherian module by i., so $M$ is Noetherian due Proposition \ref{R.3.4.P}.
914
915 Conversely, $M$ Noetherian implies $M$ finite.
916
917 item as in previous implications:\\
918 $M$ finite and $A$ Noetherian $\Longrightarrow$ $M$ is Noetherian,\\
919 $\Longrightarrow$ since $N \subseteq M$, then $N$ is Noetherian too\\
920 $\Longrightarrow$ which implies that $N$ is a finite $A$-module.
921
922 \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.
923 \end{enumerate}
924\end{proof}
925
926
927\vspace{0.5cm}
928\subsection{Hilbert basis}
929\begin{thm}{R.3.6}[Hilbert basis theorem] \label{hilbert-basis}
930 If $A$ a Noetherian ring, then so is the polynomial ring $A[x]$.
931\end{thm}
932\begin{proof}
933 Prove that any ideal $I \subset A[x]$ is fingen.
934
935 Define auxiliary sets $J_n \subset A$ by
936 $$J_n = \{ a \in A ~|~ \exists f \in I ~\text{s.th.}~ f = a x^n + b_{n-1}x^{n-1} + \ldots b_0 \}$$
937 ie. $J_n$ is the set of leading coefficients of $I$ of degree $n$.
938
939 $J_n$ is an ideal, since $I$ is an ideal.
940
941 $J_n \subset J_{n+1}$, since for $f \in I$ also $x f \in I$.
942
943 Therefore $J_1 \subset J_2 \subset \ldots \subset J_k \subset \ldots$ is an increasing chain of ideals.\\
944 Using the assumption that $A$ is Noetherian, deduce that $J_n = J_{n+1}$ for some $n$.
945
946 For each $m \leq n, ~~ J_m \subset A$ is fingen, ie.
947 $$J_m = (a_{m,1}, \ldots a_{m, r_m})$$
948
949 By definition of $J_m$, for each $a_{m,j}$ with $1 \leq j \leq r_m$,\\
950 $\exists$ a polynomial $f_{m, j} \in I$ of degree $m$ having the leading coefficient $a_{m, j}$.
951
952 $$\Longrightarrow~~ \{ f_{m,j} \}_{m<n; 1 \leq j \leq r_m}$$
953 the set of elements of $I$.
954
955 Claim: this finite set ($\{ f_{m,j} \}$) generates $I$.
956
957 $\forall f \in I$, if $\deg f =m$, then its leading coefficient is $a \in J_m$,
958
959 hence if $m \geq n$, then $a \in J_m=J_n$, so that
960 $$a = \sum b_i a_{n,i} ~~ \text{with}~ b_i \in A$$
961 and
962 $$f - \sum b_i X^{m-n} \cdot f_{n, i}$$
963 has degree $<m$.
964
965 Similarly, if $m \leq n$, then $a \in J_m$, so that
966 $$a = \sum b_i a_{m, i} ~~\text{with}~ b_i \in A$$
967 and
968 $$f - \sum b_i f_{n, i}$$
969 has degree $<m$.
970
971
972 \vspace{0.3cm}
973 By induction on $m$, $f$ can be written as a linear combination of finitely many elements.
974
975 Thus, any ideal of $A[x]$ is finitely generated.
976\end{proof}
977
978\vspace{0.5cm}
979\begin{cor}{R.3.6.C}
980 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.
981
982 In particular, any fingen algebra over $\mathbb{Z}$ or over a field $K$ is Noetherian.
983\end{cor}
984\begin{proof}
985 the assumption is that $B$ is a quotient of a polynomial ring,
986 $$B \cong A[x_1, \ldots, x_n] / I$$
987 for some ideal $I$.
988
989 By the Hilbert basis theorem \ref{hilbert-basis} and induction,\\
990 $A$ being Noetherian implies that $A[x_1, \ldots, x_n]$ is Noetherian.
991
992 And by Corollary \ref{R.3.5}(iv),\\
993 $A[x_1, \ldots, x_n]$ being Noetherian implies that $A[x_1, \ldots, x_n]/I$ is Noetherian.
994\end{proof}
995
996
997
998\vspace{1cm}
999\section{Finite ring extensions and Noether normalization}
1000
1001\subsection{A-algebras and integral domains}
1002
1003\begin{defn}{}[A-algebra]
1004 An $A$-algebra is a ring $B$ with a ring homomorphism $\psi: A \longrightarrow B$.
1005
1006 $B$ is an $A$-module with multiplication defined by $\psi(a) \cdot b~~~ (a \in A, b \in B)$.
1007
1008 When $A \subset B$, $B$ is an extenaion ring of $A$; denoted $\psi(A) = A' \subset B$.
1009\end{defn}
1010
1011\begin{defn}{R.4.1}\label{R.4.1}
1012 Let $B$ be an $A$-algebra.
1013
1014 \begin{enumerate}[i.]
1015 \item $B$ is a \emph{finite} $A$-algebra (\emph{finite over $A$}) if it is finite as an $A$-module.
1016 \item $y \in B$ is \emph{integral over} $A$ if $\exists$ a monic polynomial
1017 $$f(Y) = Y^n + a_{n-1} Y^{n-1} + \ldots + a_0 ~\in A'[Y]$$
1018 such that $f(y)=0:$
1019 $$f(y) = y^n + a_{n-1} y^{n-1} + \ldots + a_0 = 0$$
1020
1021 The algebra $B$ is \emph{integral over} $A$ if $\forall~ b \in B$ is integral.
1022 \end{enumerate}
1023\end{defn}
1024
1025
1026\begin{prop}{R.4.2}\label{R.4.2}
1027 Let $\psi: A \longrightarrow B$ be an $A$-algebra, and $y \in B$. Three equivalent conditions:
1028
1029 \begin{enumerate}[i.]
1030 \item $y$ is integral over $A$
1031 \item subring $A'[y] \subset B$ generated by $A' = \psi(A)$ and $y$ is finite over $A$
1032 \item $\exists$ an $A$-subalgebra $C \subset B$ such that $A'[y] \subset C$ and $C$ is finite over $A$
1033 \end{enumerate}
1034
1035 Notes: $A'$ is the image of $A$ in $B$, ie. $A' = \psi(A)$.\\
1036 $A'[y]$ is the smallest subring of $B$ containing both coefficients from $A$ and the element $y$.
1037\end{prop}
1038\begin{proof} .\\
1039 \begin{itemize}
1040 \item[(i to ii):]
1041 since $y$ integral over $A$ $~~\Longrightarrow~$ by \ref{R.4.1} (ii), $y$ satisfies
1042 $$f(y) = y^n + a_{n-1} y^{n-1} + \ldots + a_0 = 0$$
1043
1044 So any power $y^k~~ (k\geq n)$ can be expressed in terms of $\{1, y, y^2, \ldots, y^{n-1} \}$.
1045
1046 Thus the set $\{1, y, y^2, \ldots, y^{n-1} \}$ spans $A'[y]$ as an $A$-module.
1047
1048 \item[(iii to i):]
1049 since $A'[y] \subset C ~~\Longrightarrow~ y \in C$\\
1050 since $C$ finite over $A$, $C$ has finite generators $\{ c_1, \ldots, c_n \}$ such that $C = A \cdot c_1 + A \cdot c_2 + \ldots + A \cdot c_n$
1051
1052 Thus $y \cdot c_i \in C$,
1053 $$y \cdot c_i = \sum_{j=1}^n a_{ij} c_j$$
1054 with $a_{ij} \in A$.
1055
1056 By the Cayley-Hamilton theorem (\ref{2.4}),
1057 $$y^n + a_{n-1} y^{n-1} + \ldots + a_1 y + a_0 = 0$$
1058 Therefore, $y$ is integral (by \ref{R.4.1} (ii)).
1059 \end{itemize}
1060\end{proof}
1061
1062
1063\begin{prop}{R.4.3}[Tower Laws]\label{R.4.3}
1064 \\Let $B$ be an $A$-algebra.
