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\title{Math 421-621\\Fall 2013\\Homework 2}
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\centerline{\it Due Wednesday, January 23, 2013.}


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\noindent 1. (5 pt) We say that the $R-$module $M$ is {\it simple} if its only submodules are $M$ and $0$. Show that if $M$ is simple and $\phi:M\longrightarrow N$ is an $R-$module homomorphism, then $\phi$ is either one-to-one or the $0$ map.


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\noindent 2. Let $R$ be a commutative ring with identity.
\begin{itemize}
\item[a)] (5 pt) Show that $\text{Hom}_R(M,N)$ is an $R-$module.
\item[b)] (5 pt) Show that if $N=M$ then $\text{Hom}_R(M,N)$ also has a ring structure (is this ring necessarily commutative?).
\end{itemize}

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\noindent 3. (5 pt) Show that there is an $R-$module isomorphism

\[
\text{Hom}_R(R,M)\cong M.
\]

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\noindent 4. Let $n>0$ be a natural number. Compute the following.
\begin{itemize}
\item[a)] (5 pt) $\text{Hom}_{\mathbb{Z}}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z})$.
\item[b)] (5 pt) $\text{Hom}_{\mathbb{Z}}(\mathbb{Z}/n\mathbb{Z},\mathbb{Z})$.
\end{itemize}

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\noindent 5. We say that the $R-$module $M$ is {\it cyclic} if there is an $m\in M$ such that $M=Rm$. 
\begin{itemize}
\item[a)] (5 pt) Show that if $M$ is cyclic and $\phi:M\longrightarrow N$ is an $R-$module homomorphism, then $\phi$ is completely determined by $\phi(m)$.
\item[b)] (5 pt) Show that the homomorphic image of a cyclic module is cyclic.
\item[c)] (5 pt) Is it true that a submodule of a cyclic module is necessarily cyclic? Prove or give a counterexample.
\end{itemize}



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