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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.

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The answer is negative.

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.

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