Corollary

With the hypotheses and notation Of Theorem 2, let $f_1, f_2$ be two homomorphisms of $A$ to a field $L$ with the same restriction to $A$. Then there exists $\sigma \in \mathscr{G}$ such that

for all $x \in A'$.

Let $p'_i$ be the kernel of $f_i$ ($i = 1, 2$) which is a prime ideal of $A'$; by hypothesis $p'_1 \cap A = p'_2 \cap A$ and this intersection is a prime ideal $p$ of $A$; there therefore exists $\tau \in \mathscr{G}$ such that $\tau.p'_2 = p'_1$ (Theorem 2 (i)); replacing $f_1$ by the homomorphism $x \mapsto f_1(\tau.x)$ we may then assume that $p'_2 = p'_1$ (an ideal which we shall denote by $p'$). By taking the quotient we then derive from $f_1$ and $f_2$ two injective homomorphisms $f'_1, f'_2$ from $A'/p'$ to $L$ which therefore extend to two injective homomorphisms $f''_1, f''_2$ from the field of fractions $k'$ of $A'/p'$ to $L$. As $k'$ is a quasi-Galois extension of $k$, so is $k''_1 = f''_1(k')$ and $k''_2 = f''_2(k')$ (k being identified with a subfield of $L$) and, as there is a k-isomorphism of $k''_1$ onto $k''_2, k''_1 = k''_2$ (Algebra, Chapter V, § 6, Proposition 6). Thus $f''_1 \circ {f''_2}^{-1}$ is a k-automorphism of $k'$; by Theorem 2 (ii) it is therefore of the form 6, where $\sigma \in \mathscr{G}^z(p')$. In particular, for all $x \in A'$ the elements $f_2(x)$ and $f_1(\sigma.x)$ are equal.

Remarks

(1) Note that under the hypotheses of Theorem 2 $k'$ may be infinite over $k$ if $k'$ is not separable over $k$ (Exercise 9).

(2) Clearly $k$ is a Galois extension of $k$ if the field $k$ is perfect. It is then finite over $k$.