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\begin{document}

\begin{center}
{\bf \Large \underline{Honors Algebra 4, MATH 371 Winter 2010}}\\ 
	{\large Assignment 5 Solutions}
\end{center}

For the problems 1--7, we fix a principal ideal domain $R$.

\begin{enumerate}
	
	
	\item Let $M$ be any $R$-module.
	\begin{enumerate}
		\item For $m\in M$, the {\em annhilator of $m$ in $R$}
		is defined to be
		$$\ann_R(m):=\{r\in R\ :\ rm=0\}.$$
		Prove that $\ann_R(m)$ is an ideal of $R$.  
		
\noindent{\bf Solution:} It is clear that $\ann_R(m)$ is closed under $R$-multiplication.  If $r,s\in\ann_R(m)$
then $(r+s)m=rm+sm=0$ so $\ann_R(M)$ is closed under addition as well, and is hence an ideal.		
		
		\item We say that $m\in M$ is {\em torsion} if $\ann_R(m)\neq 0$ and we define  
		the {\em torsion submodule of $M$} to be
		$$\Tor(M):=\{m\in M\ :\ \ann_R(m)\neq 0\}.$$
		We say that an $R$-module $N$ is {\em torsion free} if $\Tor(N)=0$.
		Prove that $\Tor(M)$ really is a submodule of $M$ and that the quotient $M/\Tor(M)$
		is torsion free.
		
\noindent{\bf Solution:} Suppose $m,n\in \Tor(M)$ and that $r,s\in R\setminus\{0\}$
are elements of $\ann_R(m)$ and $\ann_R(n)$, respectively.  Then $rs$ is nonzero as $R$ is
a domain and $rs\in \ann_R(um+n)$ for any $u\in R$.  Thus, $\Tor(M)$ is a submodule of $M$.
Suppose $m\in M$ and let $\overline{m}$ be the image of $m$ in $M/\Tor(M)$.  If $\overline{m}$ 
is torsion, there exists $r\in R\setminus\{0\}$ such that $r\overline{m}=0$, or equivalently $rm\in \Tor(M)$.
Thus, there exists $s\in R\setminus\{0\}$ such that $srm =0$ so since $sr\neq 0$ we conclude that $m\in \Tor(M)$
and hence $\overline{m}=0$. 
		
	\end{enumerate}
		
	\item Let $M$ be any submodule of a free module $R^n$.  Show 
		that $M$ is itself a free module, of rank at most $n$ as follows:
		\begin{enumerate}
			\item Let $\pi_i:M\rightarrow R$ be the composition of
			the inclusion $M\hookrightarrow R^n$ with projection $R^n\rightarrow R$
			on to the $i$th factor; it is an $R$-module homomorphism.  
			If $\pi_1(M)=0$, show that $M$ is a submodule of $R^{n-1}$ in a natural way.\label{free1}
			 
\noindent{\bf Solution:} The projection $R^n\rightarrow R$ on to the $i$th factor clearly has kernel
isomorphic to $R^{n-1}$.  Thus, if $\pi_1(M)=0$ then $M = \ker(\pi_1)\subseteq R^{n-1}$.			 
			 
			\item If $\pi_1(M)\neq 0$ then it is an ideal of $R$, necessarily principal, 
			say $\pi_1(M)=(d).$ For $m\in M$ with $\pi_1(m)=d$, show that $M\simeq Rm \oplus \ker \pi_1$
			and that $\ker \pi_1$ is naturally a submodule of a free module of rank $n-1$.\label{free2}			

\noindent{\bf Solution:} Let $x\in M$ be arbitrary.  By assumption, $\pi_1(x)=rd$ for some $r\in R$
and hence $\pi_1(x-rm)=0$.  Thus, $x=rm + (x-rm)$ is in $Rm + \ker\pi_1$.  This sum is nesessarily
direct, since if $\pi_1(rm) = rd=0$ then $r=0$.  Now $\ker \pi_1$ is a submodule of $R^{n-1}$
in a natural way as in part (\ref{free1}).

			\item Conclude by induction on $n$.
			
