I have a question about a step in the proof of 5.3.17 in Liu's "Algebraic Geometry" (page 201):
Why is $B = A[T] / (P(T))$ flat over $A$?
Does anybody see a clever base change to a flat module?
My attempt was to consider the flat map $k(y) \to k(y)[T]/P(T)$ but the can quotient map $A \to A/M_y =k(y)$ doesn't provide the base change in "correct" direction, namely if $R \to S$ is flat and $R \to R'$ is a ring map then $R' \to R' \otimes_R S$ is flat,
so $A \to k(y)$ provides a map "in a wrong direction" for this criterion.
Can anybody see another flat map $R \to S$ with $R \to A$ such that $B = A \otimes_R S$?
First note that $B\cong A^{\oplus n}$ is a free $A$-module of rank $n=\deg(f)$, because $P$ is monic (you do not need it to be irreducible, the isomorphism sending a free generator to the class of the corresponding power of the variable still works).
Now you can show that free modules of finite rank are flat. Let $$ 0\to M_1\to M_2\to M_3\to 0$$ be a short exact sequence of $A$-modules. Tensoring with $A^{\oplus n}$ we obtain $$ 0\to M_1^{\oplus n}\to M_2^{\oplus n}\to M_3^{\oplus n}\to 0$$ which is exact, because it is a finite direct sum of short exact sequences.