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]
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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{innerremark}{Remark}
62\newenvironment{remark}[1]
63{\renewcommand\theinnerremark{#1}\innerremark}
64{\endinnerremark}
65
66\newtheorem{innereg}{Example}
67\newenvironment{eg}[1]
68{\renewcommand\theinnereg{#1}\innereg}
69{\endinnereg}
70
71\newtheorem{innerex}{Exercise}
72\newenvironment{ex}[1]
73{\renewcommand\theinnerex{#1}\innerex}
74{\endinnerex}
75
76\newcommand{\aA}{\mathfrak{a}} % TODO: use goth font
77\newcommand{\mM}{\mathfrak{m}}
78
79\title{Commutative Algebra notes}
80\author{arnaucube}
81\date{2026}
82
83\begin{document}
84
85\maketitle
86
87\begin{abstract}
88 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}.
89
90 Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
91
92 The proofs may slightly differ from the ones from the books, since I try to extend them for a deeper understanding.
93\end{abstract}
94
95\tableofcontents
96
97\section{Ideals}
98
99\subsection{Definitions}
100
101\begin{defn}{}[ideal]
102 $I \subset R$ ($R$ ring) such that $0 \in I$ and $\forall x \in I,~ r \in R,~ xr, rx \in I$.\\
103 \hspace*{2em} ie. $I$ absorbs products in $R$.
104\end{defn}
105
106\begin{defn}{}[prime ideal]
107 if $a, b \in R$ with $ab \in P$ and $P \neq R$ ($P$ a prime ideal), implies $a \in P$ or $b \in P$.
108\end{defn}
109
110\begin{defn}{}[principal ideal]
111 generated by a single element, $(a)$.
112
113 $(a)$: principal ideal, the set of all multiples $xa$ with $x \in R$.
114\end{defn}
115
116\begin{defn}{}[maximal ideal]
117 $\mM \subset A$ ($A$ 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 $I=A$.
118\end{defn}
119
120
121\begin{defn}{}[unit]
122 $x \in A$ such that $xy=1$ for some $y \in A$. ie. element \emph{which divides 1}.
123\end{defn}
124
125\begin{cor}{1.8} \label{1.8}
126 $A = A^{\times} \sqcup \bigcup m$ (where $\sqcup$ denotes ``disjoint union"), ie. $f \in A$ is either a unit or it is contained in a maximal ideal, not both.
127\end{cor}
128
129\begin{defn}{}[zerodivisor]
130 $x \in A$ such that $\exists~ 0 \neq y \in A$ such that $xy=0 \in A$. ie. $x$ \emph{divides 0}..
131
132
133 If a ring does not have zerodivisors is an integral domain.
134\end{defn}
135
136\begin{defn}{}[prime spectrum - $Spec(A)$]
137 set of prime ideals of $A$. ie.
138
139 $$Spec(A) = \{ P ~|~ P \subset A~ \text{is a prime ideal} \}$$
140\end{defn}
141
142\begin{defn}{}[integral domain]
143 Ring in which the product of any two nonzero elements is nonzero.
144
145 ie. no zerodivisors.
146
147 ie. $\forall~ 0 \neq a,~ 0 \neq b \in A,~ ab \neq 0 \in A$.
148
149 Every field is an integral domain, not the converse.
150\end{defn}
151
152\begin{defn}{}[principal ideal domain - PID]
153 integral domain in which every ideal is principal. ie.
154 ie. $\forall I \subset R,~ \exists~ a \in I$ such that $I = (a) = \{ ra ~|~ r \in R \}$.
155\end{defn}
156
157\begin{defn}{}[nilpotent]
158 $a \in A$ such that $a^n=0$ for some $n>0$.
159\end{defn}
160\begin{defn}{}[nilrad A]
161 set of all nilpotent elements of $A$; is an ideal of $A$.
162
163 if $nilrad A = 0 ~\Longrightarrow$ $A$ has no nonzero nilpotents.
164
165
166 $$nilrad A = \bigcap_{P \in Spec(A)} P$$
167\end{defn}
168
169\begin{defn}{}[idempotent]
170 $e \in A$ such that $e^2=e$.
171\end{defn}
172
173\begin{defn}{}[radical of an ideal]
174 $$rad I = \{ f \in A | f^n \in I~ \text{for some}~ n \}$$
175
176 $rad I$ is an ideal.
177
178 $nilrad A = rad 0$
179
180 $rad I = \bigcap_{\substack{P \in \operatorname{Spec}(A)\\ P \supset I}} P$
181\end{defn}
182
183\begin{defn}{1.13}[local ring] \label{1.13}
184 A ring is \emph{local} if it has a unique maximal ideal.
185
186 Notation: local ring $A$, its maximal ideal $\mM$, residue field $K=A/\mM$:
187 $$A \supset \mM ~\text{or}~ (A, \mM) ~\text{or}~ (A, \mM, K)$$
188
189 \vspace{0.3cm}
190 By Corollary \ref{1.8}, $A$ is local\\
191 \hspace*{4em}$\Longleftrightarrow$ $A$ has only one maximal ideal.\\
192 \hspace*{4em}$\Longleftrightarrow$ all the nonunits of $A$ form an ideal.
193\end{defn}
194
195\begin{prop}{(R/m=Field)}\label{maximal-quotient-field}
196 Let $R$ be a commutative ring with identity, and let $m \subsetneq R$ be a maximal ideal. Then the quotient ring $R/m$ is a field.
197\end{prop}
198\begin{proof}
199 Let $x + m$ be a nonzero element of $R/m$. Since
200 $$x + m \neq m$$
201 we have $x \not\in m$.
202
203 Consider the ideal $m + (x)$, where $(x)$ is the principal ideal generated by $x$.
204 Since $x \not\in m$, this ideal strictly contains $m$:
205 $$m \subsetneq m + (x)$$
206
207 Because $m$ is maximal, there are no proper ideals strictly between $m$ and $R$. Therefore,
208 $m + (x) = R$
209
210 In particular, $1 \in m + (x)$. Hence, there exist
211 $a \in m$ and $y \in R$ such that $a + yx = 1$.
212
213 Passing to the quotient modulo $m$, we obtain
214 $$
215 (y+m)(x+m)
216 = yx+m
217 = 1+m,
218 $$
219 because $a \in m$ implies $a+m=m$.
220
221 Thus, $y+m$ is the multiplicative inverse of
222 $x+m$. Therefore, every nonzero element of
223 $R/m$ is invertible, and hence $R/m$ is a field.
224\end{proof}
225
226
227
228\subsection{Z and K[X], two Principal Ideal Domains}
229
230
231\begin{lemma}{}
232 $\mathbb{Z}$ is a PID.
233\end{lemma}
234\begin{proof}
235 Let $I$ a nonzero ideal of $\mathbb{Z}$.
236
237 Since $I \neq \{0\}$, there is at least one nonzero integer in $I$. Choose the smallest element of $I$, namely $d$.
238
239 Observe that $(d) \subseteq I$, since $d \in I$. Then, every multiple $nd \in I$, since $I$ is an ideal.
240
241 Take $a \in I$. By the Euclidean division algorithm in $\mathbb{Z}$, $a=qd+r$, with $q,r \in \mathbb{Z}$ and $0 \leq r \leq d$.
242
243 Then $r = a - qd \in I$, but $d$ was chosen to be the smallest positive element of $I$, so the only possibility is $r=0$.
244
245 Hence, $a=qd$, so $a \in (d)$, giving $I \subseteq (d)$.
246
247 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).
248\end{proof}
249
250\begin{lemma}{}
251 $K[X]$ is a PID.
252\end{lemma}
253\begin{proof}
254 This proof follows very similarly to the previous proof.\\
255
256 Let $K$ be a field, $K[X]$ a polynomial ring.
257
258 Take $\{0\} \neq I \subseteq K[X]$.
259
260 Since $I \neq \{0\}$, there is at least one non-zero polynomial in $I$.
261
262 Let $p(X) \in I$ be of minimal degree among nonzero elements of $I$.
263
264 Observe that $(p(X)) \subseteq I$, because $p(X) \in I$ and $I$ is an ideal.
265
266 Let $f(X) \in I$. By Euclidean division algorithm in $K[X]$, $\exists q, r \in K[X]$ such that $f(X) = q(X) \cdot p(X) + r(X)$ with either $r(X)=0$ or $deg(r) < deg(p)$.
267
268 Since $f,p \in I$, then $r(X) = f(X) - q(X)\cdot p(X) \in I$
269
270 If $r(X) \neq 0$, then $deg(r) < deg(p)$, which contradicts the minimality of $deg(p)$ in $I$.
271
272 Therefore, $r(X)=0$, thus $f(X)=q(X)\cdot p(X)$, hence $f(X) \in (p(X))$.
273 Henceforth, $I \subseteq (p(X))$.
274
275 Then, since $(p(X)) \subseteq I$ and $I \subseteq (p(X))$, we have that $I = (p(X))$.
276
277 So every ideal of $K[X]$ is principal; thus $K[X]$ is a PID.
278
279\end{proof}
280
281
282
283\subsection{Zorn's lemma and Jacobson radicals}
284
285Let $\Sigma$ be a partially orddered set. Given subset $S \subset \Sigma$, an \emph{upper bound} of $S$ is an element $u \in \Sigma$ such that $s<u \forall s \in S$.
286
287A \emph{maximal element} of $\Sigma$, is $m \in \Sigma$ such that $m<s$ does not hold for any $s \in \Sigma$.
288
289A 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 $s_2 \leq s_1$.
290
291\begin{lemma}{R.1.7}[Zorn's lemma] \label{zorn}
292 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 $\Sigma$.
293
294 Then $\Sigma$ has a maximal element.
295\end{lemma}
296
297\begin{thm}{AM.1.3} \label{1.3}
298 Every ring $A \neq 0$ has at least one maximal ideal.
299\end{thm}
300\begin{proof}
301 By Zorn's lemma \ref{zorn}.
302\end{proof}
303
304\begin{cor}{AM.1.4} \label{1.4}
305 if $I \neq (1)$ an ideal of $A$, $\exists$ a maximal ideal of $A$ containing $I$.
306\end{cor}
307
308\begin{cor}{AM.1.5} \label{1.5}
309 Every non-unit of $A$ is contained in a maximal ideal.
310\end{cor}
311
312\begin{defn}{}[Jacobson radical]
313 The \emph{Jacobson radical} of a ring $A$ is the intersection of all the maximal ideals of $A$.
314
315 Denoted $Jac(A)$.
316
317 $Jac(A)$ is an ideal of $A$.
318\end{defn}
319
320\begin{prop}{AM.1.9} \label{1.9}
321 $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$.
322\end{prop}
323\begin{proof}
324 Suppose $1-xy$ not a unit.
325
326 By \ref{1.5}, $1-xy \in \mM$ for $\mM$ some maximal ideal.
327
328 But $x \in Jac(A) \subseteq \mM$, since $Jac(A)$ is the intersection of all maximal ideals of $A$.
329
330 Hence $xy \in \mM$, and therefore $1 \in \mM$, which is absurd, thus $1-xy$ is a unit.
331
332 Conversely:\\
333 Suppose $x \not\in \mM$ for some maximal ideal $\mM$.
334
335 Then $\mM$ and $x$ generate the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$.
336
337 Hence $1 -xy \in \mM$, and is therefore not a unit.
338\end{proof}
339
340\section{Modules}
341
342\subsection{Modules concepts}
343
344Let $A$ be a ring. An $A$-module is an Abelian group $M$ with a multiplication
345map
346\begin{align*}
347 A \times M &\longrightarrow M\\
348 (f, m) &\longmapsto fm
349\end{align*}
350satisfying $\forall~ f,g \in A,~~ m, n \in M$.
351\begin{enumerate}[i.]
352 \item $f(m \pm n)=fm \pm fn$
353 \item $(f \pm g) m = fm \pm gm$
354 \item $(fg) m = f(gm)$
355 \item $1_A m = m$
356\end{enumerate}
357
358Let $\psi: M \longrightarrow M$ an $A$-linear endomorphism of $M$.\\
359$A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi$.
360\begin{itemize}
361\item since $\psi$ is $A$-linear, $A[\psi]$ is a commutative ring.
362\item $M$ is a module over $A[\psi]$, so $\psi$ beomes multiplication by a ring element.
363\end{itemize}
364
365\begin{remark}{}
366 Vector spaces are defined over fields, a Module is the generalization of a vector space where the scalars come from a ring instead of a field.
367\end{remark}
368
369\subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries}
370
371\begin{prop}{AM.2.4}(Cayley-Hamilton Theorem) \label{2.4}
372 Let $M$ a fingen $A$-module. Let $\aA$ an ideal of $A$, let $\psi$ an
373 $A$-module endomorphism of $M$ such that $\psi(M) \subseteq \aA M$.
374
375 Then $\psi$ satisfies
376
377 $$\psi^n + a_1 \psi^{n-1} + \ldots + a_{n-1} \psi + a_n = 0$$
378
379 with $a_i \in \aA$.
380\end{prop}
381\begin{proof}
382 Since $M$ fingen, let $\{ x_1, \ldots, x_n \}$ be generators of $M$.\\
383 By hypothesis, $\psi(M) \subseteq \aA M$; so for any generator $x_i$, it's image $\psi(x_i) \in \aA M$.
384
385 Any element in $\aA M$ is a linear combination of the generators with coefficients in the ideal $\aA$, thus
386 $$\psi(x_i)= \sum_{j=1}^n a_{ij} x_j$$
387 with $a_{ij} \in \aA$.
388
389 Thus, for a module with $n$ generators, we have $n$ different $\psi(x_i)$ equations:
390
391 $$
392 \left.
393 \begin{aligned}
394 \psi(x_1) &= a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n\\
395 \psi(x_2) &= a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n\\
396 \ldots\\
397 \psi(x_n) &= a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n
398 \end{aligned}
399 \right\}
400 \begin{aligned}
401 &\text{n elements $\psi(x_i) \in \aA M$ which}\\
402 &\text{are linear combinations of the}\\
403 &\text{$n$ generators of $M$}
404 \end{aligned}
405 $$
406
407 Next step: rearrange in order to use matrix algebra.
408
409 Observe that each row equals $0$, and rearranging the elements at each row we get
410
411 $$
412 \left.
413 \begin{aligned}
414 &\psi(x_1) - (a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n) = 0\\
415 &\psi(x_2) - (a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n) = 0\\
416 &\ldots\\
417 &\psi(x_n) - (a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n) = 0
418 \end{aligned}
419 \right\}
420 $$
421
422 Then, group the $x_i$ terms together; as example, take the row $i=1$:
423 $$(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0$$
424
425 $$
426 \left.
427 \begin{aligned}
428 &~~~~(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0\\
429 &-a_{2,1} x_1 + (\psi - a_{2,2}) x_2 - \ldots - a_{2,n} x_n = 0\\
430 &\ldots\\
431 &-a_{1,1} x_1 - a_{1,2} x_2 - \ldots + (\psi - a_{1,n}) x_n = 0\\
432 \end{aligned}
433 \right\}
434 $$
435
436 So, $\forall i \in [n]$, as a matrix:
437
438 $$
439 \begin{pmatrix}
440 \psi - a_{1,1} & -a_{1,2} & \ldots & -a_{1,n}\\
441 -a_{2,1} & \psi-a_{2,2} & \ldots & -a_{2,n}\\
442 \vdots\\
443 -a_{n,1} & -a_{n,2} & \ldots & \psi-a_{n,n}\\
444 \end{pmatrix}
445 \begin{pmatrix}
446 x_1\\ x_2\\ \vdots\\ x_n
447 \end{pmatrix}
448 =
449 \begin{pmatrix}
450 0\\ 0\\ \vdots\\ 0
451 \end{pmatrix}
452 $$
453
454
455
456 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 $M$).\\
457 \hspace*{4em}Then we can write the previous equality as
458 \begin{equation}
459 \Phi \cdot m = 0
460 \label{eq:2.4.1}
461 \end{equation}
462
463 We know that
464 \begin{equation}
465 adj(\Phi) \Phi = det(\Phi) I
466 \label{eq:2.4.2}
467 \end{equation}
468 (aka. \emph{fundamental identity for the adjugate matrix}).
469
470 So if at \eqref{eq:2.4.1} we multiply both sides by $adj(\Phi)$,
471 \begin{align*}
472 adj(\Phi) \cdot \Phi \cdot &m = 0\\
473 (\text{recall from \eqref{eq:2.4.2}:}~ &adj(\Phi)\Phi=det(\Phi)\cdot I ~)\\
474 =det(\Phi) \cdot I \cdot &m = 0
475 \end{align*}
476
477 Thus,
478 \begin{align*}
479 det(\Phi) \cdot I \cdot &m = 0:\\
480 \begin{pmatrix}
481 det(\Phi) & 0 & \ldots & 0\\
482 0 & det(\Phi) & \ldots & 0\\
483 \vdots\\
484 0 & 0 & \ldots & det(\Phi)
485 \end{pmatrix}
486 \cdot
487 &\begin{pmatrix}
488 x_1\\ x_2\\ \vdots\\ x_n
489 \end{pmatrix}
490 =
491 \begin{pmatrix}
492 0\\ 0\\ \vdots\\ 0
493 \end{pmatrix}
494 \end{align*}
495
496 $\Longrightarrow$
497 \begin{equation}
498 det(\Phi) \cdot x_i = 0 ~~\forall i \in [n]
499 \label{eq:2.4.3}
500 \end{equation}
501
502 ie. $det(\Phi)$ is an \emph{annihilator} of the generators $x_i$ of $M$, thus
503 is an annihilator of the entire module $M$.
