Lemma 15.3.4. Let $R$ be a ring. Let $I \subset R$ be an ideal. Assume that every element of $1 + I$ is a unit (in other words $I$ is contained in the Jacobson radical of $R$). Let $M$ be a finite flat $R$-module such that $M/IM$ is a projective $R/I$-module. Then $M$ is a finite projective $R$-module.

Proof. By Algebra, Lemma 10.78.5 we see that $M_\mathfrak p$ is finite free for all prime ideals $\mathfrak p \subset R$. By Algebra, Lemma 10.78.2 it suffices to show that the function $\rho _ M : \mathfrak p \mapsto \dim _{\kappa (\mathfrak p)} M \otimes _ R \kappa (\mathfrak p)$ is locally constant on $\mathop{\mathrm{Spec}}(R)$. Because $M/IM$ is finite projective, this is true on $V(I) \subset \mathop{\mathrm{Spec}}(R)$. Since every closed point of $\mathop{\mathrm{Spec}}(R)$ is in $V(I)$ and since $\rho _ M(\mathfrak p) = \rho _ M(\mathfrak q)$ whenever $\mathfrak p \subset \mathfrak q \subset R$ are prime ideals, we conclude by an elementary argument on topological spaces which we omit. $\square$

Comment #3653 by Brian Conrad on

Comment #3749 by on

OK, I replaced "radical" by "Jacobson radical" in all places where appropriate and whenever we used the notation $\text{rad}(R)$ I have added text saying this denotes the Jacobson radical. See the corresponding changes here.

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