diff --git a/.gitignore b/.gitignore index 552c4ca..d45d44d 100644 --- a/.gitignore +++ b/.gitignore @@ -12,3 +12,4 @@ *.snm *.vrb galois-theory-notes.bib +commutative-algebra-notes.bib diff --git a/README.md b/README.md index bc33bcd..e383c72 100644 --- a/README.md +++ b/README.md @@ -7,6 +7,7 @@ Notes, code and documents done while reading books and papers. - [Notes on "Abstract Algebra" book, by Charles C. Pinter](abstract-algebra-charles-pinter-notes.pdf) - [Notes on Weil pairing](weil-pairing.pdf) - [Notes on Galois Theory](galois-theory-notes.pdf) +- [Notes on Commutative Algebra](commutative-algebra-notes.pdf) In-between math & crypto: diff --git a/commutative-algebra-notes.pdf b/commutative-algebra-notes.pdf new file mode 100644 index 0000000..0ab13d0 Binary files /dev/null and b/commutative-algebra-notes.pdf differ diff --git a/commutative-algebra-notes.tex b/commutative-algebra-notes.tex new file mode 100644 index 0000000..47a1c16 --- /dev/null +++ b/commutative-algebra-notes.tex @@ -0,0 +1,543 @@ +\documentclass{article} +\usepackage[utf8]{inputenc} +\usepackage{amsfonts} +% \usepackage{yfonts} % WIP +\usepackage{amsthm} +\usepackage{amsmath} +\usepackage{enumerate} +\usepackage{hyperref} +\usepackage{amssymb} +\usepackage{tikz} % diagram + +\begin{filecontents}[overwrite]{commutative-algebra-notes.bib} +@misc{am, + author = {M. F. Atiyah and I. G. MacDonald}, + title = {{Introduction to Commutative Algebra}}, + year = {1969} +} +@misc{reid, + author = {Miles Reid}, + title = {{Undergraduate Commutative Algebra}}, + year = {1995} +} +@misc{mit-course, + author = {Steven Kleiman}, + title = {{Commutative Algebra - MIT OpenCourseWare}}, + year = {2008}, + note = {\url{https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/}}, + url = {https://ocw.mit.edu/courses/18-705-commutative-algebra-fall-2008/} +} +\end{filecontents} +\nocite{*} + + +\theoremstyle{definition} + +\newtheorem{innerdefn}{Definition} +\newenvironment{defn}[1] +{\renewcommand\theinnerdefn{#1}\innerdefn} +{\endinnerdefn} + +\newtheorem{innerthm}{Theorem} +\newenvironment{thm}[1] +{\renewcommand\theinnerthm{#1}\innerthm} +{\endinnerthm} + +\newtheorem{innerlemma}{Lemma} +\newenvironment{lemma}[1] +{\renewcommand\theinnerlemma{#1}\innerlemma} +{\endinnerlemma} + +\newtheorem{innerprop}{Proposition} +\newenvironment{prop}[1] +{\renewcommand\theinnerprop{#1}\innerprop} +{\endinnerprop} + +\newtheorem{innercor}{Corollary} +\newenvironment{cor}[1] +{\renewcommand\theinnercor{#1}\innercor} +{\endinnercor} + +\newtheorem{innereg}{Example} +\newenvironment{eg}[1] +{\renewcommand\theinnereg{#1}\innereg} +{\endinnereg} + +\newtheorem{innerex}{Exercise} +\newenvironment{ex}[1] +{\renewcommand\theinnerex{#1}\innerex} +{\endinnerex} + +\newcommand{\aA}{\mathfrak{a}} % TODO: use goth font +\newcommand{\mM}{\mathfrak{m}} + +\title{Commutative Algebra notes} +\author{arnaucube} +\date{} + +\begin{document} + +\maketitle + +\begin{abstract} + Notes taken while studying Commutative Algebra, mostly from Atiyah \& MacDonald book \cite{am} and Reid's book \cite{reid}. + + Usually