Lemma 88.17.6. Let $\varphi : A \to B$ be an arrow of $\textit{WAdm}^{Noeth}$. If $\varphi $ is rig-smooth, then $\varphi $ is rig-flat, and for any presentation $B = A\{ x_1, \ldots , x_ n\} /J$ and prime $J \subset \mathfrak q \subset A\{ x_1, \ldots , x_ n\} $ not containing an ideal of definition the ideal $J_\mathfrak q \subset A\{ x_1, \ldots , x_ n\} _\mathfrak q$ is generated by a regular sequence.
Proof. Let $f \in A$. To prove that $\varphi $ is rig-flat we have to show that $\varphi _{\{ f\} } : A_{\{ f\} } \to B_{\{ f\} }$ is naively rig-flat. Now either by viewing $\varphi _{\{ f\} }$ as a base change of $\varphi $ and using Lemma 88.17.4 or by using the fact that being rig-smooth is a local property (Lemma 88.17.3) we see that $\varphi _{\{ f\} }$ is rig-smooth. Hence it suffices to show that $\varphi $ is naively rig-flat.
Choose a presentation $B = A\{ x_1, \ldots , x_ n\} /J$. In order to check the second part of the lemma it suffices to check $J_\mathfrak q \subset A\{ x_1, \ldots , x_ n\} _\mathfrak q$ is generated by a regular sequence for $J \subset \mathfrak q$ for $\mathfrak q$ maximal with respect to not containing an ideal of definition, see Algebra, Lemma 10.68.6 (which shows that the set of primes in $V(J)$ where there is a regular sequence generating $J$ is open). In other words, we may assume $\mathfrak q$ is rig-closed in $A\{ x_1, \ldots , x_ n\} $. And to check that $B$ is naively rig-flat, it also suffices to check that the corresponding localizations $B_\mathfrak q$ are flat over $A$.
Let $\mathfrak q \subset A\{ x_1, \ldots , x_ n\} $ be rig-closed with $J \subset \mathfrak q$. By Lemma 88.14.9 we may choose an $f \in A$ mapping to a unit in $A\{ x_1, \ldots , x_ n\} /\mathfrak q$ and such that the prime ideal $\mathfrak p'$ in $A_{\{ f\} }$ induced is rig-closed. Below we will use that $A_{\{ f\} }\{ x_1, \ldots , x_ n\} = A\{ x_1, \ldots , x_ n\} _{\{ f\} }$; details omitted. Consider the diagram
\[ \xymatrix{ A\{ x_1, \ldots , x_ n\} _{\mathfrak q} / J_\mathfrak q \ar[r] & A_{\{ f\} }\{ x_1, \ldots , x_ n\} _{\mathfrak q'}/ J A_{\{ f\} }\{ x_1, \ldots , x_ n\} _{\mathfrak q'} \\ A\{ x_1, \ldots , x_ n\} _{\mathfrak q} \ar[r] \ar[u] & A_{\{ f\} }\{ x_1, \ldots , x_ n\} _{\mathfrak q'} \ar[u] \\ A \ar[r] \ar[u] & A_{\{ f\} } \ar[u] } \]
The middle horizontal arrow is faithfully flat as it is a local homomorphism of local rings and flat as $A_{\{ f\} }\{ x_1, \ldots , x_ n\} $ is the completion of a localization of the Noetherian ring $A\{ x_1, \ldots , x_ n\} $. Similarly the bottom horizontal arrow is flat. Hence to show that $J_\mathfrak q$ is generated by a regular sequence and that $A \to A\{ x_1, \ldots , x_ n\} _{\mathfrak q} / J_\mathfrak q$ is flat, it suffices to prove the same things for $J A_{\{ f\} }\{ x_1, \ldots , x_ n\} _{\mathfrak q'}$ and $A_{\{ f\} } \to A_{\{ f\} }\{ x_1, \ldots , x_ n\} _{\mathfrak q'}/ J A_{\{ f\} }\{ x_1, \ldots , x_ n\} _{\mathfrak q'}$. See Algebra, Lemma 10.68.5 or More on Algebra, Lemma 15.32.4 for the statement on regular sequences. Finally, we have already seen that $A_{\{ f\} } \to B_{\{ f\} }$ is rig-smooth. This reduces us to the case discussed in the next paragraph.
Let $\mathfrak q \subset A\{ x_1, \ldots , x_ n\} $ be rig-closed with $J \subset \mathfrak q$ such that moreover $\mathfrak p = A \cap \mathfrak q$ is rig-closed as well. By the characterization of rig-smooth algebras given in Lemma 88.4.2 after reordering the variables $x_1, \ldots , x_ n$ we can find $m \geq 0$ and $f_1, \ldots , f_ m \in J$ such that
$J_\mathfrak q$ is generated by $f_1, \ldots , f_ m$, and
$\det _{1 \leq i, j \leq m}(\partial f_ j/ \partial x_ i)$ maps to a unit in $A\{ x_1, \ldots , x_ n\} _\mathfrak q$.
By Lemma 88.14.12 the fibre ring
\[ F = A\{ x_1, \ldots , x_ n\} \otimes _ A \kappa (\mathfrak p) \]
is regular. Observe that the $A$-derivations $\partial / \partial x_ i$ extend (uniquely) to derivations $D_ i : F \to F$. By More on Algebra, Lemma 15.48.3 we see that $f_1, \ldots , f_ m$ map to a regular sequence in $F_\mathfrak q$. By flatness of $A \to A\{ x_1, \ldots , x_ n\} $ and Algebra, Lemma 10.99.3 this shows that $f_1, \ldots , f_ m$ map to a regular sequence in $A\{ x_1, \ldots , x_ m\} _\mathfrak q$ and the quotient by these elements is flat over $A$. This finishes the proof. $\square$
Comments (0)