1065 \begin{enumerate}[a.]
1066 \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,\\
1067 then $C$ is finite over $A$.
1068 \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 $A$.\\
1069 In particular, every $f \in A[y_1, \ldots, y_m]$ is integral over $A$.
1070 \item Transitivity of integrality: if $A \subset B \subset C$ with $C$ integral over $B$, and $B$ integral over $A$,\\
1071 then $C$ is integral over $A$.
1072 \item Integral closure as a subring: the subset
1073 $$\tilde{A} = \{ y \in B ~|~ y ~\text{is integral over}~ A \} \subset B$$
1074 is a subring of $B$.
1075
1076 Moreover, if $y \in B$ is integral over $\tilde{A}$ then $y \in \tilde{A}$, so that $\tilde{\tilde{A}} = \tilde{A}$.
1077 \end{enumerate}
1078\end{prop}
1079\begin{proof}.\\
1080 \begin{enumerate}[a.]
1081 \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,\\
1082 then the set of products $\{ \beta_i \gamma_j \}$ generates $C$ as an $A$-module.
1083
1084 Since there are $n \times m$ generators (ie. finite), $C$ is finite over $A$.
1085
1086 \item proof by induction:
1087
1088 base case: if $y_1$ integral over $A ~\Longrightarrow~$ it satisfies a monic polynomial.
1089
1090 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.
1091
1092 inductive step: let $R_k = A[y_1, \ldots, y_k]$. Assume $R_k$ is finite over $A$.
1093
1094 Since $y_{k+1}$ is integral over $A~~ \Longrightarrow~$ it is also integral over $R_k$.
1095
1096 Thus $R_{k+1} = R_k[y_{k+1}]$ is finite over $R_k$.
1097
1098 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 $A$.
1099
1100 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).
1101
1102 \item let $x \in C$, since $x$ integral over $B$, it satisfies:
1103 $$x^n + b_{n-1} x^{n-1} + \ldots + b_1 x + b_0 = 0,~~~~~b_i \in B$$
1104
1105 Let $B''=A[b_0, b_1, \ldots, b_{n-1}]$. Since each $b_i \in B$ and $B$ is integral over $A$\\
1106 \hspace*{2em} $\Longrightarrow~$ each $b_i$ is integral over $A$.
1107
1108 Since all $b_i$ are integral over $B'$ $~~\Longrightarrow~ B'[x]$ is a finite $B'$-algebra.
1109
1110 By part (a) (transitivity of finiteness), $B'[x]$ is a finite $A$-algebra.
1111
1112 Therefore, $x$ is integral over $A$.
1113
1114 \item
1115 \begin{enumerate}[I.]
1116 \item subring:\\
1117 let $x,y \in \tilde{A}$. Want to show $x+y, xy \in \tilde{A}$:
1118
1119 by part (b), the algebra $A[x,y]$ is finite over $A$.
1120
1121 Since $x+y,xy \in A[x,y]$, they are integral over $A$.
1122
1123 Thus $x+y, xy \in \tilde{A}$, since $\tilde{A} = \{ b \in B ~|~ b ~\text{integral over}~ A \}$.
1124
1125 \item idempotence\\
1126 let $z \in B$ be integral over $\tilde{A}$
1127
1128 we have a chain $A \subseteq \tilde{A} \subseteq \tilde{A}[x]$.
1129
1130 By definition, $\tilde{A}$ is integral over $A$, and $z$ is integral over $\tilde{A}$\\
1131 thus by part (c), $z$ is integral over $A$.
1132
1133 Therefore, $z \in \tilde{A}$.
1134 \end{enumerate}
1135 \end{enumerate}
1136\end{proof}
1137
1138\begin{lemma}{4.3.Aux}[Integrality implies finiteness] \label{integral-implies-finite}
1139 If $y$ integral over $A$ then $A[y]$ is finite over $A$.
1140
1141 This extends on point (b) from the previous proposition \ref{R.4.3}.
1142\end{lemma}
1143\begin{proof}
1144 Suppose $y$ is integral over $A$. By definition $\exists~~ f \in A[T]$, with
1145 $f$ monic, such that $f(y)=0$.
1146
1147 Let $deg(f)=d$, so that for $f(y)=0$ we have
1148 $$y^d + a_{d-1} y^{d-1} + \ldots + a_1 y + a_0 = 0 ~~~~ a_i \in A$$
1149
1150 Since it is monic (leading coefficient is $1$), we can rearrange it to isolate the highest power:
1151 \begin{equation}
1152 y^d = -(a_{d-1} y^{d-1} + \ldots + a_1 y + a_0)
1153 \label{eq:yn}
1154 \end{equation}
1155 Thus $y^d$ can be written using lower powers of $y$ with coefficients in $A$.
1156
1157 \vspace{0.5cm}
1158 Consider any element $p \in A[y]$, $p = c_m y^m + c_{m-1} y^{m-1} + \ldots + c_0$.
1159
1160 if $m<d$, leave it as it is.\\
1161 if $m \geq d$, use the monic equation \eqref{eq:yn} to replace $y^d$ with lower powers.
1162
1163 Repeating this process, can reduce any power of $y$ down to a linear
1164 combination of $\{1, y, y^2, \ldots, y^{d-1} \}$.
1165
1166 Thus every element in $A[y]$ can be expressed as
1167 $$\lambda_{d-1} y^{d-1} + \ldots + \lambda_2 y^2 + \lambda_1 y + \lambda_0 \cdot 1~~~~ \lambda_i \in A$$
1168
1169 Henceforth, the set $\{1, y, y^2, \ldots, y^{d-1} \}$ generates $A[y]$ as a finite $A$-module.
1170\end{proof}
1171
1172
1173
1174\vspace{0.5cm}
1175\begin{defn}{4.4}[Integral closure]
1176 Given the ring $\tilde{A}$ from \ref{R.4.3}.(d), ie. $\tilde{A} = \{ y \in B ~|~ y ~\text{integral over}~ A \} \subset B$,
1177 $\tilde{A}$ is the \emph{integral closure} of $A$ in $B$.
1178
1179 If $A=\tilde{A}$, then $A$ is \emph{integrally closed} in $B$.
1180
1181 An integral domain $A$ is \emph{normal} if it is \emph{integrally closed in its field of fractions}, that is if
1182 $$A = \tilde{A} \subset K = Frac(A)$$
1183
1184 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 $A$.
1185\end{defn}
1186
1187
1188\vspace{0.5cm}
1189
1190\subsection{Noether normalization}
1191
1192\begin{defn}{4.6}[Algebraically independent] \label{R.4.6.D}
1193 $y_1, \ldots, y_n \in A$ 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.
1194
1195 $\Longrightarrow~~ \nexists~ F(y_1, \ldots, y_n)=0$ ($F$ nonzero) with coefficients in $K$.
1196\end{defn}
1197
1198Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some finite set $y_1, \ldots, y_n$.
1199
1200
1201\vspace{0.5cm}
1202\begin{lemma}{R.4.6.L} \label{R.4.6.L}
1203 Let $A = K[y_1, \ldots, y_n]$ and $0 \neq F \in K[Y_1, \ldots, Y_n]$ such that $F(y_1, \ldots, y_n)=0$.
1204
1205 Then $\exists~ y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over
1206 $$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~~\text{and}~~ A=A^*[y_n]$$
1207\end{lemma}
1208\begin{proof}
1209 Set $y_i^* = y_i - y_n^{r_i}$ for $i \in [n-1]$ and $r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.\\
1210 \hspace*{3em}(ie. $y_i=y_i^* + y_n^{r_i}$)
1211
1212 Define $G \in A$ by
1213 $$G(y_1^*, \ldots, y_{n-1}^*, y_n) = F(y_i^* + y_n^{r_i}, y_n)=0$$
1214 viewed as a relation for $y_n$ over $K[y_1^*, \ldots, y_{n-1}^*]$.
