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\author{Bryden R. Cais}
\title{Math 593 Assignment \# 5 Solutions}
\begin{document}

\maketitle 
\begin{enumerate}

\item The $k[x]$ module
$$M=\frac{k[x]}{(x^3+x+1)}\oplus \frac{k[x]}{x^2}$$
is given to us in primary decomposition form.  As such, we can read off the rational canonical form of the
matrix $T$: it has two blocks corresponding to the two summands, and each block is the companion matrix
of the polynomial in the ``denominator'' of the summand.  That is, we have
$$T=\begin{pmatrix}0 & 0 & -1 & 0 & 0\\ 1 & 0 & -1 & 0 & 0\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1 & 0\end{pmatrix}$$
up to a change of basis of $V=k^5$.

What is going on is simple.  Let $V=\k^5$.  Then we know that $V$ has standard basis $e_i$ for $1\leq i\leq 5$.
Observe that $M$ has basis
$$f_1=(1,0),\ f_2=(x,0),\ f_3=(x^2,0),\ f_4=(0,1),\ f_5=(0,x).$$  We then have an isomorphism (because we are sending a basis to a basis)
$$M\longrightarrow V$$ given by
$$f_i\mapsto e_i.$$
To determine $T$ we simply need to see what multiplication by $x$ does to the basis $f_i$.  We have
\begin{align*}
    xf_1=(x,0)&=f_2\\
    xf_2=(x^2,0)&=f_3\\
    xf_3=(-x-1,0)&=-f_2-f_1\\
    xf_4=(0,x)&=f_5\\
    xf_5=(0,x^2)&=0.\\
\end{align*}
It follows that multiplication by $x$ on $M$ has the matrix $T$ above.

 
    \item \begin{enumerate}

        \item Let $\r$ be a representation of $G$ on $V$.  We make $V$ into a $k[G]$ module by defining
        $$gv:=\r(g)v$$ for any $g\in G$ and $v\in V$ and extending by linearity to all of $k[G]$.
        Clearly $V$ is an abelian group under $+$.  It is easy to
        verify that
        \begin{enumerate}
            \item $1v=\r(1)v=v $ since $\r(1)$ is the identity matrix (since $\r$ is a homomorphism).
            \item $(gh)v=\r(gh)v=\r(g)\r(h)v=\r(g)(hv)=g(hv)$, since $\r$ is a homomorphism.
            \item $(g+h)v=(\r(g)+\r(h))v=\r(g)v+\r(h)v=gv+hv$
            \item $g(v+w)=\r(g)(v+w)=\r(g)v+\r(g)w=gv+gw$, since $\r(g)$ is a linear transformation.
        \end{enumerate}
        Hence, we have made $V$ into a $k[G]$ module.

        \item Now suppose that $V$ has the structure of a $k[G]$ module where scalars $\lambda\in k[G]$ act by scalar
        (vector space) multiplication.  Define $\r:G\rightarrow \GL(V)$ by
        $$\r(g)v:=gv$$ for any $g\in G, v\in V$.  We must check that $\r$ defines a homomorphism $\r:G\rightarrow \GL(V)$.
        Observe that
        $$\r(gh)v:=(gh)v=g(hv)=g(\r(h)v)=\r(g)\r(h)v$$ since $V$ is a $k[G]$ module.  Moreover, since
        $$\r(1g)v=\r(1)\r(g)v=\r(g)\r(1)v=\r(g)v$$ for all $v\in V$, we have $\r(1)=1$.  It follows that
        $\r$ is a homomorphism.  The usual axioms for a $k[G]$ module imply that $\r(g)$ is a linear transformation
        of $V$.  Since
        $$\r(g)\r(g^{-1})=\r(1)=1$$ we see that $\r(g)$ is an invertible linear transformation.  It follows that $\r$
        is a representation of $G$ on $V$.  The fact that $V\simeq V_{\r}$ as $k[G]$ modules follows from our definition
        in part a).
    \end{enumerate}

