$\text{V}$ is semisimple as a $k[X]$-module if and only if $A\in\operatorname{End}_k(\text{V})$ is diagonalizable over the algebraic closure of $k$ 
Suppose $\text{V}$ is a finite-dimensional $k$-vector space and let $A\in \operatorname{End}_k\left(\text{V}\right)$.  Regard $\text{V}$ as a $k[X]$-module via $f(X)v=f(A)v$ for $f(X)\in k[X]$, $v\in \text{V}$.  Show that $\text{V}$ is semisimple as a $k[X]$-module if and only if $A$ is diagonalizable over the algebraic closure of $k$.

From the literature, it seems like the crucial observation is that the minimal polynomial of $A$ splits into linear factors if it is diagonalizable. But I do not understand how this relates to semisimplicity.
 A: Let $V$ be a $k[X]$-module, where $X$ acts as $A\in{\rm End}_k(V)$. Since $k[X]$ is a PID, the fundamental theorem of finitely-generated modules over PIDs applies. This gaurantees an elementary factor decomposition $\bigoplus k[X]/(\pi(X)^e)$ for some set of irreducibles $\pi(X)\in k[X]$ and exponents $e\in\Bbb N$.
The following are equivalent:


*

*$V$ is semisimple (is a direct sum of simple $k[X]$-modules)

*$V$'s elementary factors' exponents are all equal to one

*$V\otimes_k\overline{k}$'s elementary factors as a $\overline{k}[X]$-module are all equal to $1$

*$A$'s Jordan blocks over $\overline{k}$ are all $1\times 1$

*$A$ is diagonalizable over $\overline{k}$


I've arranged them in the order I think is easiest to prove, $(1)\Leftrightarrow(2)\Leftrightarrow(3)\Leftrightarrow(4)\Leftrightarrow(5)$.
A: The minimal polynomial $\mu$ of $A$ will remain the same when extending the base field$~k$ to its algebraic closure$~\bar k$ (in fact the splitting field of the minimal polynomial will suffice). Then by the result you cited "$A$ is diagonalisable over $\bar k$" means "$\mu$ splits into distinct (linear) factors over$~\bar k$", and this is equivalent to "$\mu$ decomposes into distinct irreducible factors over$~k$". (For the latter one uses that distinct (monic) irreducible polynomials over$~k$ cannot have a common root over$~\bar k$, for instance because their $\gcd$ is $1$.)
Now first suppose $\mu$ is square-free: it is indeed a product of distinct irreducible factors $P_1,\ldots,P_l$ over$~k$. Then by the kernel decomposition theorem, the $k[X]$-module $V$ decomposes as as a (direct) sum of its submodules annihilated by each of the $P_i$. (One does not need the structure theorem of finitely generated $k[X]$-modules for this, just the Chinese remainder theorem for $k[X]$; in particular this depends only on the $P_i$ being pairwise relatively prime.) Each submodule is a vector space over a field $k[X]/(P_i)$, and choosing a basis for it decomposes it into a sum of simple $k[X]$-submodules; therefore $V$ is semisimple.
Conversely suppose $\mu$ contains some irreducible factor $P$ with multiplicity $m>1$. Let $~S$ be the submodule of $V$ annihilated by$~P^m$; since $P^m$ is relatively prime to $\mu/P^m$, by the  same argument as before $S$ is a direct summand of$~V$. Also not all of $S$ is annihilated by$~P$, or else $\mu$ would not be the minimal polynomial. Then the submodule $S_1$ of $S$ that is annihilated by$~P$ has nonzero intersection with the image of multiplying by$~P$, showing that $S_1$ cannot be a direct summand. So $V$ is not semisimple.

The above uses exclusively canonical decompositions, coming from the kernel decomposition theorem. One can get the result a bit faster using a non-canonical decomposition from the structure theory of finitely generated modules over a PID. Its main theorem says that an invariant factor decomposition of $V$ into cyclic modules exists, of which the last summand is isomorphic to $k[X]/(\mu)$, and any other summands are of the form $k[X]/(d_i)$ for some divisor $d_i$ of$~\mu$. Now if $\mu$ is square-free then so are all the$~d_i$, and by decomposing each of the summands by the kernel decomposition theorem one gets a direct sum of simple modules. If on the other hand $\mu$ is divisible by the square of some irreducible polynomial$~P$, then $k[X]/(\mu)$ has a submodule isomorphic to $k[X]/(P^m)$ with $m>1$ (again by kernel decomposition), which shows that $V$ is not a semisimple module.
Alternatively one can go straight for the elementary divisor decomposition of $V$ (as a $k[X]$-module). This will give a decomposition into simple modules if one exists, which happens iff all elementary divisors are primes (no higher powers of primes). To make the connection with the minimal polynomial, one must use the easy fact that $\mu$ is the product over the primes (irreducible polynomials) of their highest powers that occur as elementary divisors. Then $\mu$ is clearly square-free if and only if no powers higher than $1$ occur.