1215
1216 Since $F$ is a polynomial in $y_1, \ldots, y_{n-1}^*$, can write it as a sum of monomials
1217 $$F= \sum_m a_m y^m = \sum_m a_m \prod y_i^{m_i}$$
1218 where $m=(m_1, \ldots, m_n)$ and each $a_m \neq 0$.
1219
1220
1221 Therefore,
1222 $$G= \sum a_m \prod (y_i^* + y_n^{r_i})^{m_i}$$
1223 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 $a_m y_n^{(\sum r_i m_i)}$.
1224
1225 Suppose we can arrange so that
1226 $$m \neq m' ~\Longrightarrow~ \sum r_i m_i \neq \sum r_i m_i'$$
1227
1228 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 $y_n$).
1229\end{proof}
1230
1231
1232\vspace{0.5cm}
1233\begin{thm}{R.4.6}[Noether normalization lemma] \label{noether-normalization}
1234 Let $K$ a field, $A$ a fingen $K$-algebra.
1235
1236 Then $\exists~ z_1, \ldots, z_m \in A$ such that
1237 \begin{enumerate}[i.]
1238 \item $z_1, \ldots, z_m$ are algebraically independent over $K$
1239 \item $A$ is finite over $B=K[z_1, \ldots, z_m]$
1240 \end{enumerate}
1241
1242 That is, a fingen extension $K \subset A$ can be written as a composite
1243 $$K \subset B = K[z_1, \ldots, z_m] \subset A$$
1244 where $K \subset B$ is a polynomial extension, and $B \subset A$ is finite.
1245\end{thm}
1246\begin{proof}
1247 % (by induction on the number of generators ($n$) of $A$).
1248 Let $y_1, \ldots, y_n$ be generators of $A = K[y_1, \ldots, y_n]$.
1249
1250 if $n=0$, nothing to prove since $A$ is generated by $0$ elements $~\Longrightarrow~ A=K$, and $K$ is finite.
1251
1252 if $n>0$ we have two cases:
1253 \begin{itemize}
1254 \item[-] $y_1, \ldots, y_n$ are algebraically independent over $K$, then by
1255 definition \ref{R.4.6.D} $A \cong K[y_1, \ldots, y_n]$, so that $A$ is a finite module over itself, with $m=n$.
1256 \item[-] $y_1, \ldots, y_n$ are algebraically dependent over $K$,
1257 $$\exists~ 0 \neq f \in K[y_1, \ldots, y_n] ~\text{s.th}~ f(y_1, \ldots, y_n)=0$$
1258 \end{itemize}
1259
1260 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
1261 $$y_n^d+ a_{d-1} y_n^{d-1} + \ldots + a_1 y_n + a_0 = 0~~~~~~~a_i \in K[y_1^*,
1262 \ldots, y_{n-1}^*]$$
1263 \hspace*{2em} ie. monic, so that by definition (\ref{R.4.1}), $y_n$ is
1264 integral over $K[y_1^*, \ldots, y_{n-1}^*]$.
1265
1266 $~\longrightarrow~$ 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.
1267
1268
1269 \vspace{0.3cm}
1270 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
1271 $$A^* = K[y^*_1, \ldots, y^*_{n-1}], ~~\text{and}~ A=A^*[y_n]$$
1272
1273 Setting $y_i^* = y_i - y_n^{r_i}$, so that $y_i = y_i^* + y_n^{r_i}$ $\forall i \in [n-1],~~ r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.
1274
1275 Use those new variables at $f(y_1, \ldots, y_n)=0$:
1276 $$f(y_1^* + y_n^{r_1}, y_2^* + y_n^{r_2}, \ldots, y_{n-1}^* + y_n^{r_{n-1}}, y_n) = 0$$
1277
1278 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 $y_n$.
1279
1280 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 $A^* = K[y_1^*, \ldots, y_{n-1}^*]$.
1281
1282 \vspace{0.3cm}
1283 Induction:\\
1284 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}).
1285
1286 By inductive hypothesis on $A^*,~~ \exists~ z_1, \ldots, z_m \in A^*$ algebraically independent over $K$ and with $A^*$ finite over $B=K[z_1, \ldots, z_m]$.
1287
1288 Therefore, each step of $B \subset A^* \subset A^*[y_n]=A$ is finite, and $A$ is finite over $B$ as required.
1289\end{proof}
1290
1291
1292\vspace{0.5cm}
1293\begin{eg}{ }
1294 $A = K[X,Y]/(XY-1)$. $Y$ is algebraic over $K[X]$, but not integral over $K[Y]$.
1295
1296 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 $X=0$).
1297
1298 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 $\epsilon \neq 0$.
1299
1300 This corresponds geometrically to tilting the hyperbola a little before projecting, so that no longer has a vertical asymptotic line.
1301\end{eg}
1302
1303\vspace{1cm}
1304
1305\subsection{Weak Nullstellensatz}
1306
1307\begin{prop}{R.4.9}\label{R.4.9}
1308 let $A \subset B$ be an integral extension of integral domain,\\
1309 then $A$ is a field $\Longleftrightarrow~ B$ is a field.
1310\end{prop}
1311\begin{proof}
1312 $\Longrightarrow$:\\
1313 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.
1314
1315 Since $A$ is a field, $\exists$ inverse, observe that:
1316 \begin{align*}
1317 x^n &+ a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0\\
1318 x(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = - a_0\\
1319 -a_0^{-1}(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = x^{-1} \in B
1320 \end{align*}
1321 thus there exists inverse in $B$, so $B$ is a field too.
1322
1323 $\Longleftarrow$:\\
1324 if $B$ is a field and $0 \neq x \in A$, then $x^{-1} \in B$, so $x^{-1}$ is integral over $A$.
1325
1326 So there is a relation of the form
1327 $$(x^{-1})^n + a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 =0$$
1328
1329 Therefore
1330 \begin{align*}
1331 (x^{-1})^n &+ a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 = 0\\
1332 (x^{-1})^n &= -a_{n-1} (x^{-1})^{n-1} - \ldots - a_0\\
1333 x^{-n} &= -a_{n-1} x^{-n+1} - \ldots - a_0 ~~\text{(mult by $x^{n-1}$)}\\
1334 x^{-n+(n-1)} &= -a_{n-1} x^{-n+1+(n-1)} - \ldots - a_0 x^{n-1}\\
1335 x^{-1} &= -a_{n-1} - \ldots - a_0 x^{n-1} \in A
1336 \end{align*}
1337
1338 thus there exists inverse in $A$, so $A$ is a field too.
1339\end{proof}
1340
1341\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski}
1342 let $k$ a field, $K$ a $k$-algebra which
1343 \begin{enumerate}
1344 \item is finitely generated as a $k$-algebra
1345 \item is a field
1346 \end{enumerate}
1347 Then $K$ is algebraic over $k$, so that $k \subset K$ is a finite field
1348 extension. That is, $[K:k] < \infty$.
1349\end{thm}
1350\begin{proof}
1351 $K=k[z_1, \ldots, z_m]$ a field; want to show that $K$ is algebraic over $k$.
1352 \vspace{0.3cm}
1353
1354 Since $K$ is a fingen $k$-algebra, by Noether normalization lemma (\ref{noether-normalization}),\\
1355 $\exists~ z_1, \ldots, z_m \in K$ such that
1356 \begin{itemize}
1357 \item are algebraically independent
1358 \item $K$ is integral over the polynomial ring $A=k[z_1, \ldots, z_m]$ (which by \ref{integral-implies-finite} is finite)
1359 \end{itemize}
1360\\
1361 Now we're at the situation of \ref{R.4.9}:
1362
1363 $A \subset K$ is integral, $K$ is a field $~~\Longrightarrow~$ therefore $A$ is a field.
1364
1365 \vspace{0.2cm}
1366 Since $z_1, \ldots, z_m \in K$ are algebraically independent,\\
1367 \hspace*{2em}$\Longrightarrow~ A=k[z_1, \ldots, z_m]$ is a polynomial ring in $m$ indeterminates, and this is a field only if $m=0$
1368
1369 (since in $k[z_1]$ the element $z_1$ is not invertible, since $1/z_1$ is a rational function, not a polynomial).
