If $M$ is a flat $R$-module, is $M/IM$ a flat $R/I$-module?

Question

Let $R$ be a Noetherian (local) ring, and let $M$ be a finitely generated, flat $R$-module. Further, let $I$ be an ideal of $R$.

Question: Is $M/IM$ flat over $R/I$?

Answer 1

See Matsumura's Commutative Ring Theory, Theorem 22.3.

If $A$ is a ring and $I$ an ideal of $A$, and either

(a) $I$ is nilpotent ideal or

(b) $A$ is Noetherian ring and $M$ is $I$-adically ideal-separated,

then we have a bunch of equivalent conditions:

(1) $M$ is flat over $A$;

(3) $M/IM$ is flat over $A/I$ and $I\otimes_AM=IM$.

Edit: Implication (1) to (3) holds without any assumption on $M$ (see Matsumura's note and proof).

Answer 2

Edit: This always holds, and no assumptions are needed. If $R$ is any ring, and $M$ any $R$-module, $I$ an $R$-ideal, then $M$ flat over $R$ $\implies M/IM$ flat over $R/I$ and the natural map $I \otimes M \to IM$ is an isomorphism.

Answer 3

$M/IM \simeq M \otimes_A A/I$. Flatness is preserved under any extension of scalars $M \leadsto M\otimes_A B$, so yes.