I give you two matrices over a field. How would you tell if they are conjugate or not? What theorem are you using? State it. How does it apply to this situation? Why is \(k[x]\) a PID? If two matrices are conjugate over the algebraic closure of a field, does that mean that they are conjugate over the base field too?
Solution
For \(A\in M_n(k)\), make \(k^n\) a \(k[x]\)-module by \(x\cdot v=Av\). Then \(A\) and \(B\) are similar over \(k\) iff the corresponding \(k[x]\)-modules are isomorphic. Since \(k[x]\) is Euclidean (degree is a Euclidean function), it is a PID, and the structure theorem for finitely generated modules over a PID gives the invariant factors. Equivalently, these give the rational canonical form, which is a complete similarity invariant.
Thus one tests similarity by comparing the invariant factors (or rational canonical forms) of \(A\) and \(B\).
Moreover, if \(A,B\in M_n(k)\) become similar over an extension field \(K/k\)—in particular over \(\bar k\)—then they were already similar over \(k\). A proof uses ranks: for every \(f\in k[x]\) and every \(r\ge1\), similarity over \(K\) gives \[ \operatorname{rank}_K f(A)^r=\operatorname{rank}_K f(B)^r. \] A matrix with entries in \(k\) has the same rank after scalar extension, so the corresponding ranks over \(k\) are equal. Applying this to powers of each irreducible factor recovers the elementary divisors, hence the invariant factors, of \(A\) and \(B\). Therefore their rational canonical forms over \(k\) coincide.