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\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{}

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\begin{document}
\maketitle
\vspace{-50pt}
\begin{center}
{\huge \bf Homework 3}\\\vspace{8pt}
{\Large Due Thursday night, October 12} {\footnotesize(technically 5am Oct.\ 13)}\\
\end{center}

Given $R$-module homomorphisms $A\xrightarrow{\alpha} B$ and $B\xrightarrow{\beta} C$ with $\beta\circ \alpha = 0$, we say they form a\newline \emph{short exact sequence}, and write $\qquad\qquad0\to A\xrightarrow{\alpha} B\xrightarrow{\beta} C\to 0,\qquad$ if
\begin{compactenum}[label=(\roman*)]
\item $\alpha\colon A\to B$ is injective;
\item $\beta\colon B\to C$ is surjective; and
\item $\im(\alpha) = \ker(\beta)$.
\end{compactenum}
{\small (The content is essentially the same as saying that $C\cong B/A$, except that it allows us the freedom to consider e.g.\ that $A$ might not actually be a \emph{subset} of $B$.)}
We might sometimes leave the labels $\alpha$ and $\beta$ off, if we don't need names for the maps at the moment, but these maps are still an essential part of the SES.

\begin{question}
Consider a short exact sequence $0\to A\xrightarrow{\alpha} B\xrightarrow{\beta} C\to 0$.\newline Prove that the following are equivalent.
\begin{enumerate}[label=(\Alph*)]
\item There exists a homomorphism $\sigma\colon C\to B$ such that $\beta\circ \sigma=\id_C$.
\item There exists a homomorphism $\tau\colon B\to A$ such that $\tau\circ \alpha=\id_A$.
\item There exists an isomorphism $\varphi\colon B\to A\oplus C$ under which $\alpha$ corresponds to the inclusion $A\into A\oplus C$ and $\beta$ corresponds to the projection $A\oplus C\onto C$.
\end{enumerate}
When these equivalent conditions hold, we say  the short exact sequence $0\to A\xrightarrow{\alpha} B\xrightarrow{\beta} C\to 0$~\emph{splits}.
 {\footnotesize We can also equivalently say ``$\beta\colon B\to C$ splits'' (since by (i) this only depends on $\beta$) or  ``$\alpha\colon A\to B$ splits'' (by~(ii)).}
\end{question}

%\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{5pt}
\begin{question}
\label{q:projective}
Given an $R$-module, prove that the following are equivalent.
\begin{enumerate}[label=(\Alph*)]
\item Every short exact sequence $0\to A\to B\to M\to 0$ splits.
\item There exists some $R$-module $N$ such that $M\oplus N$ is free.
\end{enumerate}
When these equivalent conditions hold, we say that the $R$-module $M$ is \emph{projective}.
\end{question}
\vspace{5pt}
{\small Given $R$-modules $M$ and $N$, recall that $\Hom_R(M,N)$ is an $R$-module where $(rf+g)(m)=r\cdot f(m)+g(m)$. We can leave off the subscript if the ring $R$ is unambiguous from context.}
\begin{questionst}
{\footnotesize (Make sure you understand this, but don't write it up.)} 

Consider a short exact sequence $0\to A\xrightarrow{\alpha} B\xrightarrow{\beta} C\to 0$.
\begin{enumerate}[label=(\alph*)]
\item Prove for any $X$ that $\Hom(C,X)$ can be identified with \[\{f\in \Hom(B,X)\,|\, f\circ \alpha= 0\}\ \  =\ \  \ker \Hom(B,X)\xrightarrow{-\circ \alpha}\Hom(A,X).\]
\item Prove for any $Y$ that $\Hom(Y,A)$ can be identified with \[\{f\in \Hom(Y,B)\,|\, \beta\circ f = 0\}\ \ =\ \ \ker \Hom(Y,B)\xrightarrow{\beta\circ -}\Hom(Y,C).\]
\end{enumerate}
\end{questionst}
\pagebreak

\begin{questionst} (NO LONGER ASSIGNED)