504
505
506 So, we're interested into calculating the $det(\Phi)$.
507
508 By the Leibniz formula,
509 $$\det(A) = \sum_{\sigma \in S_n} sgn(\sigma) \prod_{i=1}^n a_{i, \sigma(i)}$$
510
511 thus,
512 $$det(\Phi) = \underbrace{(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})}_{\text{diagonal of $\Phi$, leading term of the determinant}} - \ldots$$
513
514 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
515 \begin{itemize}
516 \item highest power is $1 \cdot \psi^n$
517 \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\
518 where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$
519 \item the rest of coefficients of $\psi^k$ are also elements in $\aA$
520 \end{itemize}
521
522 Therefore we have
523 $$det(\Phi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$
524 with $a_i \in \aA$.
525
526 \vspace{0.5cm}
527
528 Now, notice that we had $det(\Phi) \cdot x_i = 0 ~\forall~ i\in [n]$.
529
530 The matrix $\Phi$ is the \emph{characteristic matrix}, $xI-A$, viewed as an operator. Then,
531 $$det(\Phi) = det(xI-A) = p(x)$$
532 where $p(x)$ is the \emph{characteristic polynomial}.
533
534 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 $det(\Phi)=0$.
535 Therefore,
536 $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$
537
538
539 %%%%%% OLD START
540
541 % \vspace{3cm}
542 %
543 % Kronecker delta:
544 % $\delta_{ij} =
545 % \begin{cases}
546 % 1 & \text{if } i = j,\\
547 % 0 & \text{otherwise}
548 % \end{cases}$
549 %
550 % With the Kronecker delta, $\psi(x_i)$ can be expressed as
551 % $$\psi(x_i) = \sum_{j=1}^n \delta_{ij} \psi(x_j)$$
552 % so the previous matrix can be characterized as
553 % $$\sum_{j=1}^n (\delta_{ij} \psi - a_{ij}) x_j = 0$$
554 %
555 % The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$)
556 % \begin{itemize}
557 % \item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$
558 % \item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$
559 % \end{itemize}
560 %
561 % 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 $x$.
562 %
563 % Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$,
564 % the ring of operators $A[\psi]$ is a commutative ring.
565 %
566 % $\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant.
567 %
568 %
569 % \vspace{0.75cm}
570 % This is called \emph{"the determinant trick"}.\\
571 % 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 $\psi$.\\
572 % $\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\
573 % $\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$
574 %
575 % $$det(M) \cdot x_i = 0~~ \forall i$$
576 % where $x_i$ are the generators of $M$.
577 %
578 % 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: $adj(\Phi)\cdot \Phi=det(\Phi) \cdot \aA$.
579 % \vspace{0.5cm}
580 %
581 % Since $det(M)$ kills every generator, it must kill every element in $M$\\
582 % $\Longrightarrow~~ det(M)$ is the zero map.
583 %
584 % Leibniz formula of the determinant of an $n \times n$ matrix:
585 % $$
586 % det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)}
587 % $$
588 %
589 % so,
590 % $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$
591 % expanding it,
592 % \begin{itemize}
593 % \item highest power is $1 \cdot \psi^n$
594 % \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\
595 % where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$
596 % \item the rest of coefficients of $\psi^k$ are also elements in $\aA$
597 % \end{itemize}
598 %
599 % So we have
600 % $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$
601 % with $a_i \in \aA$.
602 %
603 % 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:
604 % $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$
605 % with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem.
606
607 %%%%% OLD END
608\end{proof}
609
610
611\vspace{0.5cm}
612
613\begin{cor}{AM.2.5} \label{2.5}
614 Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$.
615
616 Then, $\exists~ x \equiv 1 \pmod \aA$ such that $xM = 0$.
617\end{cor}
618\begin{proof}
619 take $\psi = \text{identity}$. Then in Cayley-Hamilton (\ref{2.4}):
620 \begin{align*}
621 &\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0\\
622 \Longrightarrow~ &id_M + a_1 id_M + a_2 id_M + \ldots + a_{n-1} id_M + a_n = 0\\
623 \Longrightarrow~ &(1 + a_1 + \ldots + a_n) id_M = 0
624 \end{align*}
625 apply it to $m \in M$, where since $id_M(m)=m$ (by definition of the identity), we then have
626 $$(1 + a_1 + \ldots + a_n) \cdot m = 0$$
627 with $a_i \in \aA$.
628
629 \begin{enumerate}[\text{part} i.]
630 \item $xM=0$:\\
631 Thus the scalar $x = (1 + a_1 + \ldots + a_n)$ annihilates every $m \in M$, ie. the entire module $M$.
632
633 \item $x \equiv 1 \pmod \aA$:\\
634 $x \equiv 1 \pmod \aA ~~ \Longleftrightarrow (x-1) \in \aA$\\
635 then from $x = (1 + \underbrace{a_1 + \ldots + a_n}_b) \in \aA$, set $b=a_1 + \ldots + a_n$,\\
636 so that $x=(1+b) \in \aA$.\\
637
638 Then $x-1 = (1+b)-1 = b \in \aA$\\
639 so $x-1 \in \aA$, thus $x \equiv 1 \pmod \aA$ as stated.
640 \end{enumerate}
641\end{proof}
642
643
644
645\begin{prop}{AM.2.6}[Nakayama's lemma] \label{2.6}
646 Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$.
647
648 Then $\aA M = M$ implies $M=0$.
649\end{prop}
650\begin{proof}
651 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 $\aA \subseteq Jac(A)$).
652
653 (recall \ref{1.9}: $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$).\\
654 By \ref{1.9}, $x$ is a unit in $A$ (thus $x^{-1}\cdot x=1$).
655
656 Hence $M = x^{-1} \cdot \underbrace{x~ \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$.
657
658 Thus, if $\aA M = M$ then $M=0$.
659\end{proof}
660
661
662
663\begin{cor}{AM.2.7} \label{2.7}
664 Let $M$ a fingen $A$-module, let $N \subseteq M$ a submodule of $M$, let $\aA \subseteq Jac(A)$ an ideal.
665
666 Then $M = \aA M + N \stackrel{\text{implies}}{\Longrightarrow} M=N$.
667\end{cor}
668\begin{proof}
669 The idea is to apply Nakayama (\ref{2.6}) to $M/N$.
670
671 Since $M$ fingen $\Longrightarrow~~ M/N$ is fingen and an $A$-module.
672
673 Since $\aA \subseteq Jac(A) ~\Longrightarrow~$ Nakayama applies to $M/N$ too.
674
675 By definition,
676 $$\aA M = \left\{ \sum a_i \cdot m_i ~~|~~ a_i \in \aA, m_i \in M \right\}$$
677 where $m_i$ are the generators of $M$.
678
679 Then, for $M/N$,
680 $$\aA (\frac{M}{N}) = \left\{ \sum a_i \cdot (m_i + N) ~~|~~ a_i \in \aA, m_i \in M \right\}$$
681
682 observe that $a_i(m_i+N)= a_i m_i +N$, thus
683 $$\sum_i a_i \cdot (m_i + N) = \underbrace{(\sum_i a_i \cdot m_i)}_{\in \aA M} + N \in \aA M + N$$
684
685 Hence,
686
687 \begin{equation}
688 \aA (\frac{M}{N}) = \left\{ x + N ~~|~~ x \in \aA M \right\} = \aA M + N
689 \label{eq:2.7.1}
690 \end{equation}
691
692 By definition, if we take $\frac{\aA M + N}{N}$, then
693 $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$
694
695 thus every $y \in \aA M +N$ can be written as
696 $$y=x+n,~~ \text{with}~ x \in \aA M,~ n\in N$$
697 which comes from \eqref{eq:2.7.1}.
698
699 Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient.
700
701 Hence, every element of $\frac{\aA M +N}{N}$ has the form
702 $$\frac{\aA M + N}{N} = \left\{ x + N ~~|~~ x \in \aA M \right\}$$
703 as in \eqref{eq:2.7.1}.
704
705 Thus
706 \begin{equation}
707 \aA (\frac{M}{N}) = \aA M + N = \frac{\aA M +N}{N}
708 \label{eq:2.7.2}
709 \end{equation}
710
711 By the Collorary assumption, $M = \aA M + N$; quotient it by $N$:
712 \begin{equation}
713 \frac{M}{N} = \frac{\aA M +N}{N}
714 \label{eq:2.7.3}
715 \end{equation}
716
717 So, from \eqref{eq:2.7.2} and \eqref{eq:2.7.3}:
718 $$\aA (\frac{M}{N}) = \aA M +N = \frac{\aA M +N}{N} = \frac{M}{N}$$
719 thus, $\aA (\frac{M}{N}) = \frac{M}{N}$.
720
721 By Nakayama's lemma \ref{2.6}, if $\aA (\frac{M}{N}) = \frac{M}{N} ~\stackrel{implies}{\Longrightarrow}~ \frac{M}{N}=0$
722
723 Note that
724 $$\frac{M}{N} = \{ m + N ~|~ m \in M \}$$
725 (the zero element in $\frac{M}{N}$ is the coset $N=0+N$)
726
727 Then, $\frac{M}{N}=0$ means that the quotient has exactly one element, the zero coset $N$.
728
729 Thus, every coset $m + N$ equals the zero coset $N$, so $m-0 \in N ~\Longrightarrow~ m \in N$.
730
731 Hence every $m \in M$ lies in $N$, ie. $\forall m \in M,~ m \in N$.
732
733 So $M \subseteq N$. But notice that by the Corollary, we had $N \subseteq M$, therefore $M = N$.
734
735 Thus, if $M = \aA M + N \stackrel{implies}{\Longrightarrow} M = N$.
736\end{proof}
737
738
739
740\begin{prop}{AM.2.8} \label{2.8}
741 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 $M$.
742\end{prop}
743\begin{proof}
744 Let $N$ submodule $M$, generated by the $x_i$.
745
746 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 $N = M$.
747\end{proof}
748
749
750\subsection{Sequences}
751
752\begin{defn}{R.2.9.a}[Exact Sequence]
753 Let a sequence of homomorphisms
754$$L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N$$
755It is \emph{exact} at $M$ if $im(\alpha)=ker(\beta)$.
756
757ie. $\beta \circ \alpha = 0$ and $\alpha$ maps surjectively to
758$ker(\beta)$.
759\end{defn}
760
761\begin{defn}{R.2.9.b}[Short Exact Sequence (s.e.s.)] \label{2.9}
762$$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$$
763is exact $\Longleftrightarrow~ L \subset M$ and $N=M / L$.
764
765Properties:
766\begin{itemize}
767 \item $\alpha$ injective
768 \item $\beta$ surjective
769 \item $\alpha:~ L \Longrightarrow ker \beta$
770 \item $\beta$ induces $M/\alpha(L) \longrightarrow N$
771\end{itemize}
772\end{defn}
773
774\begin{prop}{R.2.10}[Split exact sequence] \label{2.10}
775 For the previous s.e.s., 3 equivalent conditions:
776 \begin{enumerate}[i.]
777 \item $\exists$ isomorphism $M \cong L \oplus N$, with
778 \begin{align*}
779 \alpha:~ &m \longmapsto (m,0)\\
780 \beta:~ &(m, n) \longmapsto n
781 \end{align*}
782
783 \item $\exists$ a \emph{section} of $\beta$, that is, a map $s: N \longrightarrow M$ such that $\beta \circ s = id_N$
784 \item $\exists$ a \emph{retraction} of $\alpha$, that is, a map $r: M \longrightarrow L$ such that $r \circ \alpha = id_L$
785 \end{enumerate}
786
787 If all i, ii, iii are satisfied, it is a split exact sequence.
788\end{prop}
789\begin{proof}
790 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. $M = L \oplus N$.
791
792 \begin{itemize}
793 \item[(i to ii, iii)]
794 if $M \cong L \oplus N$ such that $\alpha:~ m \longmapsto (m,0),~~ \beta:~ s(m, n) \longmapsto n$, we can define the maps
795
796 for ii:
797 \begin{align*}
798 s:~ N &\longrightarrow L \oplus N\\
799 s(n) &\longmapsto (0, n)
800 \end{align*}
801
802 Then $\beta(s(n))=\beta(0,n)$, so $\beta \circ s = id_N$.
803
804 for iii:
805 \begin{align*}
806 r:~ L \oplus N &\longrightarrow L\\
807 r(m,n) &\longmapsto m
808 \end{align*}
809
810 Then $r(\alpha(m))=r(m,0)$, so $r \circ \alpha = id_L$.
811
812 \item[(ii to i)]
813 assume $s: N \longrightarrow M$ such that $\beta \circ s = id_M$
814
815 Want to show $M \cong im(\alpha) \oplus im(s)$.
816
817 $\forall m \in M$, consider $m - s(\beta(m))$, apply $\beta$ to it:\\
818 $\beta(m - s(\beta(m))) = \beta(m) - (\beta \circ s)(\beta(m)) = \beta(m) - \beta(m) = 0$
819
820 Since $ker(\beta) = im(\alpha),~~ \exists! l \in L ~\text{such that}~ \alpha(l) = m - s(\beta(m))$.
821
822 Thus $m = \alpha(l) + s(\beta(m))$.
823
824 \vspace{0.3cm}
825 Now, suppose $x \in im(\alpha) \cap im(s)$, then $x = \alpha(l)=s(n)$, apply $\beta$ to it: $\beta(\alpha(l)) = \beta(s(n)) ~\Longrightarrow~ 0=n$.
826
827 If $n=0$, then $s(n)=0$, so the intersection is $\{0\}$.
828
829 \vspace{0.3cm}
830 Define
831 \begin{align*}
832 \phi: L \oplus N &\longrightarrow M\\
833 \phi(l,n) &\longmapsto \alpha(l)+s(n)
834 \end{align*}
835 This isomorphism satisfies the required conditions.
836
837 \item[(iii to i)] similar to the previous one.
838 \end{itemize}
839
840 \vspace{0.3cm}
841 Overview:\\
842
843$$
8440 \longrightarrow L
845 \substack{
846 \stackrel{\alpha}{\longrightarrow}\\
847 \stackrel{\longleftarrow}{r}
848 }
849 \substack{M \\[0.5ex] \cong L \oplus N}
850 \substack{
851 \stackrel{\beta}{\longrightarrow}\\
852 \stackrel{\longleftarrow}{s}
853 }
854 N \longrightarrow 0
855$$
856\begin{align*}
857 \alpha:~ &l \longmapsto (l,0)\\
858 r:~ &(m,n) \longmapsto m\\
859 \alpha &\circ r = id_L\\
860 \beta:~ &(l,n) \longmapsto n\\
861 s:~ &n \longmapsto (0,n)\\
862 \beta &\circ s = id_N
863\end{align*}
864
865\end{proof}
866
867\section{Noetherian rings (and modules)}
868
869\begin{defn}{}[Ascending Chain Condition]
870 A partially orddered set $\Sigma$ has the \emph{ascending chain condition} (a.c.c.) if every chain
871 $$s_1 \leq s_2 \leq \ldots \leq s_k \leq \ldots$$
872 eventually breaks off, that is, $s_k = s_{k+1} = \ldots$ for some $k$.
873\end{defn}
874
875$\Longrightarrow~ \Sigma$ has the a.c.c. iff every non-empty subset $S \subset \Sigma$ has a maximal element.\\
876\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.
877
878\subsection{Noetherian rings and modules}
879\begin{defn}{R.3.2}[Noetherian ring]\\
880 Let $A$ a ring; 3 equivalent conditions:
881 \begin{enumerate}[i.]
882 \item the set $\Sigma$ of ideals of $A$ has the a.c.c.; in other words, every increasing chain of ideals
883 $$I_1 \subset I_2 \subset \ldots \subset I_k \subset \ldots$$
884 eventually stops, that is $I_k = I_{k+1}=\ldots$ for some $k$.
885 \item every nonempty set $S$ of ideals has a maximal element
886 \item every ideal $I \subset A$ is finitely generated
887 \end{enumerate}
888 If these conditions hold, then $A$ is \emph{Noetherian}.
889\end{defn}
890\begin{proof}
891 TODO
892\end{proof}
893
894\begin{defn}{R.3.4.D}[Noetherian modules]\\
895 An $A$-module $M$ is Noetherian if the submoles of $M$ have the a.c.c.,\\
896 that is, ay increasing chain
897 $$M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$$
898 of submodules eventually stops.
899\end{defn}
900
901As in with rings, it is equivalent to say that
902\begin{enumerate}[i.]
903 \item any nonempty set of modulesof $M$ has a maximal element
904 \item every submodule of $M$ is finite
905\end{enumerate}
906
907
908
909\begin{prop}{R.3.4.P}\label{R.3.4.P}
910 Let $0 \longrightarrow L \xrightarrow{\ \alpha \ } M \xrightarrow{\ \beta \ } N \longrightarrow 0$ be a s.e.s. (split exact sequence, \ref{2.10}).
911
912 Then, $M$ is Noetherian $\Longleftrightarrow~ L$ and $N$ are Noetherian.
913\end{prop}
914\begin{proof}
915 $\Longrightarrow$: trivial, since ascending chains of submodules in $L$ and $N$ correspond one-to-one to certain chains in $M$.