    \item Let $k$ be a field and $f,g\in k[x]$ nonzero polynomials.  For any $p,q\in k[x]$, we let $\bar{p}_q$ denote the image of
    $p$ in $k[x]/(q)$ under the natural quotient map $\pi_q$.  Let
    $$R=\Hom_{k[x]}\left(k[x]/f,k[x]/g\right),$$
    and suppose that $\phi\in R$.  Let $\tilde{\phi}$ be any lift of $\phi$ to an element of $\Hom_{k[x]}\left(k[x],k[x]\right)$;
    that is, $\tilde{\phi}$ is any $k[x]$ module homomorphism making the following diagram commute:
    \begin{eqnarray*}
            \begin{CD}
          k[x]   @>\tilde{\phi}>>  k[x]\\
        @V\pi_fVV      @VV\pi_gV\\
          k[x]/(f)   @>>\phi>     k[x]/(g)
            \end{CD}
    \end{eqnarray*}
    Then we have
    \begin{align*}
        \tilde{\phi}(f)&=f\tilde{\phi}(1),
    \end{align*}
    but since
    $$\phi\circ\pi_f (f)=\phi(\bar{0}_f)=\bar{0}_g,$$
    we must have
    $$\pi_g(f\tilde{\phi}(1))=\bar{0}_g,$$
    or what is the same, $g|f\tilde{\phi}(1)$.
    Since $k$ is a field, $k[x]$ is a PID (therefore also a UFD), so putting
    $$(g,f)=(h)$$ we see that
    $$g_1=g/h$$ divides $\tilde{\phi}(1)$.
    Now define the map
    $$\Pi:k[x]\longrightarrow R$$
    by $$p\mapsto \phi_p$$
    where $\phi_p$ is defined by
    $$\phi_p(\bar{1}_f)=\bar{p}_g\bar{g_1}_g.$$
    (Realize that an element of $R$ is completely determined by where it sends $\bar{1}_f$).
    It is not difficult to check that $\Pi$ is a $k[x]$ module homomorphism.
    Indeed,
    \begin{enumerate}
        \item \begin{align*}
        \Pi(p+q)(\bar{1}_f)&=\phi_{p+q}(\bar{1}_f)\\
        &=(\bar{p}_g+\bar{q}_g)\bar{g_1}_g\\
        &=\bar{p}_g\bar{g_1}_g+\bar{q}_g\bar{g_1}_g\\
        &=(\Pi(p)+\Pi(q))(\bar{1}_f),
        \end{align*}
        so that $\Pi(p+q)=\Pi(p)+\Pi(q)$.

        \item
        \begin{align*}
            \Pi(pq)(\bar{1}_f)&=\phi_{pq}(\bar{1}_f)\\
            &=\bar{p}_g\bar{q}_g\bar{g_1}_g\\
            &=\bar{p}_g\Pi(q)(\bar{1}_f)
        \end{align*}
        so that $\Pi(pq)=p\Pi(q)$ (modulo $g$, of course, but that is all we care about).
    \end{enumerate}
    That $\Pi$ is surjective follows from our opening remarks.  Now $\Pi(p)=0$ if and only if
    $\Pi(p)(\bar{1}_f)=\bar{0}_g,$ that is, if and only if
    $g|pg_1,$ or equivalently, if and only if $h|p$.  Therefore, $\ker\Pi=(h)$ and we have
    $$k[x]/(h)\simeq\Hom_{k[x]}\left(k[x]/(f),k[x]/(g)\right).$$