\noindent{\bf Solution:} The base case is $n=1$: then $M$ is an ideal of $R$ and hence is equal to
$Rd$ for some $d\in R$, which is a free $R$-module of rank at most 1 as $R$ is a domain.  Assuming
that the statement holds for $n-1$, if $M$ is a submodule of $R^n$ then by (\ref{free2})
we have $M=Rm\oplus \ker \pi_1$.  Necessarily $Rm$ is a free rank 1 $R$-module and by the induction
hypothesis, $\ker\pi_1\subseteq R^{n-1}$ is also free over $R$ of rank at most $n-1$.			
			
		\end{enumerate}	
		
		\item Prove that any finitely generated and torsion free $R$-module $M$ is 
		a submodule of a free module, and hence free as follows:
		\begin{enumerate}
			\item Let $\{m_1,\ldots, m_s\}$ be a minimal set of generators of $M$,
			and let $M_i$ be the submodule of $M$ generated by $\{m_1,\ldots,m_i\}$.
			Show that $M_1$ a free $R$-module.
			
\noindent{\bf Solution:} The map $R\rightarrow M_1$ given by $1\mapsto m_1$
is a surjective map of $R$-modules.  It is injective as $M_1\subseteq M$ is torsion free.

			\item Let $j\ge 1$ be the greatest integer such that $M_j$ is free.
			If $j=s$ we are done.  Otherwise, $M_{j+1}$ is not free so there exists
			a relation 
			$$xm_{j+1} + \sum_{1\le i\le j} r_i m_i=0$$
			with $x\in R$ nonzero.  Show that multiplication by $x$ on $M_{j+1}$ is
			an $R$-module homomorphism whose image is contained in a free $R$-module.

\noindent{\bf Solution:} It is obvious that multiplication by $x$ is an $R$-module homomorphism.
It's image is contained in $M_j$ (since $xm_{j+1}\in M_j$), which is a free $R$-module by hypothesis.

			\item Show that the kernel of multiplication by $x$ is zero, and deduce
			that $M_{j+1}$ is free after all.  Conclude that $M$ is free.
			
\noindent{\bf Solution:} This kernel is zero since $M$ is torsion-free.  Thus, $M_{j+1}$
is isomorphic to the image of multiplication by $x$, which
is a submodule of the free $R$-module $M_j$ and is hence free by the previous problem.  By our choice of $j$, this is
a contradiction to our assumption that $j<s$ and hence $M$ itself must be free.
			
			
					
		\end{enumerate}
		
		\item Let $M$ be a finitely generated $R$-module.  Prove that the short exact sequence
		$$\xymatrix{0\ar[r] & {\Tor(M)} \ar[r] & M \ar[r] & {M/\Tor(M)} \ar[r] & 0}$$
		splits, so $M\simeq \left(M/\Tor(M)\right)\oplus \Tor(M)$ is the direct sum 
		of its torsion submodule and a free module.  Show that this 
		decomposition of $M$ as a direct sum of a torsion module and a free module is unique
		up to isomorphism.
		
\noindent{\bf Solution:} By the previous problem, $M/\Tor(M)$ is finitely generated and torsio-free
so is a free $R$-module.  The desired splitting then comes from the universal mapping property of free
$R$-modules.  Now if $T$ is any torsion module and $F$ is any free module,
then the torsion submodule of $M:=T\oplus F$ is clearly $T$, and $M/T\simeq F$.  
This gives the desired uniqueness up to isomorphism.

		\item Let $T$ be a {\em torsion} $R$-module, i.e. the inclusion $\Tor(T)\hookrightarrow T$
		is an isomorphism.  For each nonzero (principal) ideal 
		$(r)\subseteq R$, we define the {\em $(r)$-primary submodule of $T$} to be
		$$T_{(r)}:=\{t\in T\ :\ r^nt=0\ \text{for some}\ n\ge 0\}.$$
		\begin{enumerate}
			\item Show that $T_{(r)}$ is a submodule of $T$ that depends only on the ideal $(r)$
			(and not on a specific choice of generator for this ideal).
			
\noindent{\bf Solution:} If $t,s\in T_{(r)}$ then $r^nt=0$ and $r^ms=0$ for some $m,n\ge 0$.
It follows that $r^{m+n}(ut+s)=0$ for any $u\in R$ so $T_{(r)}$ is an $R$-submodule of $T$.
Since $r^nt=0$ if and only if $ur^nt=0$ for any unit $u\in R^{\times}$, it follows immediately
that $T_{(r)}$ depends only on the ideal $(r)$	.		
			