while reading books and papers I take handwritten notes in a notebook, this document contains some of them re-written to $LaTeX$. + + The proofs may slightly differ from the ones from the books, since I try to extend them for a deeper understanding. +\end{abstract} + +\tableofcontents + +\section{Ideals} + +\subsection{Definitions} +% WIP + +\subsection{Lemmas, propositions and corollaries} +\begin{thm}{1.X}{Zorn's lemma} \label{zorn} + TODO +\end{thm} + +\begin{thm}{1.3} \label{1.3} + Every ring $A \neq 0$ has at lleast one maximal ideal. +\end{thm} +\begin{proof} + By Zorn's lemma \ref{zorn}. +\end{proof} + +\begin{cor}{1.4} \label{1.4} + if $I \neq (1)$ an ideal of $A$, $\exists$ a maximal ideal of $A$ containing $I$. +\end{cor} + +\begin{cor}{1.5} \label{1.5} + Every non-unit of $A$ is contained in a maximal ideal. +\end{cor} + +\begin{defn}{Jacobson radical} + The \emph{Jacobson radical} of a ring $A$ is the intersection of all the maximal ideals of $A$. + + Denoted $Jac(A)$. + + $Jac(A)$ is an ideal of $A$. +\end{defn} + +\begin{prop}{1.9} \label{1.9} + $x \in Jac(A)$ iff $(1 - xy)$ is a unit in $A$, $\forall y \in A$. +\end{prop} +\begin{proof} + Suppose $1-xy$ not a unit. + + By \ref{1.5}, $1-xy \in \mM$ for $\mM$ some maximal ideal. + + But $x \in Jac(A) \subseteq \mM$, since $Jac(A)$ is the intersection of all maximal ideals of $A$. + + Hence $xy \in \mM$, and therefore $1 \in \mM$, which is absurd, thus $1-xy$ is a unit. + + Conversely:\\ + Suppose $x \not\in \mM$ for some maximal ideal $\mM$. + + Then $\mM$ and $x$ generte the unit ideal $(1)$, so that we have $u + xy = 1$ for some $u \in \mM$ and some $y \in A$. + + Hence $1 -xy \in \mM$, and is therefore not a unit. +\end{proof} + +\section{Modules} + +\subsection{Modules} + +\subsection{Cayley-Hamilton theorem, Nakayama lemma, and corollaries} + +\begin{prop}{2.4}(Cayley-Hamilton Theorem) \label{2.4} + Let $M$ a fingen $A$-module. Let $\aA$ an ideal of $A$, let $\psi$ an + $A$-module endomorphism of $M$ such that $\psi(M) \subseteq \aA M$. + + Then $\psi$ satisfies + + $$\psi^n + a_1 \psi^{n-1} + \ldots + a_{n-1} \psi + a_n = 0$$ + + with $a_i \in \aA$. +\end{prop} +\begin{proof} + Since $M$ fingen, let $\{ x_1, \ldots, x_n \}$ be generators of $M$.\\ + By hypothesis, $\psi(M) \subseteq \aA M$; so for any generator $x_i$, it's image $\psi(x_i) \in \aA M$. + + Any element in $\aA M$ is a linear combination of the generators with coefficients in the ideal $\aA$, thus + $$\psi(x_i)= \sum_{j=1}^n a_{ij} x_j$$ + with $a_{ij} \in \aA$. + + Thus, for a module with $n$ generators, we have $n$ different $\psi(x_i)$ equations: + + $$ + \left. + \begin{aligned} + \psi(x_1) &= a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n\\ + \psi(x_2) &= a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n\\ + \ldots\\ + \psi(x_n) &= a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n + \end{aligned} + \right\} + \begin{aligned} + &\text{n elements $\psi(x_i) \in \aA M$ which}\\ + &\text{are linear combinations of the}\\ + &\text{$n$ generators of $M$} + \end{aligned} + $$ + + Next step: rearrange in order to use matrix algebra. + + Observe that each row equals $0$, and rearranging the elements at each row we get + + \begin{align*} + &\psi(x_1) - (a_{1,1} x_1 + a_{1,2} x_2 + \ldots + a_{1,n} x_n) = 0\\ + &\psi(x_2) - (a_{2,1} x_1 + a_{2,2} x_2 + \ldots + a_{2,n} x_n) = 0\\ + &\ldots\\ + &\psi(x_n) - (a_{n,1} x_1 + a_{n,2} x_2 + \ldots + a_{n,n} x_n) = 0 + \end{align*} + + Then, group the $x_i$ terms together; as example, take the row $i=1$: + $$(\psi - a_{1,1})x_1 - a_{1,2} x_2 - \ldots - a_{1,n} x_n = 0$$ + + for $i=2$: + $$-a_{2,1} x_1 + (\psi - a_{2,2}) x_2 - \ldots - a_{2,n} x_n = 0$$ + + So, $\forall i \in [n]$, as a matrix: + + $$ + \begin{pmatrix} + \psi - a_{1,1} & -a_{1,2} & \ldots & -a_{1,n}\\ + -a_{2,1} & \psi-a_{2,2} & \ldots & -a_{2,n}\\ + \vdots\\ + -a_{n,1} & -a_{n,2} & \ldots & \psi-a_{n,n}\\ + \end{pmatrix} + \begin{pmatrix} + x_1\\ x_2\\ \vdots\\ x_n + \end{pmatrix} + = + \begin{pmatrix} + 0\\ 0\\ \vdots\\ 0 + \end{pmatrix} + $$ + + Kronecker delta: + $\delta_{ij} = + \begin{cases} + 1 & \text{if } i = j,\\ + 0 & \text{otherwise} + \end{cases}$ + + With the Kronecker delta, $\psi(x_i)$ can be expressed as + $$\psi(x_i) = \sum_{j=1}^n \delta_{ij} \psi(x_j)$$ + so the previous matrix ccan be characterized as + $$\sum_{j=1}^n (\delta_{ij} \psi - a_{ij}) x_j = 0$$ + + The entries of the matrix are \emph{endomorphisms} (elements of the ring $A[\psi]$) + \begin{itemize} + \item the term $(\psi - a_{11})$ is an operator that acts on $x_1$; as $(\psi(x_1)-a_{11}\cdot x_1)$ + \item the term $(-a_{12})$ is an operator that acts on $x_2$; as multiplication by it, ie. $(-a_{12} \cdot x_2)$ + \end{itemize} + + Since $A$ is a commutative ring, and $\psi$ commutes with any $a \in A$, + the ring of operators $A[\psi]$ is a commutative ring. + + $\Longrightarrow~$ so we can treat the matrix as a matrix of real numbers and calculate its determinant. + + 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 $\psi$.\\ + $\rightarrow$ Because in module theory, we lack of "division", so can not "solve for $\psi$" the system of equations.\\ + $\rightarrow$ The determinant provides a way to find a polynomial that \emph{annihilates} the module; the \emph{characteristic polynomial}, which related $\psi$ to the ideal $\aA$ + + $$det(M) \cdot x_i = 0~~ \forall i$$ + where $x_i$ are the generators of $M$. + + Since $det(M)$ kills every generator, it must kill every element in $M$\\ + $\Longrightarrow~~ det(M)$ is the zero map. + + Leibniz formula of the determinant of an $n \times n$ matrix: + $$ + det(M) = \sum_{\sigma \in S_n} sign(\sigma) \prod_{i=1}^n M_{i, \sigma(i)} + $$ + + so, + $$(\psi - a_{11}) (\psi - a_{22}) \ldots (\psi - a_{nn})$$ + expanding it, + \begin{itemize} + \item highest power is $1 \cdot \psi^n$ + \item coefficient of $\psi^{n-1}$ is $-( \underbrace{ a_{11} + a_{22} + \ldots + a_{nn} }_{a_1})$,\\ + where, since each $a_{ii} \in \aA,~~ a_1 \in \aA$ + \item the rest of coefficients of $\psi^k$ are also elements in $\aA$ + \end{itemize} + + So we have + $$p(\psi) = \psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n$$ + with $a_i \in \aA$. + + 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: + $$\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0$$ + with $a_i \in \aA$, as stated in the Cayley-Hamilton theorem. +\end{proof} + + + +\begin{cor}{2.5} \label{2.5} + Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA M = M$. + + Then, $\exists~ x \equiv 1 \pmod \aA$ such that $xM = 0$. +\end{cor} +\begin{proof} + take $\psi = \text{identity}$. Then in Cayley-Hamilton (\ref{2.4}): + \begin{align*} + &\psi^n + a_1 \psi^{n-1} + a_2 \psi^{n-2} + \ldots + a_{n-1} \psi + a_n = 0\\ + \Longrightarrow~ &id_M + a_1 id_M + a_2 id_M + \ldots + a_{n-1} id_M + a_n = 0\\ + \Longrightarrow~ &(1 + a_1 + \ldots + a_n) id_M = 0 + \end{align*} + apply it to $m \in M$, where since $id_M(m)=m$ (by definition of the identity), we then have + $$(1 + a_1 + \ldots + a_n) \cdot m = 0$$ + with $a_i \in \aA$. + + \begin{enumerate}[\text{part} i.] + \item $xM=0$:\\ + Thus the scalar $x = (1 + a_1 + \ldots + a_n)$ annihilates every $m \in M$, ie. the entire module $M$. + + \item $x \equiv 1 \pmod \aA$:\\ + $x \equiv 1 \pmod \aA ~~ \Longleftrightarrow (x-1) \in \aA$\\ + then from $x = (1 + \underbrace{a_1 + \ldots + a_n}_b) \in \aA$, set $b=a_1 + \ldots + a_n$,\\ + so that $x=(1+b) \in \aA$.\\ + + Then $x-1 = (1+b)-1 = b \in \aA$\\ + so $x-1 \in \aA$, thus $x \equiv 1 \pmod \aA$ as stated. + \end{enumerate} +\end{proof} + + + +\begin{prop}{2.6}{Nakayama's lemma} \label{2.6} + Let $M$ a fingen $A$-module, let $\aA$ an ideal of $A$ such that $\aA \subseteq Jac(A)$. + + Then $\aA M = M$ implies $M=0$. +\end{prop} +\begin{proof} + 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 $\aA \subseteq Jac(A)$). + + By \ref{1.9}, $x$ is a unit in $A$. + + Hence $M = x^{-1} \cdot \underbrace{x \cdot M}_{=0~ \text{(by \ref{2.5})}} = 0$. + + Thus, if $\aA M = M$ then $M=0$. +\end{proof} + + + +\begin{cor}{2.7} \label{2.7} + Let $M$ a fingen $A$-module, let $N \subseteq M$ a submodule of $M$, let $\aA \subseteq Jac(A)$ an ideal. + + Then $M = \aA M + N \stackrel{\text{implies}}{\Longrightarrow} M=N$. +\end{cor} +\begin{proof} + The idea is to apply Nakayama (\ref{2.6}) to $M/N$. + + Since $M$ fingen $\Longrightarrow~~ M/N$ is fingen and an $A$-module. + + Since $\aA \subseteq Jac(A) ~\Longrightarrow~$ Nakayama applies to $M/N$ too. + + By definition, + $$\aA M = \left\{ \sum a_i \cdot m_i ~~|~~ a_i \in \aA, m_i \in M \right\}$$ + where $m_i$ are the generators of $M$. + + Then, for $M/N$, + $$\aA (\frac{M}{N}) = \left\{ \sum a_i \cdot (m_i + N) ~~|~~ a_i \in \aA, m_i \in M \right\}$$ + + observe that $a_i(m_i+N)= a_i m_i +N$, thus + $$\sum_i a_i \cdot (m_i + N) = \underbrace{(\sum_i a_i \cdot m_i)}_{\in \aA M} + N \in \aA M + N$$ + + Hence, + + \begin{equation} + \aA (\frac{M}{N}) = \left\{ x + N ~~|~~ x \in \aA M \right\} = \aA M + N + \label{eq:2.7.1} + \end{equation} + + By definition, if we take $\frac{\aA M + N}{N}$, then + $$\frac{\aA M + N}{N} = \left\{ y + N ~~|~~ y \in \aA M +N \right\} = \aA M + N$$ + + thus every $y \in \aA M +N$ can be written as + $$y=x+n,~~ \text{with} x \in \aA M,~ n\in M$$ + which comes from \eqref{eq:2.7.1}. + + Thus, $y + N = (x+n)+N = x+N$, since $n \in N$ is zero in the quotient. + + Hence, every element of $\frac{\aA M +N}{N}$ has the form + $$\frac{\aA M + N}{N} = \left\{ x + N ~~|~~ x \in \aA M \right\}$$ + as in \eqref{eq:2.7.1}. + + Thus + \begin{equation} + \aA (\frac{M}{N}) = \aA M + N = \frac{\aA M +N}{N} + \label{eq:2.7.2} + \end{equation} + + By the Collorary assumption, $M = \aA M + N$; quotient it by $N$: + \begin{equation} + \frac{M}{N} = \frac{\aA M +N}{N} + \label{eq:2.7.3} + \end{equation} + + So, from \eqref{eq:2.7.2} and \eqref{eq:2.7.3}: + $$\aA (\frac{M}{N}) = \aA M +N = \frac{\aA M +N}{N} = \frac{M}{N}$$ + thus, $\aA (\frac{M}{N}) = \frac{M}{N}$. + + By Nakayama's lemma \ref{2.6}, if $\aA (\frac{M}{N}) = \frac{M}{N} ~\stackrel{implies}{\Longrightarrow}~ \frac{M}{N}=0$ + + Note that + $$\frac{M}{N} = \{ m + N ~|~ m \in M \}$$ + (the zero element in $\frac{M}{N}$ is the coset $N=0+N$) + + Then, $\frac{M}{N}=0$ means that the quotient has exactly one element, the zero coset $N$. + + Thus, every coset $m + N$ equals the zero coset $N$, so $m-0 \in N ~\Longrightarrow~ m \in N$. + + Hence every $m \in M$ lies in $N$, ie. $\forall m \in M,~ m \in N$. + + So $M \subseteq N$. But notice that by the Corollary, we had $N \subseteq M$, therefore $M = N$. + + Thus, if $M = \aA M + N \stackrel{implies}{\Longrightarrow} M = N$. +\end{proof} + + + +\begin{prop}{2.8} \label{2.8} + 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 $M$. +\end{prop} +\begin{proof} + Let $N$ submodule $M$, generated by the $x_i$. + + 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 $N = M$. +\end{proof} + + +\subsection{Noetherean rings} + + + + + +\section{Exercises} + +For the exercises, I follow the assignements listed at \cite{mit-course}. + +The 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}. + +\subsection{Exercises Chapter 1} + +\begin{ex}{R.1.1} + Ring $A$ and ideals $I, J$ such that $I \cup J$ is not an ideal. What's the smallest ideal containing $I$ and $J$? +\end{ex} +\begin{proof} + Take ring $A= \mathbb{Z}$. Set $I = 2 \mathbb{Z},~ J=3 \mathbb{Z}$. + + $I,~J$ are ideals of $A$ ($=\mathbb{Z}$). And $I \cup J = 2 \mathbb{Z} \cup 3 \mathbb{Z}$.\\ + Observe that for $2 \in I,~ 3 \in J ~\Longrightarrow~ 2,3 \in I \cup J$, but $2+3 = 5 \not\in I \cup J$. + + Thus $I \cup J$ is not closed under addition; thus is not an ideal. + + + Smallest ideal of $\mathbb{Z}$ ($=A$) containing $I$ and $J$ is their sum: + + $$I+J = \{ a+b | a \in I, b \in J \}$$ + + $gcd(2,3)=1$, so $I+J = \mathbb{Z}$. + + Therefore, smallest ideal containing $I$ and $J$ is the whole ring $\mathbb{Z}$. +\end{proof} + +\begin{ex}{R.1.5} + let $\psi: A \longrightarrow B$ a ring homomorphism. Prove that $\psi^{-1}$ takes prime ideals of $B$ to prime ideals of $A$.