# Intersection of flat submodule with direct summand

Let $$R$$ be a (commutative) domain, $$M$$ a flat $$R$$-module which decomposes as $$M=A\oplus B$$ and $$N$$ a (not necessarily pure) flat submodule of $$M$$. Is it the case that $$N \cap A$$ is always a pure submodule of $$N$$? Equivalently, is the image of $$N$$ in the projection of $$M$$ onto $$B$$ always flat?

Let us try to check the Wikipedia's equational criterion for purity: Writing the elements of $$A \oplus B$$ as pairs $$(a,b)$$, we are given a matrix $$X$$ of elements of $$R$$ and two vectors of pairs $$(\overline a, \overline b) = ((a_1, b_1), \dots, (a_n, b_n))^{\text T} \in N^n$$, $$(\overline y, 0) \in (N \cap A)^n$$ such that $$X (\overline a, \overline b) = (\overline y, 0).$$ We seek a vector of the form $$(\overline{a'}, 0) \in (N \cap A)^n$$ which could replace $$(\overline a, \overline b)$$ in the equation. Now the only thing I can do is to use the flatness of $$B$$ via the Theorem 3.90 (a) from Pete Clark's notes (pages 74–75), yielding a matrix $$Y$$ and a vector $$\overline c \in B^k$$ such that $$XY = 0 \quad \text{and} \quad Y\overline c = \overline b.$$ However, I don't see any way to use these new data, so I got stuck here. I couldn't use the hypothesis of $$N$$ being flat either.

Currently my guess is that there's a counterexample.

A bit more modest goal would be to determine if $$N \cap A$$ is always flat or not. I'm aware of the fact that intersection of two flat submodules need not be flat in general, but here we have the extra condition of $$A$$ being a direct summand.

Of course, the answer is trivially positive for modules over Prüfer domains, where flat = torsion-free.

Let $R$ be a Noetherian UFD such that there is a rank 2 non-free finitely generated projective module $M$; various examples of such can be found in the answers to this MO question. Further, let $Q$ be the field of fractions of $R$. As $M$ is of rank 2, it embeds into $Q\oplus Q$.
Let $N$ be the intersection of $M$ with the direct summand $Q \oplus 0$: If $N$ were pure in $M$, it would imply $M/N$ being a finitely generated flat—hence projective—module, thus $M$ would be the direct sum of two rank 1 projective modules. Since $R$ is a UFD, rank 1 projectives are free, thus forcing $M$ to be free as well—a contradiction.