1\documentclass{article}
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 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
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 73
 74\title{Commutative Algebra notes}
 75\author{arnaucube}
 76\date{}
 77
 78\begin{document}
 79
 80\maketitle
 81
 82\begin{abstract}
 83    Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}.
 84
 85    Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$.
 86
 87    The proofs may slightly differ from the ones from the books, since I try to extend them for a deeper understanding.
 88\end{abstract}
 89
 90\tableofcontents
 91
 92\section{Ideals}
 93
 94\subsection{Definitions}
 95% WIP
 96
 97\subsection{Lemmas, propositions and corollaries}
 98\begin{thm}{1.X}{Zorn's lemma} \label{zorn}
 99  TODO
100\end{thm}
101
102\begin{thm}{1.3} \label{1.3}
103  Every ring $A \neq 0$ has at lleast one maximal ideal.
104\end{thm}
105\begin{proof}
106  By Zorn's lemma \ref{zorn}.
107\end{proof}
108
109\begin{cor}{1.4} \label{1.4}
110  if $I \neq (1)$ an ideal of $A$, $\exists$ a maximal ideal of $A$ containing $I$.
111\end{cor}
112
113\begin{cor}{1.5} \label{1.5}
114  Every non-unit of $A$ is contained in a maximal ideal.
115\end{cor}
116
117\begin{defn}{Jacobson radical}
118  The \emph{Jacobson radical} of a ring $A$ is the intersection of all the maximal ideals of $A$.
119
120  Denoted $Jac(A)$.
121  
122  $Jac(A)$ is an ideal of $A$.
123\end{defn}
124
125\begin{prop}{1.9} \label{1.9}
126  $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$.
127\end{prop}
128\begin{proof}
129  Suppose $1-xy$ not a unit.
130
131  By \ref{1.5}, $1-xy \in \mM$ for $\mM$ some maximal ideal.
132
133  But $x \in Jac(A) \subseteq \mM$, since $Jac(A)$ is the intersection of all maximal ideals of $A$.
134
135  Hence $xy \in \mM$, and therefore $1 \in \mM$, which is absurd, thus $1-xy$ is a unit.
136
137  Conversely:\\
138  Suppose $x \not\in \mM$ for some maximal ideal $\mM$.
139
140  Then $\mM$ and $x$ generte the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$.
141
142  Hence $1 -xy \in \mM$, and is therefore not a unit.
143\end{proof}
144
145\section{Modules}
146
147\subsection{Modules}
148
149\subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries}
150
151\begin{prop}{2.4}(Cayley-Hamilton Theorem) \label{2.4}
152  Let $M$ a fingen $A$-module. Let $\aA$ an ideal of $A$, let $\psi$ an
153  $A$-module endomorphism of $M$ such that $\psi(M) \subseteq \aA M$.
154
155  Then $\psi$ satisfies
156  
157  $$\psi^n + a_1 \psi^{n-1} + \ldots + a_{n-1} \psi + a_n = 0$$
158
159  with $a_i \in \aA$.
160\end{prop}
161\begin{proof}
162  Since $M$ fingen, let $\{ x_1, \ldots, x_n \}$ be generators of $M$.\\
163  By hypothesis, $\psi(M) \subseteq \aA M$; so for any generator $x_i$, it's image $\psi(x_i) \in \aA M$.
164
165  Any element in $\aA M$ is a linear combination of the generators with coefficients in the ideal $\aA$, thus
166  $$\psi(x_i)= \sum_{j=1}^n a_{ij} x_j$$
167  with $a_{ij} \in \aA$.
168  
169  Thus, for a module with $n$ generators, we have $n$ different $\psi(x_i)$ equations:
170
171  $$
172  \left.