1370
1371 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 $k$.
1372\end{proof}
1373
1374
1375\vspace{1cm}
1376
1377\section{Nullstellensatz}
1378
1379Note: 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 $k$.
1380
1381\vspace{0.3cm}
1382\begin{cor}{5.2} \label{5.2}
1383 $k$ algebraically closed. Then every maximal ideal of $A = k[X_1, \ldots,
1384 X_n]$ is of the form
1385 $$m = (X_1 - a_1, \ldots, X_n -a_n),~~ a_i \in k$$
1386
1387 The map $k[X_1, \ldots, X_n] \longrightarrow k[X_1, \ldots, X_n]/m=k$ is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$.
1388
1389 Thus
1390 \begin{align*}
1391 k^n &\longleftrightarrow m-Spec A\\
1392 (a_1, \ldots, a_n) &\longleftrightarrow f(a_1, \ldots, a_n)
1393 \end{align*}
1394\end{cor}
1395\begin{proof}
1396 let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.
1397
1398 By fundamental property of maximal ideals, $K=A/m$ is a field.
1399
1400 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 $x_i' = x_i +m$.
1401
1402 By Zariski's lemma (\ref{zariski}), $K=A/m$ is algeraic over $k$.
1403
1404 Since by hypothesis $k$ is algebraically closed, it has no proper algebraic extensions\\
1405 \hspace*{2em} $\Longrightarrow~~ K=k~~ \Longrightarrow~~ k \cong A/m$.
1406
1407 So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$.
1408 $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$
1409 $$\Longrightarrow x_i - a_i \in m$$
1410
1411 The ideal generated by these terms is a subset of $m$:
1412 $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
1413
1414 Since $J$ is the kernetl of the evaluation map at point $(a_1, \ldots, a_n)$,
1415 then $J$ is a maximal ideal. Together with $J \subseteq m$, then we have
1416 $J=m$, ie. $$m = (X_1 -a_1, \ldots, X_n -a_n)$$
1417
1418 \vspace{0.3cm}
1419 Let
1420 \begin{align*}
1421 \psi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\
1422 \psi: x_i &\longmapsto a_i
1423 \end{align*}
1424
1425 Since $\psi$ is a $k$-algebra homomorphism, then $\forall f \in A$:
1426 $$\psi(f(X_1, \ldots, X_n)) = f(\psi(x_1), \ldots, \psi(x_n))= f(a_1, \ldots, a_n)$$
1427
1428 Thus there is a one-to-one correspondence:
1429
1430 points in $k^n ~~~ \longleftrightarrow~~~ m-Spec A$ (maximal ideals in $k[X_1, \ldots, X_n]$
1431
1432 $(a_1, \ldots, a_n) ~~~\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)$
1433\end{proof}
1434
1435
1436\vspace{0.5cm}
1437\subsection{Variety}
1438
1439\begin{defn}{5.3}[Variety]
1440 A \emph{variety} $V \subset k^n$:
1441 $$ V = V(J) = \{ P=(a_1, \ldots, a_n) \in k^n | f(P)=0 ~\forall~ f \in J \}$$
1442
1443 \begin{itemize}
1444 \item[$\rightarrow$] $V$ is defined by $f_1(P)= \ldots = f_m(P) = 0$
1445 \item[$\rightarrow$] $V$ is defined as the simultaneous solutions of a number of polynomial equations.
1446 \end{itemize}
1447\end{defn}
1448
1449\begin{prop}{5.3} \label{5.3}
1450 $k$ 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 $k[X_1, \ldots, X_n]$.
1451 (notation: $x_i = X_i \pmod J$)
1452
1453 Then every maximal ideal of $A$ is of the form
1454 $$(x_1 -a_1, \ldots, x_n - a_n)$$
1455 for some point $(a_1, \ldots, a_n) \in V(J)$.
1456
1457 Therefore, $\exists$ a one-to-one correspondence
1458 \begin{align*}
1459 V(X) &\longleftrightarrow m-Spec A\\
1460 \text{given by}~~~(a_1, \ldots, a_n) &\longleftrightarrow (x_1 -a_1, \ldots, x_n - a_n)
1461 \end{align*}
1462\end{prop}
1463\begin{proof}
1464 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.
1465 \begin{align*}
1466 m \subset A \longleftrightarrow &m' \subseteq k[X_1, \ldots, X_n]\\
1467 &\text{s.th.}~ J \subseteq m'
1468 \end{align*}
1469
1470 Thus, every maximal ideal of $A$ is of the form
1471 $$\underbrace{(x_1 -a_1, \ldots, x_n - a_n)}_{i.} ~\text{such that}~ \underbrace{J \subset (X_1 - a_1, \ldots, X_n - a_n)}_{ii.}$$
1472
1473 \begin{enumerate}[i.]
1474 \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 $(a_1, \ldots, a_n) \in k^n$.
1475
1476 Which when projected to the quotient ring $A$, $X_i \longmapsto x_i$ (residue class), giving $(x_1 - a_1, \ldots, x_n - a_n)$.
1477
1478 \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.
1479 \end{enumerate}
1480
1481 \vspace{0.3cm}
1482 However, since $(X_1 - a_1, \ldots, X_n - a_n)$ is the kernel of the evaluation map $f \longmapsto f(a_1, \ldots, a_n)$
1483
1484 $\Longrightarrow~$ means that $m'=(X_1 - a_1, \ldots, X_n - a_n)$ consists of all polynomials that vanish at point $P=(a_1, \ldots, a_n)$.
1485
1486 If $J \subseteq m'$, then $\forall~ f \in J$ must vanish at $P$.
1487
1488 By definition, the set of points where all polynomials in $J$ vanish is the \emph{variety}, $V(J)$.
1489
1490 \vspace{0.4cm}
1491 Thus,\\
1492 every maximal ideal in $A$ corresponds to a point $(a_1, \ldots, a_n) \in k^n$, ie.
1493 $$m-Spec A \longleftrightarrow k^n$$
1494
1495 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
1496 \begin{align*}
1497 m-Spec A &\longleftrightarrow V(J)\\
1498 \text{maximal spectrum} &\longleftrightarrow \text{variety}
1499 \end{align*}
1500\end{proof}
1501
1502\begin{prop}{5.5}[Correspondeces $V$ and $I$] \label{5.5}
1503 A variety $X \subset k^n$ is by definition $X=V(J)$ (J an ideal of $k[X_1, \ldots, X_n]$).
1504
1505 So $V$ gives a map:\\
1506 $$\{ \text{ideals of}~ k[X_1, \ldots, X_n] \} \stackrel{V}{\longrightarrow} \{ \text{subsets}~ X ~\text{of}~ k^n \}$$
1507
1508 correspondence going the other way:
1509 $$\{ \text{subsets}~ X ~\text{of}~ k^n \} \stackrel{I}{\longrightarrow} \{ \text{ideals of} ~k[X_1, \ldots, X_n] \}$$
1510
1511 defined by taking a subset $X \subset k^n$ into the ideal
1512 $$I(X) = \{ f \in k[X_1, \ldots, X_n] | f(P)=0 ~\forall~ P \in X \}$$
1513
1514 \vspace{0.4cm}
1515
1516 $V,~I$ satisfy reverse inclusions:
1517 $$J \subset J' ~\Longrightarrow~ V(J) \supset V(J') ~~~~\text{and}~~~~ X \subset Y ~\Longrightarrow~ I(X) \supset I(Y)$$
1518\end{prop}
1519
1520
1521\vspace{0.4cm}
1522\subsection{Nullstellensatz}
1523
1524\vspace{0.5cm}
1525\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
1526 Let $k$ algebraically closed field.
1527
1528 \begin{enumerate}[a.]
1529 \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
1530 \item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$,
1531 $$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$
1532 \end{enumerate}
1533\end{thm}
1534\begin{proof}
1535 \begin{enumerate}[a.]
