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\begin{document}

\noindent Math 742, Spring 2016\\
Homework 2\\
Due: February 5

\section{Exercises}

\begin{enumerate}[(1)]

\item Let $R$ be a ring and let $0 \to M_1 \to M_2 \to M_3 \to 0$ be a short exact sequence of $R$-modules. Show that $M_2$ is finitely generated if $M_1$ and $M_3$ are finitely generated.

Furthermore, show that if $M_2$ is finitely generated, then $M_3$ is finitely generated. Give an example where $M_2$ is finitely generated but $M_1$ is not.

\bigskip

\item Let $R$ be a ring and let $\cC$ be the category of $R$-modules. Let $\Lambda$ be a set and suppose we are given modules indexed by $\Lambda$: $\{M_\lambda\}_{\lambda \in \Lambda}$.

  \begin{enumerate}
  \item Show that the assignment $N \mapsto \prod_{\lambda \in \Lambda} \Hom_R(M_\lambda, N)$ defines a functor $\cC \to \Set$. Show that it is representable, i.e., isomorphic to $h^M$ where $M = \bigoplus_{\lambda \in \Lambda} M_\lambda$.

  \item Consider the inclusions $i_\lambda \colon M_\lambda \to M$ which sends $m \in M_\lambda$ to the sequence which is $m$ in position $\lambda$ and $0$ elsewhere. Show that this satisfies the following universal mapping property: given any other module $N$ and homomorphisms $f_\lambda \colon M_\lambda \to N$, there is a {\it unique} homomorphism $M \to N$ such that the following diagram commutes for all $\lambda \in \Lambda$:
\[
\xymatrix{ M_\lambda \ar[r]^-{i_\lambda} \ar[rd]_-{f_\lambda} & \bigoplus_{\lambda \in \Lambda} M_\lambda \ar[d] \\ & N }.
\]

If we're given another module $P$ and homomorphisms $j_\lambda \colon M_\lambda \to P$ which satisfy the same universal mapping property, then show, just using the definition, that $M$ and $P$ are canonically isomorphic. Reinterpret (a) using Yoneda's lemma to give another proof of this fact.

A special case is when $M_\lambda=R$ for all $\lambda \in \Lambda$, in which case we get the universal mapping property for a free module $R^{\oplus \Lambda}$.

  \item Similarly, show that the assignment $N \mapsto \prod_{\lambda \in \Lambda} \Hom_R(N,M_\lambda)$ defines a functor $\cC^\op \to \Set$. Show that it is representable, i.e., isomorphic to $h_{M'}$ where $M' = \prod_{\lambda \in \Lambda} M_\lambda$. Translate this into a universal mapping property for the direct product as we did for the direct sum (you don't have to reprove anything).
  \end{enumerate}

\bigskip

\item Let $R$ be a ring and let $m,n,p$ be positive integers.
  \begin{enumerate}
  \item Show that a homomorphism between free $R$-modules $R^n \to R^m$ is the same thing as an $m \times n$ matrix whose entries come from $R$, and that composition $R^n \to R^m \to R^p$ can be computed using matrix multiplication.
  \end{enumerate}
  
In particular, an endomorphism $\alpha \colon R^n \to R^n$ is the same as a square matrix, so we can define its determinant in the usual way:
\[
\det(\alpha) = \sum_{\sigma \in S_n} {\rm sgn}(\sigma) \alpha_{1,\sigma(1)} \cdots \alpha_{n,\sigma(n)}.
\]
It satisfies all of the familiar properties from linear algebra: it's independent of choice of basis for $R^n$, there's a Laplace expansion formula, and $\det(\alpha\beta) = \det(\alpha)\det(\beta)$. (You don't need to prove this.)
\begin{enumerate}
\setcounter{enumii}{1}
  \item Show that $\alpha$ is bijective if and only if $\det(\alpha)$ is a unit.
  \item Show that $\alpha$ is injective if and only if $\det(\alpha)$ is not a zerodivisor.
  \item In particular, injective need not imply bijective. However, show that if $\alpha$ is surjective, then $\alpha$ is bijective.
  \end{enumerate}

\bigskip

\item Let $R$ be a principal ideal domain. Here are two facts you may use without proof:
  \begin{itemize}
  \item A submodule of a finitely generated $R$-module is also finitely generated\footnote{This is better studied in the general context of noetherian rings, which comes later in the course.}.
  \item (Smith normal forms): Given an $n \times m$ matrix $A$ with entries in $R$, there exists an $n \times n$ matrix $S$ and an $m \times m$ matrix $T$ such that $SAT$ is diagonal (i.e., only has entries in the $(i,i)$ positions), and each element is of the form $up^d$ where $u$ is a unit, $p$ is a prime element, and $d \ge 0$ is an integer.
  \end{itemize}

Use them to show the following:

  \begin{enumerate}
  \item Show that every finitely generated $R$-module $M$ is isomorphic to the direct sum of a free $R$-module and its torsion submodule\footnote{$m$ is in the torsion submodule if there exists nonzero $x \in R$ such that $xm=0$} and that, furthermore, the torsion submodule is isomorphic to a direct sum of modules $R/(p^d)$ where $p$ is prime and $d > 0$ is an integer.
  \item Let $k$ be a field and let $X$ be an $n \times n$ matrix. Describe a $k[x]$-module structure on $k^n$ where $x$ acts by $X$. Show that this is a finitely generated torsion module. 
  
Now assume $k$ is algebraically closed. Interpret the decomposition for the module $M$ from (a) in terms of a normal form for $X$ (this is the {\bf Jordan canonical form}).
  \end{enumerate}

\bigskip

\item Let $R$ be a ring and suppose we have the following commutative diagram of $R$-modules:
\[
\xymatrix{ A \ar[r] \ar[d]^-f & B \ar[r] \ar[d]^-g & C \ar[r] \ar[d]^-h & D \ar[d]^-i \\
A' \ar[r] & B' \ar[r] & C' \ar[r] & D' }
\]
Assume the top row and bottom row are both exact.
\begin{enumerate}
\item If $g,i$ are injective and $f$ is surjective, show that $h$ is injective.
\item If $f,h$ are surjective and $i$ is injective, show that $g$ is surjective.
\end{enumerate}


\end{enumerate}

\section{Suggested exercises (don't submit)}

From Altman--Kleiman:
\begin{itemize}
\item Chapter 4: 3, 17, 18, 19
\item Chapter 5: 14, 16, 22, 29
\item Chapter 6: 5, 9
\item Chapter 7: 2, 3, 9
\end{itemize}

\end{document}
