The Stacks project

Example 86.15.3. By Examples, Lemma 108.17.1 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 $\mathfrak p = (0)$.

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