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Let $R$ be a ring with 1, not necessarily commutative, with no zero divisors. Suppose $S$ is a flat extension of $R$. What additional assumptions, if any, would allow us to assert that a subring $R \subseteq T \subseteq S$ must also be flat over $R$?

I'm interested in seeing relevant (counter-)examples, as well as any necessary or sufficient conditions you can think of.

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    $\begingroup$ If $R$ is commutative, as a localization of $R$, its quotient field $K$ is flat. So the additional assumptions would have to be strong enough to guarantee that every intermediate ring between $R$ and $K$ is flat over $R$. $\endgroup$
    – moonlight
    May 31, 2017 at 10:40
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    $\begingroup$ At least in the commutative case, where being a principal ideal domain is equivalent to "every sub-module of a free module is free", a sufficient condition is $R$ a PID and $S$ free as $R$-module. $\endgroup$
    – Ben
    Jun 1, 2017 at 6:11

1 Answer 1

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I made some progress under the assumption that $R$ is commutative. We say $R$ satisfies "Condition 1" iff the following holds:

Given arbitrary ring extensions $R \subseteq T \subseteq S$, if $S$ is a flat $R$-module then so is $T$.

It turns out that:

$R$ satisfies Condition 1 if and only if $R$ is a Prufer domain.

$\textit{Proof:}$

($\Leftarrow$) If $R$ is a Prufer domain, then every submodule of a flat $R$-module is flat. Condition 1 follows immediately.

($\Rightarrow$) Suppose $R$ is not a Prufer domain. Then there is an ideal $I \subseteq R$ such that $I$ is not flat as an $R$-module. As additive abelian groups, define: $E \doteq R \oplus I$ and $F \doteq R \oplus R$. Define a multiplication on $F$ as follows: $(a,x) \cdot (b,y) \doteq (ab, ay+bx)$. Endow $E$ with the same multiplication. It is easily checked that this turns $E,F$ into commutative, unital rings, and we have $R \subseteq E \subseteq F$ naturally.

We note that as R-modules, we have $E \cong R \oplus I$ and $F \cong R \oplus R$. Then $F$ is flat, but $E$ is not since $I$ isn't flat. Then Condition 1 fails. $\Box$

So if we are considering your question for $R$ not a Prufer domain, then we must impose some restriction on $S$ to guarantee flatness of $T$.

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    $\begingroup$ Actually the converse holds for flat overrings; see Richman, Generalized quotient rings, Theorem 4. $\endgroup$
    – user26857
    Jun 4, 2017 at 19:44
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    $\begingroup$ @user26857 that result actually implies the one above quite readily, too. Thanks for the source $\endgroup$
    – M10687
    Jun 4, 2017 at 20:34

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