1536 \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$:\\
1537 Let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.\\
1538 Then $L=k[X_1, \ldots, X_n]/m$ is a field (by TODO ref).
1539
1540 By Zariski's lemma (\ref{zariski}), since $L$ is generated as a $k$-algebra by the images of the variables $x_i$, and $k$ is algebraically closed.
1541
1542 Then the only algebraic extension of $k$ is $k$ itself. Thus $L \cong k$.
1543
1544 \vspace{0.3cm}
1545 Then $\exists$ a surjective homomorphism $\psi: k[X_1, \ldots, X_n] \longrightarrow k$.
1546
1547 Let $a_i = \psi(x_i)$. Then $x_i - a \in ker(\psi) = m ~\forall~ i$.
1548
1549 Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$.
1550
1551 Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$.
1552
1553 Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$.\\
1554 $\Longrightarrow~$ thus $P \in V(J)$, and thus $V(J) \neq \emptyset$.
1555
1556\item $I(V(J)) = rad J$:\\
1557 \begin{align*}
1558 I(V(J)) &= rad J\\
1559 \text{vanishing ideal of a variety} &= \text{radical of the ideal defining the variety}
1560 \end{align*}
1561 where $rad~J = \{ f \in R ~|~ f^n \in J ~\text{for some}~ n>0 \}$.
1562
1563 Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$.
1564
1565 Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$
1566
1567 Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie.
1568 $$\exists~ (a_1, \ldots, a_n, a_{n+1}) \in V(J')$$
1569
1570 Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires
1571 $$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$
1572 a contradiction.
1573
1574 Therefore, $V(J') = \emptyset$.
1575
1576 \vspace{0.4cm}
1577
1578 Since $V(J')=\emptyset$, by the Weak Nullstellensatz/Zariski (\ref{zariski}),\\
1579 \hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$.
1580
1581 Every element in an ideal is a linear combination of its generators: $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
1582
1583 $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)} g_i) + \text{(polynomial)} \cdot (1 - Yf)$$
1584
1585 which, since $1 \in J'$,
1586 $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$
1587
1588 substitute $Y=1/f$,
1589 $$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$
1590 thus
1591 $$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$
1592 multiply by $f^n$,
1593 $$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$
1594
1595 thus $f^n$ is a linear combination of $g_i$.\\
1596 Thus $f^n \in J$, so $f \in rad~J$.
1597\end{enumerate}
1598\end{proof}
1599
1600
1601\vspace{0.4cm}
1602\subsection{Irreducible varieties}
1603
1604\begin{defn}{5.7}[Irreducible variety]
1605 a variety $X \subset k^n$ is \emph{irreducible} if it is nonempty and not the union of two proper subvarieties; that is, if
1606 $$X = X_1 \cup X_2 ~~\text{for varieties}~ X_1, X_2 ~\Longrightarrow~ X= X_1 ~\text{or}~ X_2$$
1607\end{defn}
1608
1609\begin{prop}{5.7} \label{5.7}
1610 a variety $X$ is irreducible iff $I(X)$ is prime.
1611\end{prop}
1612\begin{proof}
1613 set $I=I(X)$.
1614
1615 if $I$ not prime, then $f,g \in A \setminus I$ be such that $fg \in I$.
1616
1617 Define new ideals
1618 $$J_1=(I, f) ~~\text{and}~~ J_2=(I, g)$$
1619
1620 Then, since $f \not\in I(X)$, it follows that $V(J_1) \subsetneq X$
1621
1622 \hspace*{2em}$\Longrightarrow$ so $X=V(J_1) \cup V(J_2)$ is reducible.
1623
1624 The converse is similar.
1625\end{proof}
1626
1627\begin{cor}{5.8} \label{cor.5.8}
1628 let $k$ algebraically closed field. Then $V$ and $I$ induce one-to-one correspondences
1629
1630 \begin{align*}
1631 \{ \text{radical ideals}~J~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{varieties}~ X \subset k^n \}\\
1632 \{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}
1633 \end{align*}
1634
1635 Therefore,
1636 $$Spec k[X_1, \ldots, X_n] = \{ \text{irreducible varieties}~ X \subset k^n \}$$
1637\end{corollary}
1638\end{cor}
1639
1640\begin{prop}{5.8} \label{5.8}
1641 let $A= k[x_1, \ldots, x_n]$ a fingen $k$-algebra ($k$ an algebraically closed field).
1642
1643 Write $J$ for the ideal of relations holding between $x_1, \ldots, x_n$, so that $A=k[X_1, \ldots, X_n]/J$.
1644
1645 Then there is a one-to-one correspondence
1646 $$Spec A \longleftrightarrow \{ \text{irreducible subvarieties}~ X \subset V(J) \}$$
1647\end{prop}
1648\begin{proof}
1649 By definition, $Spec A = \{ P ~|~ P \subset A ~\text{is prime ideal} \}$.
1650
1651 By Corollary \ref{cor.5.8}:
1652 $$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$
1653
1654 About varieties:
1655 \begin{itemize}
1656 \item $I(X)$ in $R=k[X_1, \ldots, X_n]$ is the ideal of the variety $X$\\
1657 \hspace*{2em}ie. the set of all polynomials that vanish on every point of $X$.
1658 \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.
1659 \end{itemize}
1660
1661 If $P$ is prime in $A~~\Longrightarrow~$ it must correspond to some prime ideal $\mathfrak{P}$ in $R$ (also $J \subset R$).
1662
1663 Then from Nullstellensatz (\ref{nullstellensatz}), every prime ideal $\mathfrak{P}$ in $R$ is the ideal of some irreducible variety $X$.\\
1664 \hspace*{2em}$\Longrightarrow~~ \mathfrak{P}=I(X)$.
1665
1666
1667 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\\
1668 $$\Longrightarrow~~ P = \{ f+J ~|~ f \in I(X) \}~~ = I(X) ~\text{mod}~J$$
1669
1670 Now, $J$ is the set of equations defining our "universe" $V(J)$.
1671
1672 Since $A=R/J ~~ \Longrightarrow~$ we thus have $J \subseteq R$.
1673
1674 Also we have a correspondence between $A=R/J$ and $R$.
1675
1676 Let $\mathfrak{P}$ be the preimage of $P$ in $R=k[X_1, \ldots, X_n]$, ie.
1677 \begin{align*}
1678 R &\longrightarrow A=R/J\\
1679 \mathfrak{P} &\longmapsto P\\
1680 I(X) &\longmapsto I(X) ~\text{mod}~J
1681 \end{align*}
1682
1683 $\Longrightarrow~ \text{thus}~ J \subseteq \mathfrak{P}$.
1684
1685
1686 Now, $J \susbset \mathfrak{P} = I(X)$;\\
1687 by \ref{5.5}, the set of points where $\mathfrak{P}$ vanishes must be inside the set of points where $J$ vanishes, so
1688 $$V(J) \supseteq V(\mathfrak{P}) = V(I(X)) = X$$
1689 $\Longrightarrow~~ X \subseteq V(J)$, ie. the irreducible variety $X$ must be a subvariety of $V(J)$.
1690
1691 Therefore,
1692 $$\mathfrak{P} \in Spec A \longleftrightarrow X \subseteq V(J)$$
1693 where $X = V(\mathfrak{P})$.
1694\end{proof}
1695
1696\newpage
1697
1698\section{Exercises}
1699
1700For the exercises, I follow the assignments listed at \cite{mit-course}.
1701
1702The 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}.
1703
1704\subsection{Exercises Chapter 1}
1705
1706\begin{ex}{R.1.1}
1707 Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$?
1708\end{ex}
1709\begin{proof}
1710 Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$.
1711
1712 $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\
1713 Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$.
1714
1715 Thus $I \cup J$ is not closed under addition; thus is not an ideal.
1716
1717
1718 Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum:
1719
1720 $$I+J = \{ a+b | a \in I, b \in J \}$$
1721
1722 $gcd(2,3)=1$, so $I+J = \mathbb{Z}$.
