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I have a question about a step in the proof of 5.3.17 in Liu's "Algebraic Geometry" (page 201):

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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 $B = A \otimes_R S$?

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    $\begingroup$ Why need a base change? $B$ is a free $A$-module (of finite rank equal to $\deg P$). $\endgroup$
    – user26857
    Feb 17, 2019 at 9:40
  • $\begingroup$ @user26857:For this argument we then also need that $P$ irreducible? But every lift $P$ of $\tilde{P}$ is irreducible since a factorisation of $P$ would provide a factorisation of $\tilde{P}$? does this argument work? $\endgroup$
    – user267839
    Feb 17, 2019 at 15:03
  • $\begingroup$ No, we don't need $P$ irreducible: monic it's enough. $\endgroup$
    – user26857
    Feb 17, 2019 at 15:28
  • $\begingroup$ Please upvote and accept the answer. (If consider it incomplete feel free to ask for more details.) $\endgroup$
    – user26857
    Feb 18, 2019 at 21:50

1 Answer 1

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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.

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