Flatness of subring 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.
 A: 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 $T$. 
