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\author{Bryden R. Cais}
\title{Math 593 Assignment \# 6 Solutions}
\begin{document}

\maketitle
\begin{enumerate}

\item Let $k$ be a field and $A_i$ a finite dimensional vector space over $k$.  Consider the exact sequence:
$$
\begin{CD}
    0 @>f_{-1}>> A_0 @>f_0>> A_1 @>f_1>> A_2 @>f_2>> \ldots @>f_{n-2}>> A_{n-1}@>f_{n-1}>> A_n@>f_{n}>> 0
\end{CD}
$$
The familiar rank nullity theorem tells us that
$$\dim \im f_i+\dim\ker f_i=\dim A_i.$$
However, since the sequence is exact we have $\im f_i=\ker f_{i+1}$ so that
$$\dim A_i=\dim \im f_{i}+\dim\im f_{i-1}$$ for $0\leq i\leq n$.
Hence
\begin{align*}
    \sum_{i=0}^n (-1)^i \dim A_i &=\sum_{i=0}^n (-1)^i \left(\dim \im f_{i}+\dim\im f_{i-1}\right)\\
    &=\sum_{i=0}^n (-1)^i \dim \im f_{i}+\sum_{i=0}^n (-1)^i \dim \im f_{i-1}\\
        &=\sum_{i=0}^n (-1)^i \dim \im f_{i}-\sum_{i=0}^n (-1)^i \dim \im f_{i}+(\dim\im f_{-1}-\dim\im f_n)\\
        &=(\dim\im f_{-1}-\dim\im f_n)\\
        &=0,
\end{align*}
since the image of $0$ under $f_{-1}$ is $0$ and the image of $A_n$ under $f_n$ is 0.

\item \begin{enumerate}
    \item Suppose that the sequence
    \begin{align}
    \begin{CD}
    0 @>>> N @>u>> M @>v>> L @>>> 0
    \end{CD}\label{seq}
    \end{align}
    splits, that is, we have the following commutative diagram
    \begin{align}
    \begin{CD}
    0 @>>> N @>u>> M @>v>> L @>>> 0\\
      &&     @V\phi_N VV @V\varphi VV @VV\phi_L V  \\
    0 @>>> N @>>\iota_N>  N\oplus L @>>\pi_L> L @>>> 0
    \end{CD}\label{seq2}
    \end{align}
    where $\iota_N(n)=(n,0)$ is the natural inclusion map, $\pi_L(n,l)=l$ the natural projection map, and $\phi_N,\phi_L,\varphi$
    are isomorphisms.  Define
    $$j:M\longrightarrow N$$ by
    $$j:=\phi_N^{-1}\circ\pi_N\circ\varphi,$$
    where $\pi_N(n,l)=n$ is the natural projection map (and $\phi_N^{-1}$ is well defined since $\phi_N$ is an isomorphism).
    It is obvious that $j$ is a homomorphism since it is the composition of homomorphisms.
    Then we have
    $$j\circ u=\phi_N^{-1}\circ\pi_N\circ\varphi\circ u=\id_N$$ since the diagram above is commutative.  Observe that
    we need not assume that $\phi_N=\id_N$.

    Similarly, define
    $$s:L\longrightarrow M$$ by
    $$s:=\varphi^{-1}\circ\iota_L\circ\phi_L,$$
    where $\iota_L(l)=(0,l)$ is the natural inclusion map.  Again, it is obvious that $s$ is a homomorphism and we have
    $$v\circ s=v\circ\varphi^{-1}\circ\iota_L\circ\phi_L=\id_L$$ since the diagram commutes.
    This shows that (a) implies (b) and (c).

