A similar argument holds if u is algebraic, with p(u) = 0 and q(v) = 0, where q is the image of p. That is, the coefficients of p are mapped, via c, to the coefficients of q. Since c is an isomorphism from K onto L, p is irreducible iff q is irreducible. Thus v is a root of an irreducible polynomial, and L(v) is indeed a finite field extension of L. Again, the extended isomorphism is completely determined by the image of K and u.
As a corollary, the elements u and v are roots of the same irreducible polynomial p(x) over K iff K(u) and K(v) are isomorphic with K fixed. The previous paragraph allows us to start with the trivial isomorphism that fixes K and extend it to u → v, provided u and v are both roots of p(x). Conversely, assume the isomorphism, and raise u and v to successive powers, until they are spanned by lower powers of u and v respectively. Since the map is an isomorphism that fixes K, the resulting polynomials are identical, and u and v are both roots of p(x).