1723
1724 Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$.
1725\end{proof}
1726
1727\begin{ex}{R.1.5}
1728 let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\
1729 In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$.
1730\end{ex}
1731\begin{proof}
1732 (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$).
1733
1734 Let
1735 $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
1736 The claim is that $\psi^{-1}(P)$ is prime ideal of $A$.
1737
1738 \begin{enumerate}[i.]
1739 \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
1740 $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$).
1741
1742 If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so
1743 $$\psi(a-b)= \psi(a) - \psi(b) \in P$$
1744 hence $a-b \in \psi^{-1}(P)$.
1745
1746 If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\
1747 Thus $ra \in \psi^{-1}(P)$.
1748
1749 $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$.
1750
1751 \item show that $\psi^{-1}(P)$ is prime:\\
1752 $\psi^{-1}(P) \neq A$, 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 $B$.
1753
1754 Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$.
1755
1756 Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\
1757 Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$.
1758
1759 Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$.
1760 \end{enumerate}
1761\end{proof}
1762
1763
1764\begin{ex}{R.1.6}
1765 prove or give a counter example:
1766 \begin{enumerate}[a.]
1767 \item the intersection of two prime ideals is prime
1768 \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
1769 \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
1770 \item the map $\psi^{-1}$ of Proposition 1.2 takes maximal ideals of $A/I$
1771 to maximal ideals of $A$
1772 \end{enumerate}
1773\end{ex}
1774\begin{proof}
1775 \begin{enumerate}[a.]
1776 \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime.
1777
1778 Then $I \cap J = (2) \cap (3) = (6)$.
1779
1780 The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$.
1781
1782 Thus the intersection of two primes can not be prime.
1783
1784 \item $P_1=(2),~ P_2=(3)$, both prime.
1785
1786 Then,
1787 $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$
1788
1789 $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$.
1790
1791 (recall: principal= generated by a single element)
1792
1793 So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$.
1794
1795 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.
1796
1797 Henceforth, the sum of two prime ideals is not necessarily prime.
1798
1799 \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$.
1800
1801 Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$.
1802
1803 Then
1804 \begin{align*}
1805 \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\
1806 \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A
1807 \end{align*}
1808
1809 But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
1810
1811 Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
1812
1813 \item $\psi: A \longrightarrow A/I$ quotient homomorphism, $I \subseteq A$ an ideal.
1814
1815 Let $M$ a maximal ideal of $A/I$, then $\frac{(A/I)}{M}$ is a field (Proposition 1.3).
1816
1817 By the isomorphism theorems,
1818 $$\frac{(A/I)}{M} \cong \frac{A}{\psi^{-1}(M)}$$
1819
1820
1821 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 $A$.
1822
1823 $\Longrightarrow~$ under $\psi$, preimages of maximal ideals are maximal.
1824 \end{enumerate}
1825\end{proof}
1826
1827\begin{ex}{R.1.12.a}
1828 if $I,J$ ideals and $P$ prime ideal, prove that
1829 $$IJ \subset P ~\Longleftrightarrow~ I \cap J \subset P ~\Longleftrightarrow~ I ~\text{or}~ J \subset P$$
1830\end{ex}
1831\begin{proof}
1832 assume $I \subseteq P$ (for $J \subseteq P$ will be the same, symmetric), take $x \in IJ$,
1833
1834 then
1835 $$x = \sum_{k=1}^n a_k b_k$$
1836 with $a_k \in I,~ b_k \in J$.
1837
1838 Each $a_k \in I \subseteq P$. Since $P$ an ideal,
1839 $$\sum_{k=1}^n a_k b_k \in P$$
1840
1841 thus $x \in P$, hence $IJ \subseteq P$.
1842
1843 So $I \subseteq P$ or $J \subseteq P$ $~\Longrightarrow IJ \subseteq P$.
1844
1845 \vspace{0.5cm}
1846 Conversely,\\
1847 assume $P$ prime and $IJ \subseteq P$.
1848
1849 Suppose by contradiction that $I \not\subseteq P$ and $J \not\subseteq P$.
1850 \begin{itemize}
1851 \item[-] since $I \not\subseteq P,~ \exists a \in I$ with $a \not\in P$
1852 \item[-] since $J \not\subseteq P,~ \exists b \in J$ with $b \not\in P$
1853 \end{itemize}
1854 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 $P$.
1855
1856 Thus $I \not\subseteq P$ and $J \not\subseteq P$ are false.
1857
1858 \vspace{0.3cm}
1859 So both directions are proven, hence
1860 $$IJ \subseteq P ~\Longrightarrow~ I \subseteq P ~\text{or}~ J \subseteq P$$
1861
1862\end{proof}
1863
1864\begin{ex}{R.1.18}
1865 Use Zorn's lemma to prove that any prime ideal $P$ contains a minimal prime ideal.
1866\end{ex}
1867\begin{proof}
1868 Let $P$ prime ideal of $R$.
1869
1870 $$S = \{ Q \subseteq R ~|~ Q ~\text{a prime ideal AND}~ Q \subseteq P \}$$
1871
1872 Goal: show that $S$ has a minimal element, the minimal ideal contained in $P$.
1873
1874 $P \subset S$, so $S$ is nonempty.
1875
1876 Let $C \subseteq S$ be a chain (= totally ordered subset) with respect to inclusion.
1877
1878 Define
1879 $$Q_C = \bigcap_{Q \in C} Q$$
1880
1881 Clearly $Q_C \subseteq P$, since each $Q \in C$ is $Q \subseteq P$.
1882
1883 Since $C$ is ordered by inclusion, it is a decreasing chain of prime ideals.
1884
1885 Intersection of a decreasing chain of prime ideals is again a prime ideal:
1886 \begin{itemize}
1887 \item[-] if $ab \in Q_C$, then $ab \in Q ~\forall Q \in C$
1888 \item[-] since $Q$ prime, $\forall Q \in C$ either $a \in Q$ or $b \in Q$
1889 \end{itemize}
1890
1891 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 $Q_2 \subseteq Q_1$.
1892
1893 In either case: contradiction, since the smaller one would have to contain the element that was assumed to be excluded.
1894
1895 Thus $\forall Q \in C$ the same element $a, b$ must lie in all $Q$. $\Longrightarrow~$ lies in the intersection of them, $Q_C$.
1896
1897 Henceforth, $Q_C$ is a prime ideal and lies in $S$, and its a lower bound of $C$ in $S$.
1898
1899 Now, $S$ is nonempty, and every chain in $S$ has a lower bound in $S$ (its intersection).\\
1900 Therefore, $S$ has a minimal element $P_{min}$.
1901
1902 By construction, $P_{min}$ is a prime ideal $P_{min} \subseteq P$, and by minimality there are no strictly smaller prime ideals inside $P$.
1903
1904 So $P_{min}$ is a minimal prime ideal, contained in $P$.
1905\end{proof}
1906
1907\begin{ex}{R.1.10}
1908\end{ex}
1909\begin{proof}
1910\end{proof}
1911
1912\begin{ex}{R.1.11}
1913\end{ex}
1914\begin{proof}
1915\end{proof}
1916
1917\begin{ex}{R.1.4}
1918\end{ex}
1919\begin{proof}
1920\end{proof}
1921
1922\subsection{Exercises Chapter 2}
1923
1924\begin{ex}{R.2.9}
1925$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ is a s.e.s. of $A$-modules. Prove that if $N, L$ are finite over $A$, then $M$ is finite over $A$.