			
			\item If $r,s\in R$ have $gcd(r,s)=1$, show that $T_{(r)}\cap T_{(s)} = 0$.
			Conclude that for such $r,s$ we have $T_{(rs)}\simeq T_{(r)}\oplus T_{(s)}$.
			Hint: use the fact that for any nonnegative integers $n,m$, there exist $u,v\in R$
			with $ur^n+vs^m = 1$.\label{injpri}
			
\noindent{\bf Solution:} Let $t\in T_{(r)}\cap T_{(s)}$ and suppose that $r^nt=0$ and $s^mt=0$.
Since $(r^n,s^m)$ is the unit ideal, there exist $u,v\in R$ such that $ur^n+vs^m=1$ and hence
$$t=(ur^n+vs^m)t = u(r^nt) + v(s^mt) = 0+ 0 =0.$$
Now the natural map $T_{(r)} \oplus T_{(s)}\rightarrow T_{(rs)}$ sending $(t,t')$ to $t+t'$
is surjective since if $(rs)^n t =0$ then choosing $u,v\in R$ with $ur^n+vs^n=1$ we have
$t=ur^nt+vs^nt$ and $ur^nt\in T_{(s)}$ while $vs^nt\in T_{(r)}$.  This map has kernel $T_{(r)}\cap T_{(s)}=0$
so is an isomorphism.
			
			\item Prove that there is a canonical isomorphism of $R$-modules
			$$T\simeq \bigoplus_{p} T_p$$
			where $p$ ranges over the distinct prime ideals of $R$.
			
\noindent{\bf Solution:} For each $p$, we have a canonical inclusion $T_p\hookrightarrow T$, which gives
a canonical mapping 
$$\varphi: \bigoplus_{p} T_p\rightarrow T.$$
By (\ref{injpri}), we see that $\varphi$ is injective.  If $t\in T$ then $\ann_R(t)$ is a principal
ideal of $R$, say generated by $(d)$.  If $d=\pi_1^{e_1}\cdots \pi_k^{e_k}$ is the prime factorization of 
$d$ in $R$ and $p_i=(\pi_i)$ the prime ideal corresponding to $\pi_i$, the an easy induction argument
using (\ref{injpri}) shows that $t$ is in the image of the restriction of $\varphi$ to the submodule
$\oplus_{i=1}^k T_{p_i}$ of $\bigoplus_{p} T_p$.  It follows that $\varphi$ is surjective as well, hence
an isomorphism.



		\end{enumerate}
		
		\item Let $p$ be a prime ideal of $R$ and $T_p$ a finitely generated, nonzero $p$-primary
		$R$-module. 
		\begin{enumerate}
			\item Show that any quotient or sub-module of $T_p$ is again $p$-primary
			and finitely generated (be careful for finite generation of submodules!)
	
\noindent{\bf Solution:} Clearly any quotient of a finitely generated module is again finitely generated,
as the images of the generators are a generating set.  Since $R$ is a PID, it is noetherian and any submodule
of a finitely generated $R$-module is again finitely generated (proof?).  That any quotient or submodule
of a $p$-primary module is $p$-primary is obvious.	
	
			
			\item Let $\ann(T_p):=\{ r\in R\ :\ rt=0\ \text{for all}\ t\in T_p\}$ be the
			{\em annihilator of $T_p$ in $R$}.  Prove that $\ann(T_p)$ is a proper ideal of $R$,
			and is equal to $p^N$ for some positive integer $N$.  
			
\noindent{\bf Solution:} Let $t\in T_p$ and suppose $\ann_R(t)=(r)$ for some $r\in R$. 
If $\pi$ is a principal generator of $p$, we know that $\pi^nt=0$ for some $n$ which 
without loss of generality we can assume is minimal with this property.
Then $\pi^n$ is a multiple of $r$, so we must have that $r=\pi^du$ for a unit $u$ and some $d\le n$.
We conclude that $\ann_R(t)=p^n$.
By definition, $\ann(T_p)$ is equal to the intersection 
$$\bigcap_{t\in T_p} \ann_R(t),$$  
and hence has the form $p^N$ for some $N$.  It is a proper ideal since $T_p$ is nonzero, and
for $t\in T_p$ nonzero $\ann_R(t)$ is a proper ideal.