916
917 $\Longleftarrow$: suppose $M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$ is an increasing chain of submodules of $M$.
918
919 Then identifying $\alpha(L)$ with $L$ and taking intersection gives a chain
920 $$L \cap M_1 \subset L \cap M_2 \subset \ldots \subset L \cap M_k \subset \ldots$$
921 of submodules of $L$, and applying $\beta$ gives a chain
922 $$\beta(M_1) \subset \beta(M_2) \subset \ldots \beta(M_k) \subset \ldots$$
923 of submodules of $N$.
924
925 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.
926\end{proof}
927
928\begin{lemma}{R.3.4.L}
929 for submodules $M_1 \subset M_2 \subset M$,
930 $$L \cap M_1 = L \cap M_2 ~\text{and}~ \beta(M_1) = \beta(M_2) ~\Longrightarrow~ M_1 = M_2$$
931\end{lemma}
932\begin{proof}
933 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 $\beta(m)=\beta(n)$.
934
935 Then $\beta(m-n)=0$, so that
936 $$m - n \in M_2 \cap ker(\beta)=M_1 \cap ker(\beta)$$
937
938 Hence $m \in M_1$, thus $M_1 = M_2$.
939\end{proof}
940
941
942
943\begin{cor}{R.3.5}[Properties of Noetherian modules]\label{R.3.5}
944 \begin{enumerate}[i.]
945 \item if $\forall i \in [r],~~M_i$ are Noetherian modules, then
946 $\bigoplus_{i=1}^r M_i$ is Noetherian.
947 \item if $A$ a Noetherian ring, then an $A$-module $M$ is Noetherian iff it is finite over $A$.
948 \item if $A$ a Noetherian ring, $M$ a finite module, then any submodule $N \subset M$ is again finite.
949 \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.
950 \end{enumerate}
951\end{cor}
952\begin{proof}
953 \begin{enumerate}[i.]
954 \item a direct sum $M_1 \oplus M_2$ is a particular case of an exact sequence.
955
956 Then, Proposition \ref{R.3.4.P} proves this statement when $r=2$. The case $r>2$ follows by induction.
957
958 \item if $M$ finite, then $\exists~$ surjective homomorphism
959 $$A^r \longrightarrow M \longrightarrow 0$$
960 for some $r$, so that $M$ is a quotient
961 $$M \cong A^r / N$$
962 for some submodule $N \subset A^r$.
963
964 $A^r$ is a Noetherian module by i., so $M$ is Noetherian due Proposition \ref{R.3.4.P}.
965
966 Conversely, $M$ Noetherian implies $M$ finite.
967
968 item as in previous implications:\\
969 $M$ finite and $A$ Noetherian $\Longrightarrow$ $M$ is Noetherian,\\
970 $\Longrightarrow$ since $N \subseteq M$, then $N$ is Noetherian too\\
971 $\Longrightarrow$ which implies that $N$ is a finite $A$-module.
972
973 \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.
974 \end{enumerate}
975\end{proof}
976
977
978\vspace{0.5cm}
979\subsection{Hilbert basis}
980\begin{thm}{R.3.6}[Hilbert basis theorem] \label{hilbert-basis}
981 If $A$ a Noetherian ring, then so is the polynomial ring $A[x]$.
982
983 ie. Every ideal in a polynomial ring over a field has a finite set of
984 generators (is fingen).
985\end{thm}
986\begin{proof}
987 Prove that any ideal $I \subset A[x]$ is fingen.
988
989 Define auxiliary sets $J_n \subset A$ by
990 $$J_n = \{ a \in A ~|~ \exists f \in I ~\text{s.th.}~ f = a x^n + b_{n-1}x^{n-1} + \ldots b_0 \}$$
991 ie. $J_n$ is the set of leading coefficients of $I$ of degree $n$.
992
993 $J_n$ is an ideal, since $I$ is an ideal.
994
995 $J_n \subset J_{n+1}$, since for $f \in I$ also $x f \in I$.
996
997 Therefore $J_1 \subset J_2 \subset \ldots \subset J_k \subset \ldots$ is an increasing chain of ideals.\\
998 Using the assumption that $A$ is Noetherian, deduce that $J_n = J_{n+1}$ for some $n$.
999
1000 For each $m \leq n, ~~ J_m \subset A$ is fingen, ie.
1001 $$J_m = (a_{m,1}, \ldots a_{m, r_m})$$
1002
1003 By definition of $J_m$, for each $a_{m,j}$ with $1 \leq j \leq r_m$,\\
1004 $\exists$ a polynomial $f_{m, j} \in I$ of degree $m$ having the leading coefficient $a_{m, j}$.
1005
1006 $$\Longrightarrow~~ \{ f_{m,j} \}_{m<n; 1 \leq j \leq r_m}$$
1007 the set of elements of $I$.
1008
1009 Claim: this finite set ($\{ f_{m,j} \}$) generates $I$.
1010
1011 $\forall f \in I$, if $\deg f =m$, then its leading coefficient is $a \in J_m$,
1012
1013 hence if $m \geq n$, then $a \in J_m=J_n$, so that
1014 $$a = \sum b_i a_{n,i} ~~ \text{with}~ b_i \in A$$
1015 and
1016 $$f - \sum b_i X^{m-n} \cdot f_{n, i}$$
1017 has degree $<m$.
1018
1019 Similarly, if $m \leq n$, then $a \in J_m$, so that
1020 $$a = \sum b_i a_{m, i} ~~\text{with}~ b_i \in A$$
1021 and
1022 $$f - \sum b_i f_{n, i}$$
1023 has degree $<m$.
1024
1025
1026 \vspace{0.3cm}
1027 By induction on $m$, $f$ can be written as a linear combination of finitely many elements.
1028
1029 Thus, any ideal of $A[x]$ is finitely generated.
1030\end{proof}
1031
1032\vspace{0.5cm}
1033\begin{cor}{R.3.6.C}
1034 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.
1035
1036 In particular, any fingen algebra over $\mathbb{Z}$ or over a field $K$ is Noetherian.
1037\end{cor}
1038\begin{proof}
1039 the assumption is that $B$ is a quotient of a polynomial ring,
1040 $$B \cong A[x_1, \ldots, x_n] / I$$
1041 for some ideal $I$.
1042
1043 By the Hilbert basis theorem \ref{hilbert-basis} and induction,\\
1044 $A$ being Noetherian implies that $A[x_1, \ldots, x_n]$ is Noetherian.
1045
1046 And by Corollary \ref{R.3.5}(iv),\\
1047 $A[x_1, \ldots, x_n]$ being Noetherian implies that $A[x_1, \ldots, x_n]/I$ is Noetherian.
1048\end{proof}
1049
1050
1051
1052\vspace{1cm}
1053\section{Finite ring extensions and Noether normalization}
1054
1055\subsection{A-algebras and integral domains}
1056
1057\begin{defn}{}[A-algebra / k-algebra]
1058 An $A$-algebra is a ring $B$ with a ring homomorphism $\psi: A \longrightarrow B$.
1059
1060 $B$ is an $A$-module with multiplication defined by $\psi(a) \cdot b~~~ (a \in A, b \in B)$.
1061
1062 When $A \subset B$, $B$ is an extension ring of $A$; denoted $\psi(A) = A' \subset B$.
1063\end{defn}
1064
1065\begin{defn}{R.4.1}\label{R.4.1}
1066 Let $B$ be an $A$-algebra.
1067
1068 \begin{enumerate}[i.]
1069 \item $B$ is a \emph{finite} $A$-algebra (\emph{finite over $A$}) if it is finite as an $A$-module.
1070 \item $y \in B$ is \emph{integral over} $A$ if $\exists$ a monic polynomial
1071 $$f(Y) = Y^n + a_{n-1} Y^{n-1} + \ldots + a_0 ~\in A'[Y]$$
1072 such that $f(y)=0:$
1073 $$f(y) = y^n + a_{n-1} y^{n-1} + \ldots + a_0 = 0$$
1074
1075 The algebra $B$ is \emph{integral over} $A$ if $\forall~ b \in B$ is integral.
1076 \end{enumerate}
1077\end{defn}
1078
1079
1080\begin{prop}{R.4.2}\label{R.4.2}
1081 Let $\psi: A \longrightarrow B$ be an $A$-algebra, and $y \in B$. Three equivalent conditions:
1082
1083 \begin{enumerate}[i.]
1084 \item $y$ is integral over $A$
1085 \item subring $A'[y] \subset B$ generated by $A' = \psi(A)$ and $y$ is finite over $A$
1086 \item $\exists$ an $A$-subalgebra $C \subset B$ such that $A'[y] \subset C$ and $C$ is finite over $A$
1087 \end{enumerate}
1088
1089 Notes: $A'$ is the image of $A$ in $B$, ie. $A' = \psi(A)$.\\
1090 $A'[y]$ is the smallest subring of $B$ containing both coefficients from $A$ and the element $y$.
1091\end{prop}
1092\begin{proof} .\\
1093 \begin{itemize}
1094 \item[(i to ii):]
1095 since $y$ integral over $A$ $~~\Longrightarrow~$ by \ref{R.4.1} (ii), $y$ satisfies
1096 $$f(y) = y^n + a_{n-1} y^{n-1} + \ldots + a_0 = 0$$
1097
1098 So any power $y^k~~ (k\geq n)$ can be expressed in terms of $\{1, y, y^2, \ldots, y^{n-1} \}$.
1099
1100 Thus the set $\{1, y, y^2, \ldots, y^{n-1} \}$ spans $A'[y]$ as an $A$-module.
1101
1102 \item[(iii to i):]
1103 since $A'[y] \subset C ~~\Longrightarrow~ y \in C$\\
1104 since $C$ finite over $A$, $C$ has finite generators $\{ c_1, \ldots, c_n \}$ such that $C = A \cdot c_1 + A \cdot c_2 + \ldots + A \cdot c_n$
1105
1106 Thus $y \cdot c_i \in C$,
1107 $$y \cdot c_i = \sum_{j=1}^n a_{ij} c_j$$
1108 with $a_{ij} \in A$.
1109
1110 By the Cayley-Hamilton theorem (\ref{2.4}),
1111 $$y^n + a_{n-1} y^{n-1} + \ldots + a_1 y + a_0 = 0$$
1112 Therefore, $y$ is integral (by \ref{R.4.1} (ii)).
1113 \end{itemize}
1114\end{proof}
1115
1116
1117\begin{prop}{R.4.3}[Tower Laws]\label{R.4.3}
1118 \\Let $B$ be an $A$-algebra.
1119 \begin{enumerate}[a.]
1120 \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,\\
1121 then $C$ is finite over $A$.
1122 \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 $A$.\\
1123 In particular, every $f \in A[y_1, \ldots, y_m]$ is integral over $A$.
1124 \item Transitivity of integrality: if $A \subset B \subset C$ with $C$ integral over $B$, and $B$ integral over $A$,\\
1125 then $C$ is integral over $A$.
1126 \item Integral closure as a subring: the subset
1127 $$\tilde{A} = \{ y \in B ~|~ y ~\text{is integral over}~ A \} \subset B$$
1128 is a subring of $B$.
1129
1130 Moreover, if $y \in B$ is integral over $\tilde{A}$ then $y \in \tilde{A}$, so that $\tilde{\tilde{A}} = \tilde{A}$.
1131 \end{enumerate}
1132\end{prop}
1133\begin{proof}.\\
1134 \begin{enumerate}[a.]
1135 \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,\\
1136 then the set of products $\{ \beta_i \gamma_j \}$ generates $C$ as an $A$-module.
1137
1138 Since there are $n \times m$ generators (ie. finite), $C$ is finite over $A$.
1139
1140 \item proof by induction:
1141
1142 base case: if $y_1$ integral over $A ~\Longrightarrow~$ it satisfies a monic polynomial.
1143
1144 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.
1145
1146 inductive step: let $R_k = A[y_1, \ldots, y_k]$. Assume $R_k$ is finite over $A$.
1147
1148 Since $y_{k+1}$ is integral over $A~~ \Longrightarrow~$ it is also integral over $R_k$.
1149
1150 Thus $R_{k+1} = R_k[y_{k+1}]$ is finite over $R_k$.
1151
1152 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 $A$.
1153
1154 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).
1155
1156 \item let $x \in C$, since $x$ integral over $B$, it satisfies:
1157 $$x^n + b_{n-1} x^{n-1} + \ldots + b_1 x + b_0 = 0,~~~~~b_i \in B$$
1158
1159 Let $B''=A[b_0, b_1, \ldots, b_{n-1}]$. Since each $b_i \in B$ and $B$ is integral over $A$\\
1160 \hspace*{2em} $\Longrightarrow~$ each $b_i$ is integral over $A$.
1161
1162 Since all $b_i$ are integral over $B'$ $~~\Longrightarrow~ B'[x]$ is a finite $B'$-algebra.
1163
1164 By part (a) (transitivity of finiteness), $B'[x]$ is a finite $A$-algebra.
1165
1166 Therefore, $x$ is integral over $A$.
1167
1168 \item
1169 \begin{enumerate}[I.]
1170 \item subring:\\
1171 let $x,y \in \tilde{A}$. Want to show $x+y, xy \in \tilde{A}$:
1172
1173 by part (b), the algebra $A[x,y]$ is finite over $A$.
1174
1175 Since $x+y,xy \in A[x,y]$, they are integral over $A$.
1176
1177 Thus $x+y, xy \in \tilde{A}$, since $\tilde{A} = \{ b \in B ~|~ b ~\text{integral over}~ A \}$.
1178
1179 \item idempotence\\
1180 let $z \in B$ be integral over $\tilde{A}$
1181
1182 we have a chain $A \subseteq \tilde{A} \subseteq \tilde{A}[x]$.
1183
1184 By definition, $\tilde{A}$ is integral over $A$, and $z$ is integral over $\tilde{A}$\\
1185 thus by part (c), $z$ is integral over $A$.
1186
1187 Therefore, $z \in \tilde{A}$.
1188 \end{enumerate}
1189 \end{enumerate}
1190\end{proof}
1191
1192\begin{lemma}{4.3.Aux}[Integrality implies finiteness] \label{integral-implies-finite}
1193 If $y$ integral over $A$ then $A[y]$ is finite over $A$.
1194
1195 This extends on point (b) from the previous proposition \ref{R.4.3}.
1196\end{lemma}
1197\begin{proof}
1198 Suppose $y$ is integral over $A$. By definition $\exists~ f \in A[T]$, with
1199 $f$ monic, such that $f(y)=0$. (here $T$ is a formal variable).
1200
1201 Let $deg(f)=d$, so that for $f(y)=0$ we have
1202 $$y^d + a_{d-1} y^{d-1} + \ldots + a_1 y + a_0 = 0 ~~~~ a_i \in A$$
1203
1204 Since it is monic (leading coefficient is $1$), we can rearrange it to isolate the highest power:
1205 \begin{equation}
1206 y^d = -(a_{d-1} y^{d-1} + \ldots + a_1 y + a_0)
1207 \label{eq:yn}
1208 \end{equation}
1209 Thus $y^d$ can be written using lower powers of $y$ with coefficients in $A$.
1210
1211 \vspace{0.5cm}
1212 Consider any element $p \in A[y]$, $p = c_m y^m + c_{m-1} y^{m-1} + \ldots + c_0$.
1213
1214 if $m<d$, leave it as it is.\\
1215 if $m \geq d$, use the monic equation \eqref{eq:yn} to replace $y^d$ with lower powers.
1216
1217 Repeating this process, can reduce any power of $y$ down to a linear
1218 combination of $\{1, y, y^2, \ldots, y^{d-1} \}$.
1219
1220 Thus every element in $A[y]$ can be expressed as
1221 $$\lambda_{d-1} y^{d-1} + \ldots + \lambda_2 y^2 + \lambda_1 y + \lambda_0 \cdot 1~~~~ \lambda_i \in A$$
1222
1223 Henceforth, the set $\{1, y, y^2, \ldots, y^{d-1} \}$ generates $A[y]$ as a finite $A$-module.
1224\end{proof}
1225
1226
1227
1228\vspace{0.5cm}
1229\begin{defn}{4.4}[Integral closure]
1230 Given the ring $\tilde{A}$ from \ref{R.4.3}.(d), ie. $\tilde{A} = \{ y \in B ~|~ y ~\text{integral over}~ A \} \subset B$,
1231 $\tilde{A}$ is the \emph{integral closure} of $A$ in $B$.
1232
1233 If $A=\tilde{A}$, then $A$ is \emph{integrally closed} in $B$.
1234
1235 An integral domain $A$ is \emph{normal} if it is \emph{integrally closed in its field of fractions}, that is if
1236 $$A = \tilde{A} \subset K = Frac(A)$$
1237
1238 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 $A$.
1239\end{defn}
1240
1241
1242\vspace{0.5cm}
1243
1244\subsection{Noether normalization}
1245
1246\begin{defn}{4.6}[Algebraically independent] \label{R.4.6.D}
1247 $y_1, \ldots, y_n \in A$ 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.
1248
1249 $\Longrightarrow~~ \nexists~ F(y_1, \ldots, y_n)=0$ ($F$ nonzero) with coefficients in $K$.
1250\end{defn}
1251
1252Recall: a $K$-algebra $A$ is fingen over $K$ if $A=K[y_1, \ldots, y_n]$ for some finite set $y_1, \ldots, y_n$.
