I have a question about an identification used in Morishita's "Knots and Primes" (page 25):
Let $A$ be an integral domain. We consider a finite Galois algebra $B$ over $A$ (the latter means that $K=Frac(B)$ is Galois over $F=Frac(A)$.
The author defines $Gal(B/A):= Aut(B/A)$.
Now the question(*): Why is $Aut(B/A)= Aut(K/F)$?
(the latter equals $Gal(K/F)$ by definition)
The problem is the following: I know that by universal property of fraction constructions every $A$-automorphism $\phi:B \to B$ can be uniquely extended to $\bar{\phi}: K \to K$.
Therefore we obtain the inclusion $Aut(B/A) \subset Aut(K/F)$.
But is every automorphism from $Aut(K/F)= Gal(K/F)$ already determined by an automorphism from $Aut(B/A)$?
The main problem is that generally an automorphism from $Gal(K/F)$ don't need to preserve the ring $B$, right? (see this former thread:Do automorphisms of quotient fields preserve the underlying ring?)
The only restriction is that it preserves $F=Frac(A)$. Does it suffice in this special case to obtain the equality (*)?
For the definition of "galois algebra" given in the text, this equality of automorphism groups follows since $B$ is the integral closure of $A$ in $L$.
Explicitly, we have $\sigma(B)=B$ for any $\sigma$ that preserves $A$, since applying $\sigma$ to a monic polynomial with $\alpha\in B$ as a root yields a monic polynomial with $\sigma(\alpha)$ as a root.
This is false in general however, if your definition of "galois algebra" is that $B/A$ is an extension of integral domains such that $Quot(B)=L$, $Quot(A)=K$, with $L/K$ galois. An example of this occurs in $\mathbb{Q}(i)/\mathbb{Q}$, take the subrings $\mathbb{Z}(i)_{(2+i)}/\mathbb{Z}_{(2)}$. Here the subscripts denote localisation at the prime ideals $(2+i)$ and $(2)$.
Complex conjugation does not preserve this subring, $2+i$ is not a unit in $\mathbb{Z}(i)_{(2+i)}$, since it generates a prime ideal, but $2-i$ is a unit, since it has inverse $\frac{1+i}{3+i}$ which has denominator not in $(2+i)$.