Does flatness ascend through a free ring map?

Question

Suppose I have a composition of ring maps (rings commutative with identity)

$$R \xrightarrow{f} S \xrightarrow{g} T$$

where $f$ and $g$ are monomorphisms, $S$ is $R$-free, and $T$ is (faithfully) $R$-flat.

Is it also true that $T$ is $S$-flat?

For example, if $R[[x]]$ is flat over $R$, is $R[[x]]$ flat over $R[x]$? What if we replace $R[[x]]$ by an overring of $R[x]$?

I know that faithfulness will not generally lift like this (if $T$ is a localization of $S$ it certainly won't) but it seems intuitive to me that flatness would ascend through a free ring extension.

Answer 1

Bourbaki's counter-example:

Let $K$ be a field. Then $K[X]$ is free, $K(X)$ is faithfully flat over $K$, but is certainly not faithfully flat over $K[X]$ since this would imply $fK(X)\ne K(X)$ for any irreducible polynomial in $K[X]$.