1253
1254
1255\vspace{0.5cm}
1256\begin{lemma}{R.4.6.L} \label{R.4.6.L}
1257 Let $A = K[y_1, \ldots, y_n]$ and $0 \neq F \in K[Y_1, \ldots, Y_n]$ such that $F(y_1, \ldots, y_n)=0$.
1258
1259 Then $\exists~ y^*_1, \ldots, y^*_{n-1} \in A$ such that $y_n$ is integral over
1260 $$A^* = K[y^*_1, \ldots, y^*_{n-1}] ~~\text{and}~~ A=A^*[y_n]$$
1261\end{lemma}
1262\begin{proof}
1263 Set $y_i^* = y_i - y_n^{r_i}$ for $i \in [n-1]$ and $r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.\\
1264 \hspace*{3em}(ie. $y_i=y_i^* + y_n^{r_i}$)
1265
1266 Define $G \in A$ by
1267 $$G(y_1^*, \ldots, y_{n-1}^*, y_n) = F(y_i^* + y_n^{r_i}, y_n)=0$$
1268 viewed as a relation for $y_n$ over $K[y_1^*, \ldots, y_{n-1}^*]$.
1269
1270 Since $F$ is a polynomial in $y_1, \ldots, y_{n-1}^*$, can write it as a sum of monomials
1271 $$F= \sum_m a_m y^m = \sum_m a_m \prod y_i^{m_i}$$
1272 where $m=(m_1, \ldots, m_n)$ and each $a_m \neq 0$.
1273
1274
1275 Therefore,
1276 $$G= \sum a_m \prod (y_i^* + y_n^{r_i})^{m_i}$$
1277 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 $a_m y_n^{(\sum r_i m_i)}$.
1278
1279 Suppose we can arrange so that
1280 $$m \neq m' ~\Longrightarrow~ \sum r_i m_i \neq \sum r_i m_i'$$
1281
1282 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 $y_n$).
1283\end{proof}
1284
1285
1286\vspace{0.5cm}
1287\begin{thm}{R.4.6}[Noether normalization lemma] \label{noether-normalization}
1288 Let $K$ a field, $A$ a fingen $K$-algebra.
1289
1290 Then $\exists~ z_1, \ldots, z_m \in A$ such that
1291 \begin{enumerate}[i.]
1292 \item $z_1, \ldots, z_m$ are algebraically independent over $K$
1293 \item $A$ is finite over $B=K[z_1, \ldots, z_m]$
1294 \end{enumerate}
1295
1296 That is, a fingen extension $K \subset A$ can be written as a composite
1297 $$K \subset B = K[z_1, \ldots, z_m] \subset A$$
1298 where $K \subset B$ is a polynomial extension, and $B \subset A$ is finite.
1299\end{thm}
1300\begin{proof}
1301 % (by induction on the number of generators ($n$) of $A$).
1302 Let $y_1, \ldots, y_n$ be generators of $A = K[y_1, \ldots, y_n]$.
1303
1304 if $n=0$, nothing to prove since $A$ is generated by $0$ elements $~\Longrightarrow~ A=K$, and $K$ is finite.
1305
1306 if $n>0$ we have two cases:
1307 \begin{itemize}
1308 \item[-] $y_1, \ldots, y_n$ are algebraically independent over $K$, then by
1309 definition \ref{R.4.6.D} $A \cong K[y_1, \ldots, y_n]$, so that $A$ is a finite module over itself, with $m=n$.
1310 \item[-] $y_1, \ldots, y_n$ are algebraically dependent over $K$,
1311 $$\exists~ 0 \neq f \in K[y_1, \ldots, y_n] ~\text{s.th}~ f(y_1, \ldots, y_n)=0$$
1312 \end{itemize}
1313
1314 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
1315 $$y_n^d+ a_{d-1} y_n^{d-1} + \ldots + a_1 y_n + a_0 = 0~~~~~~~a_i \in K[y_1^*,
1316 \ldots, y_{n-1}^*]$$
1317 \hspace*{2em} ie. monic, so that by definition (\ref{R.4.1}), $y_n$ is
1318 integral over $K[y_1^*, \ldots, y_{n-1}^*]$.
1319
1320 $~\longrightarrow~$ 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.
1321
1322
1323 \vspace{0.3cm}
1324 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
1325 $$A^* = K[y^*_1, \ldots, y^*_{n-1}], ~~\text{and}~ A=A^*[y_n]$$
1326
1327 Setting $y_i^* = y_i - y_n^{r_i}$, so that $y_i = y_i^* + y_n^{r_i}$ $\forall i \in [n-1],~~ r_1, \ldots, r_{n-1} \geq 1 \in \mathbb{Z}$.
1328
1329 Use those new variables at $f(y_1, \ldots, y_n)=0$:
1330 $$f(y_1^* + y_n^{r_1}, y_2^* + y_n^{r_2}, \ldots, y_{n-1}^* + y_n^{r_{n-1}}, y_n) = 0$$
1331
1332 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 $y_n$.
1333
1334 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 $A^* = K[y_1^*, \ldots, y_{n-1}^*]$.
1335
1336 \vspace{0.3cm}
1337 Induction:\\
1338 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}).
1339
1340 By inductive hypothesis on $A^*,~~ \exists~ z_1, \ldots, z_m \in A^*$ algebraically independent over $K$ and with $A^*$ finite over $B=K[z_1, \ldots, z_m]$.
1341
1342 Therefore, each step of $B \subset A^* \subset A^*[y_n]=A$ is finite, and $A$ is finite over $B$ as required.
1343\end{proof}
1344
1345
1346\vspace{0.5cm}
1347\begin{eg}{ }
1348 $A = K[X,Y]/(XY-1)$. $Y$ is algebraic over $K[X]$, but not integral over $K[Y]$.
1349
1350 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 $X=0$).
1351
1352 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 $\epsilon \neq 0$.
1353
1354 This corresponds geometrically to tilting the hyperbola a little before projecting, so that no longer has a vertical asymptotic line.
1355\end{eg}
1356
1357\vspace{1cm}
1358
1359\subsection{Weak Nullstellensatz}
1360
1361\begin{prop}{R.4.9}\label{R.4.9}
1362 let $A \subset B$ be an integral extension of integral domain,\\
1363 then $A$ is a field $\Longleftrightarrow~ B$ is a field.
1364\end{prop}
1365\begin{proof}
1366 $\Longrightarrow$:\\
1367 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.
1368
1369 Since $A$ is a field, $\exists$ inverse, observe that:
1370 \begin{align*}
1371 x^n &+ a_{n-1} x^{n-1} + \ldots + a_1 x + a_0 = 0\\
1372 x(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = - a_0\\
1373 -a_0^{-1}(x^{n-1} &+ a_{n-1} x^{n-2} + \ldots + a_1) = x^{-1} \in B
1374 \end{align*}
1375 thus there exists inverse in $B$, so $B$ is a field too.
1376
1377 $\Longleftarrow$:\\
1378 if $B$ is a field and $0 \neq x \in A$, then $x^{-1} \in B$, so $x^{-1}$ is integral over $A$.
1379
1380 So there is a relation of the form
1381 $$(x^{-1})^n + a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 =0$$
1382
1383 Therefore
1384 \begin{align*}
1385 (x^{-1})^n &+ a_{n-1} (x^{-1})^{n-1} + \ldots + a_0 = 0\\
1386 (x^{-1})^n &= -a_{n-1} (x^{-1})^{n-1} - \ldots - a_0\\
1387 x^{-n} &= -a_{n-1} x^{-n+1} - \ldots - a_0 ~~\text{(mult by $x^{n-1}$)}\\
1388 x^{-n+(n-1)} &= -a_{n-1} x^{-n+1+(n-1)} - \ldots - a_0 x^{n-1}\\
1389 x^{-1} &= -a_{n-1} - \ldots - a_0 x^{n-1} \in A
1390 \end{align*}
1391
1392 thus there exists inverse in $A$, so $A$ is a field too.
1393\end{proof}
1394
1395\begin{thm}{4.10.prev}[Zariski's lemma] \label{zariski}
1396 $k$ a field, $L$ fingen $k$-algebra and a field. Then $L$ is a finite algebraic extension of $k$.
1397
1398 If $k$ is algebraically closed, then $L=k$.
1399\end{thm}
1400
1401\vspace{0.4cm}
1402From Zariski's lemma we can see that:\\
1403let maximal ideal $m \subset k[X_1, \ldots, X_n]$, then $k[X_1, \ldots, X_n]/m$ is a (quotient) field.
1404
1405It's a fingen $k$-algebra, so by Zariski's lemma, $k[X_1, \ldots, X_n]/m$ is algebraic over $k$.
1406
1407Since $k[X_1, \ldots, X_n]/m$ is algebraically closed, it must equal $k$; ie.
1408$$\frac{k[X_1, \ldots, X_n]}{m}=k$$
1409
1410
1411Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, we have the natural map
1412$$\psi: k \longrightarrow \frac{k[X_1, \ldots, X_n]}{m}$$
1413so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $k[X_1, \ldots, X_n]/m$ must be an element of $k$).\\
1414Then, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$.
1415Thus,
1416$$a_i = X_i +m$$
1417$$\Longrightarrow X_i - a_i \in m$$
1418
1419 % So, $\forall x_i \in k$, its image in the quotient field $A/m$ must be an element of $k$.
1420 % $$\Longrightarrow x_i'=a_i \in k, ~\forall i \in [n]$$
1421 % $$\Longrightarrow x_i - a_i \in m$$
1422
1423The ideal generated by these terms is a subset of $m$:
1424$$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
1425
1426Thus $m=(X_1 - a_1, \ldots, X_n - a_n)$ (shown also at \ref{5.2}).
1427
1428\vspace{0.4cm}
1429Then:
1430For $k$ algebraically closed, every maximal ideal of $k[X_1, \ldots, X_n]$ has the form
1431$(X_1 - a_1, \ldots, X_n - a_n)$ for some $a_i \in k$.
1432
1433Equivalently, $V(I) = \emptyset ~~\Longleftrightarrow 1 \in I$.
1434
1435\vspace{0.4cm}
1436
1437\begin{thm}{R.4.10}[Weak Nullstellensatz - Zariski's lemma] \label{zariski2}
1438 let $k$ a field, $K$ a $k$-algebra which
1439 \begin{enumerate}
1440 \item is finitely generated as a $k$-algebra
1441 \item is a field
1442 \end{enumerate}
1443 Then $K$ is algebraic over $k$, so that $k \subset K$ is a finite field
1444 extension. That is, $[K:k] < \infty$.
1445\end{thm}
1446\begin{proof}
1447 $K=k[z_1, \ldots, z_m]$ a field; want to show that $K$ is algebraic over $k$.
1448 \vspace{0.3cm}
1449
1450 Since $K$ is a fingen $k$-algebra, by Noether normalization lemma (\ref{noether-normalization}),\\
1451 $\exists~ z_1, \ldots, z_m \in K$ such that
1452 \begin{itemize}
1453 \item are algebraically independent
1454 \item $K$ is integral over the polynomial ring $A=k[z_1, \ldots, z_m]$ (which by \ref{integral-implies-finite} is finite)
1455 \end{itemize}
1456\\
1457 Now we're at the situation of \ref{R.4.9}:
1458
1459 $A \subset K$ is integral, $K$ is a field $~~\Longrightarrow~$ therefore $A$ is a field.
1460
1461 \vspace{0.2cm}
1462 Since $z_1, \ldots, z_m \in K$ are algebraically independent,\\
1463 \hspace*{2em}$\Longrightarrow~ A=k[z_1, \ldots, z_m]$ is a polynomial ring in $m$ indeterminates, and this is a field only if $m=0$
1464
1465 (since in $k[z_1]$ the element $z_1$ is not invertible, since $1/z_1$ is a rational function, not a polynomial).
1466
1467 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 $k$.
1468\end{proof}
1469
1470
1471\vspace{1cm}
1472
1473\section{Nullstellensatz}
1474
1475\subsection{Corollary from Weak Nullstellensatz}
1476Note: 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 $k$.
1477
1478\vspace{0.3cm}
1479\begin{cor}{5.2} \label{5.2}
1480 $k$ algebraically closed. Then every maximal ideal of $A = k[X_1, \ldots,
1481 X_n]$ is of the form
1482 $$m = (X_1 - a_1, \ldots, X_n -a_n),~~ a_i \in k$$
1483
1484 The map $k[X_1, \ldots, X_n] \longrightarrow k[X_1, \ldots, X_n]/m=k$ is the natural evaluation map $f(X_1, \ldots, X_n) \longmapsto f(a_1, \ldots, a_n)$.
1485
1486 Thus
1487 \begin{align*}
1488 k^n &\longleftrightarrow m-Spec~ A\\
1489 (a_1, \ldots, a_n) &\longleftrightarrow f(a_1, \ldots, a_n)
1490 \end{align*}
1491\end{cor}
1492\begin{proof}
1493 let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.
1494
1495 By fundamental property of maximal ideals, $K=A/m$ is a field.
1496
1497 Since $A$ is a fingen $k$-algebra (generated by $X_1, \ldots, X_n$), then $K=A/m$ is also a fingen $k$-algebra, generated by residues $\psi(x_i) = x_i' = x_i +m$ (where $\psi: k \longrightarrow A/m$).
1498
1499 By Zariski's lemma (\ref{zariski}), $K=A/m$ is algeraic over $k$.
1500
1501 Since by hypothesis $k$ is algebraically closed, it has no proper algebraic extensions\\
1502 \hspace*{2em} $\Longrightarrow~~ K=k~~ \Longrightarrow~~ k \cong A/m$.
1503
1504 \vspace{0.3cm}
1505 Recall that since $A$ is a $k$-algebra, we have the natural map $\psi: k \longrightarrow A/m$, so that $\forall x_i \in k,~~\psi(x_i)=a_i \in k$ (its image in the quotient field $A/m$ must be an element of $k$).\\
1506 Also, by definition of a quotient ring, $\psi(x_i)$ is the coset $\psi(x_i)=X_i +m$.
1507 Thus,
1508 $$a_i = X_i +m$$
1509 $$\Longrightarrow X_i - a_i \in m$$
1510
1511 The ideal generated by these terms is a subset of $m$:
1512 $$J=(X_1 -a_1, \ldots, X_n -a_n) \subseteq m$$
1513
1514 Since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$,
1515 then $J$ is a maximal ideal. Together with $J \subseteq m$, then we have
1516 $J=m$, ie. $$m = (X_1 -a_1, \ldots, X_n -a_n)$$
1517
1518 \vspace{0.3cm}
1519 Let
1520 \begin{align*}
1521 \phi: k[X_1, \ldots, X_n] &\longrightarrow k[X_1, \ldots, X_n]/m\\
1522 \phi: x_i &\longmapsto a_i
1523 \end{align*}
1524
1525 Which is the natural evaluation map $f(X_1, \ldots, X_n) \longrightarrow f(a_1, \ldots, a_n)$.
1526 \vspace{0.2cm}
1527
1528 Since $\phi$ is a $k$-algebra homomorphism, then $\forall f \in A$:
1529 $$\phi(f(X_1, \ldots, X_n)) = f(\phi(x_1), \ldots, \phi(x_n))= f(a_1, \ldots, a_n)$$
1530
1531 Thus there is a one-to-one correspondence:
1532
1533 points in $k^n ~~~ \longleftrightarrow~~~ m-Spec~ A$ (maximal ideals in $k[X_1, \ldots, X_n]$
1534
1535 $(a_1, \ldots, a_n) ~~~\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)$
1536\end{proof}
1537
1538
1539\vspace{0.5cm}
1540\subsection{Variety}
1541
1542\begin{defn}{5.3}[Variety]
1543 A \emph{variety} $V \subset k^n$:
1544 $$ V = V(J) = \{ P=(a_1, \ldots, a_n) \in k^n | f(P)=0 ~\forall~ f \in J \}$$
1545
1546 \begin{itemize}
1547 \item[$\rightarrow$] $V$ is defined by $f_1(P)= \ldots = f_m(P) = 0$
1548 \item[$\rightarrow$] $V$ is defined as the simultaneous solutions of a number of polynomial equations.
1549 \end{itemize}
1550\end{defn}
1551
1552\begin{prop}{5.3} \label{5.3}
1553 $k$ 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 $k[X_1, \ldots, X_n]$.
1554 (notation: $x_i = X_i \pmod J$)
1555
1556 Then every maximal ideal of $A$ is of the form
1557 $$(x_1 -a_1, \ldots, x_n - a_n)$$
1558 for some point $(a_1, \ldots, a_n) \in V(J)$.
1559
1560 Therefore, $\exists$ a one-to-one correspondence
1561 \begin{align*}
1562 V(X) &\longleftrightarrow m-Spec~ A\\
1563 \text{given by}~~~(a_1, \ldots, a_n) &\longleftrightarrow (x_1 -a_1, \ldots, x_n - a_n)
1564 \end{align*}
1565\end{prop}
1566\begin{proof}
1567 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.
1568 \begin{align*}
1569 m \subset A \longleftrightarrow &m' \subseteq k[X_1, \ldots, X_n]\\
1570 &\text{s.th.}~ J \subseteq m'
1571 \end{align*}
1572
1573 Thus, every maximal ideal of $A$ is of the form
1574 $$\underbrace{(x_1 -a_1, \ldots, x_n - a_n)}_{i.} ~\text{such that}~ \underbrace{J \subset (X_1 - a_1, \ldots, X_n - a_n)}_{ii.}$$
1575
1576 \begin{enumerate}[i.]
