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\parbox{2.25in}{\bf Math 4720/5210}
\hfill {\Large \bf Homework \#8} \hfill
\parbox{2.25in}{\bf \hfill Due: Mon., Apr. $29^\text{th}$, 2019}

\vspace{0.2in}

\begin{description}
\item[\#1 All Torsion-ed up Inside] Let $R$ be a ring with $1$ and $M$ an $R$-module. We call $m \in M$ a {\it torsion element} if $r \cdot m=0$ for some $0 \not= r \in R$ (i.e., $m$ is annihilated by some nonzero element of our ring). 

Define: $\mathrm{Tor}_R(M) = \{ m \in R \;|\; \exists\; 0\not= r \in R \text{ such that } r \cdot m =0\}$.
\begin{enumerate}[(a)]
    \item Compute $\mathrm{Tor}_{\mathbb{Z}}(\mathbb{Z}_3)$ (considering $\mathbb{Z}_3$ as a $\mathbb{Z}$-module) and $\mathrm{Tor}_{\mathbb{Z}_3}(\mathbb{Z}_3)$ (considering $\mathbb{Z}_3$ as a $\mathbb{Z}_3$-module).
    \item Show that if $R$ is an integral domain, then $\mathrm{Tor}(M)$ is a submodule of $M$.
    \item Show that if $R$ has zero divisors and $M \not= \{0\}$, then $\mathrm{Tor}(M) \not= \{0\}$.
    \item Let $\varphi:M \to N$ be a $R$-module homomorphism. Show that $\varphi(\mathrm{Tor}_R(M)) \subseteq \mathrm{Tor}_R(N)$.
    \item {\bf [Grad. Students]} Give an example of a ring $R$ and $R$-module $M$ where $\mathrm{Tor}(M)$ is not a submodule of $M$. 
    
             {\it Hint:} This can be done with some $\mathbb{Z}_n$ as a $\mathbb{Z}_n$-module.
\end{enumerate}

\item[\#2 Am I Extendable?] Let $G$ be a {\bf finite} abelian group. Then $G$ is a $\mathbb{Z}$-module. Can the action of $\mathbb{Z}$ be extended to an action of $\mathbb{Q}$ making $G$ into a $\mathbb{Q}$-module? Why or why not?

\item[\#3 Homies] {\bf [Grad. Students]} Let $R$ be a commutative ring with $1$ and let $M$ be an $R$-module. 

Show that $\mathrm{Hom}_R(R,M) \cong M$ as $R$-modules.

{\it Hint:} A homomorphism from a free module to some other module is determined by its action on a basis.

\item[\#4 Irreducible Fun!] Let $R$ be a ring with $1$ and $M$ an $R$-module and $N$ a submodule of $M$. Then $N$ is called a {\it cyclic} $R$-module if there exists some $n \in N$ such that $N=\langle n \rangle = R\cdot n = \{ r \cdot n \;|\; R \}$. An $R$-module $M$ is called {\it irreducible} if $M \not=\{0\}$ and the only submodules of $M$ are $\{0\}$ and $M$ itself.

\begin{enumerate}[(a)]
\item Let $x \in M$. Show that $\langle x \rangle$ is a submodule of $M$.
\item Show that $M$ is irreducible if and only if $M \not=\{0\}$ and $M$ is cyclic such that $M=\langle x \rangle$ for all $0\not=x \in M$.
\item {\bf [Grad. Students]} Let $R$ be commutative with $1$. Show that $M$ is irreducible if and only if $R/I \cong M$ (as $R$-modules) for some maximal ideal $I \triangleleft R$.

{\it Suggestion:} First, consider why there must be some $0\not=\varphi:R \to M$ and how this gives $R/I \cong M$ for some ideal. Then consider why this only happens when $I$ is maximal.
\item Determine all of the irreducible $\mathbb{Z}$-modules. 

{\it Hint:} Consider part (b)'s result.
\item Let $M_1$ and $M_2$ be irreducible $R$-modules and $\varphi:M_1 \to M_2$ a $R$-module homomorphism. Show that either $\varphi=0$ or $\varphi$ is an isomorphism. 

{\it Note:} This result is called {\it Schur's Lemma}. As a consequence, if $M=M_1=M_2$ is irreducible, we have that every $\varphi \in \mathrm{End}_R(M)=\mathrm{Hom}_R(M,M)$ is either the zero map or an isomorphism. Now, $\mathrm{End}_R(M)$ is a ring under addition of homomorphisms and function composition, it is not the zero ring since the identity map (a homomorphism) is not the zero map (since $M \not= \{0\}$ which is part of the definition of irreducibility), and we have that every non-zero homomorphism is invertible (i.e., a unit in $\mathrm{End}(M)$). Thus $\mathrm{End}_R(M)$ is a {\it division ring} when $M$ is an irreducible $R$-module. 
\end{enumerate}

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