    \item Suppose that there is a homomorphism $j:M\longrightarrow N$ such that $j\circ u=\id_N$.  Define
    $$\varphi:M\longrightarrow N\oplus L$$ by
    $$\varphi(m)=(j(m),v(m)).$$
    Since $j,v$ are homomorphisms, it is clear that $\varphi$ is a homomorphism.  Now suppose that $\varphi(m)=0$.
    Then $j(m)=(0,0)$ so that $m\in\ker j=\im u$ by exactness.  It follows that $m=u(n)$ for some $n\in N$, and hence that
    $j(m)=j\circ u (n)=n=0$.  Therefore, $m=u(0)=0$ so that $\varphi$ is injective.  Let $(n,l)\in N\oplus L$ be arbitrary.
    Since $v$ is surjective, there exists $m^{\prime}\in M$ such that $v(m^{\prime})=l$.  Set
    $$m=u(n)+m^{\prime}-u\circ j (m^{\prime}).$$
    Then
    $$v(m)=v(u(n)+m^{\prime}-u\circ j (m^{\prime}))=v(m^{\prime})=l$$
    since $\im u=\ker j$, and
    $$j(m)=j(u(n)+m^{\prime}-u\circ j (m^{\prime}))=j\circ u (n)+j(m^{\prime})-j\circ u\circ j (m^{\prime})=n$$
    since $j\circ u=\id_N$.  Surjectivity of $\varphi$ follows.  Thus, $\varphi$ is an isomorphism.  Taking the obvious
    maps $\phi_N=\id_N$ and $\phi_L=\id_L$, it is easy to see that the diagram (\ref{seq2}) is commutative and hence that
    the sequence (\ref{seq}) splits.

    Similarly, suppose that we have a homomorphism $s:L\longrightarrow M$ such that $v\circ s=\id_L$.
    Define
    $$\psi: N\oplus L\longrightarrow M$$ by
    $$\psi(n,l)=u(n)+s(l).$$
    Since $u,s$ are homomorphisms, it is clear that $\psi$ is a homomorphism.  Now suppose that $\psi(n,l)=0$.
    Then certainly $v(u(n)+s(l))=l=0$ since $\ker v=\im u$ and $v\circ s=\id_L$ so that $u(n)=0$.  Since $u$ is injective,
    it follows that $(n,l)=(0,0)$.
    Now let $m\in M$ be arbitrary.
    Observe that
    $$m=(m-s\circ v(m))+s\circ v(m)$$
    and that
    $$v(m-s\circ v(m))=v(m)-v\circ s\circ v(m)=v(m)-v(m)=0$$ since $v\circ s=\id_L$.  It follows that
    $(m-s\circ v(m))\in\ker v=\im u$ so there exists $n\in N$ such that $u(n)=(m-s\circ v(m))$.  Then set $l=v(m)$
    whence $m=\psi(n,l)$ and $\psi$ is surjective.
    Thus, $\psi$ is an isomorphism and we let $\varphi=\psi^{-1}$, $\phi_N=\id_N$ and $\phi_L=\id_L$.
    As before, it is easy to see that the diagram (\ref{seq2}) is commutative and hence that
    the sequence (\ref{seq}) splits.

    We have thus shown that (b) and (c) imply (a).  This completes the proof.
\end{enumerate}

\item \begin{enumerate}
    \item Let $F$ be a free $R$ module with basis $\{x_i\}_{i\in I}$, and $N$ an arbitrary $R$ module.  Let $n_i\in N$ for $i\in I$.
    Because $F$ is free, every element $f\in F$ has a {\em unique} representation
    \begin{eqnarray}
    f=\sum_{i\in I} a_ix_i \label{uniq}
    \end{eqnarray}
     for $a_i\in R$.  Define
    $$\phi:F\longrightarrow N$$ by $\phi(x_i)=n_i$ and extend $R$-linearity to all of $F$.  By definition, $\phi$ is a $R$ module
    homomorphism.  Moreover, the uniqueness of (\ref{uniq}) guarantees that $\phi$ is well defined.  Since $\{x_i\}$ is a basis for $F$,
    any homomorphism $F\rightarrow N$ agreeing with $\phi$ on $\{x_i\}$ agrees with $\phi$ on all of $F$.  Hence $\phi$ is unique.

    \item By part (a), there is a unique homomorphism
    $$\phi:F\rightarrow \bigoplus_{i\in I} R$$ given by
    $\phi(x_i)=(0,0,\ldots,0,1,0,\ldots)$ where the 1 is $i^{\text{th}} $ position.  (OK, we are supposing that $I$ is countable.)
    If
    $$\phi\left(\sum_{i\in I}a_ix_i\right)=(0,0\ldots),$$
    then we clearly have $a_i=0$ for all $i$.  This shows that $\phi$ is injective.  Surjectivity is also clear:
    if $a=(a_i)_{i\in I}\in \oplus_{i\in I} R$ then $\phi(\sum_{i\in I} a_ix_i)=a$.  Thus $\phi$ is an isomorphism.