1577 \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 $(a_1, \ldots, a_n) \in k^n$.
1578
1579 Which when projected to the quotient ring $A$, $X_i \longmapsto x_i$ (residue class), giving $(x_1 - a_1, \ldots, x_n - a_n)$.
1580
1581 \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.
1582 \end{enumerate}
1583
1584 \vspace{0.3cm}
1585 However, since $(X_1 - a_1, \ldots, X_n - a_n)$ is the kernel of the evaluation map $f \longmapsto f(a_1, \ldots, a_n)$
1586
1587 $\Longrightarrow~$ means that $m'=(X_1 - a_1, \ldots, X_n - a_n)$ consists of all polynomials that vanish at point $P=(a_1, \ldots, a_n)$.
1588
1589 If $J \subseteq m'$, then $\forall~ f \in J$ must vanish at $P$.
1590
1591 By definition, the set of points where all polynomials in $J$ vanish is the \emph{variety}, $V(J)$.
1592
1593 \vspace{0.4cm}
1594 Thus,\\
1595 every maximal ideal in $A$ corresponds to a point $(a_1, \ldots, a_n) \in k^n$, ie.
1596 $$m-Spec~ A \longleftrightarrow k^n$$
1597
1598 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
1599 \begin{align*}
1600 m-Spec~ A &\longleftrightarrow V(J)\\
1601 \text{maximal spectrum} &\longleftrightarrow \text{variety}
1602 \end{align*}
1603\end{proof}
1604
1605\begin{prop}{5.5}[Correspondeces $V$ and $I$] \label{5.5}
1606 A variety $X \subset k^n$ is by definition $X=V(J)$ (J an ideal of $k[X_1, \ldots, X_n]$).
1607
1608 So $V$ gives a map:\\
1609 $$\{ \text{ideals of}~ k[X_1, \ldots, X_n] \} \stackrel{V}{\longrightarrow} \{ \text{subsets}~ X ~\text{of}~ k^n \}$$
1610
1611 correspondence going the other way:
1612 $$\{ \text{subsets}~ X ~\text{of}~ k^n \} \stackrel{I}{\longrightarrow} \{ \text{ideals of} ~k[X_1, \ldots, X_n] \}$$
1613
1614 defined by taking a subset $X \subset k^n$ into the ideal
1615 $$I(X) = \{ f \in k[X_1, \ldots, X_n] | f(P)=0 ~\forall~ P \in X \}$$
1616
1617 \vspace{0.4cm}
1618
1619 $V,~I$ satisfy reverse inclusions:
1620 $$J \subset J' ~\Longrightarrow~ V(J) \supset V(J') ~~~~\text{and}~~~~ X \subset Y ~\Longrightarrow~ I(X) \supset I(Y)$$
1621\end{prop}
1622
1623
1624\vspace{0.4cm}
1625\subsection{Nullstellensatz}
1626
1627\vspace{0.5cm}
1628\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
1629 Let $k$ algebraically closed field.
1630
1631 \begin{enumerate}[a.]
1632 \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
1633 \item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$,
1634 $$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$
1635 \end{enumerate}
1636\end{thm}
1637\begin{proof}
1638 \begin{enumerate}[a.]
1639 \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$:\\
1640 Let $m \subset k[X_1, \ldots, X_n]$ be a maximal ideal.\\
1641
1642 Now,
1643 \begin{itemize}
1644 \item[-] since $m$ maximal, $L= k[X_1, \ldots, X_n]/m$ is a field.
1645 \item[-] since $k[X_1, \ldots, X_n]$ is a fingen $k$-algebra, $L$ is a fingen $k$-algebra.
1646 \end{itemize}
1647 $\Longrightarrow~$ thus $L$ is a finite field extension of $k$.
1648
1649 (Recall: if $k$ algebraically closed and $L$ a fingen field extension of $k$,
1650 then $L \cong k$.)
1651
1652 Therefore,
1653 $$L=\frac{k[X_1, \ldots, X_n]}{m} \cong k$$
1654
1655 Since $k[X_1, \ldots, X_n]$ is a $k$-algebra, $\exists$ a surjective homomorphism
1656 $$\psi: k[X_1, \ldots, X_n] \longrightarrow k$$
1657
1658 Let $a_i = \psi(x_i)$. Then $x_i - a_i \in ker(\psi) = m ~\forall~ i \in [n]$.
1659
1660 Since the ideal $(X_1 - a_1, \ldots, X_n - a_n)$ is maximal and contained in $m$, they must be equal, ie. $m = (X_1 - a_1, \ldots, X_n - a_n)$. (as in \ref{5.2})
1661
1662 Therefore, $P=(a_1, \ldots, a_n) \in k^n$ is a zero for every polynomial in $m$,
1663 $$f \in m ~\Longleftrightarrow~ f(P)=0$$
1664
1665 Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$, ie. every element of $J$ vanishes at $P$.\\
1666 $\Longrightarrow~$ therefore $P \in V(J)$, and thus $V(J) \neq \emptyset$.
1667
1668\item $I(V(J)) = rad J$:\\
1669 \begin{align*}
1670 I(V(J)) &= rad J\\
1671 \text{vanishing ideal of a variety} &= \text{radical of the ideal defining the variety}
1672 \end{align*}
1673 where $rad~J = \{ f \in R ~|~ f^n \in J ~\text{for some}~ n>0 \}$.
1674
1675 Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$.
1676
1677 Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$
1678
1679 Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie.
1680 $$\exists~ (a_1, \ldots, a_n, a_{n+1}) \in V(J')$$
1681
1682 Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires
1683 $$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$
1684 a contradiction.
1685
1686 Therefore, $V(J') = \emptyset$.
1687
1688 \vspace{0.4cm}
1689
1690 Since $V(J')=\emptyset$, by the Weak Nullstellensatz/Zariski (\ref{zariski}),\\
1691 \hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$.
1692
1693 Every element in an ideal is a linear combination of its generators: $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
1694
1695 $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf)$$
1696
1697 which, since $1 \in J'$,
1698 $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$
1699
1700 substitute $Y=1/f$,
1701 $$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$
1702 thus
1703 $$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$
1704 multiply by $f^n$,
1705 $$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$
1706
1707 thus $f^n$ is a linear combination of $g_i$.\\
1708 Thus $f^n \in J$, so $f \in rad~J$.
1709\end{enumerate}
1710\end{proof}
1711
1712
1713\vspace{0.4cm}
1714\subsection{Irreducible varieties}
1715
1716\begin{defn}{5.7}[Irreducible variety]
1717 a variety $X \subset k^n$ is \emph{irreducible} if it is nonempty and not the union of two proper subvarieties; that is, if
1718 $$X = X_1 \cup X_2 ~~\text{for varieties}~ X_1, X_2 ~\Longrightarrow~ X= X_1 ~\text{or}~ X_2$$
1719\end{defn}
1720
1721\begin{prop}{5.7} \label{5.7}
1722 a variety $X$ is irreducible iff $I(X)$ is prime.
1723\end{prop}
1724\begin{proof}
1725 set $I=I(X)$.
1726
1727 if $I$ not prime, then $f,g \in A \setminus I$ be such that $fg \in I$.
1728
1729 Define new ideals
1730 $$J_1=(I, f) ~~\text{and}~~ J_2=(I, g)$$
1731
1732 Then, since $f \not\in I(X)$, it follows that $V(J_1) \subsetneq X$
1733
1734 \hspace*{2em}$\Longrightarrow$ so $X=V(J_1) \cup V(J_2)$ is reducible.
1735
1736 The converse is similar.
1737\end{proof}
1738
1739\begin{cor}{5.8} \label{cor.5.8}
1740 let $k$ algebraically closed field. Then $V$ and $I$ induce one-to-one correspondences
1741
1742 \begin{align*}
1743 \{ \text{radical ideals}~J~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{varieties}~ X \subset k^n \}\\
1744 \{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} &\longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}
1745 \end{align*}
1746
1747 Therefore,
1748 $$Spec~ k[X_1, \ldots, X_n] = \{ \text{irreducible varieties}~ X \subset k^n \}$$
1749\end{cor}
1750
1751\begin{prop}{5.8} \label{5.8}
1752 let $A= k[x_1, \ldots, x_n]$ a fingen $k$-algebra ($k$ an algebraically closed field).
1753
1754 Write $J$ for the ideal of relations holding between $x_1, \ldots, x_n$, so that $A=k[X_1, \ldots, X_n]/J$.
1755
1756 Then there is a one-to-one correspondence
1757 $$Spec~ A \longleftrightarrow \{ \text{irreducible subvarieties}~ X \subset V(J) \}$$
1758\end{prop}
1759\begin{proof}
1760 By definition, $Spec~ A = \{ P ~|~ P \subset A ~\text{is prime ideal} \}$.
1761
1762 By Corollary \ref{cor.5.8}:
1763 $$\{ \text{prime ideals}~P~\text{of}~k[X_1, \ldots, X_n] \} \longleftrightarrow \{ \text{irreducible varieties}~ X \subset k^n \}$$
1764
1765 About varieties:
1766 \begin{itemize}
1767 \item $I(X)$ in $R=k[X_1, \ldots, X_n]$ is the ideal of the variety $X$\\
1768 \hspace*{2em}ie. the set of all polynomials that vanish on every point of $X$.
1769 \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.
1770 \end{itemize}
1771
1772 If $P$ is prime in $A~~\Longrightarrow~$ it must correspond to some prime ideal $\mathfrak{P}$ in $R$ (also $J \subset R$).
1773
1774 Then from Nullstellensatz (\ref{nullstellensatz}), every prime ideal $\mathfrak{P}$ in $R$ is the ideal of some irreducible variety $X$.\\
1775 \hspace*{2em}$\Longrightarrow~~ \mathfrak{P}=I(X)$.
1776
1777
1778 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\\
1779 $$\Longrightarrow~~ P = \{ f+J ~|~ f \in I(X) \}~~ = I(X) ~\text{mod}~J$$
1780
1781 Now, $J$ is the set of equations defining our "universe" $V(J)$.
1782
1783 Since $A=R/J ~~ \Longrightarrow~$ we thus have $J \subseteq R$.
1784
1785 Also we have a correspondence between $A=R/J$ and $R$.
1786
1787 Let $\mathfrak{P}$ be the preimage of $P$ in $R=k[X_1, \ldots, X_n]$, ie.
1788 \begin{align*}
1789 R &\longrightarrow A=R/J\\
1790 \mathfrak{P} &\longmapsto P\\
1791 I(X) &\longmapsto I(X) ~\text{mod}~J
1792 \end{align*}
1793
1794 $\Longrightarrow~ \text{thus}~ J \subseteq \mathfrak{P}$.
1795
1796
1797 Now, $J \subset \mathfrak{P} = I(X)$;\\
1798 by \ref{5.5}, the set of points where $\mathfrak{P}$ vanishes must be inside the set of points where $J$ vanishes, so
1799 $$V(J) \supseteq V(\mathfrak{P}) = V(I(X)) = X$$
1800 $\Longrightarrow~~ X \subseteq V(J)$, ie. the irreducible variety $X$ must be a subvariety of $V(J)$.
1801
1802 Therefore,
1803 $$\mathfrak{P} \in Spec~ A \longleftrightarrow X \subseteq V(J)$$
1804 where $X = V(\mathfrak{P})$.
1805\end{proof}
1806
1807\subsection{Again: Zariski, Weak Nullstellensatz, Nullstllensatz}
1808This subsection proves again Zariski's lemma, Weak Nullstellensatz and Nullstellensatz, in a more interconnected way than the proofs described previously.
1809
1810\begin{lemma}{}[Zariski's lemma]
1811 $k$ a field, $K$ a fingen $k$-algebra and a field.
1812
1813 Then $K$ is a finite algebraic extension of $k$.
1814
1815 If $k$ algebraically closed, then $K=k$.
1816\end{lemma}
1817\begin{proof}
1818 $K$ a fingen $k$-algebra and a field $\Longleftrightarrow$ $K=k[X_1, \ldots, X_n]$ a field.\\
1819 Want to show that $K$ is algebraic over $k$.
1820
1821 Since $K$ is a fingen $k$-algebra, by Noether normalization \ref{noether-normalization},\\
1822 $\exists~ \{y_1, \ldots, y_r\} \subset K$ such that
1823 \begin{itemize}
1824 \item the set $\{y_1, \ldots, y_r\}$ is algebraically independent over $k$
1825 \item $K$ is integral over the polynomial ring $R=k[y_1, \ldots, y_r] \subset K$.
1826 \end{itemize}
1827
1828 By \ref{R.4.9}:\\
1829 if $A \subset B$ integral extension of integral domain,\\
1830 then $A$ is a field $\Longleftrightarrow~ B$ is a field.
1831
1832 which applied to our case:
1833 $$B:=K ~\text{(a field)},~~ A:= R=k[y_1, \ldots, y_r]$$
1834 since $K$ is integral over $R$ and $K$ a field $\Longrightarrow$ by \ref{R.4.9} $R$ must be a field.
1835
1836 \vspace{0.3cm}
1837 Since $y_i \in K$ are algebraically independent\\
1838 $\Longrightarrow~ R=k[y_1, \ldots, y_r]$ is a polynomial ring in $r$ indeterminates.
1839
1840 Then, if $r \geq 1$, we have at least one variable ($k[y_1]$);\\
1841 in $k[y_1]$, the element $y_1$ is not invertible, since $1/y_1$ is a rational function, not a polynomial.\\
1842 (alternatively: the only invertible elements in a polynomial ring over a field are the non-zero constants of the field $k$).
1843
1844 Therefore, $R=k[y_1, \ldots, y_r]$ can be a field iff $r=0$, and we know that is a field, therefore $r=0$ and $R=k$.
1845
1846 By Noether normalization, at the previous step we knew that $K$ is integral over $R$ ($=k$).
1847
1848 \vspace{0.3cm}
1849 By \ref{integral-implies-finite}, a fingen algebra that is integral over its base field is a finite vector space over that field.
1850
1851 Thus $[K:k]<\infty$, and $K$ is algebraic over $k$.
1852\end{proof}
1853
1854\vspace{0.5cm}
1855This brings us to the \emph{Weak Nullstellensatz}:
1856
1857\begin{thm}{4.10}[Weak Nullstellensatz]\label{weak-nullstellensatz-again}
1858 $k$ a field, $K$ a $k$-algebra with
1859 \begin{itemize}
1860 \item $K$ fingen as a $k$-algebra
1861 \item $K$ a field
1862 \end{itemize}
1863 Then $K$ is algebraic over $k$, so that $k \subset K$ is a finite field extension; that is $[K:k] < \infty$.
1864\end{thm}
1865\begin{proof}
1866 proven at previous proof.
1867\end{proof}
1868
1869
1870
1871\vspace{0.8cm}
1872Now, let $m\subset k[X_1, \ldots, X_n]$ a maximal ideal. Let $K=k[X_1, \ldots, X_n]$.
1873
1874Then $\frac{k[X_1, \ldots, X_n]}{m} = \frac{K}{m}$ is a field (by basic ideal quotient theory \ref{maximal-quotient-field}).
1875
1876If it's a fingen $k$-algebra $\Longrightarrow$ by Zariski's lemma, $\frac{K}{m}$ is algebraic over $k$.
1877
1878And since $k$ is algebraically closed $\Longrightarrow~ \frac{K}{m}=k$.
1879
1880By definition of $K$ being a $k$-algebra,
1881$$\exists~ \psi: k \longrightarrow \frac{K}{m} ~~~\Longrightarrow~ k \stackrel{iso}{\cong} \frac{K}{m}$$
1882
1883Thus, $\forall x_i \in k,~ \psi(x_i)=a_i \in k$.
1884And by definition of a quotient ring,
1885
1886$\psi(x_i)$ is the coset
1887\begin{align*}
1888 \psi(x_i)&=x_i+m\\
1889 &\Longrightarrow a_i = x_i +m\\
1890 &x_i - a_i \in m
1891\end{align*}
1892
1893Since $x_i - a_i \in m ~~\Longrightarrow~~ J=(X_1 - a_1, \ldots, X_n -a_n) \subseteq m$,
1894
1895since $J$ is the kernel of the evaluation map at point $(a_1, \ldots, a_n)$, then $J$ is a maximal ideal.
1896
1897(Corollary \ref{5.2}) Thus $J=m$, ie. $m=(X_1 - a_1, \ldots, X_n -a_n)$, with $a_i \in k$.
1898
1899\vspace{0.4cm}
1900We had $k \stackrel{iso}{\cong} \frac{K}{m} = \frac{k[X_1, \ldots, X_n]}{m}$.
1901
1902We have the quotient map
1903\begin{align*}
1904 \phi: k[X_1, \ldots, X_n] &\longrightarrow \frac{k[X_1, \ldots, X_n]}{m} \cong k\\
1905 x_i &\mapsto a_i \in k
1906\end{align*}
1907which is the natural evaluation map $f(X_1, \ldots, X_n) \longrightarrow f(a_1, \ldots, a_n)$.
