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\title{Math studies algebra Spring 2018: Homework 5}

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\author{JC}

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\maketitle

   This homework is due by class time on Friday 2 March. It must be typeset
   (preferably in\ \LaTeX) and submitted as a PDF file on the Canvas site,
   with a filename of the form \begin{verbatim} andrewID_alg_homeworknumber.pdf \end{verbatim}   
   For each minute that it is late, the grade will be reduced by 10 percent. 

\medskip

\begin{enumerate}


\item  Let $V$ be a finite dimensional VS over an algebraically closed field $k$.
   A linear map $T: V \rightarrow V$ is said to be {\em diagonalisable} if there is 
   a basis of $V$ consisting of eigenvectors of $T$ (equivalently there is a basis 
   wrt which the matrix of $T$ is diagonal). $T$ is said to be {\em nilpotent}
   if $T^k = 0$ for some $k > 0$. 

  Prove that every linear $T$ can be written as the sum of a diagonalisable map and
    a nilpotent map in exactly one way.

\item  Let $R$ be a commutative ring with $1$ and let $M$ be an $R$-module such that
   for all modules $N_1$ and $N_2$ and all surjective $\alpha: N_1 \rightarrow N_2$, the
   map $Hom(M, \alpha)$ is surjective. Prove that there exists a module $N$ such that
   $M \oplus N$ is free. Hint: $M$ is a quotient of a free module.     


 \item  We saw in class that if $M$ is an fg module over a PID $R$ then
   $M \simeq R/(a_1) \oplus \ldots \oplus R/(a_m) \oplus R^t$ where the $a_i$ are
   nonzero nonunits with $a_i \vert a_{i+1}$. 
      Fill in the steps of the following outline of
   a proof that this representation is essentially unique.

   \begin{enumerate}

   \item If  $T(M)$ is the submodule consisting of torsion elements then
   $T(M) \simeq R/(a_1) \oplus \ldots \oplus R/(a_m)$ and $M/T(M) \simeq R^t$. 

   \item $t$ is the rank of $M$.   

   \item If $p$ is a prime element of $R$ then $R/(p)$ is a field.
     
   \item If $p$ is a prime of $R$, $s$ is an integer greater than zero, $N$ is an $R$-module
     and we define $N[p^s] = \{ n \in N : p^s n = 0 \}$ then $N[p^s]/N[p^{s-1}]$ becomes
     an $R/(p)$-module under the obvious scalar multiplication.

   \item  If $M \simeq R/(a_1) \oplus \ldots \oplus R/(a_m)$ as above and $p$ is prime, then
     the dimension of $M[p^s]/M[p^{s-1}]$ as a $R/(p)$-VS can be determined from the
     sequence $t_1 \ldots t_m$ where $t_i$ is the power to which $p$ appears in the prime
     factorisation of $a_i$.

   \item  If $M \simeq R/(a_1) \oplus \ldots \oplus R/(a_m)$ then the ideals $(a_i)$ can be determined from
     the dimensions of $M[p^s]/M[p^{s-1}]$ as  $R/(p)$-VS's for $p$ prime and $s > 0$.
       


   \end{enumerate}

   
\end{enumerate}  

     
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