Question about Fibers $X_y$ I have some questions about the proof of 5.3.17 in Liu's "Algebraic Geometry" (page 201):

Hypotheses of Thm. 3.15:



*

*Why is $B = A[T] / (P(T))$ a flat A-module? That's clear that $A[T]$ should be as a free A-module flat, but I'm not sure if flatness is stable under quotient structure.

*That's not clear how the Prop. 3.1.24 provides the isomorphism $O_{Spec B} \cong q_* O_{X_B}$.
Here the statement of 3.1.24:



*Why is fiber of $q$ over a closed point of $Spec B$ isomorphical to $X_y \times_{Spec k(y)} Spec E$? 

 A: *

*Note that $A[T]/(P(T))$ is a free module and hence flat. Just check using the fact that $P$ is monic and irreducible. Note that $P$ is irreducible since $\tilde{P}$ is irreducible.  

*Apply the flat base change theorem for the cartesian diagram
$\require{AMScd}
\begin{CD}
X \times_Y Spec(B) @>{g}>> X\\
@V{q}VV @V{f}VV \\
Spec(B) @>{h}>> Spec(A)
\end{CD}
$
to get $h^*f_* \mathcal{O}_{X} \cong q_*g^*\mathcal{O}_X \cong q_*\mathcal{O}_{X_B}$ also note that you have been given $f_*\mathcal{O}_X \cong \mathcal{O}_{Spec(A)}$. Now plug this in the left hand side of the isomorphism above to get the desired isomorphism. Note that one should check that this is same as the isomorphism mentioned in proposition 1.24 as there is a lot of maps coming into picture.


*The fiber of $q$ over a closed point in $Spec(B)$ is given by the fiber diagram


$\require{AMScd}
\begin{CD}
X' @>{}>> X \times_Y Spec(B) @>{g}>> X\\
@VVV @V{q}VV @V{f}VV \\
Spec(E) @>>> Spec(B) @>{h}>> Spec(A)
\end{CD}
$
The outer square is cartesian. Note that the bottom arrow is composition of two morphism $Spec(E) \rightarrow Spec(k(y)) \rightarrow Spec(A)$. Then consider the following
$\require{AMScd}
\begin{CD}
X_y \times_{Spec(k(y))} Spec(E) @>{}>> X_y @>{g}>> X\\
@VVV @VVV @V{f}VV \\
Spec(E) @>>> Spec(k(y)) @>{h}>> Spec(A)
\end{CD}
$
This is a diagram with both the squares cartesian and hence the the whole big square is cartesian. Note that the bottom map in the two big squares in the diagrams under section 3 is same and hence the fiber product must be the same(universal property of fiber diagrams). Hence fiber of $q$ over the closed point that is $X'$ is nothing but $X_y \times_{Spec(k(y))} Spec(E)$.
