Let c() be a field isomorphism from K onto L, and let c(u) = v, where u is transcendental over K and v is transcendental over L. It is easy to extend the isomorphism to map K(u) onto L(v). Since isomorphisms preserve addition and multiplication, the extended isomorphism is completely determined by the image of K and u.
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).