\documentclass[11pt]{article}
\usepackage{amsmath,amssymb,amsthm,amsfonts,verbatim}
\usepackage{enumitem}
\usepackage{microtype}
\usepackage{url}
\usepackage{hyperref}
\usepackage[margin=1in]{geometry}
\usepackage{xypic}
\theoremstyle{definition}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{fact}[theorem]{Fact}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{question}{Question}
\newtheorem{innerquestionst}{Question}
\newenvironment{questionst}{\renewcommand\theinnerquestionst{{\textrm \thequestion\st}}
\addtocounter{question}{1}\innerquestionst}{\endinnerquestionst}


\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{challenge}[question]{Challenge Problem}

\title{\vspace{-50pt}Math 210A: Modern Algebra\\
\large Thomas Church ({\tt tfchurch@stanford.edu})\\
\href{http://math.stanford.edu/~church/teaching/210A-F17/}{\nolinkurl{http://math.stanford.edu/~church/teaching/210A-F17}}}
\author{}
\date{}

\newcommand{\Q}{\mathbb{Q}}
\newcommand{\R}{\mathbb{R}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\N}{\mathbb{N}}
\newcommand{\C}{\mathbb{C}}
\newcommand{\F}{\mathbf{F}}
\renewcommand{\L}{\mathcal{L}}
\renewcommand{\P}{\mathcal{P}}
\newcommand{\abs}[1]{\left\vert #1\right\vert}
\renewcommand{\phi}{\varphi}
\newcommand{\m}{\mathfrak{m}}


\DeclareMathOperator{\im}{range}  

\newcommand{\iso}{\cong}

\newcommand{\spc}{\ \  \ \ \ }
\newcommand{\st}{$^{\ast}$}

\renewcommand{\baselinestretch}{1.1}
\begin{document}
\maketitle
\vspace{-50pt}
\begin{center}
{\huge \bf Homework 2}\\\vspace{8pt}
{\Large Due Thursday night, October 5} {\footnotesize(technically 5am Oct.\ 6)}\\
\end{center}

\vspace{28pt}

\noindent Note that Q\ref{S1complex} says ``optional, replaces Q\ref{S1real}''. This means that this question is optional (really!); however, if you write up and submit this optional question Q\ref{S1complex}, you do not have to submit Q\ref{S1real}.\\[22pt]

\noindent {\small Recall that:\begin{itemize}
\item a domain is a commutative ring in which $xy=0\implies (x=0\text{ or }y=0)$;
\item a prime ideal $P\subsetneq R$ is an ideal for which $xy\in P\implies(x\in P\text{ or }y\in P)$ \newline (or equivalently, for which the quotient $R/P$ is a domain); and
\item a maximal ideal $\m\subsetneq R$ is one for which the quotient ring $R/\m$ is a field.
\end{itemize}}\vspace{12pt}
\begin{question}
Let $R$ be a commutative ring, and let $I=\{r\in R\,|\, \exists k>0\text{ such that } r^k=0\}$.
\begin{enumerate}[label=(\alph*)]
\item Prove  $I$ is an ideal.
\item Prove $I$ is the intersection of all the prime ideals of $R$.\newline \small You may use without proof the following fact, a consequence of Zorn's lemma: if $S$ is a subset of $R$ satisfying $0\notin S$ and $S\cdot S\subset S$, then the set of ideals $J\subset R$ for which $J\cap S=\emptyset$ has a maximal element.
\end{enumerate}
\end{question}

\vspace{12pt}
\begin{question}
\label{S1real}
Let $R=C^0([0,1])$ be the ring of real-valued continuous functions on the closed interval $[0,1]$.
For every point $p\in [0,1]$, we obtain a maximal ideal $\m_p=\{f\in R\,|\, f(p)=0\}$.

Prove that \emph{every} maximal ideal of $R$ is of the form $\m_p$ for a unique $p\in [0,1]$.\\[-12pt]

\noindent(Hint: You may wish to recall that $[0,1]$ is compact, which means that for any collection of open intervals covering it, there is some finite subcollection that still covers it.)\\[10pt]



\noindent Note that this means that you can recover the set $[0,1]$ just from the \emph{ring} $R$.\newline (This actually works for any compact Hausdorff space, not just $[0,1]$; the proof is the same.)\\[8pt] {\small (Optional, to think about: can you also recover the \emph{topology} on $[0,1]$ from the ring $R$?)}
\end{question}
\vfill
\phantom{x}\hfill(cont)
\newpage
\pagebreak

\begin{question}[optional, replaces Q\ref{S1real}]
\label{S1complex}
[This question is very hard, \textbf{100\% optional}, and cannot be done without material from outside this course.]

Let $R=C^\infty(S^1;\C)$ be the ring of complex-valued smooth functions on the circle $S^1$, which for concreteness I will realize as smooth 1-periodic functions on $\R$:
\[R\iso \{f\in C^\infty(\R;\C)\,|\,f(x+1)=f(x)\}.\]
The proof of Q\ref{S1real} applies in exactly the same way to $R$, showing that every maximal ideal of $R$ is of the form $\m_p=\{f\in R\,|\,f(p)=0\}$ for a unique $p\in [0,1)\approx S^1$; you do not have to prove this.\newline
{\footnotesize (The complex-valued vs real-valued is not an important point, it just simplifies the following.)}\\

For any $f\in R$ we can define complex numbers $a_n\in \C$ for all $n\in \Z$ by $a_n=\int_0^1 f(x)e^{-2\pi i n x}\,dx$.\newline (Remark: It is a fact that these numbers decay rapidly as $n\to \infty$,\newline
in the sense that for all $k\geq 0$ we have $n^k \abs{a_n}\to 0$ and $n^k \abs{a_{-n}}\to 0$ as $n\to +\infty$.)