			
			\item Suppose $\{t_1,\ldots,t_s\}$ is a set of generators for $T_p$ as an $R$-module.
			Show that there exists a set of $s$ generators $\{y_1,\ldots,y_s\}$ of $T_p$ 
			with 
			$$\ann_R(y_1) = \ann_R(T_p)=p^N$$\label{mingenfix}
			
\noindent{\bf Solution:} We have seen that $\ann_R(t_i)=p^{n_i}$ for some integers $n_i$,
and hence that these ideals form a chain.
It follows that $\ann(T_p)$, which is the intersection of the $\ann_R(t_i)$,
coincides with $\ann_R(t_i)$ for some $i$ (specificaly, for any value of $i$
with maximal $n_i$).  By reordering the $t_i$'s if need be, we conclude as desired.  

			
			\item Let $\langle y_1\rangle$ be the $R$-submodule of $T_p$ generated by $y_1$
			and set $T_p':=T_p/\langle y_1\rangle$; it is again a finitely generated 
			$p$-primary $R$-module.  Suppose $y'\in T_p'$ has $\ann_R(y')=p^m$ for some
			positive integer $m$.  Show that $m\le N$ and that there exists $y\in T_p$
			projecting to $y'$ with $\ann_R(y)=p^m$.\label{lift} 
			
\noindent{\bf Solution:} That $m\le N$ is clear.  Let $x$ be any lift of $y'$ to $T_p$,
so $p^mx \in \langle y_1 \rangle$, say $p^mx = ry_1$.  Then $p^{N-m}ry_1 = p^Nx=0$
so we conclude that $p^m | r$, say $r=p^m z$.  Then $y:=x-zy_1$ lifts $y'$ and
is killed by $p^m$.  We conclude that $\ann_R(y)=p^d$ for some $d\le m$.  But 
since $p^d$ then kills $y'$, we must have $d=m$. 
			
			\item Prove that there exist integers positive $m_1\le m_2\le \ldots \le m_s$
			with $\ann_R(T_p)=p^{m_s}$	and an isomorphism of $R$-modules
			\begin{equation}
				T_p\simeq R/p^{m_1} \oplus R/p^{m_2}\oplus \cdots \oplus R/p^{m_s} \label{primarydecomp}
			\end{equation}
			as follows:
			Fix a minimal set of generators $\{y_1,\ldots,y_s\}$ of $T_p$ and proceed
			by induction on $s$.  
			\begin{enumerate}
				\item By (\ref{mingenfix}), we may assume $\ann_R(y_1)=\ann_R(T_p)$.
				If $s=1$ conclude.
				
	\noindent{\bf Solution:} If $s=1$, the map $R\rightarrow T_p$ sending $r$ to $ry_1$
	is surjective with kernel $\ann_R(y_1)= p^{m_1}$.			
				
				\item If $s > 1$, consider the short exact sequence of $R$-modules
				\begin{equation}
				  	\xymatrix{
						0\ar[r] & {\langle y_1 \rangle} \ar[r] & T_p \ar[r] & T_p' \ar[r] & 0
					}\label{tosplit}
				 \end{equation} 
				 Using the induction hypothesis and part (\ref{lift}) to appropriately
				 lift generators of the direct summands occurring in the decomposition
				 of $T_p'$ as in (\ref{primarydecomp}), show that this sequence
				 is split, and conclude as desired.
				 
		\noindent{\bf Solution:} Since the images of $y_2,\ldots,y_s$ generate $T_p'$
		as an $R$-module, the induction hypothesis implies that we have an isomorphism
		$$T_p'\simeq  R/p^{m_2} \oplus R/p^{m_3}\oplus \cdots \oplus R/p^{m_s}.$$
		Choose $z_i\in T_p'$  mapping to a generator of the submodule $R/p^{m_i}$ in the
		above direct sum decomposition, and let $x_i$ be any lift of $z_i$ to $T_p$
		satisfying $\ann_R(x_i)=p^{m_i}$; this is possible by (\ref{lift}).
		Consider the $R$-linear mapping $T_p'\rightarrow T_p$ sending $z_i$ to $x_i$.
		This is well-defined by our choice of $x_i$ and gives a splitting of (\ref{tosplit}).
		We conclude as desired, since $\langle y_1\rangle\simeq R/p^{m_1}$.
					 
				 
			\end{enumerate}
			
			\item Show that the $m_i$ as in (\ref{primarydecomp}) are uniquely determined
			by $T_p$.  Hint: for each $j\ge 0$, consider the $R$-module $p^jT_p/p^{j+1}T_p$.
			Show that this is a vector space over the field $R/p$ and compute its dimension.
			
