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\title{Math 728\\Fall 2004\\Homework 4}
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\centerline{\it Due Monday September 27, 2004.}

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\noindent 1. (5 pt) Suppose that $M$ and $N$ are $R-$modules and that there are one to one $R-$module homomorphisms $f:M\longrightarrow N$ and $g:N\longrightarrow M$ (so one can identify $M$ as a submodule of $N$ and $N$ as a submodule of $M$). Does it follow that $M\cong N$? Prove this or give a counterexample.

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\noindent 2. (5 pt) Let $M$ and $N$ be $R-$modules. Suppose that you have $R-$module homomorphisms $f:M\longrightarrow N$ and $g:N\longrightarrow M$ such that $fg=1_N$ and $gf=1_M$. Show that $N=\text{im}(f)\oplus\text{ker}(g)$ (and hence $M=\text{im}(g)\oplus\text{ker}(f)$).

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\noindent 3. (5 pt) Consider the following commutative diagram of $R-$module homomorphisms
\begin{center}
$\xymatrix{ & 0\ar[d] & 0\ar[d] & 0\ar[d] & \\
0\ar[r] & A_1\ar[r]\ar[d] & A_2\ar[r]\ar[d] & A_3\ar[r]\ar[d] & 0\\
0\ar[r] & B_1\ar[r]\ar[d] & B_2\ar[r]\ar[d] & B_3\ar[r]\ar[d] & 0\\
0\ar[r] & C_1\ar[r]\ar[d] & C_2\ar[r]\ar[d] & C_3\ar[r]\ar[d] & 0\\
 & 0 & 0 & 0 & } $
\end{center}

Show that if the columns and the top two rows are exact, then the bottom row is exact.

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\noindent 4. (5 pt) Show that if $R$ is commutative with identity and $M$ is an $R-$module then there is an $R-$module isomorphism

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

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\noindent 5. (5 pt) Let $R$ be commutative with identity. Show that every $R-$module is free if and only if $R$ is a field.



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