# Base Change Preserves Flatness

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 $$B = A \otimes_R S$$?

• Why need a base change? $B$ is a free $A$-module (of finite rank equal to $\deg P$). Feb 17, 2019 at 9:40
• @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? Feb 17, 2019 at 15:03
• No, we don't need $P$ irreducible: monic it's enough. Feb 17, 2019 at 15:28
• Please upvote and accept the answer. (If consider it incomplete feel free to ask for more details.) Feb 18, 2019 at 21:50

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.