\\ + In particular if $A \subset B$ and $P$ a prime ideal of $B$, then $A \cap P$ is a prime ideal of $A$. +\end{ex} +\begin{proof} + (Recall: prime ideal is if $a,b \in R$ and $a \cdot b \in P$ (with $R \neq P$), implies $a \in P$ or $b \in P$). + + Let + $$\psi^{-1}(P) = \{ a \in A | \psi(a) \in P \} = A \cap P$$ + The claim is that $\psi^{-1}(P)$ is prime iddeal of $A$. + + \begin{enumerate}[i.] + \item show that $\psi^{-1}(P)$ is an ideal of $A$:\\ + $0_A \in \psi^{-1}(P)$, since $\psi(0_A)=0_B \in P$ (since every ideal contains $0$). + + If $a,b \in \psi^{-1}(P)$, then $\psi(a), \psi(b) \in P$, so + $$\psi(a-b)= \psi(a) - \psi(b) \in P$$ + hence $a-b \in \psi^{-1}(P)$. + + If $a \in \psi^{-1}(P)$ and $r \in A$, then $\psi(ra) = \psi(r) \psi(a) \in P$, since $P$ is an ideal.\\ + Thus $ra \in \psi^{-1}(P)$. + + $\Longrightarrow$ so $\psi^{-1}$ is an ideal of $A$. + + \item show that $\psi^{-1}(P)$ is prime:\\ + $\psi^{-1}(P) \neq A$, 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 $B$. + + Take $a,b \in A$ with $ab \in \psi^{-1}(P)$; then $\psi(ab) \in P$, and since $\psi$ is a ring homomorphism, $\psi(ab) = \psi(a)\psi(b)$. + + Since $P$ prime ideal, then $\psi(a)\psi(b) \in P$ implies either $\psi(a) \in P$ or $\psi(b) \in P$.\\ + Thus $a \in \psi^{-1}(P)$ or $b \in \psi^{-1}(P)$. + + Hence $\psi^{-1}(P)$ ($=A \cap P$) is a prime ideal of $A$. + \end{enumerate} +\end{proof} + + +\begin{ex}{R.1.6} + prove or give a counter example: + \begin{enumerate}[a.] + \item the intersection of two prime ideals is prime + \item the ideal $P_1+P_2$ generated by $2$ prime ideals $P_1,P_2$ is prime + \item if $\psi: A \longrightarrow B$ ring homomorphism, then $\psi^{-1}$ takes maximal ideals of $B$ to maximal ideals of $A$ + \end{enumerate} +\end{ex} +\begin{proof} + \begin{enumerate}[a.] + \item let $I = 2 \mathbb{Z} = (2)$, $J = 3 \mathbb{Z} = (3)$ be ideals of $\mathbb{Z}$, both prime. + + Then $I \cap J = (2) \cap (3) = (6)$. + + The ideal $(6)$ is not prime in $\mathbb{Z}$, since $2 \cdot 3 \in (6)$, but $2 \neq (6)$ and $3 \neq (6)$. + + Thus the intersection of two primes can not be prime. + + \item $P_1=(2),~ P_2=(3)$, both prime. + + Then, + $$P_1 + P_2 = (2)+(3)=\{ a+b | a \in P_1, b \in P_2 \}$$ + + $\longrightarrow~$ in a principal ideal domain (like $\mathbb{Z}$), the sum of two principal ideals is again principal, and given by $(m)+(n)=(gcd(m,n))$. + + (recall: principal= generated by a single element) + + So, $P_1+P_2= (2)+(3) = (gcd(2,3))=(1)=\mathbb{Z}$. + + 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. + + Henceforth, the sum of two prime ideals is not necessarily prime. + + \item let $A=\mathbb{Z},~ B=\mathbb{Q},~ \psi: A \longrightarrow B$. + + Since $\mathbb{Q}$ is a field, its only maximal ideal is $(0)$. + + Then + \begin{align*} + \psi^{-1}( (0) ) &= (0) \subset \mathbb{Z}\\ + \text{ie.}~ \psi^{-1}( m_B ) &= (m_B) \subset A + \end{align*} + + But $(0)$ is not maximal in $\mathbb{Z}$, because $\mathbb{Z}/(0) \cong \mathbb{Z}$ is not a field. + + Thus the preimages of maximal ideals under arbitrary ring homomorphisms need not be maximal. + \end{enumerate} +\end{proof} + +\subsection{Exercises Chapter 2} + +\bibliographystyle{unsrt} +\bibliography{commutative-algebra-notes.bib} + +\end{document}