Let $S\subset R$ be the subring consisting of those functions for which $a_{-1}=a_{-2}=\cdots=0$, i.e.
\[S=\left\{f\in R\,\left\vert \int_0^1 f(x)e^{-2\pi i n x}\,dx = 0\text{ for all } n<0\,\right.\right\}\]
{\small (You do not have to prove that $S$ is a subring of $R$, though you might benefit from thinking about why it is.)}\newline
For every $p\in [0,1)$, we still have a maximal ideal $\m_p\subset S$ given by $\m_p=\{f\in S\,|\,f(p)=0\}$.\\[10pt]

Exhibit a maximal ideal of $S$ that is \emph{not} of this form, and ideally exhibit \emph{two} such maximal ideals. {\small (If you really want a challenge: can you classify \emph{all} maximal ideals of $S$? Warning: I do not know that this is possible using things you know. But you could at least come up with a guess, even if you can't completely prove it's correct.)}
\end{question}
\vfill

\begin{question}
Let $a\in \Z$ and $b\in \Z$ be coprime. Let $C$ be any abelian group, and let \[f\colon (\Z/a\Z)\times (\Z/b\Z)\to C\] be a $\Z$-bilinear map. Prove that $f=0$.
\end{question}
\vfill

\begin{question}
Let $N$ be a submodule of the $R$-module $M$. Prove that if $N$ is finitely generated and $M/N$ is finitely generated, then $M$ is finitely generated.
\end{question}
\vfill

\begin{question}
Let $M$ be a finitely generated $R$-module. Let $\pi\colon M\twoheadrightarrow R^n$ be a surjective homomorphism, and let $K=\ker(\pi)$. Prove that the $R$-module $K$ is finitely generated.
\end{question}

{\small Localization of modules, which features in the next question, will be covered on Monday.}
\vspace{-8pt}
\begin{question}
Let $R$ be a commutative ring, and let $S\subset R$ be a multiplicative set ($S\cdot S\subset S$). Let $M$ be a finitely generated $R$-module. Prove that the localization $S^{-1}M$ satisfies
\[S^{-1}M=0\qquad\qquad\iff\qquad\qquad \exists s\in S\text{ with }s\cdot M=0.\]
\end{question}

\vfill
\phantom{x}\hfill(cont)
\newpage




\begin{question}
Let $f\colon X\to Y$ be a homomorphism of $R$-modules.
\begin{enumerate}[label=(\alph*),wide]
\item Consider all pairs $(A,\alpha)$ of an $R$-module $A$ and a homomorphism $\alpha\colon A\to X$ with $f\circ \alpha=0$.
\[\xymatrix{A \ar[r]^{\alpha} \ar[dr]_{0}& X\ar[d]^{f} \\ & Y}\]

 You will prove that these exists a ``universal'' such pair. Specifically, you must construct\newline some $(M,\mu\colon M\to X)$ with $f\circ \mu=0$ with the property that:\newline for any $(A,\alpha\colon A\to X)$ with $f\circ \alpha=0$, there exists a unique $a\colon A\to M$ such that $\alpha=\mu\circ a$. 
%\[\xymatrix{A \ar[r]^{\alpha} \ar[dr]_{0}& X\ar[d]^{f} & & X\ar[dr]^{0}\ar[d]_{f} &\\ & Y&  & Y\ar[r]_{\beta} & B }\]

\[\xymatrix{A \ar[r]_{a}\ar[ddrr]_0 \ar@/^1pc/[rr]^{\alpha}& M \ar[ddr]^0\ar[r]_\mu & X\ar[dd] \\ && \\ & & Y}\]
{\small (We might abbreviate this property as saying roughly: ``Every $\alpha$ with $f\circ \alpha=0$ factors uniquely through $M$''.)}\\[15pt]

\item On the other side, consider all pairs $(B,\beta\colon Y\to B)$ with $\beta\circ f=0$.
\[\xymatrix{X \ar[d]_{f} \ar[dr]^{0}& \\Y\ar[r]_{\beta}  & B}\]
Prove that these exists a ``universal'' such pair, by  constructing\newline some $(N,\nu\colon Y\to N)$ with $\nu\circ f=0$ with the property that:\newline for any $(B,\beta\colon Y\to B)$ with $\beta\circ f=0$, there exists a unique $b\colon N\to B$ such that $\beta=b\circ \nu$. 
\[\xymatrix{ X \ar[dd]_f\ar[ddr]^0\ar[ddrr]^0 && \\ && \\ Y \ar[r]^{\nu} \ar@/_1pc/[rr]_{\beta}& N \ar[r]^b & B}\]
{\small (We might abbreviate this property as saying roughly:  ``Every $\beta$ with $\beta\circ f=0$ factors uniquely through $N$''.)}

\end{enumerate}
\end{question}

\end{document}