    \item Suppose that $M\supset N$ are $R$ modules and $M/N\simeq F$ is free.  Then we have an exact sequence
    \begin{eqnarray}
    \begin{CD}
    0 @>>> N @>u>> M@>v>> F@>>> 0.\label{seq3}
    \end{CD}\
    \end{eqnarray}
    Let $\{x_i\}$ be a basis for $F$.  Since $v$ is surjective, there exist $\{m_i\}_{i\in I}$ such that
    $v(m_i)=x_i.$ Since $F$ is free, by part (a) we have a unique homomorphism $s:F\rightarrow M$
    given by $s(x_i)=m_i$.  It is clear that $v\circ s=\id_L$, so that by problem 2, the sequence (\ref{seq3})
    splits and $M\simeq F\oplus N$, as desired.
\end{enumerate}

\item \begin{enumerate}
        \item Let $P$ be a free module, and suppose we have an exact sequence $M\rightarrow N\rightarrow 0$ together
        with a homomorphism $s: P\rightarrow N$.  Let $\{x_i\}$ be a generating set for $P$.  Since $u$ is surjective,
        for each $i$ there exists $m_i\in M$ such that $u(m_i)=x_i$.  By part (a) of problem 3, there exists a unique
        homomorphism
        $$\bar{s}:P\longrightarrow M$$
        given by $\bar{s}(x_i)=m_i$.  It follows that the diagram
\begin{diagram}
  &                             &    P     &      &    \\
  & \relax\ldTo^{\bar{s}}       & \dTo_{s} &      &    \\
M &         \rTo^{u}            &    N     & \rTo &  0 \\
\end{diagram}
is commutative, and hence that $P$ is projective.

        \item Suppose that $P$ is projective and we have an exact sequence
        \begin{diagram}
        0 & \rTo & M & \rTo & N & \rTo & P & \rTo 0.
        \end{diagram}
        Then since $P$ is projective, there exists a homomorphism $\bar{s}$ such that the following diagram is commutative:
\begin{diagram}
 &      &   &      &   &                             &    P         &      &    \\
 &      &   &      &   & \relax\ldTo^{\bar{s}}       & \dTo_{\id_P} &      &    \\
0& \rTo & M & \rTo & N &         \rTo^{u}            &    P         & \rTo &  0 \\
\end{diagram}
We now apply problem 2(c), from which it follows that any such exact sequence splits.  Thus (a) implies (b).

Next, assume (b) and recall that every module is the quotient of some free module.  We therefore have an exact sequence
\begin{diagram}
        0 & \rTo & Q & \rTo & F & \rTo & P & \rTo 0,
\end{diagram}
where $F$ is free.  Since we are assuming that any such sequence splits, we have an isomorphism
$$F\longrightarrow P\oplus Q,$$
that is, $P$ is a direct summand of a free module.  Thus (b) implies (c).

Finally, suppose (c) and let $\begin{CD}M @>u>> N @>>> 0\end{CD}$ be exact.  Suppose that we have a homomorphism $s:P\rightarrow N$.
Let $\pi_P:P\oplus Q\rightarrow P$ be the natural projection map.  Since there exists a module $Q$ such that $P\oplus Q$ is free,
by part (a) of this problem, we have a homomorphism $\bar{s}^{\prime}$ such that the diagram

\begin{diagram}
  &                              &  P\oplus Q   &      &   \\
  &\ldTo(2,4)^{\bar{s}^{\prime}} &  \dTo_{\pi_P}&      &   \\
  &                              &    P         &      &    \\
  &                              &     \dTo_{s} &      &    \\
 M&   \rTo^{u}                   &    N         & \rTo &  0 \\
\end{diagram}
commutes.  We therefore have a map $\bar{s}:P\longrightarrow M$ given by
$$\bar{s}=\bar{s}^{\prime}\circ\iota_P$$
where $\iota_P:P\longrightarrow P\oplus Q$ is the natural inclusion map, such that
$$s=u\circ\bar{s}.$$  It follows that $P$ is projective and hence (c) implies (a).

\end{enumerate}

\item Until next time\ldots

\end{enumerate}

\end{document}
