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I've read the claim about base change for flat modules in several sources (Lang's Algebra, Hartshorne's Algebraic Geometry, A&M), but unfortunately it isn't proven anywhere. The claim is that

Let $A$ be a commutative $R$-algebra, and $F$ a flat $R$-module. Then $A\otimes_R F$ is a flat $A$-module.

The proof is supposedly immediate, but sadly not to me. Is there a nice standard proof of the claim I could read? Thank you.

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up vote 4 down vote accepted

It amounts to the following observation: if $P$ is any $A$-module, $$P \otimes_R M \cong (P \otimes_A A) \otimes_R M \cong P \otimes_A (A \otimes_R M)$$ Hence, if we have a short exact sequence of $A$-modules $$0 \longrightarrow P'' \longrightarrow P \longrightarrow P' \longrightarrow 0$$ the tensored sequence $$0 \longrightarrow P'' \otimes_R M \longrightarrow P \otimes_R M \longrightarrow P' \otimes_R M \longrightarrow 0$$ is also a short exact sequence, by hypothesis, and via the isomorphism discussed above, $$0 \longrightarrow P'' \otimes_A (A \otimes_R M) \longrightarrow P \otimes_A (A \otimes_R M) \longrightarrow P' \otimes_A (A \otimes_R M) \longrightarrow 0$$ is also a short exact sequence, hence $(A \otimes_R M)$ is a flat $A$-module.

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Thank you Zhen Lin. – Hailie Mathieson Mar 17 '12 at 16:25

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