Can it be that $R[[x]]$ is flat over $R$ but not over $R[x]$?

Question

I have been trying to better understand the ascension of flatness. After asking a poor question in general terms (see Does flatness ascend through a free ring map?), I realized that I do not understand even the fundamental polynomial/power series situation.

Namely, consider $$R \rightarrow R[X] \rightarrow R[[X]]$$

If $R[[X]]$ is faithfully flat as an $R$-module, will flatness "ascend" to $R[[X]]$ as an $R[X]$-module?

The flatness of $R[[X]]$ over $R[X]$ has been asked about on MO but without answer or relevant comment.

I am guessing that flatness need not ascend since if we pick a countable coherent ring of large global dimension then $R[X]$ shouldn't (?) be coherent and I see no reason why it's completion should (?) be flat (but I have no confidence in this guess). However it seems I do not know enough to produce (and verify) a counterexample.

I would really appreciate it if someone could shed light on this particular situation and / or share some thoughts on the topic of "ascension of flatness" generally, i.e. what features of the rings $R \rightarrow S \rightarrow T$ with $T$ faithfully flat as an $R$-module could cause $T$ to be faithfully flat as an $S$-module.

Answer 1

Let $D$ be a domain with fraction field $K$, denote by $D'$ the integral closure of $D$ and by $D''$ the complete integral closure (that is, the ring of elements $\frac{a}{b} \in K$ such that for some $d \in D, d(\frac{a}{b})^n \in D \ \forall n$)

Lemma: If $D[[X]]$ is flat over $D[X]$, then $D'' = D'$.

Proof: Let $\frac{a}{b} \in K$ almost integral such that $d \frac{a^n}{b^n} \in D \ \forall n$. Then let $G = d\sum_{n=0}^{\infty} (\frac{aX}{b})^n \in D[[X]]$. We have $(b-aX)G = bd$. If $D[[X]]$ is flat then $$\big((b-aX)\cap(bd)\big)D[[x]] = (b-aX)D[[X]]\cap(bd)D[[X]] = bdD[[X]]$$

Thus there exists $f \in (b-aX) \cap (bd)$ such that $fF = bd$. Since $bd \mid f$, write $f = f' bd$ where $f'_0 = 1$, $f' \in D[X]$. Since $b-aX \mid f$, write $g(b-aX) = f$. Thus $\frac{g}{d}(1 - \frac{a}{b}X) = f'$ is a factorization of $f'$ in $K[X]$, and since the lowest coefficients are all $1$, it's well known that this is a factorization in $D'[X]$. $\square$

So in order to ensure $R[[X]]$ is (faithfully) flat over $R$ but not flat over $R[X]$, a coherent domain in which $D''\not= D'$ will suffice. This is easy, since valuation domains are coherent and are completely integrally closed precisely when they are rank 1. We conclude:

Any valuation domain $V$ of rank > 1 is such that $V[[x]]$ is (faithfully) flat over $V$ but $V[[X]]$ is not flat over $V[X]$.