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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 4}\\\vspace{8pt}
{\Large Due Thursday night, October 19} {\footnotesize(technically 5am Oct.\ 20)}\\
\end{center}

\begin{question}
Consider the situation of the snake lemma, where each row is exact:
\[\xymatrix{
&A\ar[r]^{f} \ar[d]^\alpha    & B\ar[r]^{g}\ar[d]^\beta  & C \ar[r]\ar[d]^\gamma & 0\\
0\ar[r]&A'\ar[r]^{f'}  & B'\ar[r]^{g'}  & C' &
}\]
\begin{enumerate}
\item[(a)] Construct a connecting homomorphism $d\colon \ker \gamma\to \coker\alpha$.
\item[(b*)] Check that this yields a complex $\ker \alpha\to \ker \beta\to \ker\gamma\xrightarrow{d} \coker \alpha\to \coker \beta\to \coker\gamma$.
\item[(c*)] Check that this sequence is exact at $\ker \beta$ and $\coker \beta$.
\item[(d)] Check that this sequence is exact at $\ker \gamma$ and $\coker \alpha$.
\item[(e*)] Check that if $f$ is injective, then $0\to \ker\alpha\to \ker \beta$ is exact at $\ker \alpha$ also;\newline
and that if $g'$ is surjective, then $\coker \beta\to \coker \gamma\to 0$ is exact at $\coker \gamma$ also.
\end{enumerate}
{\footnotesize Note you only need to write up (a) and (d).}
\end{question}
\enlargethispage{\baselineskip}

\vfill
A \emph{free resolution} of an $R$-module $M$ is a complex
\[\cdots\to F_3\to F_2\to F_1\to F_0\to M\to 0\]
which is exact everywhere and where each $F_i$ is free.\newline
{\small (Similarly, a \emph{projective resolution} of $M$ is an exact sequence $\cdots\to P_2\to P_1\to P_0\to M\to 0$\phantom{xxxxx}\linebreak where each $P_i$ is projective, and so on.)}
\begin{question}
Prove that every $R$-module $M$ has a free resolution \[\cdots\to F_2\to F_1\to F_0\to M\to 0.\]
\end{question}
\vfill
\begin{question}
Let $M$ and $N$ be $R$-modules, and suppose you have free resolutions\newline\indent $\cdots\to F_2\xrightarrow{d} F_1\xrightarrow{d} F_0\to M\xrightarrow{d} 0$ \qquad and\qquad   $\cdots\to G_2\xrightarrow{d} G_1\xrightarrow{d} G_0\xrightarrow{d} N\to 0$. 

\noindent Given a homomorphism $f\colon M\to N$, prove  there exist maps $f_i\colon F_i\to G_i$ making a commutative diagram
\[\xymatrix{
\cdots\ar[r]&F_2\ar[r]^{d} \ar[d]^{f_2} &F_1\ar[r]^{d} \ar[d]^{f_1} &F_0\ar[r] \ar[d]^{f_0}   & M\ar[d]^f\ar[r]&0\\
\cdots\ar[r]&G_2\ar[r]^{d}&G_1\ar[r]^{d} &G_0\ar[r]   & N\ar[r]&0\\
}\]
(To think about, and write up if you find a good answer:) In what sense are the maps $f_i$ unique?
\end{question}

\begin{question}
Compute an explicit free resolution for $M$ in the following situations:
\begin{enumerate}[label=(\alph*)]
\item $R=\Z$, \quad $M=\Z\oplus \Z/12\Z$
\item $R=\Z$, \quad $M=\Z/3\Z\oplus \Z/4\Z$
\item $R=\R[T]$, \quad $M=\R^2$, with $R$-module structure where $T$ acts by $\begin{pmatrix}1&2\\0&1\end{pmatrix}$
\item $R=\R[x,y]$, \quad $M=\R$, with $R$-module structure where $x$ and $y$ act by 0.
\item $R=\Z[\sqrt{-30}]$, \quad $M=\mathbb{F}_2$, with $R$-module structure where $\sqrt{-30}$ acts by 0.
\item $R=\R[x,y]$, \quad $M=\R[x,y]/I$\newline where $I$ is the ideal of all polynomials with no constant, linear, or quadratic term.\newline
{\small (in other words, $M$ consists of at-most-quadratic polynomials in $x$ and $y$)}
\item $R=\Z[t]/(t^2-1)$, \quad $M=\Z$, with $R$-module structure where $t$ acts by the identity.
\item $R=$ a ring of your choice, \quad $M=$ an interesting module of your choice.
\end{enumerate}
\end{question}
\vspace{50pt}
\begin{question}
Consider the map $f\colon M\to N$ from $M=\Z\oplus \Z/12\Z$ to $N=\Z/3\Z\oplus \Z/4\Z$ sending $(a\in \Z,b\in \Z/12\Z)$ to $(\overline{a}\in \Z/3\Z, \overline{b}\in \Z/4\Z)$.
If \[\cdots\to F_1\to F_0\to M\to 0\qquad\text{and}\qquad\cdots\to G_1\to G_0\to N\to 0\] are the free resolutions of $M$ and $N$ that you constructed in Q4(a) and Q4(b), describe explicitly the maps $f_i\colon F_i\to G_i$ as in Q3.
\end{question}

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