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{*}
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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}{AM.1.X}{Zorn's lemma} \label{zorn}
 99  TODO
100\end{thm}
101
102\begin{thm}{AM.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}{AM.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}{AM.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}{AM.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
149Let $A$ be a ring. An $A$-module is an Abelian group $M$ with a multiplication
150map
151\begin{align*}
152  A \times M &\longrightarrow M\\
153  (f, m) &\longmapsto fm
154\end{align*}
155satisfying $\forall~ f,g \in A,~~ m, n \in M$.
156\begin{enumerate}[i.]
157  \item $f(m \pm n)=fm \pm fn$
158  \item $(f \pm g) m = fm \pm gm$
159  \item $(fg) m = f(gm)$
160  \item $1_A m = m$
161\end{enumerate}
162
163Let $\psi: M \longrightarrow M$ an $A$-linear endomorphism of $M$.\\
164$A[\psi] \subset End M$ is the subring geneerated by $A$ and the action of $\psi$.
165\begin{itemize}
166\item since $\psi$ is $A$-linear, $A[\psi]$ is a commutative ring.
167\item $M$ is a module over $A[\psi]$, so $\psi$ beomes multiplication by a ring element.
168\end{itemize}
169
170\subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries}
171
172\begin{prop}{AM.2.4}(Cayley-Hamilton Theorem) \label{2.4}
173  Let $M$ a fingen $A$-module. Let $\aA$ an ideal of $A$, let $\psi$ an
174  $A$-module endomorphism of $M$ such that $\psi(M) \subseteq \aA M$.
175
176  Then $\psi$ satisfies
177  
178  $$\psi^n + a_1 \psi^{n-1} + \ldots + a_{n-1} \psi + a_n = 0$$
179
180  with $a_i \in \aA$.
181\end{prop}
182\begin{proof}
183  Since $M$ fingen, let $\{ x_1, \ldots, x_n \}$ be generators of $M$.\\
184  By hypothesis, $\psi(M) \subseteq \aA M$; so for any generator $x_i$, it's image $\psi(x_i) \in \aA M$.
185
186  Any element in $\aA M$ is a linear combination of the generators with coefficients in the ideal $\aA$, thus
187  $$\psi(x_i)= \sum_{j=1}^n a_{ij} x_j$$
188  with $a_{ij} \in \aA$.
189  
190  Thus, for a module with $n$ generators, we have $n$ different $\psi(x_i)$ equations:
191
192  $$
193  \left.
194  \begin{aligned}
195    \psi(x_1) &= a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n\\
196    \psi(x_2) &= a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n\\
197    \ldots\\
198    \psi(x_n) &= a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n
199  \end{aligned}
200  \right\}
201  \begin{aligned}
202      &\text{n elements $\psi(x_i) \in \aA M$ which}\\
203      &\text{are linear combinations of the}\\
204      &\text{$n$ generators of $M$}
205  \end{aligned}
206  $$
207  
208  Next step: rearrange in order to use matrix algebra.
209
210  Observe that each row equals $0$, and rearranging the elements at each row we get
211
212  \begin{align*}
213    &\psi(x_1) - (a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n) = 0\\
214    &\psi(x_2) - (a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n) = 0\\
215    &\ldots\\
216    &\psi(x_n) - (a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n) = 0
217  \end{align*}
218
219  Then, group the $x_i$ terms together; as example, take the row $i=1$:
220    $$(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0$$
221
222    for $i=2$:
223    $$-a_{2,1} x_1 + (\psi - a_{2,2}) x_2 - \ldots - a_{2,n} x_n = 0$$
224    
225  So, $\forall i \in [n]$, as a matrix:
226  
227  $$
228  \begin{pmatrix}
229    \psi - a_{1,1} & -a_{1,2} & \ldots & -a_{1,n}\\
230    -a_{2,1} & \psi-a_{2,2} & \ldots & -a_{2,n}\\
231    \vdots\\
232    -a_{n,1} & -a_{n,2} & \ldots & \psi-a_{n,n}\\
233  \end{pmatrix}
234  \begin{pmatrix}
235    x_1\\ x_2\\ \vdots\\ x_n
236  \end{pmatrix}
237  =
238  \begin{pmatrix}
239    0\\ 0\\ \vdots\\ 0
240  \end{pmatrix}
241  $$
242
243  Kronecker delta:
244  $\delta_{ij} =
245  \begin{cases}
246    1 & \text{if } i = j,\\
247    0 & \text{otherwise}
248  \end{cases}$
249
250  With the Kronecker delta, $\psi(x_i)$ can be expressed as
251  $$\psi(x_i) = \sum_{j=1}^n \delta_{ij} \psi(x_j)$$
252  so the previous matrix can be characterized as
253  $$\sum_{j=1}^n (\delta_{ij} \psi - a_{ij}) x_j = 0$$
254
255  The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$)
256  \begin{itemize}
257    \item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$
258    \item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$
259  \end{itemize}
260
261  Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$,
262  the ring of operators $A[\psi]$ is a commutative ring.