			\noindent{\bf Solution:} The dimension $d_j$ of the $\F_p$-vector space $p^jT_p/p^{j+1}T_p$
			is easily seen to be equal to the number of $m_i$ which are greater than $j$.  It is not hard
			to see that the $m_i$ can be recovered from the $d_j$, and hence are uniquely determined by
			$T_p$ as this is obviously true of the $d_j$.
		\end{enumerate}
		
		You have just proved:
		\begin{theorem}[Structure theorem for finitely generated modules over a PID: Primary decomposition]
			Let $R$ be a principal ideal domain and $M$ a finitely generated $R$-module.
			Then there exist direct sum decompositions
			$$R \simeq F\oplus T\quad\text{and}\quad T\simeq \bigoplus_p T_p$$
			where $F$ is a free $R$-module of finite rank, $T$ is a torsion $R$-module
			and $T_p$ is a $p$-primary torsion $R$-module.  Here, $F,T,T_p$ are 
			each uniquely determined by $M$.  Furthermore, for each $p$ there 
			exist integers $0 < m_1 \le,\ldots, \le m_s$ and a direct sum decomposition
			$$T_p\simeq R/p^{m_1}\oplus \cdots \oplus R/p^{m_s}$$
			with $s$ the minimal number of generators of $T_p$ and $p^m_s=\ann_R(T_p)$.
			The integers $\{m_i\}$ are uniquely determined by $M$.
		\end{theorem}\label{primary}
		
		\item Now prove:
		\begin{corollary}[Structure theorem for finitely generated modules over a PID: Invariant factor decomposition]
					Let $R$ be a principal ideal domain and $M$ a finitely generated $R$-module.
					Then there exists a direct sum decomposition 
					$$M\simeq F \oplus T$$
					with $F$ a free module and $T=\Tor(M)$ the torsion submodule of $M$.
					Moreover, there exists a nonnegative integer $d$ and nonzero elements $a_1,a_2,\ldots,a_m$ of $R$
			which are not units and which satisfy the divisibility relations
			$$a_1 | a_2 | \cdots | a_m$$
			such that there is a canonical isomorphism of $R$-modules
			$$T\simeq R/(a_1)\oplus R/(a_2) \oplus \cdots \oplus R/(a_d).$$
			The ideals $(a_i)$ are uniquely determined by $T$ $($and hence by $M$$)$; they are called the
			{\em invariant factors} of $M$.
		\end{corollary}
		Hint: Use Theorem \ref{primary} and the Chinese Remainder Theorem.
		
			\noindent{\bf Solution:} By the previous exercise, we have decompositions
			$$R \simeq F\oplus T\quad\text{and}\quad T\simeq \bigoplus_{i=1}^k T_{p_i}$$
			with $F$ free and $T$ torsion and uniquely determined by $M$ up to isomorphism. 
			For each $i$, we have
			$$T_{p_i}\simeq R/p_i^{m_1(i)}\oplus \cdots \oplus R/p_i^{m_{s_i}(i)}$$
			with $s_i$ and 
			$$m_1(i) \le m_2(i) \le \cdots \le m_{s_i}(i)$$ 
			uniquely determined by $T_{p_i}$ and hence by $M$.  
			Let $m=\max_i\{s _i\}$ and define $a_{m-j}\in R$ to be any generator of the principal ideal
			$$
 				\prod_{i=1}^k p_i^{m_{s_i-j}(i)}
			$$
			where we put $m_{s_i-j}(i)=0$ if $s_i-j < 0$. Then $a_1| a_2|\cdots | a_m$
			and the ideals $(a_i)$ are uniquely determined by $M$ as this is true of the $m_{j}(i)$.
		
		
		\item Now let $F$ be a field and $V$ a finite dimensional $F$-vector space equipped with
		a linear transformation $A:V\rightarrow V$.  We consider $V$ as an $F[X]$-module via
		\begin{equation*}
			(a_0 + a_1 X + \cdots + a_n X^n)v := a_0v + a_1Av + \cdots + a_nA^nv,
		\end{equation*}
		where $A^i$ denotes the composition of $A$ with itself $i$-times.  Since $F[X]$ is a PID,
		the structure theorems for $V$ as an $F[X]$-module that you proved above apply.  
		\begin{enumerate}
				\item Let $(a_i)$ for $i=1,\ldots, d$ be the invariant factors of $V$.  Show that
				each $a_i$ may be taken to be a monic polynomial, and that with this normalization 
				the $a_i$'s are uniquely determined by $V$ and $A$ (not just up to units).
				