173  \begin{aligned}
174    \psi(x_1) &= a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n\\
175    \psi(x_2) &= a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n\\
176    \ldots\\
177    \psi(x_n) &= a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n
178  \end{aligned}
179  \right\}
180  \begin{aligned}
181      &\text{n elements $\psi(x_i) \in \aA M$ which}\\
182      &\text{are linear combinations of the}\\
183      &\text{$n$ generators of $M$}
184  \end{aligned}
185  $$
186  
187  Next step: rearrange in order to use matrix algebra.
188
189  Observe that each row equals $0$, and rearranging the elements at each row we get
190
191  \begin{align*}
192    &\psi(x_1) - (a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n) = 0\\
193    &\psi(x_2) - (a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n) = 0\\
194    &\ldots\\
195    &\psi(x_n) - (a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n) = 0
196  \end{align*}
197
198  Then, group the $x_i$ terms together; as example, take the row $i=1$:
199    $$(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0$$
200
201    for $i=2$:
202    $$-a_{2,1} x_1 + (\psi - a_{2,2}) x_2 - \ldots - a_{2,n} x_n = 0$$
203    
204  So, $\forall i \in [n]$, as a matrix:
205  
206  $$
207  \begin{pmatrix}
208    \psi - a_{1,1} & -a_{1,2} & \ldots & -a_{1,n}\\
209    -a_{2,1} & \psi-a_{2,2} & \ldots & -a_{2,n}\\
210    \vdots\\
211    -a_{n,1} & -a_{n,2} & \ldots & \psi-a_{n,n}\\
212  \end{pmatrix}
213  \begin{pmatrix}
214    x_1\\ x_2\\ \vdots\\ x_n
215  \end{pmatrix}
216  =
217  \begin{pmatrix}
218    0\\ 0\\ \vdots\\ 0
219  \end{pmatrix}
220  $$
221
222  Kronecker delta:
223  $\delta_{ij} =
224  \begin{cases}
225    1 & \text{if } i = j,\\
226    0 & \text{otherwise}
227  \end{cases}$
228
229  With the Kronecker delta, $\psi(x_i)$ can be expressed as
230  $$\psi(x_i) = \sum_{j=1}^n \delta_{ij} \psi(x_j)$$
231  so the previous matrix ccan be characterized as
232  $$\sum_{j=1}^n (\delta_{ij} \psi - a_{ij}) x_j = 0$$
233
234  The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$)
235  \begin{itemize}
236    \item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$
237    \item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$
238  \end{itemize}
239
240  Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$,
241  the ring of operators $A[\psi]$ is a commutative ring.
242
243  $\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant.
244
245  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 $\psi$.\\
246  $\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\
247  $\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$
248
249  $$det(M) \cdot x_i = 0~~ \forall i$$
250  where $x_i$ are the generators of $M$.
251
252  Since $det(M)$ kills every generator, it must kill every element in $M$\\
253  $\Longrightarrow~~ det(M)$ is the zero map.
254
255  Leibniz formula of the determinant of an $n \times n$ matrix:
256  $$
257  det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)}
258  $$
259  
260  so,
261  $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$
262  expanding it,
263  \begin{itemize}
264    \item highest power is $1 \cdot \psi^n$
265    \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\
266      where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$
267    \item the rest of coefficients of $\psi^k$ are also elements in $\aA$
268  \end{itemize}
269
270  So we have
271  $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$
272  with $a_i \in \aA$.
273
274  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:
275  $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$
276  with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem.
277\end{proof}
278
279
280
281\begin{cor}{2.5} \label{2.5}
282  Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$.
283
284  Then, $\exists~ x \equiv 1 \pmod \aA$ such that $xM = 0$.
285\end{cor}
286\begin{proof}
287  take $\psi = \text{identity}$. Then in Cayley-Hamilton (\ref{2.4}):
288  \begin{align*}
289    &\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0\\
290    \Longrightarrow~ &id_M + a_1 id_M + a_2 id_M + \ldots + a_{n-1} id_M + a_n = 0\\
291    \Longrightarrow~ &(1 + a_1 + \ldots + a_n) id_M = 0
292  \end{align*}
293  apply it to $m \in M$, where since $id_M(m)=m$ (by definition of the identity), we then have
294  $$(1 + a_1 + \ldots + a_n) \cdot m = 0$$
295  with $a_i \in \aA$.
296
297  \begin{enumerate}[\text{part} i.]
