The Stacks project

Example 88.15.3. By Examples, Lemma 110.18.1 there exists a local Noetherian $2$-dimensional domain $(A, \mathfrak m)$ complete with respect to a principal ideal $I = (a)$ and an element $f \in \mathfrak m$, $f \not\in I$ with the following property: the ring $A_{\{ f\} }[1/a]$ is nonreduced. Here $A_{\{ f\} }$ is the $I$-adic completion $(A_ f)^\wedge $ of the principal localization $A_ f$. To be sure the ring $A_{\{ f\} }[1/a]$ is nonzero. Let $B = A_{\{ f\} }/ \text{nil}(A_{\{ f\} })$ be the quotient by its nilradical. Observe that $A \to B$ is adic and topologically of finite type. In fact, $B$ is a quotient of $A\{ x\} = A[x]^\wedge $ by the map sending $x$ to the image of $1/f$ in $B$. Every prime $\mathfrak q$ of $B$ not containing $a$ must lie over $(0) \subset A$1. Hence $B_\mathfrak q$ is flat over $A$ as it is a module over the fraction field of $A$. Thus $A \to B$ is naively rig-flat. On the other hand, the map

\[ A_{\{ f\} } \longrightarrow B_{\{ f\} } = (B_ f)^\wedge = B = A_{\{ f\} } /\text{nil}(A_{\{ f\} }) \]

is not flat after inverting $a$ because we get the nontrivial surjection $A_{\{ f\} }[1/a] \to A_{\{ f\} }[1/a]/\text{nil}(A_{\{ f\} }[1/a])$. Hence $A_{\{ f\} } \to B_{\{ f\} }^\wedge $ is not naively rig-flat!

[1] Namely, we can find $\mathfrak q \subset \mathfrak q' \subset B$ with $a \in \mathfrak q'$ because $B$ is $a$-adically complete. Then $\mathfrak p' = A \cap \mathfrak q'$ contains $a$ but not $f$ hence is a height $1$ prime. Then $\mathfrak p = A \cap \mathfrak q$ must be strictly contained in $\mathfrak p'$ as $a \not\in \mathfrak p$. Since $\dim (A) = 2$ we see that $\mathfrak p = (0)$.

Comments (0)

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Your email address will not be published.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 0GGN. Beware of the difference between the letter 'O' and the digit '0'.



View Example 88.15.3 as pdf