\noindent \sout{Let $R$ be a commutative ring, and let $M$ be a finitely generated $R$-module.}\newline \sout{Prove that if $\alpha\colon M\to M$ is surjective, then it is an isomorphism.}
\end{questionst}
{\small \noindent \sout{(Note the following useful consequence: any $n$ elements that generate $R^n$ are actually a basis of $R^n$.)}}

\vfill
To the equivalent definitions (A)$\iff$(B)$\iff$(C) of finite generation on HW1, we could add the following equivalent condition (you don't need to prove this).\newline An $R$-module $M$ is finitely generated if and only if:
\begin{enumerate}
\item[(D)] There exists a short exact sequence $0\to A\to F\to M\to 0$ where $F$ is a finitely generated free module.
\end{enumerate}



\begin{question}
Let $M$ be a finitely generated $R$-module. Prove that the following are equivalent.
\begin{enumerate}[label=(\Alph*)]
\item There exists a short exact sequence $0\to A\to F\to M\to 0$ where $F$ is a finitely generated free module and $A$ is finitely generated.
\item For \emph{every} short exact sequence $0\to Q\to F\to M\to 0$ where $F$ is a finitely generated free module, $Q$ is finitely generated.
\end{enumerate}
When these equivalent conditions hold, we say that the $R$-module $M$ is \emph{finitely presented}.
\end{question}

\vfill
\begin{questionst}
{\footnotesize (Make sure you understand this, but don't write it up.)} 
Let $R$ be a commutative ring, and $S\subset R$ a multiplicative set. Consider a short exact sequence $0\to A\xrightarrow{\alpha} B\xrightarrow{\beta} C\to 0$. Applying the localization functor, we obtain maps $S^{-1}A\xrightarrow{S^{-1}\alpha} S^{-1}B$ and $S^{-1}B\xrightarrow{S^{-1}\beta} S^{-1}C$.\\[5pt]

Prove that $0\to S^{-1}A\xrightarrow{S^{-1}\alpha} S^{-1}B\xrightarrow{S^{-1}\beta} S^{-1}C\to 0$ is a short exact sequence.\newline {\small(Hint: this is essentially equivalent to $\ker(S^{-1} f)=S^{-1}\ker(f)$ and $\im(S^{-1}f)=S^{-1}\im(f)$, which we saw in class. So just make sure you understand why it holds.)}
\end{questionst}
\vfill
\phantom{x}\hfill(cont)
\newpage

Let $R$ be a commutative ring, and $S\subset R$ a multiplicative set. Given an $R$-linear map $f\colon M\to N$, in class we defined\footnote{in class I called this $S^{-1}f$ instead of $\frac{f}{S}$, but that will get too hard to write.} the $R[\frac{1}{S}]$-linear map $\frac{f}{S}\colon S^{-1}M\to S^{-1}N$ given by $\frac{f}{S}(\frac{m}{s})= \frac{f(m)}{s}$.

A natural question is whether \emph{every} $R[\frac{1}{S}]$-linear map from $S^{-1}M$ to $S^{-1}N$ is of this form. 
Thinking a moment shows this can't quite be true: for example, the map $g\colon \Z[\frac{1}{2}]\to \Z[\frac{1}{2}]$ given by $g(x)=\frac{1}{2^{10}}\cdot x$ can't come from a map $\Z\to \Z$.

So the real question is whether, for every $g\colon S^{-1}M\to S^{-1}N$, there exist $s\in S$ and $f\colon M\to N$ such that $s\cdot g=\frac{f}{S}$. In this question, you will show that this holds when $M$ is finitely presented.
\vspace{15pt}
\begin{question}
\label{q:localization}If we set $L(f)=\frac{f}{S}$, 
this gives a set function
\[L\colon \Hom_R(M,N)\to \Hom_{R[\frac{1}{S}]}(S^{-1}M,S^{-1}N).\]
Observe that $L$ is actually $R$-linear (you do not need to prove this). 