    \item
    \begin{enumerate}
        \item Let $V=\R^2$ and $T_{\t}:V\rightarrow V$ be rotation by $\t$.  Then $V_{\t}$ is irreducible if and only if
        $\t\not\in \pi\Z.$  First suppose that that $\t\not\in\pi\Z.$  Then $\sin(\t)\neq 0$ so that for any nonzero
        $v\in V$, the magnitude of
        $v\times T_{\t}v$ is not zero; that is, $v,T_{\t}v$ form the sides of a parallelogram in $\R^2$ with
        nonzero area.  As such, $v,T_{\t}v$ span $\R^2$.  (There are lots of ways of saying this; the idea is that as long as $\t$
        is not a multiple of $\pi$, the vectors $v,T_{\t}v$ are nor multiples of each other).  Let $W\subset V$ be a $T_{\t}$ invariant
        nontrivial subspace and let $w\in W$ be nonzero.  Then $T_{\t}w\in W$ so that $W$ contains a basis for $V$ and hence $W=V$
        so that $V$ is irreducible.  Conversely, if $\t=n\pi$ for some $n\in\Z$ then $T_{\t}v=\pm v$ for any $v\in V$; as such
        $V_{\t}$ has the nontrivial proper submodules $M_v=\{\lambda v:\lambda\in\R\}$ for all $v\neq 0$ in $V$ and is thus reducible.

        \item A nontrivial submodule of $V_{\r}$ is a nontrivial subspace $W\subset V$ that is preserver by
        the action of $G$.  Therefore, $V$ is irreducible if and only if for every $W\subset V,$ there exists some $g\in G$
        such that $\r(g)W\nsubseteq W$.  Equivalently, $V$ is irreducible if and only if the $G$ orbit of any nonzero
        vector $v\in V$ spans $V$ (if the orbit doesn't span $V$ then it spans a proper, nontrivial subspace $W$, which
        is then preserved by the action of $G$).

        \item $V$ is reducible as a $k[S_3]$ module.  To see this, let $W\subset V$ be the subspace given
        by
        $$W=\{(a,a,a):a\in k\}.$$
        Clearly this is a nontrivial proper subspace of $V$.  Moreover, $W$ is $S_3$ invariant since
        $S_3$ acts by permuting the components of any vector $v\in V$.  It follows that $W$ is a nontrivial, proper
        $k[S_3]$ submodule of $V$.
    \end{enumerate}

    \item
    \begin{enumerate}
        \item Suppose that $M$ is an irreducible $R$ module.  Let $\phi\in\End_R(M)$ and suppose that $\phi\neq 0$.
        Then since $\ker \phi\subset M$ is a submodule, $\ker \phi=0$ or $M$; since $\phi \neq 0$, $\phi$ is then injective.
        Hence $\phi$ is an isomorphism and therefore invertible; that is, $\End_R(M)$ is a division ring.
        Notice that we really do need to show that $\phi$ is both injective and surjective: if $M$ were finite, one implies
        the other, but in general injectivity of a map $\phi:M\rightarrow M$ does \textit{not} imply surjectivity.  For example,
        the map $\phi:\Z\rightarrow\Z$ given by $\phi(n)= 2n$ is certainly injective, but definitely not surjective.

        \item There are many ways of seeing this.  The intuition is that $T$ acts just like $i=\sqrt{-1}$ and gives $R^2$
        complex structure.  Moreover, since $x$ acts by $T$, $\R[x]$ is like $\C$.  It is clear that the $\C$ linear endomorphisms
        of $\C$ is isomorphic to $\C$ since any such endomorphism is completely determined by where the identity is sent, and it can be sent
        anywhere.  To make this intuition into a proof, we proceed as follows:

        We take the standard basis $e_1,e_2$ for $V$.  In this basis, any endomorphism ov $V$ is given
	as a $2\times 2$ matrix with entries in $\R$.  Suppose $\varphi\in\End_{\R[x]}(V_T)$ and let
	$$\begin{pmatrix}a & b\\ c& d\end{pmatrix}$$ be the matrix of $\varphi$.  Since $\varphi$
	must commute with $T$, and since the matrix of $T$ is given by
	$$\begin{pmatrix} 0 & -1\\ 1 & 0\end{pmatrix},$$ a short calculation shows that we must have
	$$\begin{pmatrix}-c & -d\\ a & b\end{pmatrix}=\begin{pmatrix} b & -a\\ d & -c\end{pmatrix},$$
	that is, $b=-c,a=d.$  It follows that every $\varphi\in \End_{\R[x]}(V_T)$ has the form
	$$a+bT$$ for some $a,b\in\R.$  Therefore, we define
	$$\phi:\C\longrightarrow\End_{\R[x]}(V_T) $$
	by
	$$\phi(a+bi)=a+bT.$$
	By our remarks above, $\phi$ is surjective.  Injectivity is also clear.  That $\phi$ is a homomorphism follows
	from a short calculation:
	$$\phi((a+bi)(c+di))=\phi(ac-bd+(ad+bc)i)=(ac-bd)+(ad+bc)T,$$
	on the other hand,
	$$\phi(a+bi)\phi(c+di)=(a+bT)(c+dT)=ac+adT+bcT+bdT^2=(ac-bd)+(ad+bc)T$$
	since $T^2=-1.$  It is obvious that $\phi$ preserves addition and that $\phi(1)=1.$
	Since $\phi$ is a bijective ring homomorphism, it is an isomorphism, and we are done. 
	%Define 
        %$$\phi: \R[x]\longrightarrow \End_{R[x]}(V_T)$$
        %by
        %$$f\mapsto \varphi_f,$$
        %where
        %$$\varphi_f(v)= fv.$$
        %Observe that $\phi$ is a surjective $\R[x]$ module homomorphism:
        %\begin{enumerate}
        %    \item Surjectivity follows from the fact that every $\R[x]$ linear endomorphism of $V_T$ is
        %    multiplication by an element of $\R[x]$.  This can be shown, for example, by an explicit calculation
        %    with matrices: letting $\{e_1,e_2\}$ be the standard basis for $\R^2$, observe that
        %    $$Te_1=e_2,Te_2=-e_1$$ so that if $\varphi\in \End_{R[x]}(V_T)$ commutes with $T$ we have
        %    $$T\varphi(e_1)=\varphi(e_2),T\varphi(e_2)=-\varphi(e_1).$$  It follows that we can write
        %    $$\varphi=a\cdot 1+b T$$ for $a,b\in\R$.  Already this is more information than we need.
        %    \item $\phi(hf)v=\varphi_{hf}(v)=(hf)v=h(fv)=h\varphi_f(v)$ for any $h\in \R[x]$ and $v\in V_T$.
        %    \item $\phi(f+g)v=\varphi_{f+g}(v)=(f+g)v=fv+gv=\phi(f)v+\phi(g)v$.
       % \end{enumerate}
      % Now suppose $\phi(f)=0$.  Then $fv=0$ for all $v\in V$.  By the division algorithm, we may write
      % $$f=(x^2+1)g+h$$
      % where $g,h\in \R[x]$ and $h=0$ or $\deg(h)\leq 1$.
      % Then
      % \begin{align*}
      %      fv&=g\left((T^2+1)v\right)+hv\\
      %      &=g\left(-v+v\right)+hv\\
      %      &=hv,
      % \end{align*}
      % since $T^2$ is rotation by $\pi$.  Thus $fv=0$ for
      % all $v\in V$ entails $hv=0$ for all $v\in V$.  Write $h(x)=ax+b$.
      % Then $hv=aTv+bv.$  Observe that if $v\neq 0$, the vectors $Tv,v$ are orthogonal so that if $v$ is nonzero, so is $aTv+bv$ unless $a=b=0$.
      % We conclude that $h=0$ so that $(x^2+1)|f$.  Conversely, if $(x^2+1)|f$ then $\phi(f)=0$.  Therefore
      % $\ker \phi=(x^2+1)$ so that
      % $$\C\simeq \R/(x^2+1)\simeq \End_{\R[x]}(V_T).$$

       %Of course, a really slick way to so this would be to observe that $\End_{\R[x]}(V_T)$ is a field by 4 a) and 5 a).
      % Moreover, an explicit isomorphism with $\C$ is given by
       %$$a+bT\mapsto a+bi$$ (by part i above). 
    \end{enumerate}

\end{enumerate}

\end{document}