298    \item $xM=0$:\\
299      Thus the scalar $x = (1 + a_1 + \ldots + a_n)$ annihilates every $m \in M$, ie. the entire module $M$.
300
301    \item $x \equiv 1 \pmod \aA$:\\
302      $x \equiv 1 \pmod \aA ~~ \Longleftrightarrow (x-1) \in \aA$\\
303      then from $x = (1 + \underbrace{a_1 + \ldots + a_n}_b) \in \aA$, set $b=a_1 + \ldots + a_n$,\\
304      so that $x=(1+b) \in \aA$.\\
305
306      Then $x-1 = (1+b)-1 = b \in \aA$\\
307      so $x-1 \in \aA$, thus $x \equiv 1 \pmod \aA$ as stated.
308  \end{enumerate}
309\end{proof}
310
311
312
313\begin{prop}{2.6}{Nakayama's lemma} \label{2.6}
314  Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$.
315
316  Then $\aA M = M$ implies $M=0$.
317\end{prop}
318\begin{proof}
319  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 $\aA \subseteq Jac(A)$).
320
321  By \ref{1.9}, $x$ is a unit in $A$.
322
323  Hence $M = x^{-1} \cdot \underbrace{x \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$.
324
325  Thus, if $\aA M = M$ then $M=0$.
326\end{proof}
327
328
329
330\begin{cor}{2.7} \label{2.7}
331  Let $M$ a fingen $A$-module, let $N \subseteq M$ a submodule of $M$, let $\aA \subseteq Jac(A)$ an ideal.
332
333  Then $M = \aA M + N \stackrel{\text{implies}}{\Longrightarrow} M=N$.
334\end{cor}
335\begin{proof}
336  The idea is to apply Nakayama (\ref{2.6}) to $M/N$.
337
338  Since $M$ fingen $\Longrightarrow~~ M/N$ is fingen and an $A$-module.
339
340  Since $\aA \subseteq Jac(A) ~\Longrightarrow~$ Nakayama applies to $M/N$ too.
341
342  By definition,
343  $$\aA M = \left\{ \sum a_i \cdot m_i ~~|~~ a_i \in \aA, m_i \in M \right\}$$
344  where $m_i$ are the generators of $M$.
345
346  Then, for $M/N$,
347  $$\aA (\frac{M}{N}) = \left\{ \sum a_i \cdot (m_i + N) ~~|~~ a_i \in \aA, m_i \in M \right\}$$
348
349  observe that $a_i(m_i+N)= a_i m_i +N$, thus
350  $$\sum_i a_i \cdot (m_i + N) = \underbrace{(\sum_i a_i \cdot m_i)}_{\in \aA M} + N \in \aA M + N$$
351
352  Hence,
353
354  \begin{equation}
355    \aA (\frac{M}{N}) = \left\{ x + N ~~|~~ x \in \aA M \right\} = \aA M + N
356    \label{eq:2.7.1}
357  \end{equation}
358
359  By definition, if we take $\frac{\aA M + N}{N}$, then
360  $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$
361
362  thus every $y \in \aA M +N$ can be written as
363  $$y=x+n,~~ \text{with} x \in \aA M,~ n\in M$$
364  which comes from \eqref{eq:2.7.1}.
365
366  Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient.
367  
368  Hence, every element of $\frac{\aA M +N}{N}$ has the form
369  $$\frac{\aA M + N}{N} = \left\{ x + N ~~|~~ x \in \aA M \right\}$$
370  as in \eqref{eq:2.7.1}.
371
372  Thus
373  \begin{equation}
374    \aA (\frac{M}{N}) = \aA M + N = \frac{\aA M +N}{N}
375    \label{eq:2.7.2}
376  \end{equation}
377
378  By the Collorary assumption, $M = \aA M + N$; quotient it by $N$:
379  \begin{equation}
380    \frac{M}{N} = \frac{\aA M +N}{N}
381  \label{eq:2.7.3}
382  \end{equation}
383
384  So, from \eqref{eq:2.7.2} and \eqref{eq:2.7.3}:
385  $$\aA (\frac{M}{N}) = \aA M +N = \frac{\aA M +N}{N} = \frac{M}{N}$$
386  thus, $\aA (\frac{M}{N}) = \frac{M}{N}$.