263
264  $\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant.
265
266  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 $\psi$.\\
267  $\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\
268  $\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$
269
270  $$det(M) \cdot x_i = 0~~ \forall i$$
271  where $x_i$ are the generators of $M$.
272
273  Since $det(M)$ kills every generator, it must kill every element in $M$\\
274  $\Longrightarrow~~ det(M)$ is the zero map.
275
276  Leibniz formula of the determinant of an $n \times n$ matrix:
277  $$
278  det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)}
279  $$
280  
281  so,
282  $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$
283  expanding it,
284  \begin{itemize}
285    \item highest power is $1 \cdot \psi^n$
286    \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\
287      where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$
288    \item the rest of coefficients of $\psi^k$ are also elements in $\aA$
289  \end{itemize}
290
291  So we have
292  $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$
293  with $a_i \in \aA$.
294
295  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:
296  $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$
297  with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem.
298\end{proof}
299
300
301
302\begin{cor}{AM.2.5} \label{2.5}
303  Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$.
304
305  Then, $\exists~ x \equiv 1 \pmod \aA$ such that $xM = 0$.
306\end{cor}
307\begin{proof}
308  take $\psi = \text{identity}$. Then in Cayley-Hamilton (\ref{2.4}):
309  \begin{align*}
310    &\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0\\
311    \Longrightarrow~ &id_M + a_1 id_M + a_2 id_M + \ldots + a_{n-1} id_M + a_n = 0\\
312    \Longrightarrow~ &(1 + a_1 + \ldots + a_n) id_M = 0
313  \end{align*}
314  apply it to $m \in M$, where since $id_M(m)=m$ (by definition of the identity), we then have
315  $$(1 + a_1 + \ldots + a_n) \cdot m = 0$$
316  with $a_i \in \aA$.
317
318  \begin{enumerate}[\text{part} i.]
319    \item $xM=0$:\\
320      Thus the scalar $x = (1 + a_1 + \ldots + a_n)$ annihilates every $m \in M$, ie. the entire module $M$.
321
322    \item $x \equiv 1 \pmod \aA$:\\
323      $x \equiv 1 \pmod \aA ~~ \Longleftrightarrow (x-1) \in \aA$\\
324      then from $x = (1 + \underbrace{a_1 + \ldots + a_n}_b) \in \aA$, set $b=a_1 + \ldots + a_n$,\\
325      so that $x=(1+b) \in \aA$.\\
326
327      Then $x-1 = (1+b)-1 = b \in \aA$\\
328      so $x-1 \in \aA$, thus $x \equiv 1 \pmod \aA$ as stated.
329  \end{enumerate}
330\end{proof}
331
332
333
334\begin{prop}{AM.2.6}{Nakayama's lemma} \label{2.6}
335  Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$.
336
337  Then $\aA M = M$ implies $M=0$.
338\end{prop}
339\begin{proof}
340  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 $\aA \subseteq Jac(A)$).
341
342  By \ref{1.9}, $x$ is a unit in $A$ (thus $x^{-1}\cdot x=1$).
343
344  Hence $M = x^{-1} \cdot \underbrace{x~ \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$.
345
346  Thus, if $\aA M = M$ then $M=0$.
347\end{proof}
348
349
350
351\begin{cor}{AM.2.7} \label{2.7}
352  Let $M$ a fingen $A$-module, let $N \subseteq M$ a submodule of $M$, let $\aA \subseteq Jac(A)$ an ideal.
353
354  Then $M = \aA M + N \stackrel{\text{implies}}{\Longrightarrow} M=N$.
355\end{cor}
356\begin{proof}
357  The idea is to apply Nakayama (\ref{2.6}) to $M/N$.
358
359  Since $M$ fingen $\Longrightarrow~~ M/N$ is fingen and an $A$-module.
360
361  Since $\aA \subseteq Jac(A) ~\Longrightarrow~$ Nakayama applies to $M/N$ too.
362
363  By definition,
364  $$\aA M = \left\{ \sum a_i \cdot m_i ~~|~~ a_i \in \aA, m_i \in M \right\}$$
365  where $m_i$ are the generators of $M$.
