Lemma 110.12.6. If $R$ is a valuation ring of dimension $> 1$, then $R[[x]]$ is flat over $R$ but not flat over $R[x]$.
Proof. The arguments above show that this is true if we can show that $R$ is not completely normal (valuation rings are normal, see Algebra, Lemma 10.50.3). Let $\mathfrak p \subset \mathfrak m \subset R$ be a chain of primes. Pick nonzero $x \in \mathfrak p$ and $y \in \mathfrak m \setminus \mathfrak p$. Then $x y^{-n} \in R$ for all $n \geq 1$ (if not then $y^ n/x \in R$ which is absurd because $y \not\in \mathfrak p$). Hence $1/y$ is almost integral over $R$ but not in $R$. $\square$
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