1926\end{ex}
1927\begin{proof}
1928 Denote the generators of $L$ and $N$ respectively as
1929 \begin{align*}
1930 \{l_1, \ldots, l_k \} &\subseteq L\\
1931 \{n_1, \ldots, n_p \} &\subseteq N
1932 \end{align*}
1933
1934 By s.e.s. definition,
1935 \begin{itemize}
1936 \item[-] $\alpha$ is injective (one-to-one), so
1937 $$\forall l_i \in L,~ \exists~ x_i \in M ~\text{s.th.}~ \alpha(l_i)=x_i$$
1938
1939 \item[-] $\beta$ is surjective (onto), so
1940 $$\forall n_j \in N,~ \exists~ y_j \in M ~\text{s.th.}~ \beta(y_j)=n_j$$
1941\end{itemize}
1942
1943 We will show that $\{x_1, \ldots, x_k, y_1, \ldots, y_p \}$ generate $M$, and thus $M$ is finite:
1944
1945 Let $m \in M$, then $\beta(m) \in N$, and
1946 $$\beta(m) = \sum_{j=1}^p a_j n_j ~~~\text{with}~ a_j \in A$$
1947
1948 Take $m' \in M$, with $m' = \sum a_j y_j$, then
1949 $$\beta(m') = \sum a_j \beta(y_j) = \sum a_j n_j = \beta(m)$$
1950
1951 Then, since $\beta(m)=\beta(m')~~ \Longrightarrow~~ \beta(m-m')=0$, thus
1952 $$(m-m') \in ker(\beta)$$
1953
1954 By \emph{exactness} property, since $\alpha: L \longrightarrow ker(\beta)$, we have $ker(\beta)=im(\alpha)$.
1955
1956 Therefore, $\exists~ l \in L$ such that $\alpha(l)= m-m'$.
1957
1958 Since $\{l_i\}_k$ generate $L$,
1959 $$l = \sum^k b_i l_i$$
1960 thus
1961 $$m-m' = \alpha(l) = \alpha(\underbrace{\sum b_i l_i}_{l}) = \sum b_i \underbrace{\alpha(l_i)}_{x_i} = \sum b_i x_i$$
1962
1963 Rearrange,
1964 $$m = m' + \sum b_i x_i = \sum_{j=1}^p a_j y_j + \sum_{i=1}^k b_i x_i ~~~~~ \forall m \in M$$
1965
1966 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.\\
1967 Thus $M$ is finitely generated over $A$.
1968\end{proof}
1969
1970
1971\subsection{Exercises Chapter 3}
1972
1973\begin{ex}{R.3.2}
1974 $K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space.
1975 Prove that $A$ is Noetherian and Artinian.
1976\end{ex}
1977\begin{proof}
1978 $dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in \aA$ and $c \in K$, then $c \cdot x \in \aA$.
1979
1980 \begin{enumerate}
1981 \item Noetherian:\\
1982 let $I_1 \subseteq I_2 \subseteq \ldots$ be an ascending chain of ideals in $A$.
1983
1984 Since each $I_i$ is a subspace, we have
1985 $$dim_K(I_1) \leq dim_K(I_2) \leq \ldots \leq n$$
1986 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.
1987
1988 \item Artinian:\\
1989 Similarly, if $I_1 \supseteq I_2 \supseteq \ldots$ a descending chain of ideals in $A$.
1990
1991 then
1992 $$n \geq dim_K(I_1) \geq dim_K(I_2) \geq \ldots \geq 0$$
1993 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.
1994 \end{enumerate}
1995\end{proof}
1996
1997\begin{ex}{R.3.5}
1998 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 $M$.
1999
2000 Prove if the following holds or not:
2001 $$\beta(M_1)=\beta(M_2) ~\text{and}~ \alpha^{-1}(M_1)=\alpha^{-1}(M_2)~~\Longrightarrow~ M_1=M_2$$
2002\end{ex}
2003\begin{proof}
2004 Counterexample showing that it does not hold:
2005
2006 Let $K$ a field, $M = K \oplus K~, L=K,~N=K$.
2007
2008 Set, for $l \in L,~ (m_1, m_2) \in M$,
2009 \begin{align*}
2010 \alpha:~ &l \longmapsto (l, 0)\\
2011 \beta:~ &(m_1, m_2) \longmapsto m_2
2012 \end{align*}
2013
2014 So we have
2015 $$0 \longrightarrow K \stackrel{\alpha}{\longrightarrow} K^2 \stackrel{\beta}{\longrightarrow} K \longrightarrow 0$$
2016
2017 Then,
2018 \begin{align*}
2019 M_1 &= \{ (x, x) ~|~ x \in K\} ~~~~\sim\text{(diagonal line)} \\
2020 M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)}
2021 \end{align*}
2022
2023 (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 $M$).
2024
2025 Then,
2026 \begin{align*}
2027 \beta(M_1) &= \{ x ~|~ x \in K\} = K \\
2028 \beta(M_2) &= \{ x ~|~ x \in K\} = K
2029 \end{align*}
2030 thus, $\beta(M_1) = \beta(M_2)$.
2031
2032 For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1) = \{0\}$,\\
2033 for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2) = \{0\}$,\\
2034 thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$.
2035
2036
2037 So we've seen that
2038 \begin{align*}
2039 \beta(M_1) = \beta(M_2)\\
2040 \alpha^{-1}(M_1)=\alpha^{-1}(M_2)
2041 \end{align*}
2042 while having $M_1 \neq M_2$.
2043\end{proof}
2044
2045
2046
2047\begin{ex}{R.3.3}
2048Let $A$ a ring, $I_1, \ldots, I_k$ ideals such that each $A/I_i$ is a Noetherian ring.
2049Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bigcap I_i = 0$ then $A$ is also Noetherian.
2050\end{ex}
2051\begin{proof}
2052\begin{enumerate}[i.]
2053 \item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian.
2054 $\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian.
2055 \item Take the canonical homomorphism
2056 $$\phi: A \longrightarrow \bigoplus_{i=1}^n A/ I_i$$
2057 by $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$.
2058
2059 $\phi$ is injective: $ker(\phi)= \{ a \in A | a \in I_i \forall i \}$.
2060
2061 Since we're given $\cap I_i = 0$, then $ker(\phi)=\cap I_i$, and $\phi$ is injective.
2062
2063 Thus, $\phi$ is the isomorphism $A \cong im(\phi)$, where $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$.
2064
2065 We know that any submodule of a Noetherian module is Noetherian, thus, since
2066 \begin{itemize}
2067 \item $A/I_i$ is Noetherian by hypothesis of the exercise
2068 \item $A \cong im(\phi)$
2069 \item $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$
2070 \end{itemize}
2071 then, $A$ is Noetherian.
2072\end{enumerate}
2073\end{proof}
2074
2075\begin{ex}{R.3.4}
2076 Prove that if A is a Noetherian ring and M a finite A-module, then there
2077 exists an exact sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$.
2078 That is, M has a presentation as an A-module in terms of finitely many generators and relations.
2079\end{ex}
2080\begin{proof}
2081 since $M$ fingen $~\Longrightarrow~$ generators $\{m_1, \ldots, m_2 \} \subseteq M$ span $M$.
2082
2083 Let $\beta$ be a surjective $A$-linear map, which forms a free $A$-module of rank $p$ onto $M$:
2084 \begin{align*}
2085 \beta: A^p &\longrightarrow M\\
2086 (a_1, \ldots, a_p) &\longmapsto \sum_{i=1}^p a_i m_i
2087 \end{align*}
2088
2089 Let $K=ker(\beta)$. By the 1st Isomorphism Theorem,
2090 $$M \cong A^p / K$$
2091
2092 Since $A$ is a Noetherian ring, then every free $A$-module of finite rank (eg. $A^p$) is a Noetherian module.
2093
2094 Every submodule of a Noetherian module is fingen.
2095
2096 $\Longrightarrow~$ since $K \subseteq A^p, ~\Longrightarrow~ K ~~(=ker(\beta))$ is fingen.
2097
2098 Since $K$ fingen, let $\{k_1, \ldots, l_q\}$ be generators of $K$.
2099
2100 Define $\psi: A^q \longrightarrow K$.
2101
2102 Compose it with the inclusion map $i: K \longrightarrow A^p$,
2103 $$\alpha = i \circ \psi:~ A^q \longrightarrow A^p$$
2104
2105 So we have the whole sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$, where
2106 \begin{itemize}
2107 \item $\beta$ is surjective
2108 \item $im(\alpha)=ker(\beta)$
2109 \end{itemize}
2110 so that it is a exact sequence, thus, $M$ has a finite presentation.
