Sorry, I have found that always p0 = p.
Let F and K be fields, F subfield of K. Given a matrix T in Fn (the algebra of n x n matrices over F), let p be the minimal polynomial of T over F and p0 the minimal polynomial of T over K, considering T as an element of Kn. Then p0 | p, since p0 divides all polynomials over K (and hence all polynomials over F) which are satisfied by T. Of course p0 | p in K[x] but not necessarily in F[x]. My question is: how could I find a matrix T in Fn such that p0 is different from p?
For example, let's make F = Q, the field of rational numbers, and K = F( ), the field obtained by adjoining to F (within say the field of real numbers. With n = 2, if T = (a_ij) is the matrix with a11= 1, a12= 1/2, a21= 2, a22= -1, then p= x^2 - 2, whose roots are all in K. But then p0 = p. Perhaps with n a little larger. And perhaps, after all, p0 is always equal to p.