Regular ring maps

Recall that a map of Noetherian rings $f: A \to B$ is regular if it is flat and all fiber rings are geometrically regular. Note that even if $f$ is a flat local map of local Noetherian rings, one truly has to check geometric regularity of all fibers, not just the closed fiber. This is kind of annoying, not only from a practical point of view, but also philosophically.

The point of this short post is to note that under a mild assumption, it’s enough to look at the closed fiber only:

Lemma. Let $f:A \to B$ be a flat local map of Noetherian local rings, and assume that $A$ is quasi-excellent. Then the following are equivalent:

All fiber rings of $f$ are geometrically regular, i.e. $f$ is a regular ring map.
$\overline{B} = B/ \mathfrak{m}_A B$ is geometrically regular over $k = A/\mathfrak{m}_A$ .
$k \to \overline{B}$ is formally smooth in the $\mathfrak{m}_{\overline{B}}$ -adic topology.
$A \to B$ is formally smooth in the $\mathfrak{m}_B$ -adic topology.

Proof. 1. $\Rightarrow$ 2. is trivial. The equivalences 2. $\Leftrightarrow$ 3. $\Leftrightarrow$ 4. are proved in Stacks, Tag 07NQ. The implication 4. $\Rightarrow$ 1. is a theorem of M. Andre (Localisation de la lissite formelle, Manuscripta Math. 13, pp. 297-307); this is the only place we use the quasi-excellence of $A$ .

Corollary. Let $K \to L$ be a separable extension of fields. Then $K[[x_1,\dots,x_n]] \to L[[x_1,\dots,x_n]]$ is regular.

Proof. Flatness is trivial (e.g. by the local criterion), and condition 2. in the lemma above is trivial.

Is there a truly easy proof of this corollary? Even if you give yourself access to some big hammers like Popescu’s theorem, I don’t see a simple argument.

Corollary. Let $K \to L$ be any extension of characteristic zero nonarchimedean fields, and let $A$ be any tft $K$ -algebra. Then $A \to A \widehat{\otimes}_K L$ is regular.

Proof. It suffices to check that $A_\mathfrak{m} \to (A \widehat{\otimes}_K L)_\mathfrak{n}$ is regular, where $\mathfrak{n}$ is any maximal ideals upstairs which contracts to a maximal ideal downstairs. But now flatness is easy, and condition 2. in the lemma is easy to verify by some standard structural properties of tft algebras.

Andre’s theorem is stated on p. 260 of Matsumura’s Commutative Ring Theory, where Matsumura calls it “an extremely strong theorem.” I remember reading this sentence in grad school and completely failing to understand the point (or maybe even the statement…) of Andre’s result. But now I get it.

Update (April 14). Here’s another corollary which seems useful.

Corollary. Let $A \to B$ be a regular map of Noetherian rings, with $A$ quasi-excellent. Then

( $A[[x_1,\dots,x_n]]$ is quasi-excellent and) $A[[x_1,\dots,x_n]] \to B[[x_1,\dots,x_n]]$ is a regular ring map.
For any ideal $I \subset A$ , (the $I$ -adic completion $\hat{A}$ is quasi-excellent and) the induced map $\hat{A} \to \hat{B}$ on $I$ -adic completions is a regular ring map.

Proof. By a theorem of Gabber, if $A$ is quasi-excellent, then so are $A[[x_1,\dots,x_n]]$ and $\hat{A}$ , cf. this nice paper for a detailed discussion. Part 2. is now an easy consequence of 1., by writing $\hat{A} \to \hat{B}$ as the base change of $A[[x_1,\dots,x_n]] \to B[[x_1,\dots,x_n]]$ along a suitable surjection $A[[x_1,\dots,x_n]] \to \hat{A}$ .

For 1., we reduce by an easy induction to the case $n=1$ . Let $Q \subset B[[x]]$ be any prime ideal, with contractions $P \subset A[[x]]$ , $\mathfrak{q} \subset B$ , $\mathfrak{p} \subset A$ . Then we get a commutative square

$A[[x]]_P \to B[[x]]_Q$
$\quad\quad\quad\quad\uparrow\quad\quad\quad\quad\uparrow$
$A_{\mathfrak{p}}[[x]] \to B_{\mathfrak{q}}[[x]]$

of local Noetherian rings where all arrows are flat ring maps. Moreover, the vertical arrows are localizations, and the horizontal arrows are flat local maps. We need to check that the upper horizontal arrow is regular for any $Q$ . Since the upper row comes from the lower row by localization, it’s enough to check that the lower horizontal arrow is regular. But now we win, because $A_{\mathfrak{p}}[[x]] \to B_{\mathfrak{q}}[[x]]$ is a flat local map with quasi-excellent source, and the closed fiber of this map is the same as the closed fiber of the (local) map $A_{\mathfrak{p}} \to B_{\mathfrak{q}}$ . Since $A \to B$ is regular by assumption, $B_{\mathfrak{q}}/\mathfrak{p}B_{\mathfrak{q}}$ is geometrically regular over $\mathrm{Frac}(A/\mathfrak{p})$ , and now we’re done by applying condition 2. in the Lemma. $\square$

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