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\author{Bryden R. Cais}
\title{Math 593 Assignment \# 7 Solutions}
\begin{document}

\maketitle
\begin{enumerate}
    \item
    \item Put $R=\Q[x,y]$ and let $M=(x,y)\subset R$ (considered as an $R$ module).  We claim that we have an exact sequence
    $$\begin{CD}0@>>>R@>A>>R^2@>B>>M@>>>0\end{CD}$$
    where
    $$Ar=r\begin{pmatrix}-y \\x\end{pmatrix}$$ and
    $$B\begin{pmatrix}f\\g\end{pmatrix}=fx+gy.$$
    To verify this assertion, it suffices to check that $\im A=\ker B$ since
    it is obvious that $B$ is surjective.  Suppose that $fx+gy=0$.  Then $fx=-gy$ and $y|f$ since $R$ is a UFD
    and $y$ is irreducible.  Similarly, $x|g$ and it follows that $f=-ry,\ g=rx$ for some $r\in R$, that is, $\ker B=\im A$.
    It follows that the presentation matrix for $M$ is the matrix of multiplication by $A$ as a map $R\rightarrow R^2$.
    It is clear that this matrix is
    $$\begin{pmatrix}-y\\x\end{pmatrix}.$$

    \item The presentation matrix is
    $$\begin{pmatrix}3 & 2\\ 6 & -4 \\ 12 & 10\end{pmatrix}.$$
    We use row and column operations to simplify this matrix.  We have
    \begin{align*}
    \begin{CD}
    \begin{pmatrix}3 & 2\\ 6 & -4 \\ 12 & 10\end{pmatrix}@>\text{col 1$-$col 2}>>\begin{pmatrix}1 & 2\\ 10 & -4 \\ 2 & 10\end{pmatrix}
    @>\text{row 2$-10\cdot$row 1}>>\begin{pmatrix}1 & 2\\ 0 & -24 \\ 2 & 10\end{pmatrix}
        @>\text{row 3$-2\cdot$row 1}>>\\
        \begin{pmatrix}1 & 2\\ 0 & -24 \\ 0 & 6\end{pmatrix}@>\text{col 2$-2\cdot$col 1}>>
        \begin{pmatrix}1 & 0\\ 0 & -24 \\ 0 & 6\end{pmatrix}@>\text{row 2$+4\cdot$row 3}>>
        \begin{pmatrix}1 & 0\\ 0 & 0 \\ 0 & 6\end{pmatrix},
    \end{CD}
    \end{align*}
    from which it is evident that
    $$M\simeq \Z\oplus\Z/(6).$$

    \item  Let $A$ be an integral domain and $M$ an $A$ module.
    \begin{enumerate}
        \item Let $T(M)\subset M$ denote the set of all torsion elements and let $m_1,m_2\in T(M)$.
        Then there exist $a_1,a_2\in A\setminus\{0\}$ such that $a_im_i=0$ for $i=1,2$.
        Then for any $a\in A$ we have
        $$a_1a_2(m_1+m_2)=a_2(a_1m_1)+a_1(a_2m_2)=0$$ and
        $$a_1(am_1)=a(a_1m_1)=0.$$
        Since $A$ is an integral domain, $a_1a_2\neq 0$ and therefore $m_1+m_2,\ am_1\in T(M)$ which shows that
        $T(M)$ is a submodule of $M$.

        \item Suppose that $m\in T(M/T(M))$.  Let $\tilde{m}$ be any lift of $m$ to $M$.  Then there exists some $a\in A\setminus\{0\}$
        such that $am=0$, or equivalently, such that $a\tilde{m}\in T(M)$.  Then there exists $b\in A\setminus \{0\}$ such that
        $ba\tilde{m}=0$ in $M$.  Since $A$ is an integral domain, $ba\neq 0$ so that $\tilde{m}\in T(M)$.  Therefore, $m=0$ and
        we are done.

        \item Let $A=\C[x,y]$.  Considering $M=(x,y)$ as an $A$ module, it is not hard to see that $A$ is torsion free: this follows
        directly from the fact that $A$ is an integral domain and $M\subset A$.  However $M$ is not free since we have a nontrivial
        relation between the generators $x,y$; namely $x(y)-y(x)=0$ and since $M$ is not generated by any single element of $\C[x,y]$.
        Now consider the $A$ module
        $$M=A\oplus A/(x).$$  It is clear that $N=A/(x)$ considered as an $A$ submodule of $M$
        by the obvious inclusion is the torsion submodule of $M$ and is a proper submodule.
    \end{enumerate}

    \item Let $A$ be a Euclidean ring and let $M$ be a finitely generated torsion free $A$ module.  Let $S=\{m_i\}_{i=1}^s$ generate $M$
    and without loss of generality suppose that $m_1,\ldots,m_n$ is a maximal linearly independent subset of $S$.  Let $F$ be the free
    $A$ module generated by $m_1,\ldots,m_n$.

    \begin{enumerate}
        \item By our choice of $m_j$ for $1\leq j\leq n$, for any $n+1\leq i\leq s$ the set $m_1,\ldots,m_n,m_i$ is linearly dependent.
        We therefore have a nontrivial relation
        $$\sum_{j=1}^n a_jm_j+a_im_i=0,$$
        where not all $a_j$ are zero.  In particular, $a_i\neq 0$ since $m_1,\ldots,m_n$ are linearly independent.  It follows that
        $$a_im_i\in F.$$

        \item Set $a=a_{n+1}\cdots a_s\in A$.  Since each $a_i\neq 0$ and $A$ is an integral domain, we have $a\neq 0$.  Define
        $\phi_a:M\rightarrow M$ by $\phi_a(m)=am$.  Since $a\neq 0$ and $M$ is torsion free, $\phi_a$ is injective.  Moreover,
        any $m\in M$ has the form
        $$m=\sum_{j=1}^n b_jm_j+\sum_{i=n+1}^s b_{i}m_i$$ for some $b_l\in A$.  We have
        $$am=a\sum_{j=1}^n b_jm_j+\sum_{i=n+1}^s (b_{i}a_{n+1}\cdots \hat{a_i}\cdots a_s)(a_im_i),$$
        where $\hat{a_i}$ denotes the omission of $a_i$ from the product.  Since $a_im_i\in F$ by construction, it follows
        that $am\in F$ and hence that $\im\phi_a\subset F$.

        \item Since $\phi_a$ is injective is gives an isomorphism
        $$\phi_a:M\rightarrow \im\phi_a\subset F.$$  Since $F$ is free and finitely generated, it follows that $\im \phi_a$ and hence $M$
        is also free.
    \end{enumerate}
\end{enumerate}

\end{document}