2111\end{proof}
2112
2113\subsection{Exercises Chapter 4}
2114
2115\begin{ex}{R.4.1.a}
2116 $k[X^2] \subset k[X]$ is a finite extension, hence integral. Find the integral dependence relation for any $f \in k[X]$.
2117\end{ex}
2118\begin{proof}
2119 $\forall f(X) \in k[X]$ can be uniquely decomposed into its even and odd parts:
2120 $$f(X) = p(X^2) + X \cdot q(X^2)$$
2121
2122 with $p(X^2),~ q(X^2) \in k[X^2]$, and\\
2123 $p(X^2)$: sum of all terms with even exponents\\
2124 $q(X^2)$: sum of all terms with odd exponents, and then factoring out $X$.
2125
2126 \vspace{0.3cm}
2127 (Observation: this is used in FRI cryptographic protocol\\
2128 \href{https://github.com/arnaucube/math/blob/master/notes_fri_stir.pdf}{https://github.com/arnaucube/math/blob/master/notes\_fri\_stir.pdf})
2129
2130
2131 Rearrange it
2132 \begin{align*}
2133 f(X) &- p(X^2) = X \cdot q(X^2), ~~\text{square:}\\
2134 (f(X) &- p(X^2))^2 = X^2 \cdot q(X^2)^2\\
2135 f(X)^2-2 p(X^2) f(X) &+ p(X^2)^2 = X^2 \cdot q(X^2)^2\\
2136 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
2137 \end{align*}
2138
2139 Denote the last polynomial as $P(T) \in k[X^2]$, where $f(X)$ is a root of $P(T)$.
2140
2141 \vspace{0.2cm}
2142 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 $a_i \in k[X^2]$.
2143
2144 We have that
2145 \begin{align*}
2146 a_1 &= -2 p(X^2)\\
2147 a_0 &= p(X^2)^2 - X^2 q(X^2)^2
2148 \end{align*}
2149
2150
2151 So for example, for $f(X)= X^3 + X^2 + X + 1$:
2152 \begin{align*}
2153 f(X) = (X^2 + 1) &+ X (X^2 + 1)\\
2154 (f(X) - (X^2+1))^2 &= X^2 (X^2 + 1)^2\\
2155 (f(X) - p(X))^2 &= X^2 (q(X))^2
2156 \end{align*}
2157\end{proof}
2158
2159
2160\begin{ex}{R.4.5}
2161 Let $A=k[X,Y]/(Y^2 - X^2 - X^3)$. Prove that the normalization of $A$ is $k[t]$ where $t=Y/X$.
2162\end{ex}
2163\begin{proof}
2164$A=k[X,Y]/(Y^2 - X^2 - X^3)$, express $X$ and $Y$ in terms of $t$:
2165
2166Since $t=Y/X$ then $Y=tX$, and combined with $Y^2 = X^2 + X^3$, then
2167\begin{align*}
2168 (tX)^2 &= X^2 + X^3\\
2169 t^2 X^2 &= X^2 + X^3, ~~\text{assuming}~X \neq 0:\\
2170 t^2 &= 1+X, ~\text{thus}\\
2171 X&=t^2-1 \in k[X]
2172\end{align*}
2173
2174Then, $Y=tX=t(t^2-1)=t^3-t \in k[X]$.
2175
2176Hence $X, Y \in k[X]$.
2177
2178Therefore, $k[X,Y]/(Y^2 - X^2 - X^3) \subseteq k[t]$.
2179
2180\vspace{0.4cm}
2181By \ref{noether-normalization} (Noether normalization lemma), to show that $k[t]$ is the \emph{normalization}, must show that $k[t]$ is \emph{integral} over $A$.
2182
2183From $X=t^2-1 ~~\Longrightarrow~~ t^2-1-X=0 ~~\Longrightarrow~ t^2 - (1+X) = 0$.
2184
2185$(1+X) \in A$, so $t$ satisfies the monic polynomial
2186$$P(T) = T^2 - (1+X) \in A[T]$$
2187Thus $t$ is integral over $A$.
2188
2189Since $k[t]$ is generated by $t$ over $k$, and $k \subset A$, then the entire ring $k[t]$ is integral over $A$.
2190
2191Since $k[t]$ is a polynomial ring over a field, which is a UFD, it is integrally closed (since all UFD are integrally closed).
2192
2193$Frac A = k(X,Y)$, since $X=t^2-1, ~Y=t^3-t ~~\Longrightarrow~ k(X,Y) \subseteq k(t)$
2194
2195and $t=Y/X \in k(X,Y)$, thus $k(X,Y)=k(t)$.
2196
2197Since $k[t]$ is integrally closed and is the integral closure of $A$ in its fraction field $k(t)$, we conclude that the normalization of $A$ is $k[t]$.
2198\end{proof}
2199
2200
2201
2202\begin{ex}{R.4.9}
2203 $k$ 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.
2204\end{ex}
2205\begin{proof}
2206 (want to find a linear combination of the coordinates such that the original variables satisfy monic polynomials over the new ring $B$)
2207
2208 The relations defining $A$ are
2209 \begin{align*}
2210 X^2-Y^3-1=0 ~~&\Longrightarrow~ Y^3 = X^2 -1 ~~~(*)\\
2211 XZ -1=0 ~~&\Longrightarrow~ Z= 1/X = X^{-1}
2212 \end{align*}
2213
2214Thus $A$ can be denoted as $A = k[X,Y,X^{-1}]/(Y^3 - X^2 - 1)$.
2215
2216
2217Now, $Y$ is inegral over $k[X]$, since $Y^3 - (X^2 - 1) = 0$ is monic in $Y$ with coefficients in $k[X]$.
2218
2219$Z$ is not integral over $k[X]$, since $Z=1/X$ and $X$ is not a unit in $k[X]$.
2220
2221Choose $\alpha, \beta \in k$ such that $X$ (and thus $Z$) becomes integral over $B$:\\
2222set $\alpha=0,~\beta=1 ~~\Longrightarrow~~ B=k[X+\alpha Y + \beta Z]= k[X+Z]$.
2223
2224Let $w=X+Z$; since $XZ=1$, we have
2225$$w=X+\frac{1}{X} ~~\Longrightarrow~~ Xw=X^2+1 ~~\Longrightarrow~~ X^2 -w X +1 = 0 ~~(**)$$
2226
2227which is monic with coefficients in $k[w]$, thus $X$ is integral over $B$.
2228
2229Since $Z=w-X$, $Z$ is also integral.
2230
2231\vspace{0.4cm}
2232Generators of $A$ as a $B$-module:
2233
2234we hadd $B=k[w]$ with $w=X+Z$.
2235
2236From $(**)$ we have $X^2 - wX + 1=0$, so $X^2 = wX -1$.
2237
2238Thus 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 $\{1, X\}$.
2239
2240Fitting $X^2$ into $(*)$,
2241\begin{align*}
2242 X^2 &- Y^3 -1 =0\\
2243 Y^3 &= X^2-1\\
2244 Y^3 &= wX -2
2245\end{align*}
2246
2247thus any power of $Y$ higher than $2$ can be reduced (eg. $Y^4 = Y(wX-2) = w(XY) - 2Y$).
2248
2249So its partial basis is $\{1, Y, Y^2\}$.
2250
2251For $Z$, since $XZ=1$ and $w=X+Z \Longrightarrow Z=w-X$, thus $Z$ is a $B$-linear combination of $\{1,X\}$.
2252
2253\vspace{0.2cm}
2254Combining the previous partial basis, the generators are
2255$$\{ 1, X \} \times \{ 1,Y, Y^2 \} = \{ 1, Y, Y^2, X, XY, XY^2 \}$$
2256
2257\end{proof}
2258
2259\bibliographystyle{unsrt}
2260\bibliography{commutative-algebra-notes.bib}
2261
2262\end{document}