4
$\begingroup$

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.

$\endgroup$
5
$\begingroup$

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.