\smallskip
Prove that if $M$ is finitely presented, then $L$ is the localization map of the $R$-module $\Hom_R(M,N)$.
More precisely, for any $M$ the universal property gives a map \[L'\colon S^{-1}\Hom_R(M,N)\to \Hom_{R[\frac{1}{S}]}(S^{-1}M,S^{-1}N);\] you must prove that if $M$ is finitely presented, then $L'$ is an isomorphism.
\end{question}
\vspace{45pt}
\begin{question}
Give a counterexample to Q\ref{q:localization} when $M$ is not finitely presented, by exhibiting some $g\colon S^{-1}M\to S^{-1}N$ for which there do not exist $s\in S$ and $f\colon M\to N$ such that $s\cdot g=\frac{f}{S}$.\newline
{\small (Note: you don't have to take $R$ to be some crazy ring for this.)}
\end{question}



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


\begin{question}
Given elements $r_1,\ldots,r_k$ in a commutative ring $R$, prove  the following are equivalent.
\begin{enumerate}[label=(\Alph*)]
\item These elements generate the unit ideal: $(r_1,\ldots,r_k)=R$;\newline in other words, there exist $a_1,\ldots,a_k\in R$ such that $a_1r_1+\cdots+a_kr_k=1$.
\item An $R$-module $M$ is 0 $\iff$ the $R[\frac{1}{r_i}]$-module $M[\frac{1}{r_i}]$ is 0 for all $i=1,\ldots,k$.
\end{enumerate}
\end{question}

\vfill
\begin{question}
\label{q:equiv}
Let $R$ be a commutative ring, and let $M$ be an $R$-module.\newline Prove that if $M$ is \textbf{finitely presented}, the following are equivalent.
\begin{enumerate}[label=(\Alph*)]
\item $M$ is projective. {\small (see Q\ref{q:projective})}
\item $M$ is \emph{locally free}, meaning there exist $r_1,\ldots,r_k$ in $R$ with $(r_1,\ldots,r_k)=R$\newline such that $M[\frac{1}{r_i}]$ is a free $R[\frac{1}{r_i}]$-module for all $i=1,\ldots,n$.
\item $M_P$ is a free $R_P$-module for all prime ideals $P$.
\item $M_{\m}$ is a free $R_{\m}$-module for all maximal ideals $\m$.
\end{enumerate}
\end{question}

\vfill
\begin{question} (Hard)
Extend the equivalence in Q\ref{q:equiv} to include the following equivalent condition\newline (still under the assumption that $M$ is finitely presented):
\begin{enumerate}[label=(\Alph*)]
\item[(E)] Every linear dependence in $M$ is trivial, in the sense below.

\end{enumerate}
A linear dependence in $M$ is a list of module elements $m_1,\ldots,m_n\in M$ and ring elements~${r_1,\ldots,r_n\in R}$ such that $r_1m_1+\cdots+r_nm_n=0$ in $M$.\\[10pt]

A \emph{trivial} linear dependence is, colloquially, something like
\begin{align*}&(10v_1-3v_2)\\+2\cdot &(-3v_1+v_2)\\+&(-4v_1+v_2)\\
=&(10-6-4)v_1+(-3+2+1)v_2\\
=&0v_1+0v_2=0.
\end{align*}
 Formally, a linear dependence is \emph{trivial} if there exist module elements $v^1,\ldots,v^k\in M$ and ring elements $a_i^j\in R$ such that \[a_i^1v^1+a_i^2v^2+\cdots+a_i^kv^k = m_i\qquad\text{for all $i$}\] \[r_1a_1^j+r_2a_2^j+\cdots+r_na_n^j=0\qquad\text{for all $j$}\]
\end{question}
\end{document}