1908
1909Since $\phi$ is a $k$-algebra homomorphism, then
1910$$\forall~ f \in K,~ \phi(f(X_1, \ldots, X_n)) = f(\phi(x_1), \ldots, \phi(x_n)) = f(a_1, \ldots, a_n)$$
1911
1912Thus, $\exists$ a one-to-one correspondence:
1913\begin{align*}
1914 \text{points in}~ k^n ~~~ &\longleftrightarrow~~~ m-Spec~ A, ~\text{maximal ideals in}~ k[X_1, \ldots, X_n]\\
1915 (a_1, \ldots, a_n) ~~~&\longleftrightarrow~~~ (X_1 - a_1, \ldots, X_n - a_n)
1916\end{align*}
1917
1918\vspace{0.5cm}
1919Observe: $x_i - a_i \in ker(\psi) =m~ \forall~ i \in [n]$.
1920
1921We also have seen that $m=(X_1-a_1, \ldots, X_n-a_n)$.
1922
1923So for $P=(a_1, \ldots, a_n) \in k^n$, it is a zero for every polynomial in $m$, ie.
1924$$\forall~ f \in m ~\Longrightarrow~ f(P)=0$$
1925
1926Since $J \subseteq m$, $P$ is also a zero for every polynomial in $J$; ie. every element of $J$ vanishes at $P$.
1927
1928$\Longrightarrow~ P \in V(J)$, and thus $V(J) \neq \emptyset$. This is what the Nullstellensatz first part states:
1929\begin{thm}{5.6}[Nullstellensatz] \label{nullstellensatz}
1930 \begin{enumerate}[a.]
1931 \item if $J \subsetneq k[X_1, \ldots, X_n]$ then $V(J) \neq \emptyset$
1932 \item $I(V(J)) = rad J$, in other words, for $f \in k[X_1, \ldots, X_n]$,
1933 $$f(P)=0 ~\forall~ P \in V ~~\Longleftrightarrow~ f^n \in J ~\text{for some $n$.}$$
1934 \end{enumerate}
1935\end{thm}
1936\begin{proof}
1937 \begin{enumerate}[a.]
1938 \item already proven in the previous steps.
1939
1940 \item $I(V(J)) = rad J$\\
1941 ie. vanishing ideal of a variety = radical of the ideal defining the variety
1942
1943 where $rad~J = \{ f \in R | f^n \in J ~\text{for some}~ n>0 \}$.
1944
1945 Want to show that if a polynomial vanishes at all points where $g_1, \ldots, g_m$ vanish, then $f \in rad(g_1, \ldots, g_m)$.
1946
1947 Consider the ring $k[X_1, \ldots, X_n, Y]$ and the ideal $J'$ generated by $\{ g_1, \ldots, g_m, 1-Y f \}$
1948
1949 Suppose there is a point $(a_1, \ldots, a_n, a_{n+1})$ that is a zero of $J'$. ie.
1950 $$\exists~ \underbrace{(a_1, \ldots, a_n, a_{n+1})}_{a} \in V(J')$$
1951
1952 Since $g_i(a)=0$, our hypothesis says $f(a)=0$. However, the last generator $(1-Yf)$ requires
1953 $$1 - a_{n+1} f(a) = 0 ~~\Longrightarrow~ \text{implies}~ 1 - a_{n+1} \cdot 0 = 0 ~\Longrightarrow~ 1-0=0$$
1954 a contradiction.
1955
1956 Therefore, $V(J') = \emptyset$.
1957
1958 \vspace{0.4cm}
1959
1960 By the Weak Nullstellensatz (\ref{weak-nullstellensatz-again}),\\
1961 \hspace*{2em} if $V(J')=\emptyset$ then $J'=(1)$, so $1 \in J'=(1)$.
1962
1963 Every element in an ideal is a linear combination of its generators:\\
1964 $J'$ is generated by $\{ g_1, \ldots, g_m, 1-Yf \}$
1965
1966 $$\Longrightarrow~ \forall j \in J',~~ j=(\sum \text{(polynomial)}~ g_i) + \text{(polynomial)} \cdot (1 - Yf)$$
1967
1968 which, since $1 \in J'$,
1969 $$1= \left( \sum_{i=1}^m p_i(X,Y) g_i(X) \right) + q(X,Y) \cdot (1 - Y f(X))$$
1970
1971 substitute $Y=1/f$,
1972 $$1= \left( \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X) \right) + q(X,\frac{1}{f}) \cdot \underbrace{(1 - \frac{1}{f} f(X))}_{0}$$
1973 thus
1974 $$1= \sum_{i=1}^m p_i(X,\frac{1}{f}) g_i(X)$$
1975 multiply by $f^n$,
1976 $$f^n= \sum_{i=1}^m A_i(X) g_i(X)$$
1977
1978 thus $f^n$ is a linear combination of $g_i$.\\
1979 Therefore $f^n \in J$, so $f \in rad~J$.
1980 \end{enumerate}
1981\end{proof}
1982
1983
1984
1985\section{Rings of fractions $S^{-1}A$ and localization}
1986
1987\subsection{Rings of fractions $S^{-1}A$}
1988
1989\begin{defn}{6.1}[ring of fractions] \label{def.6.1}
1990 let $A$ a ring, $S \subset A$ a multiplicative set ($1 \in S$, and $st \in S
1991 ~\forall s, t \in S$).
1992
1993 Introduce th following relation $\sim$ on $A \times S$:
1994 $$(a, s) \sim (b, t) \Longleftrightarrow \exists y \in S ~\text{such that}~ u(at - bs) = 0$$
1995
1996 (write $a/s$ for the equivalence class of $(a, s)$.
1997
1998 Then, the \emph{ring of fractions of $A$ with respect to $S$} is
1999 $$S^{-1}A = (A \times S)/\sim$$
2000
2001 with ring op'ns defined by the usual arithmetic op'ns on fractions:
2002 $$\frac{a}{s} \pm \frac{b}{t} = \frac{(at \pm bs)}{st} ~~\text{and}~~ \frac{a}{s} \cdot \frac{b}{t} = \frac{ab}{st}$$
2003\end{defn}
2004
2005\begin{prop}{6.1} \label{prop.6.1}
2006 \begin{enumerate}[i.]
2007 \item $\sim$ is an equivalence relation
2008 \item the ring op'ns are well defined, and $S^{-1}A$ is a ring
2009 \item
2010 \begin{align*}
2011 \psi: A &\longrightarrow S^{-1}A\\
2012 a &\longmapsto a/1
2013 \end{align*}
2014 is a ring homomorphism.
2015 \end{enumerate}
2016\end{prop}
2017
2018\begin{eg}{ }
2019 TODO
2020\end{eg}
2021
2022\begin{lemma}{6.2} \label{6.2}
2023 For $f \in A$, write $S = \{ 1, f, f^2, \ldots \}$, and $A_f \cong S^{-1}A$.
2024
2025 Then
2026 $$A_f \cong \frac{A[X]}{(Xf-1)}$$
2027\end{lemma}
2028\begin{proof}
2029 Define the homomorphism
2030 \begin{align*}
2031 \psi: A[X] &\longrightarrow A_f\\
2032 a &\longmapsto a/1\\
2033 X &\longmapsto 1/f
2034 \end{align*}
2035
2036By the 1st isomorphism theorem:
2037
2038 \begin{tikzpicture}[node distance=1.5cm, auto]
2039 \node (G) {$A[X]$};
2040 \node (H) [right of=G] {$A_f$};
2041 \node (GmodK) [below of=G, xshift=0.75cm] {$A[X]/\ker(\psi)$};
2042
2043 \draw[->] (G) to node {$\psi$} (H);
2044 \draw[->] (G) to node [swap] {$\phi$} (GmodK);
2045 \draw[->] (GmodK) to node [swap] {$\eta$} (H);
2046 \end{tikzpicture}
2047
2048 Thus we want to prove that $ker(\psi) = (Xf-1)$, so that $\frac{A[X]}{ker(\psi)} = \frac{A[X]}{(Xf-1)}$, and the lemma is proven.
2049
2050 \vspace{0.3cm}
2051 First, observe that $\psi(Xf-1) = \psi(X)\psi(f)-\psi(1) = \frac{1}{f} \frac{f}{1} - 1 = 1-1=0$,
2052 so $Xf-1 \in ker(\psi)$, ie. $(Xf-1) \subseteq ker(\psi)$.
2053
2054 Now, we want to prove that $ker(\psi) \subseteq (Xf-1)$.
2055
2056 Take $h \in ker(\psi)$, will prove that $h \in (Xf-1)~ \forall~ h \in ker(\psi)$, and thus $ker(\psi) \subseteq(Xf-1)$.
2057
2058 \vspace{0.3cm}
2059 Want to prove that $h(X)$ is a multiple of $(Xf-1)$.
2060
2061 Let
2062 $$h(X)=a_n X^n + a_{n-1} X^{n-1} + \ldots + a_1 X + a_0$$
2063
2064 % Since $h \in ker(\psi) ~~\Longrightarrow~~ \psi(h)=0$, so
2065 % $$\psi(h) = \frac{a_n}{f^n} + \frac{a_{n-1}}{f^{n-1}} + \ldots \frac{a_1}{f} + \frac{a_0}{1} = 0 \in A_f$$
2066
2067 multiply $h(X)$ by $f^n$:
2068 $$f^n \cdot h(X)=a_n (f^n X^n) + a_{n-1} f (f^{n-1} X^{n-1}) + a_{n-2} f^2 (f^{n-2} X^{n-2}) \ldots$$
2069
2070 Note that since $\forall~ i \geq 1,~~ f^i X^i = (Xf -1)\cdot (f^{i-1} X^{i-1} + f^{i-2} X^{i-2} + \ldots + 1)$, then $f^i X^i \equiv 1 \pmod{Xf-1}$.
2071
2072 So,
2073 $$f^n \cdot h(X)=\underbrace{a_n (1) + a_{n-1} f (1) + a_{n-2} f^2 (1) + \ldots + a_0 f^n}_{C~~\text{(constant)}} \pmod{Xf-1}$$
2074
2075 $$\Longrightarrow~~ f^n \cdot h(X)=C \pmod{Xf-1}$$
2076 $$\Longleftrightarrow~~ f^n \cdot h(X)=Q(X) \cdot (Xf-1) + C$$
2077
2078 Want to remove $C$, but it is non-zero. Note that in $A_f$ (ring of
2079 fractions), $\frac{a}{f^n} = 0$ iff $\exists~ k$ such that $f^k \cdot a = 0$ in $A$.
2080
2081 So, multiply both sides by $f^k$:
2082 $$f^k \cdot f^n \cdot h(X)=f^k \cdot (Q(X) \cdot (Xf-1) + C)$$
2083 $$\underbrace{f^k f^n}_{f^{nk}} \cdot h(X)=\underbrace{f^k Q(X)}_{Q'(X)} \cdot (Xf-1) + \underbrace{f^k C}_{0}$$
2084
2085 $$\Longrightarrow~~ f^{n+k} \cdot h(X)=Q'(X) \cdot (Xf-1)$$
2086 $$\Longleftrightarrow~~ f^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2087
2088 multiply it by $X^{n+k}$:
2089 $$X^{n+k} \cdot f^{n+k} \cdot h(X) \equiv X^{n+k} \cdot 0 \pmod{Xf-1}$$
2090 $$(Xf)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2091
2092 Now, since we had $Xf \equiv 1$ in $\frac{A[X]}{(Xf-1)}$,
2093 $$(1)^{n+k} \cdot h(X) \equiv 0 \pmod{Xf-1}$$
2094 $$\Longrightarrow~~ h(X) \equiv 0 \pmod{Xf-1}$$
2095 By definition this is saying $h(X) \in (Xf-1) ~\forall~ k \in ker(\psi)$.
2096
2097 Thus $ker(\psi) \subseteq (Xf-1)$.
2098
2099 Initially we saw that $(Xf-1) \subseteq ker(\psi)$. Therefore $ker(\psi)=(Xf-1)$.
2100
2101 Hence,
2102 $$A_f \cong \frac{A[X]}{(Xf-1)}$$
2103\end{proof}
2104
2105
2106\vspace{0.4cm}
2107
2108Given a ring homomorphism $\psi: A \longrightarrow B$, there is a correspondence
2109
2110$$e: \{ \text{ideals of } A \} \longrightarrow \{ \text{ideals of } B \}$$
2111
2112given by $e(I) = \psi(I)B = IB$ (called \emph{extension}),
2113
2114and
2115
2116$$r: \{ \text{ideals of } B \} \longrightarrow \{ \text{ideals of } A \}$$
2117
2118given by $r(J) = \psi^{-1} J$ (called \emph{restriction}, written $A \cap J$).
2119
2120Set $B= S{-1}A$. Then $S^{-1}I = e(I) = \psi(I)B$.
2121
2122\begin{prop}{6.3} \label{prop.6.3}
2123 \begin{enumerate}[a.]
2124 \item $\forall$ ideal $J$ of $S^{-1}A$, $e(r(J))=J$
2125 \item $\forall$ ideal $I$ of $A$,
2126 $$r(e(I)) = \{ a \in A ~|~ as \in I ~\text{for some}~ s \in S \}$$
2127 \item if $P$ prime and $P \cap S = \emptyset$,\\
2128 then $e(P)=S^{-1}P$ is a prime ideal of $S^{-1}A$.
2129 \end{enumerate}
2130\end{prop}
2131\begin{proof}
2132\end{proof}
2133
2134\begin{cor}{6.3} \label{cor.6.3}
2135 for an ideal $I$ of $A$, the necessary and sufficient condition for
2136 $r(e(I))=I$ is
2137 $$as \in I \Longrightarrow a \in I~~ \forall s \in S$$
2138\end{cor}
2139
2140
2141\subsection{Localization}
2142
2143If $P$ prime ideal, then $S=A \setminus P$ is a multiplicative set.
2144Define the set $A_P=S^{-1}A$.
2145
2146\begin{prop}{6.4} \label{6.4}
2147 $a/s \in A_P$ is a unit of $A_P ~\Longleftrightarrow~ a \not\in P$.
2148
2149 Therefore $A_P$ is a \emph{local ring}, with maximal ideal $e(P)=P A_P$.
2150
2151 The local ring $(A_P, PA_P)$ is called the \emph{localization} of $A$ at $P$.
2152\end{prop}
2153\begin{proof}
2154 \begin{enumerate}
2155 \item[$\Longleftarrow$] (if $a \not\in P ~\Longrightarrow~ a/s \in A_P$ is a unit)\\
2156
2157 if $a \not\in P$, then by definition of $S,~~ a \in S$.
2158
2159 In the localization $A_P$, every element of $S$ is invertible.
2160
2161 The inverse of $\frac{a}{s}$ is $\frac{s}{a}$\\
2162 \hspace*{2em}$\longrightarrow~~ \frac{a}{s} \cdot \frac{s}{a} = 1$, thus
2163 $a/s \in A_P$ is a unit.
2164
2165 \item[$\Longrightarrow$] (if $a/s$ unit $~\Longrightarrow~ a \not\in P$)\\
2166
2167 Suppose $a/s$ a unit; then $\exists~ \frac{b}{t} \in A_P$ such that
2168 $\frac{a}{s} \cdot \frac{b}{t} = 1$.
2169
2170 By definition of equality in localization, $\frac{ab}{st}=1$ means $\exists u \in S$ such that
2171 \begin{equation}
2172 u(ab \cdot 1 - st \cdot 1) = 0~~ \Longrightarrow~ uab=ust
2173 \label{eq:6.4}
2174 \end{equation}
2175 with $u,s,t \in S$.
2176
2177 Since $S=A \setminus P$ and $P$ is a prime ideal,\\
2178 the products of elements outside $P$ must also be outside of $P$.
2179 $$u,s,t \not\in P ~\Longrightarrow~ ust \not\in P$$
2180
2181 At \eqref{eq:6.4}, we know that $uab=ust \not\in P$, thus $uab \not\in P$.
2182
2183 If $uab \not\in P$, then $a \not\in P$.
2184
2185
2186 Next we will prove that $A_P$ is a local ring.\\
2187
2188 A ring is local if it has exactly one maimal ideal (\ref{1.13}); so we
2189 show that the set of non-units forms an ideal.
2190
2191 $a/s$ is not a unit iff $a \in P$. Let $m= \{ a/s | a \in P,~ s \not\in P \} = PA_P$.
2192 Want to show that $PA_P$ is the unique maximal ideal:
2193 \begin{itemize}
2194 \item it's an ideal: let $\frac{a}{s}, \frac{b}{t} \in PA_P$ with $a,b \in P$.
2195
2196 Then $\frac{a}{b} + \frac{b}{t} \in PA_P$ with numerator in $P$.
2197
2198 If multiply a fraction in $PA_P$ by any fraction in $A_P$, the numerator stays in $P$, thus $PA_P$ is an ideal.
2199
2200 \item every element outside of $PA_P$ is a unit (proven at the beginning of this proposition's proof ($\Longleftarrow$)).
2201
2202 \item maximality: every ideal strictly larger than $PA_P$ must contain a unit.\\
2203 If an ideal contains a unit, it must be the entire ring $A_P$.\\
2204 \hspace*{2em}$\Longrightarrow$ therefore, $PA_P$ is the unique maximal ideal.
2205 \end{itemize}
2206 \end{enumerate}
2207\end{proof}
2208
2209
2210In other words:
2211
2212\emph{Localization}: formal way to 'force' certain elements to have inverses.
2213
2214It's the smallest ring that contains $A$ and makes all the elements of $S$ invertible.
