Diagonalizability and elementary divisors How to prove that an $n \times n$ matrix $A$ over a field $\mathbb F$ is diagonzalizable if and only if every elementary divisor of $A$ has degree $1$? I kind of know why this is true but I am not able to make an argument 
if $A$ is diagonalizable then it is similar to a diagonal matrix with entries $a_1,...,a_n$ where these are the eigenvalues of $A$ (with possible repetition). then $A$ has the same elementary divisors as this matrix.. but what are the elementary divisors of this matrix? i am thinking that if say $a_{n_1},...,a_{n_s}$ are the distinct ones among $a_1,...,a_n$ then the elementary divisors are $x-a_{n_1},...,x-a_{n_s}$ but I don't know how to justify this.
for the converse i am not very sure how to do this.
thanks for help
 A: It is useful to be a bit more precise by what you mean by elementary divisors of$~A$. Their definition supposes a module over a PID. Here one takes the $F[X]$ module defined by $F^n$ with the action of $X$ given by$~A$. If that decomposes as a direct sum of submodules (stable subspaces) whose annihilating polynomials are powers of irreducible polynomials, those powers are the elementary divisors. Now if $A$ is diagonalisable, you can decompose $F^n$ into $1$-dimensional modules spanned by eigenvectors, one generated by an eigenvector for eigenvalue$~a_i$ is annihilated by $X-a_i$; these are therefore the elementary divisors.
The general equivalence is in fact very straightforward. The mentioned $F[X]$-module always decomposes (non canonically) as direct sum of submodules isomorphic to $F[X]/D_i$ where $D_i$ are the elementary divisors, and these submodules are irreducible (their only submodules are the zero module and the whole submodule). The dimension of $F[X]/D_i$ as $F$-vector space is $\deg(D_i)$. So the elementary divisors are all of degree$~1$ if and only if $F^n$ admits a decomposition as a direct sum of $1$-dimensional subspaces, which happens if and only of $F^n$ admits a basis of eigenvectors for$~A$, in other words if $A$ is diagonalisable over$~F$.