			\noindent{\bf Solution:} Since the units of $F[X]$ are just $F^{\times}$, any nonzero
			ideal of $F[X]$ has a unique generator which is monic.				
				
				\item Let
				$$g:=b_0 + b_1X + \cdots +  X^k \in F[X]$$
				be any monic polynomial and consider the finite-dimensional $F$-vector space
				$$V_g:=F[X]/(g).$$
				Multiplication by $X$ gives a linear transformation of $V_g$, which we denote by 
				$m_X$.  Show that the matrix of $m_X$ with respect to the basis $1,X,X^2,\cdots, X^{k-1}$
				of $V_g$ is the {\em companion matrix} of $g$, given by
				the $k\times k$ matrix over $F$\label{companion}
				\begin{equation*}
					C_g:=\left(\begin{array}{cccccc}
						0 & 0 & \cdots & \cdots & \cdots & -b_0\\ 
						1 & 0 & \cdots & \cdots & \cdots & -b_1 \\
						0 & 1 & \cdots & \cdots & \cdots & -b_2 \\
						0 & 0 & \ddots &         &        & \vdots \\
						\vdots & \vdots && \ddots &        & \vdots \\
						0  & 0 & \cdots & \cdots &    1    & -b_{k-1}
					\end{array}\right)
				\end{equation*}
				
					\noindent{\bf Solution:} Obvious, using the identity 
					$$X\cdot X^{k-1}=X^k = -b_0-b_1X-\cdots - b_{k-1}X^{k-1}.$$		
				
				
				\item Prove that there exists a basis of $V$ with respect to which the matrix of $A$ 
				is in {\em rational canonical form}, i.e. has the block form\label{rcan}
				\begin{equation}
					\left( 
					\begin{array}{cccc}
						C_{a_1} &  &    &\\
						        & C_{a_2} &  & \\
						        &         & \ddots & \\
						        &        & & C_{a_d}
					\end{array}
					\right)\label{rcan}
				\end{equation}
				Hint: First observe that $V$ is a torsion $F[X]$-module.  Now use the invariant factor form
				of the structure theorem for modules over a PID and part (\ref{companion}).
				
	\noindent{\bf Solution:} Since $A$ satisfies its characteristic polynomial by Cayley-Hamilton,
	this characteristic polynomial annhihilates $V$ as an $F[X]$-module.  We deduce that $V$ is
	torsion and hence isomorphic to $F[X]/(a_1)\oplus \cdots\oplus F[X]/(a_d)$ for $a_1,\cdots,a_d$
	the invariant factors of $V$, normalized to be monic.  Let $B_i$ be the basis of 
	$F[X]/(a_i)$ given by $1,X,X^2,\ldots, X^{\deg a_i -1}$ and let $B$ be the basis of $V$
	corresponding to the union of the $B_i$.  Then the matrix of $A$ with respect to $B$ is
	easily seen to be as above, by part (\ref{companion}).  
	 			
				
				
				\item Prove that any square matrix $M$ over $F$ 
				is similar to a matrix in rational canonical form; i.e. there exists an invertible matrix $P$ such that
				$P^{-1}MP$ has the form (\ref{rcan}), and show moreover that $P$ and this rational canonical
				form of $M$ are uniquely determined by $M$.  Conclude that two $n\times n$ 
				matrices over $F$ are similar if and only if they have the same rational canonical form.
				
		\noindent{\bf Solution:} Two matrices are similar if and only if there is a change of basis
		bringing one matrix to the other.  It therefore follows from (\ref{rcan}) that there
		is a change of basis matrix bringing $M$ into rational canonical form.  Clearly this rational
		canonical form is uniquely determined by $M$, as it is uniquely determined by the invariant factors
		of $F^n$ as an $F[X]$-module via $M$.  If $M$ and $M'$ are similar, then $F^n$ as an $F[X]$
		module via $M$ is isomorphic to $F^n$ as an  			
		$F[X]$ module via $M'$, so the invariant factors are the same.  The converse is clear from 
		the above.
		
		  
		\end{enumerate}
		
		
\end{enumerate}
\end{document}