The Stacks project

Lemma 109.12.6. 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 10.50.10). 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 $R$. $\square$

Comments (0)

There are also:

  • 3 comment(s) on Section 109.12: Nonflat completions

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F1Z. Beware of the difference between the letter 'O' and the digit '0'.