387
388  By Nakayama's lemma \ref{2.6}, if $\aA (\frac{M}{N}) = \frac{M}{N} ~\stackrel{implies}{\Longrightarrow}~ \frac{M}{N}=0$
389
390  Note that
391  $$\frac{M}{N} = \{ m + N ~|~ m \in M \}$$
392  (the zero element in $\frac{M}{N}$ is the coset $N=0+N$)
393
394  Then, $\frac{M}{N}=0$ means that the quotient has exactly one element, the zero coset $N$.
395
396  Thus, every coset $m + N$ equals the zero coset $N$, so $m-0 \in N ~\Longrightarrow~ m \in N$.
397
398  Hence every $m \in M$ lies in $N$, ie. $\forall m \in M,~ m \in N$.
399
400  So $M \subseteq N$. But notice that by the Corollary, we had $N \subseteq M$, therefore $M = N$.
401
402  Thus, if $M = \aA M + N \stackrel{implies}{\Longrightarrow} M = N$.
403\end{proof}
404
405
406
407\begin{prop}{2.8} \label{2.8}
408  Let $x_i \forall i \in [n]$ be elements of $M$ whose images $\frac{M}{m M}$ from a basis of this vecctor space. Then the $x_i$ generate $M$.
409\end{prop}
410\begin{proof}
411  Let $N$ submodule $M$, generated by the $x_i$.
412
413  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 $N = M$.
414\end{proof}
415
416
417\subsection{Noetherean rings}
418
419
420
421
422
423\section{Exercises}
424
425For the exercises, I follow the assignements listed at \cite{mit-course}.
426
427The 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}.
428
429\subsection{Exercises Chapter 1}
430
431\begin{ex}{R.1.1}
432  Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$?
433\end{ex}
434\begin{proof}
435  Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$.
436
437  $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\
438  Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$.
439
440  Thus $I \cup J$ is not closed under addition; thus is not an ideal.
441
442
443  Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum:
444
445  $$I+J = \{ a+b | a \in I, b \in J \}$$
446
447  $gcd(2,3)=1$, so $I+J = \mathbb{Z}$.
448
449  Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$.
450\end{proof}
451
452\begin{ex}{R.1.5}
453  let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\
454  In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$.
455\end{ex}
456\begin{proof}
457  (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$).
458
459  Let
460  $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
461  The claim is that $\psi^{-1}(P)$ is prime iddeal of $A$.
462
463  \begin{enumerate}[i.]
464    \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
465      $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$).
466
467      If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so
468      $$\psi(a-b)= \psi(a) - \psi(b) \in P$$
469      hence $a-b \in \psi^{-1}(P)$.
470
471      If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\
472      Thus $ra \in \psi^{-1}(P)$.
473
474      $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$.
475
476    \item show that $\psi^{-1}(P)$ is prime:\\
477      $\psi^{-1}(P) \neq A$, 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 $B$.
478
479      Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$.
480
481      Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\
482      Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$.
483
484      Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$.
485  \end{enumerate}
486\end{proof}
487
488
489\begin{ex}{R.1.6}
490  prove or give a counter example:
491  \begin{enumerate}[a.]
492    \item the intersection of two prime ideals is prime
493    \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
494    \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
495  \end{enumerate}
496\end{ex}
497\begin{proof}
498  \begin{enumerate}[a.]
499    \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime.
500
501      Then $I \cap J = (2) \cap (3) = (6)$.
502
503      The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$.
504
505      Thus the intersection of two primes can not be prime.
506
507    \item $P_1=(2),~ P_2=(3)$, both prime.
508
509      Then,
510      $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$
511
512      $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$.
513
514      (recall: principal= generated by a single element)
515
516      So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$.
517
518      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.
519
520      Henceforth, the sum of two prime ideals is not necessarily prime.
521
522    \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$.
523
524      Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$.
525
526      Then
527      \begin{align*}
528        \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\
529        \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A
530      \end{align*}
531
532      But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
533
534      Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
535  \end{enumerate}
536\end{proof}
537
538\subsection{Exercises Chapter 2}
539
540\bibliographystyle{unsrt}
541\bibliography{commutative-algebra-notes.bib}
542
543\end{document}