2215
2216There is a natural map $\psi: A \longrightarrow S^{-1}A$, by $f(a)=\frac{a}{1},~ f(s)=\frac{1}{s}~~ \text{for}~ s \in S$.
2217
2218
2219\subsection{Localization commutes with taking quotients}
2220Let $A$ ring, $S$ multiplicative set, $I$ ideal.
2221
2222Write $T$ for the image of $S$ in $A/I$.
2223
2224Then $S^{-1}I=I \cdot S^{-1}A$ is an ideal of $S^{-1}A$, can take the quotient ring $S^{-1}A / S^{-1} I$.
2225
2226$\longleftrightarrow$ can take the quotient $A/I$ and then localize to get $T^{-1}(A/I)$.
2227
2228
2229\begin{cor}{6.7} \label{6.7}
2230 $$T^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
2231 In particular, for $P$ prime ideal,
2232 $$\frac{A_P}{PA_P} = k(P) = Frac(\underbrace{A/P}_{\substack{\text{integral}\\\text{domain}}})$$
2233
2234 From \ref{6.4}, $A_P$: local ring, $PA_P$: unique maximal ideal.
2235 $k(P)$ and $Frac(A/P)$ are field of fractions.
2236\end{cor}
2237\begin{proof}
2238 the quotient ring $A/I$ can be viewed as an $A$-module and
2239 $$\underbrace{ T^{-1}(A/I) }_{\text{ring of fractions}} \cong
2240 \underbrace{ \frac{S^{-1}A}{S^{-1}I} }_{\text{module of fractions}}$$
2241
2242 The \ref{cor.6.6}.i, gives an isomorphism of modules
2243 $$ T^{-1}(A/I) = S^{-1}(A/I) \cong \frac{S^{-1}A}{S^{-1}I}$$
2244 it's easy to see that this is a ring homomorphism.
2245\end{proof}
2246% TODO 6.6 (proposition and corollary)
2247
2248
2249
2250\newpage
2251
2252\section{Exercises}
2253
2254For the exercises, I follow the assignments listed at \cite{mit-course}.
2255
2256The 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}.
2257
2258\subsection{Exercises Chapter 1}
2259
2260\begin{ex}{R.1.1}
2261 Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$?
2262\end{ex}
2263\begin{proof}
2264 Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$.
2265
2266 $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\
2267 Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$.
2268
2269 Thus $I \cup J$ is not closed under addition; thus is not an ideal.
2270
2271
2272 Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum:
2273
2274 $$I+J = \{ a+b | a \in I, b \in J \}$$
2275
2276 $gcd(2,3)=1$, so $I+J = \mathbb{Z}$.
2277
2278 Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$.
2279\end{proof}
2280
2281\begin{ex}{R.1.5}
2282 let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\
2283 In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$.
2284\end{ex}
2285\begin{proof}
2286 (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$).
2287
2288 Let
2289 $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
2290 The claim is that $\psi^{-1}(P)$ is prime ideal of $A$.
2291
2292 \begin{enumerate}[i.]
2293 \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
2294 $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$).
2295
2296 If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so
2297 $$\psi(a-b)= \psi(a) - \psi(b) \in P$$
2298 hence $a-b \in \psi^{-1}(P)$.
2299
2300 If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\
2301 Thus $ra \in \psi^{-1}(P)$.
2302
2303 $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$.
2304
2305 \item show that $\psi^{-1}(P)$ is prime:\\
2306 $\psi^{-1}(P) \neq A$, 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 $B$.
2307
2308 Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$.
2309
2310 Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\
2311 Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$.
2312
2313 Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$.
2314 \end{enumerate}
2315\end{proof}
2316
2317
2318\begin{ex}{R.1.6}
2319 prove or give a counter example:
2320 \begin{enumerate}[a.]
2321 \item the intersection of two prime ideals is prime
2322 \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
2323 \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
2324 \item the map $\psi^{-1}$ of Proposition 1.2 takes maximal ideals of $A/I$
2325 to maximal ideals of $A$
2326 \end{enumerate}
2327\end{ex}
2328\begin{proof}
2329 \begin{enumerate}[a.]
2330 \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime.
2331
2332 Then $I \cap J = (2) \cap (3) = (6)$.
2333
2334 The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$.
2335
2336 Thus the intersection of two primes can not be prime.
2337
2338 \item $P_1=(2),~ P_2=(3)$, both prime.
2339
2340 Then,
2341 $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$
2342
2343 $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$.
2344
2345 (recall: principal= generated by a single element)
2346
2347 So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$.
2348
2349 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.
2350
2351 Henceforth, the sum of two prime ideals is not necessarily prime.
2352
2353 \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$.
2354
2355 Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$.
2356
2357 Then
2358 \begin{align*}
2359 \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\
2360 \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A
2361 \end{align*}
2362
2363 But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
2364
2365 Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
2366
2367 \item $\psi: A \longrightarrow A/I$ quotient homomorphism, $I \subseteq A$ an ideal.
2368
2369 Let $M$ a maximal ideal of $A/I$, then $\frac{(A/I)}{M}$ is a field (Proposition 1.3).
2370
2371 By the isomorphism theorems,
2372 $$\frac{(A/I)}{M} \cong \frac{A}{\psi^{-1}(M)}$$
2373
2374
2375 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 $A$.
2376
2377 $\Longrightarrow~$ under $\psi$, preimages of maximal ideals are maximal.
2378 \end{enumerate}
2379\end{proof}
2380
2381\begin{ex}{R.1.12.a}
2382 if $I,J$ ideals and $P$ prime ideal, prove that
2383 $$IJ \subset P ~\Longleftrightarrow~ I \cap J \subset P ~\Longleftrightarrow~ I ~\text{or}~ J \subset P$$
2384\end{ex}
2385\begin{proof}
2386 assume $I \subseteq P$ (for $J \subseteq P$ will be the same, symmetric), take $x \in IJ$,
2387
2388 then
2389 $$x = \sum_{k=1}^n a_k b_k$$
2390 with $a_k \in I,~ b_k \in J$.
2391
2392 Each $a_k \in I \subseteq P$. Since $P$ an ideal,
2393 $$\sum_{k=1}^n a_k b_k \in P$$
2394
2395 thus $x \in P$, hence $IJ \subseteq P$.
2396
2397 So $I \subseteq P$ or $J \subseteq P$ $~\Longrightarrow IJ \subseteq P$.
2398
2399 \vspace{0.5cm}
2400 Conversely,\\
2401 assume $P$ prime and $IJ \subseteq P$.
2402
2403 Suppose by contradiction that $I \not\subseteq P$ and $J \not\subseteq P$.
2404 \begin{itemize}
2405 \item[-] since $I \not\subseteq P,~ \exists a \in I$ with $a \not\in P$
2406 \item[-] since $J \not\subseteq P,~ \exists b \in J$ with $b \not\in P$
2407 \end{itemize}
2408 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 $P$.
2409
2410 Thus $I \not\subseteq P$ and $J \not\subseteq P$ are false.
2411
2412 \vspace{0.3cm}
2413 So both directions are proven, hence
2414 $$IJ \subseteq P ~\Longrightarrow~ I \subseteq P ~\text{or}~ J \subseteq P$$
2415
2416\end{proof}
2417
2418\begin{ex}{R.1.18}
2419 Use Zorn's lemma to prove that any prime ideal $P$ contains a minimal prime ideal.
2420\end{ex}
2421\begin{proof}
2422 Let $P$ prime ideal of $R$.
2423
2424 $$S = \{ Q \subseteq R ~|~ Q ~\text{a prime ideal AND}~ Q \subseteq P \}$$
2425
2426 Goal: show that $S$ has a minimal element, the minimal ideal contained in $P$.
2427
2428 $P \subset S$, so $S$ is nonempty.
2429
2430 Let $C \subseteq S$ be a chain (= totally ordered subset) with respect to inclusion.
2431
2432 Define
2433 $$Q_C = \bigcap_{Q \in C} Q$$
2434
2435 Clearly $Q_C \subseteq P$, since each $Q \in C$ is $Q \subseteq P$.
2436
2437 Since $C$ is ordered by inclusion, it is a decreasing chain of prime ideals.
2438
2439 Intersection of a decreasing chain of prime ideals is again a prime ideal:
2440 \begin{itemize}
2441 \item[-] if $ab \in Q_C$, then $ab \in Q ~\forall Q \in C$
2442 \item[-] since $Q$ prime, $\forall Q \in C$ either $a \in Q$ or $b \in Q$
2443 \end{itemize}
2444
2445 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 $Q_2 \subseteq Q_1$.
2446
2447 In either case: contradiction, since the smaller one would have to contain the element that was assumed to be excluded.
2448
2449 Thus $\forall Q \in C$ the same element $a, b$ must lie in all $Q$. $\Longrightarrow~$ lies in the intersection of them, $Q_C$.
2450
2451 Henceforth, $Q_C$ is a prime ideal and lies in $S$, and its a lower bound of $C$ in $S$.
2452
2453 Now, $S$ is nonempty, and every chain in $S$ has a lower bound in $S$ (its intersection).\\
2454 Therefore, $S$ has a minimal element $P_{min}$.
2455
2456 By construction, $P_{min}$ is a prime ideal $P_{min} \subseteq P$, and by minimality there are no strictly smaller prime ideals inside $P$.
2457
2458 So $P_{min}$ is a minimal prime ideal, contained in $P$.
2459\end{proof}
2460
2461\begin{ex}{R.1.10}
2462\end{ex}
2463\begin{proof}
2464\end{proof}
2465
2466\begin{ex}{R.1.11}
2467\end{ex}
2468\begin{proof}
2469\end{proof}
2470
2471\begin{ex}{R.1.4}
2472\end{ex}
2473\begin{proof}
2474\end{proof}
2475
2476\subsection{Exercises Chapter 2}
2477
2478\begin{ex}{R.2.9}
2479$0 \longrightarrow L \stackrel{\alpha}{\longrightarrow} M \stackrel{\beta}{\longrightarrow} N \longrightarrow 0$ is a s.e.s. of $A$-modules. Prove that if $N, L$ are finite over $A$, then $M$ is finite over $A$.
2480\end{ex}
2481\begin{proof}
2482 Denote the generators of $L$ and $N$ respectively as
2483 \begin{align*}
2484 \{l_1, \ldots, l_k \} &\subseteq L\\
2485 \{n_1, \ldots, n_p \} &\subseteq N
2486 \end{align*}
2487
2488 By s.e.s. definition,
2489 \begin{itemize}
2490 \item[-] $\alpha$ is injective (one-to-one), so
2491 $$\forall l_i \in L,~ \exists~ x_i \in M ~\text{s.th.}~ \alpha(l_i)=x_i$$
2492
2493 \item[-] $\beta$ is surjective (onto), so
2494 $$\forall n_j \in N,~ \exists~ y_j \in M ~\text{s.th.}~ \beta(y_j)=n_j$$
2495\end{itemize}
2496
2497 We will show that $\{x_1, \ldots, x_k, y_1, \ldots, y_p \}$ generate $M$, and thus $M$ is finite:
2498
2499 Let $m \in M$, then $\beta(m) \in N$, and
2500 $$\beta(m) = \sum_{j=1}^p a_j n_j ~~~\text{with}~ a_j \in A$$
2501
2502 Take $m' \in M$, with $m' = \sum a_j y_j$, then
2503 $$\beta(m') = \sum a_j \beta(y_j) = \sum a_j n_j = \beta(m)$$
2504
2505 Then, since $\beta(m)=\beta(m')~~ \Longrightarrow~~ \beta(m-m')=0$, thus
2506 $$(m-m') \in ker(\beta)$$
2507
2508 By \emph{exactness} property, since $\alpha: L \longrightarrow ker(\beta)$, we have $ker(\beta)=im(\alpha)$.
2509
2510 Therefore, $\exists~ l \in L$ such that $\alpha(l)= m-m'$.
2511
2512 Since $\{l_i\}_k$ generate $L$,
2513 $$l = \sum^k b_i l_i$$
2514 thus
2515 $$m-m' = \alpha(l) = \alpha(\underbrace{\sum b_i l_i}_{l}) = \sum b_i \underbrace{\alpha(l_i)}_{x_i} = \sum b_i x_i$$
2516
2517 Rearrange,
2518 $$m = m' + \sum b_i x_i = \sum_{j=1}^p a_j y_j + \sum_{i=1}^k b_i x_i ~~~~~ \forall m \in M$$
2519
2520 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.\\
2521 Thus $M$ is finitely generated over $A$.
2522\end{proof}
2523
2524
2525\subsection{Exercises Chapter 3}
2526
2527\begin{ex}{R.3.2}
2528 $K$ a field, $A \supset K$ a ring which is finite dimensional as a $K$-vector space.
2529 Prove that $A$ is Noetherian and Artinian.
2530\end{ex}
2531\begin{proof}
2532 $dim(A)=n < \infty$, so every ideal $\aA$ of $A$ is a $K$-subspace of $A$, because if $x \in \aA$ and $c \in K$, then $c \cdot x \in \aA$.
2533
2534 \begin{enumerate}
2535 \item Noetherian:\\
2536 let $I_1 \subseteq I_2 \subseteq \ldots$ be an ascending chain of ideals in $A$.
2537
2538 Since each $I_i$ is a subspace, we have
2539 $$dim_K(I_1) \leq dim_K(I_2) \leq \ldots \leq n$$
2540 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.
2541
2542 \item Artinian:\\
2543 Similarly, if $I_1 \supseteq I_2 \supseteq \ldots$ a descending chain of ideals in $A$.
2544
2545 then
2546 $$n \geq dim_K(I_1) \geq dim_K(I_2) \geq \ldots \geq 0$$
2547 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.
2548 \end{enumerate}
2549\end{proof}
2550
2551\begin{ex}{R.3.5}
2552 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 $M$.
2553
2554 Prove if the following holds or not:
2555 $$\beta(M_1)=\beta(M_2) ~\text{and}~ \alpha^{-1}(M_1)=\alpha^{-1}(M_2)~~\Longrightarrow~ M_1=M_2$$
2556\end{ex}
2557\begin{proof}
2558 Counterexample showing that it does not hold:
2559
2560 Let $K$ a field, $M = K \oplus K~, L=K,~N=K$.
2561
2562 Set, for $l \in L,~ (m_1, m_2) \in M$,
2563 \begin{align*}
2564 \alpha:~ &l \longmapsto (l, 0)\\
2565 \beta:~ &(m_1, m_2) \longmapsto m_2
2566 \end{align*}
2567
2568 So we have
2569 $$0 \longrightarrow K \stackrel{\alpha}{\longrightarrow} K^2 \stackrel{\beta}{\longrightarrow} K \longrightarrow 0$$
2570
2571 Then,
2572 \begin{align*}
2573 M_1 &= \{ (x, x) ~|~ x \in K\} ~~~~\sim\text{(diagonal line)} \\
2574 M_2 &= \{ (0, x) ~|~ x \in K\} ~~~~\sim\text{(y-axis)}
2575 \end{align*}
2576
2577 (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 $M$).
2578
2579 Then,
2580 \begin{align*}
2581 \beta(M_1) &= \{ x ~|~ x \in K\} = K \\
2582 \beta(M_2) &= \{ x ~|~ x \in K\} = K
2583 \end{align*}
2584 thus, $\beta(M_1) = \beta(M_2)$.
2585
2586 For $M_1,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_1) = \{0\}$,\\
2587 for $M_2,~~ (l,0)\in M$ iff $l=0$, thus $\alpha^{-1}(M_2) = \{0\}$,\\
2588 thus $\alpha^{-1}(M_1)=\alpha^{-1}(M_2)$.
2589
2590
2591 So we've seen that
2592 \begin{align*}
2593 \beta(M_1) = \beta(M_2)\\
2594 \alpha^{-1}(M_1)=\alpha^{-1}(M_2)
2595 \end{align*}
2596 while having $M_1 \neq M_2$.
2597\end{proof}
2598
2599
2600
2601\begin{ex}{R.3.3}
2602Let $A$ a ring, $I_1, \ldots, I_k$ ideals such that each $A/I_i$ is a Noetherian ring.
2603Prove that $\bigoplus A/I_i$ is a Noetherian $A$-module, and deduce that if $\bigcap I_i = 0$ then $A$ is also Noetherian.
2604\end{ex}
2605\begin{proof}
2606\begin{enumerate}[i.]
2607 \item by Corollary \ref{R.3.5} (i), if $M_i$ Noetherian modules, then $\bigoplus M_i$ is Noetherian.
2608 $\Longrightarrow$ thus $\bigoplus A/I_i$ is Noetherian.
2609 \item Take the canonical homomorphism
2610 $$\phi: A \longrightarrow \bigoplus_{i=1}^n A/ I_i$$
2611 by $\phi(a) = (a+I_1, a+I_2, \ldots, a+I_n)$.
2612
2613 $\phi$ is injective: $ker(\phi)= \{ a \in A | a \in I_i \forall i \}$.
2614
2615 Since we're given $\cap I_i = 0$, then $ker(\phi)=\cap I_i$, and $\phi$ is injective.
2616
2617 Thus, $\phi$ is the isomorphism $A \cong im(\phi)$, where $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$.
2618
2619 We know that any submodule of a Noetherian module is Noetherian, thus, since
2620 \begin{itemize}
2621 \item $A/I_i$ is Noetherian by hypothesis of the exercise
2622 \item $A \cong im(\phi)$
2623 \item $im(\phi)$ is an $A$-submodule of $\bigoplus A/I_i$
2624 \end{itemize}
2625 then, $A$ is Noetherian.
