Show that a matrix \(A \in M_n(K)\) is similar to a diagonal matrix over \(K\) if and only if all the elementary divisors of \(A\) are linear (degree 1 polynomials over \(K\)).
Solution
View \(V=K^n\) as a \(K[x]\)-module by \[ x\cdot v=Av. \] The elementary-divisor decomposition has the form \[ V\cong\bigoplus_j K[x]/(p_j^{e_j}), \] where each \(p_j\) is monic irreducible.
If \(A\) is diagonalizable over \(K\), choose an eigenbasis with eigenvalues \(\lambda_1,\dots,\lambda_n\). Then \[ V\cong\bigoplus_{i=1}^n K[x]/(x-\lambda_i), \] so every elementary divisor has degree \(1\).
Conversely, if every elementary divisor has degree \(1\), then each is necessarily \[ x-\lambda \] for some \(\lambda\in K\); in particular no higher power \((x-\lambda)^e\) with \(e>1\) occurs. Hence \[ V\cong\bigoplus_i K[x]/(x-\lambda_i), \] a direct sum of one-dimensional eigenspaces. Therefore \(V\) has an eigenbasis and \(A\) is diagonalizable.
Thus \[ \boxed{A\text{ is diagonalizable over }K \iff\text{ all elementary divisors have degree }1}. \]