366
367  Then, for $M/N$,
368  $$\aA (\frac{M}{N}) = \left\{ \sum a_i \cdot (m_i + N) ~~|~~ a_i \in \aA, m_i \in M \right\}$$
369
370  observe that $a_i(m_i+N)= a_i m_i +N$, thus
371  $$\sum_i a_i \cdot (m_i + N) = \underbrace{(\sum_i a_i \cdot m_i)}_{\in \aA M} + N \in \aA M + N$$
372
373  Hence,
374
375  \begin{equation}
376    \aA (\frac{M}{N}) = \left\{ x + N ~~|~~ x \in \aA M \right\} = \aA M + N
377    \label{eq:2.7.1}
378  \end{equation}
379
380  By definition, if we take $\frac{\aA M + N}{N}$, then
381  $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$
382
383  thus every $y \in \aA M +N$ can be written as
384  $$y=x+n,~~ \text{with} x \in \aA M,~ n\in M$$
385  which comes from \eqref{eq:2.7.1}.
386
387  Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient.
388  
389  Hence, every element of $\frac{\aA M +N}{N}$ has the form
390  $$\frac{\aA M + N}{N} = \left\{ x + N ~~|~~ x \in \aA M \right\}$$
391  as in \eqref{eq:2.7.1}.
392
393  Thus
394  \begin{equation}
395    \aA (\frac{M}{N}) = \aA M + N = \frac{\aA M +N}{N}
396    \label{eq:2.7.2}
397  \end{equation}
398
399  By the Collorary assumption, $M = \aA M + N$; quotient it by $N$:
400  \begin{equation}
401    \frac{M}{N} = \frac{\aA M +N}{N}
402  \label{eq:2.7.3}
403  \end{equation}
404
405  So, from \eqref{eq:2.7.2} and \eqref{eq:2.7.3}:
406  $$\aA (\frac{M}{N}) = \aA M +N = \frac{\aA M +N}{N} = \frac{M}{N}$$
407  thus, $\aA (\frac{M}{N}) = \frac{M}{N}$.
408
409  By Nakayama's lemma \ref{2.6}, if $\aA (\frac{M}{N}) = \frac{M}{N} ~\stackrel{implies}{\Longrightarrow}~ \frac{M}{N}=0$
410
411  Note that
412  $$\frac{M}{N} = \{ m + N ~|~ m \in M \}$$
413  (the zero element in $\frac{M}{N}$ is the coset $N=0+N$)
414
415  Then, $\frac{M}{N}=0$ means that the quotient has exactly one element, the zero coset $N$.
416
417  Thus, every coset $m + N$ equals the zero coset $N$, so $m-0 \in N ~\Longrightarrow~ m \in N$.
418
419  Hence every $m \in M$ lies in $N$, ie. $\forall m \in M,~ m \in N$.
420
421  So $M \subseteq N$. But notice that by the Corollary, we had $N \subseteq M$, therefore $M = N$.
422
423  Thus, if $M = \aA M + N \stackrel{implies}{\Longrightarrow} M = N$.
424\end{proof}
425
426
427
428\begin{prop}{AM.2.8} \label{2.8}
429  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 $M$.
430\end{prop}
431\begin{proof}
432  Let $N$ submodule $M$, generated by the $x_i$.
433
434  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 $N = M$.
435\end{proof}
436
437
438\begin{prop}{AM.2.10} \label{2.10}
439  Split exact sequence. TODO
440\end{prop}
441
442\section{Noetherean rings}
443
444\begin{defn}{}{Ascending Chain Condition}
445  A partially orddered set $\Sigma$ has the \emph{ascending chain condition} (a.c.c.) if every chain
446  $$s_1 \leq s_2 \leq \ldots \leq s_k \leq \ldots$$
447  eventually breaks off, that is, $s_k = s_{k+1} = \ldots$ for some $k$.
448\end{defn}
449
450$\Longrightarrow~ \Sigma$ has the a.c.c. iff every non-empty subset $S \subset \Sigma$ has a maximal element.\\
451\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.
452
453\begin{defn}{R.3.2}{Noetherian ring}
454  Let $A$ a ring; 3 equivalent conditions:
455  \begin{enumerate}[i.]
456    \item the set $\Sigma$ of ideals of $A$ has the a.c.c.; in other words, every increasing chain of ideals
457      $$I_1 \subset I_2 \subset \ldots \subset I_k \subset \ldots$$
458      eventually stops, that is $I_k = I_{k+1}=\ldots$ for some $k$.
459    \item every nonempty set $S$ of iddeals has a maximal element
460    \item every iddeal $I \subset A$ is finitely generated
461  \end{enumerate}
462  If these conditions hold, then $A$ is \emph{Noetherian}.
463\end{defn}
464\begin{proof}
465  TODO
466\end{proof}
467
468\begin{defn}{R.3.4.D}{Noetherian modules}
469  An $A$-module $M$ is Noetherian if the submoles of $M$ have the a.c.c.,\\
470  that is, ay increasing chain
471  $$M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$$
472  of submodules eventually stops.
473\end{defn}
474
475As in with rings, it is equivalent to say that
476\begin{enumerate}[i.]
