Question concerning a notation in Vakil‘s notes

Question

This is Exercise 5.4.H. of Vakil‘s notes on algebraic geometry. We want to show that if $A$ is a UFD where $2$ is invertible, and $z^2-f$ is irreducible in $A[z]$, where $f\in A$ is squarefree, then $\operatorname{Spec}A[z]/(z^2-f)$ is normal. In the hint he sets $B=A[z]/(z^2-f)$ and I think we want to show that $B$ is integrally closed. For this, we assume there is $F(T)\in B(T)$ monic with a root $\alpha$ in $\operatorname{Quot}(B)\setminus\operatorname{Quot}(A)$.
First question: why do we assume $\alpha\notin\operatorname{Quot}(A)$, rather then $\alpha\notin B$? Second Question: He goes on to say that „replacing $F(T)$ by $\bar{F}(T)F(T)$ we may assume $F(T)\in A[T]$.“ What does he mean with $\bar{F}(T)$?

If you have the feeling that I might be able to answer my first question by myself doing this exercise, please ignore it. However, thanks in advance for any clarification concerning the second question.

Answer 1

To simplify the notation, let $K,L$ be the fraction fields of $A,B$, respectively.

I agree that it is a mistake to take $\alpha \in L-K$ and it should say $\alpha \in L-B$. In fact, in practice one might even worry that some element in $K$ (not in $A$) becomes integral over $B$ (of course this can't happen here because $A$ is integrally closed). In reality it is not even necessary to assume $\alpha \in L-K$. If you drop that assumption, nothing changes about the proof. What the exercise shows is that if $\alpha$ is integral over $B$ then it was in $B$ in the first place. I guess that assumption makes the proof one by contradiction, but it becomes a direct proof if you drop it.

The polynomial $\bar F(T)$ is the polynomial obtained from $F(T)$ by replacing each of its coefficients by their conjugate in $L$. It's clear that $$\overline{F(T)\bar F(T)} = \bar F(T) \bar{\bar{F}}(T) = \bar F(T) F(T)$$ which is to say that all of the coefficients of $\bar F(T) F(T)$ are fixed by Galois, hence in $A$.