2626\end{enumerate}
2627\end{proof}
2628
2629\begin{ex}{R.3.4}
2630 Prove that if A is a Noetherian ring and M a finite A-module, then there
2631 exists an exact sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$.
2632 That is, M has a presentation as an A-module in terms of finitely many generators and relations.
2633\end{ex}
2634\begin{proof}
2635 since $M$ fingen $~\Longrightarrow~$ generators $\{m_1, \ldots, m_2 \} \subseteq M$ span $M$.
2636
2637 Let $\beta$ be a surjective $A$-linear map, which forms a free $A$-module of rank $p$ onto $M$:
2638 \begin{align*}
2639 \beta: A^p &\longrightarrow M\\
2640 (a_1, \ldots, a_p) &\longmapsto \sum_{i=1}^p a_i m_i
2641 \end{align*}
2642
2643 Let $K=ker(\beta)$. By the 1st Isomorphism Theorem,
2644 $$M \cong A^p / K$$
2645
2646 Since $A$ is a Noetherian ring, then every free $A$-module of finite rank (eg. $A^p$) is a Noetherian module.
2647
2648 Every submodule of a Noetherian module is fingen.
2649
2650 $\Longrightarrow~$ since $K \subseteq A^p, ~\Longrightarrow~ K ~~(=ker(\beta))$ is fingen.
2651
2652 Since $K$ fingen, let $\{k_1, \ldots, l_q\}$ be generators of $K$.
2653
2654 Define $\psi: A^q \longrightarrow K$.
2655
2656 Compose it with the inclusion map $i: K \longrightarrow A^p$,
2657 $$\alpha = i \circ \psi:~ A^q \longrightarrow A^p$$
2658
2659 So we have the whole sequence $A^q \stackrel{\alpha}{\longrightarrow} A^p \stackrel{\beta}{\longrightarrow} M \longrightarrow 0$, where
2660 \begin{itemize}
2661 \item $\beta$ is surjective
2662 \item $im(\alpha)=ker(\beta)$
2663 \end{itemize}
2664 so that it is a exact sequence, thus, $M$ has a finite presentation.
2665\end{proof}
2666
2667\subsection{Exercises Chapter 4}
2668
2669\begin{ex}{R.4.1.a}
2670 $k[X^2] \subset k[X]$ is a finite extension, hence integral. Find the integral dependence relation for any $f \in k[X]$.
2671\end{ex}
2672\begin{proof}
2673 $\forall f(X) \in k[X]$ can be uniquely decomposed into its even and odd parts:
2674 $$f(X) = p(X^2) + X \cdot q(X^2)$$
2675
2676 with $p(X^2),~ q(X^2) \in k[X^2]$, and\\
2677 $p(X^2)$: sum of all terms with even exponents\\
2678 $q(X^2)$: sum of all terms with odd exponents, and then factoring out $X$.
2679
2680 \vspace{0.3cm}
2681 (Observation: this is used in FRI cryptographic protocol\\
2682 \href{https://github.com/arnaucube/math/blob/master/notes_fri_stir.pdf}{https://github.com/arnaucube/math/blob/master/notes\_fri\_stir.pdf})
2683
2684
2685 Rearrange it
2686 \begin{align*}
2687 f(X) &- p(X^2) = X \cdot q(X^2), ~~\text{square:}\\
2688 (f(X) &- p(X^2))^2 = X^2 \cdot q(X^2)^2\\
2689 f(X)^2-2 p(X^2) f(X) &+ p(X^2)^2 = X^2 \cdot q(X^2)^2\\
2690 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
2691 \end{align*}
2692
2693 Denote the last polynomial as $P(T) \in k[X^2]$, where $f(X)$ is a root of $P(T)$.
2694
2695 \vspace{0.2cm}
2696 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 $a_i \in k[X^2]$.
2697
2698 We have that
2699 \begin{align*}
2700 a_1 &= -2 p(X^2)\\
2701 a_0 &= p(X^2)^2 - X^2 q(X^2)^2
2702 \end{align*}
2703
2704
2705 So for example, for $f(X)= X^3 + X^2 + X + 1$:
2706 \begin{align*}
2707 f(X) = (X^2 + 1) &+ X (X^2 + 1)\\
2708 (f(X) - (X^2+1))^2 &= X^2 (X^2 + 1)^2\\
2709 (f(X) - p(X))^2 &= X^2 (q(X))^2
2710 \end{align*}
2711\end{proof}
2712
2713
2714\begin{ex}{R.4.5}
2715 Let $A=k[X,Y]/(Y^2 - X^2 - X^3)$. Prove that the normalization of $A$ is $k[t]$ where $t=Y/X$.
2716\end{ex}
2717\begin{proof}
2718$A=k[X,Y]/(Y^2 - X^2 - X^3)$, express $X$ and $Y$ in terms of $t$:
2719
2720Since $t=Y/X$ then $Y=tX$, and combined with $Y^2 = X^2 + X^3$, then
2721\begin{align*}
2722 (tX)^2 &= X^2 + X^3\\
2723 t^2 X^2 &= X^2 + X^3, ~~\text{assuming}~X \neq 0:\\
2724 t^2 &= 1+X, ~\text{thus}\\
2725 X&=t^2-1 \in k[X]
2726\end{align*}
2727
2728Then, $Y=tX=t(t^2-1)=t^3-t \in k[X]$.
2729
2730Hence $X, Y \in k[X]$.
2731
2732Therefore, $k[X,Y]/(Y^2 - X^2 - X^3) \subseteq k[t]$.
2733
2734\vspace{0.4cm}
2735By \ref{noether-normalization} (Noether normalization lemma), to show that $k[t]$ is the \emph{normalization}, must show that $k[t]$ is \emph{integral} over $A$.
2736
2737From $X=t^2-1 ~~\Longrightarrow~~ t^2-1-X=0 ~~\Longrightarrow~ t^2 - (1+X) = 0$.
2738
2739$(1+X) \in A$, so $t$ satisfies the monic polynomial
2740$$P(T) = T^2 - (1+X) \in A[T]$$
2741Thus $t$ is integral over $A$.
2742
2743Since $k[t]$ is generated by $t$ over $k$, and $k \subset A$, then the entire ring $k[t]$ is integral over $A$.
2744
2745Since $k[t]$ is a polynomial ring over a field, which is a UFD, it is integrally closed (since all UFD are integrally closed).
2746
2747$Frac A = k(X,Y)$, since $X=t^2-1, ~Y=t^3-t ~~\Longrightarrow~ k(X,Y) \subseteq k(t)$
2748
2749and $t=Y/X \in k(X,Y)$, thus $k(X,Y)=k(t)$.
2750
2751Since $k[t]$ is integrally closed and is the integral closure of $A$ in its fraction field $k(t)$, we conclude that the normalization of $A$ is $k[t]$.
2752\end{proof}
2753
2754
2755
2756\begin{ex}{R.4.9}
2757 $k$ 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.
2758\end{ex}
2759\begin{proof}
2760 (want to find a linear combination of the coordinates such that the original variables satisfy monic polynomials over the new ring $B$)
2761
2762 The relations defining $A$ are
2763 \begin{align*}
2764 X^2-Y^3-1=0 ~~&\Longrightarrow~ Y^3 = X^2 -1 ~~~(*)\\
2765 XZ -1=0 ~~&\Longrightarrow~ Z= 1/X = X^{-1}
2766 \end{align*}
2767
2768Thus $A$ can be denoted as $A = k[X,Y,X^{-1}]/(Y^3 - X^2 - 1)$.
2769
2770
2771Now, $Y$ is inegral over $k[X]$, since $Y^3 - (X^2 - 1) = 0$ is monic in $Y$ with coefficients in $k[X]$.
2772
2773$Z$ is not integral over $k[X]$, since $Z=1/X$ and $X$ is not a unit in $k[X]$.
2774
2775Choose $\alpha, \beta \in k$ such that $X$ (and thus $Z$) becomes integral over $B$:\\
2776set $\alpha=0,~\beta=1 ~~\Longrightarrow~~ B=k[X+\alpha Y + \beta Z]= k[X+Z]$.
2777
2778Let $w=X+Z$; since $XZ=1$, we have
2779$$w=X+\frac{1}{X} ~~\Longrightarrow~~ Xw=X^2+1 ~~\Longrightarrow~~ X^2 -w X +1 = 0 ~~(**)$$
2780
2781which is monic with coefficients in $k[w]$, thus $X$ is integral over $B$.
2782
2783Since $Z=w-X$, $Z$ is also integral.
2784
2785\vspace{0.4cm}
2786Generators of $A$ as a $B$-module:
2787
2788we hadd $B=k[w]$ with $w=X+Z$.
2789
2790From $(**)$ we have $X^2 - wX + 1=0$, so $X^2 = wX -1$.
2791
2792Thus 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 $\{1, X\}$.
2793
2794Fitting $X^2$ into $(*)$,
2795\begin{align*}
2796 X^2 &- Y^3 -1 =0\\
2797 Y^3 &= X^2-1\\
2798 Y^3 &= wX -2
2799\end{align*}
2800
2801thus any power of $Y$ higher than $2$ can be reduced (eg. $Y^4 = Y(wX-2) = w(XY) - 2Y$).
2802
2803So its partial basis is $\{1, Y, Y^2\}$.
2804
2805For $Z$, since $XZ=1$ and $w=X+Z \Longrightarrow Z=w-X$, thus $Z$ is a $B$-linear combination of $\{1,X\}$.
2806
2807\vspace{0.2cm}
2808Combining the previous partial basis, the generators are
2809$$\{ 1, X \} \times \{ 1,Y, Y^2 \} = \{ 1, Y, Y^2, X, XY, XY^2 \}$$
2810
2811\end{proof}
2812
2813
2814\subsection{Exercises Chapter 6}
2815
2816\begin{ex}{R.6.3.a}
2817 Let $A = A' \times A''$; prove that $A'$ and $A''$ are rings of fractions of $A$.
2818\end{ex}
2819\begin{proof}
2820 ring of fractions of $A$ = $S^{-1}A$, so want to prove that $S^{-1}A \cong A'$.
2821
2822 Localization map
2823 \begin{align*}
2824 \psi: A &\longrightarrow A'\\
2825 (a', a'') &\longmapsto a'
2826 \end{align*}
2827
2828 note that $\psi$ is surjective, since $\forall~ a' \in A',~ \psi(a', 0)=a'$.
2829
2830 Let the multiplicative set $S$ be $S=\{ e_1 \}$ with $e_1 = (1,0)$.
2831
2832 Want $\underbrace{S^{-1}A}_{\text{localization}} \cong A'$.
2833
2834 In $S^{-1}A$, $x \in A$ maps to $0$ iff $s x = 0$ for some $s \in S$.
2835
2836 Let $x=(a', a'')\in A$; then $sx=0 ~\Longrightarrow~ s \cdot (a', a'') =0$,
2837
2838 with $s \in S$, so $s=(1,0)$, hence
2839 $$(1,0)\cdot(a',a'')=(0,0)$$
2840 $$\Longrightarrow~ (a', 0) = (0, 0)$$
2841 \hspace*{4em} which implies $a' = 0$, so that $(a', a'')=(0, a')$.
2842
2843 $\Longrightarrow$ the elements that become zero in the localization are of the form $(0, a'')$
2844
2845 $$ker(\psi) = \{ (a',a'')\in A ~|~ \psi(a', a'')=0 \} = \{(0, a'') ~|~ a'' \in A'' \}$$
2846
2847 By the 1st isomorphism theorem,
2848
2849 \begin{tikzpicture}[node distance=1.5cm, auto]
2850 \node (G) {$A$};
2851 \node (H) [right of=G] {$A'$};
2852 \node (GmodK) [below of=G, xshift=0.75cm] {$\frac{A}{ker(\psi)}$};
2853
2854 \draw[->] (G) to node {$\psi$} (H);
2855 \draw[->] (G) to node [swap] {$\phi$} (GmodK);
2856 \draw[<->] (GmodK) to node [swap] {$\eta$} (H);
2857 \end{tikzpicture}
2858
2859 $\Longrightarrow~ \frac{A}{ker(\psi)} \cong A'$
2860
2861
2862 \vspace{0.5cm}
2863 Take the localization map
2864 \begin{align*}
2865 \phi: A &\longrightarrow S^{-1}A\\
2866 a &\longmapsto a/1
2867 \end{align*}
2868
2869 $$ker(\phi) = \{ x \in A ~|~ sx=0 ~\text{for some}~ s \in S \}$$
2870
2871 and we've seen that $a'=0$, so $x \in A ~\Rightarrow~ x=(a', a'') = (0, a'')$
2872
2873 $$ker(\phi) = \{ (0, a'') ~|~ a'' \in A'' \}$$
2874
2875 which is the same as $ker(\psi)$; $ker(\psi) = ker(\phi)$.
2876
2877 \vspace{0.2cm}
2878 $\Longrightarrow~~ A'$ and $S^{-1}A$ are both surjective images of $A$ with exact same kernel,
2879 thus $A' \cong S^{-1}A$.
2880\end{proof}
2881
2882\begin{ex}{R.6.4}
2883 \begin{enumerate}[a.]
2884 \item Give an example of a ring $A$ and distinct multiplicative sets $S,~ T$ such that $S^{-1}A = T^{-1}A$.
2885 \item Prove that for fixed $S$, there is a maximal multiplicative set $T$ with this property defined by
2886 $$T = \{ t \in A ~|~ at \in S ~\text{for some}~ a \in A \}$$
2887 \end{enumerate}
2888\end{ex}
2889\begin{proof}
2890 \begin{enumerate}[a.]
2891 \item
2892 (\emph{saturated sets})\\
2893
2894 General rule of saturation:\\
2895 two multiplicative sets $S,~T$ yield the same localization, $S^{-1}A = T^{-1}A$, iff the have the same saturation.
2896
2897 Saturation of $S$: $\hat{S}$, set of all elements in $A$ that divide some element of $S$:
2898 $$\hat{S} = \{ a \in A ~|~ \exists~ b \in A ~\text{s.th.}~ ab \in S \}$$
2899
2900 \vspace{0.4cm}
2901 Example 1: $A=\mathbb{Z}$,\\
2902
2903 $$S = \{ 2^n ~|~ n \geq 0 \} = \{ 1, 2, 4, 8, 16, \ldots \}$$
2904 localization:
2905 $$S^{-1} \mathbb{Z} = \{ \frac{a}{2^n} ~|~ a \in \mathbb{Z},~ n \in \mathbb{N} \}$$
2906
2907 $$T = \{ \pm 1, \pm 2, \pm 4, \pm 8, \pm 16, \ldots \}$$
2908
2909 Notice that despite $S \neq T$, we have $S^{-1}\mathbb{Z} = T^{-1} \mathbb{Z}$,\\
2910 since once $n \in \mathbb{Z}$ is invertible, $-n$ is automatically invertible: $(-n)^{-1} = - (n^{-1})$.
2911
2912 \vspace{0.4cm}
2913 Example 2: $A= k[x]$,
2914 $$S = \{ x^n ~|~ n \geq 0 \}$$
2915 $$T = \{ (2x)^n ~|~ n \geq 0 \}$$
2916 $$\Longrightarrow~ S^{-1}A = T^{-1} A = k[x, x^{-1}]$$
2917
2918 \item
2919 Show that $T$ is a multiplicative set:
2920
2921 (must contain $1$ and be closed under multiplication)\\
2922 Since $1 \in A$ and $1 \cdot 1 = 1 \in S$, then $1 \in T$.
2923
2924 Let $t_1,~ t_2 \in T$; by definition of $T$,
2925 $$\exists~ a_1, a_2 \in A ~\text{such that}~ a_1 t_1,~ a_2 t_2 \in S$$
2926
2927 Since $S$ a multiplicative set, $(a_1 t_1)(a_2 t_2) \in S$\\
2928 rearrange it: $(a_1 a_2)(t_1 t_2) \in S$.\\
2929 Since $a_1 a_2 \in A,~~ t_1 t_2 \in T$.
2930
2931 Thus $T$ is a multiplicative set.
2932
2933 Next, we show that $T$ is maximal:
2934
2935 Suppose $U$ is the maximal instead of $T$, such that $U^{-1}A \cong S^{-1}A$.
2936
2937 Let $u \in U$; the image of $u$ in $S^{-1}A$ must be a unit.
2938
2939 Units in $S^{-1}A$ are the fractions $\frac{s}{a}$ such that their inverse exists.\\
2940 \hspace*{2em} $\exists~ x=\frac{a}{s}$ such that $u \frac{a}{s} = \frac{1}{1}$\\
2941 \hspace*{2em} $\Longrightarrow~ \frac{ua}{s} = \frac{1}{1} ~\Longrightarrow~ \exists s' \in S$ such that $s'(ua-s)=0$\\
2942 $\Longrightarrow~ s'ua=s's \in S$
2943
2944 since $s,s' \in S$, their product is also in $S$.
2945
2946 Let $b=s'a$, then $ub \in S$.
2947
2948 By definition of $T$, $u \in T$.
2949
2950 Therefore, $U \subseteq T$; thus $T$ is maximal.
2951\end{enumerate}
2952\end{proof}
2953
2954
2955\bibliographystyle{unsrt}
2956\bibliography{commutative-algebra-notes.bib}
2957
2958\end{document}