477  \item any nonempty set of modulesof $M$ has a maximal element
478  \item every submodule of $M$ is finite
479\end{enumerate}
480
481
482
483\begin{prop}{R.3.4.P}
484  Let $0 \longrightarrow L \xrightarrow{\ \alpha \ } M \xrightarrow{\ \beta \ } N \longrightarrow 0$ be a s.e.s. (split exact sequence, \ref{2.10}).
485
486  Then, $M$ is Noetherian $\Longleftrightarrow~ L$ and $N$ are Noetherian.
487\end{prop}
488\begin{proof}
489  $\Longrightarrow$: trivial, since ascending chains of submodules in $L$ and $N$ correspond one-to-one to certain chains in $M$.
490
491  $\Longleftarrow$: suppose $M_1 \subset M_2 \subset \ldots \subset M_k \subset \ldots$ is an increasing chain of submodules of $M$.
492
493  Then identifying $\alpha(L)$ with $L$ and taking intersection gives a chain
494  $$L \cap M_1 \subset L \cap M_2 \subset \ldots \subset L \cap M_k \subset \ldots$$
495  of submodules of $L$, and applying $\beta$ gives a chain
496  $$\beta(M_1) \subset \beta(M_2) \subset \ldots \beta(M_k) \subset \ldots$$
497  of submodules of $N$.
498
499  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.
500\end{proof}
501
502\begin{lemma}{R.3.4.L}
503  for submodules $M_1 \subset M_2 \subset M$,\\
504
505  $L \cap M_1 = L \cap M_2$ and $\beta(M_1) = \beta(M_2) ~\Longrightarrow~ M_1 = M_2$.
506\end{lemma}
507\begin{proof}
508  if $m\in M_2$, then $\beta(m) \in \beta(M_1) = \beta(M_2)$, so that there is an $n \in M_1$ such that $\beta(m)=\beta(m)$.
509
510  Then $\beta(m-n)=0$, so that
511  $$m - n \in M_2 \cap ker(\beta)=M_1 \cap ker(\beta)$$
512
513  Hence $m \in M_1$, thus $M_1 = M_2$.
514\end{proof}
515
516\newpage
517
518\section{Exercises}
519
520For the exercises, I follow the assignements listed at \cite{mit-course}.
521
522The 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}.
523
524\subsection{Exercises Chapter 1}
525
526\begin{ex}{R.1.1}
527  Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$?
528\end{ex}
529\begin{proof}
530  Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$.
531
532  $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\
533  Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$.
534
535  Thus $I \cup J$ is not closed under addition; thus is not an ideal.
536
537
538  Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum:
539
540  $$I+J = \{ a+b | a \in I, b \in J \}$$
541
542  $gcd(2,3)=1$, so $I+J = \mathbb{Z}$.
543
544  Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$.
545\end{proof}
546
547\begin{ex}{R.1.5}
548  let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\
549  In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$.
550\end{ex}
551\begin{proof}
552  (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$).
553
554  Let
555  $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$
556  The claim is that $\psi^{-1}(P)$ is prime iddeal of $A$.
557
558  \begin{enumerate}[i.]
559    \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\
560      $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$).
561
562      If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so
563      $$\psi(a-b)= \psi(a) - \psi(b) \in P$$
564      hence $a-b \in \psi^{-1}(P)$.
565
566      If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\
567      Thus $ra \in \psi^{-1}(P)$.
568
569      $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$.
570
571    \item show that $\psi^{-1}(P)$ is prime:\\
572      $\psi^{-1}(P) \neq A$, 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 $B$.
573
574      Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$.
575
576      Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\
577      Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$.
578
579      Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$.
580  \end{enumerate}
581\end{proof}
582
583
584\begin{ex}{R.1.6}
585  prove or give a counter example:
586  \begin{enumerate}[a.]
587    \item the intersection of two prime ideals is prime
588    \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime
589    \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$
590  \end{enumerate}
591\end{ex}
592\begin{proof}
593  \begin{enumerate}[a.]
594    \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime.
595
596      Then $I \cap J = (2) \cap (3) = (6)$.
597
598      The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$.
599
600      Thus the intersection of two primes can not be prime.
601
602    \item $P_1=(2),~ P_2=(3)$, both prime.
603
604      Then,
605      $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$
606
607      $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$.
608
609      (recall: principal= generated by a single element)
610
611      So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$.
612
613      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.
614
615      Henceforth, the sum of two prime ideals is not necessarily prime.
616
617    \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$.
618
619      Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$.
620
621      Then
622      \begin{align*}
623        \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\
624        \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A
625      \end{align*}
626
627      But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field.
628
629      Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal.
630  \end{enumerate}
631\end{proof}
632
633\subsection{Exercises Chapter 2}
634
635\bibliographystyle{unsrt}
636\bibliography{commutative-algebra-notes.bib}
637
638\end{document}