## Things to do during a pandemic

• Try and fail to buy toilet paper at Edeka.
• Make ketchup from scratch.
• Watch 30 episodes of Iron Chef.
• Annoy your wife by following her around the apartment, or by listening to old blues songs too loudly.
• Try and fail to buy toilet paper at Aldi’s and at Lidl.
• Meet with your masters and PhD students on Zoom.
• Gain a proper appreciation for Auslander-Gorenstein rings.
• Buy 72 rolls of Polish toilet paper on Amazon and pay extra for it to be absurdly shipped from Britain, because that’s the only option for some reason.
• Start a cool new joint project with BB and spend way too much time thinking about it (more on this soon!).
• Promise CJ you’ll finish writing the proof of a certain result in a certain nearly final version of a certain paper and then somehow don’t finish doing it (yet). See previous item.
• Move one kilometer to a much cheaper and much nicer apartment.
• Write silly blog posts that (hopefully) no one will read.

*The Lithuanian VP, not the French VP.

## More commutative algebra

Let $B$ be a Gorenstein local ring, and let $M$ be a finitely generated $B$-module. It is a standard fact that $M$ is Cohen-Macaulay iff $\mathrm{Ext}^{i}_B (M,B)$ is zero for all $i \neq \mathrm{codim}(M)\overset{def}{=}\dim B - \dim M$. Moreover, the functor $M \mapsto \breve{M}=\mathrm{Ext}^{d}_{B}(M,B)$ induces an involutive anti-equivalence on the category of Cohen-Macaulay $B$-modules of codimension $d$.

Suppose now that we have a flat local map of local Noetherian rings $f: A \to B$ with $B$ Gorenstein, and that $M$ is a CM $B$-module which is flat over $A$. It is natural to ask whether $\breve{M}$ is also flat over $A$. The goal of this post is to prove the following partial result towards this question.

Theorem. Notation and setup as above, $\breve{M}$ is flat over $A$ if $A$ is regular or if $\mathrm{projdim}_B (M) < \infty$.

Before continuing, note that $A$ and $B/\mathfrak{m}_A B$ are automatically Gorenstein. This will be useful later.

The first key lemma is the following.

Lemma 1. Let $B \leftarrow A \rightarrow A'$ be a diagram of Noetherian commutative rings, and let $M$ be a finitely generated $B$-module flat over $A$. Suppose that the maps $B \leftarrow A \rightarrow A'$ are tor-independent (e.g. one of them is flat), and that at least one of the following conditions holds:
1. $\mathrm{projdim}_{B}(M) < \infty$;
2. $A \to A'$ is of finite tor-dimension.
Then there is a convergent spectral sequence
$\mathrm{Tor}_{-i}^{A}(\mathrm{Ext}^j_{B}(M,B),A') \Rightarrow \mathrm{Ext}^{i+j}_{B'}(M',B'),$ where $B' = B\otimes_A A'$ and $M'=M \otimes_B B'=M \otimes_A A'$.

This is not so hard to prove. The only real input is Stacks Tag 0A6A – the auxiliary conditions 1. and 2. above corresponds to cases (1) and (4) of that lemma.

Granted this lemma, the auxiliary hypotheses in the theorem let us apply this spectral sequence with $A,B,M$ chosen as in the theorem and with $A' = k=A/\mathfrak{m}_A$. Since $M$ is Cohen-Macaulay over $B$ of some codimension $d$, only $j = d$ contributes, and the spec. seq. degenerates to isomorphisms

$\mathrm{Tor}_{n}^{A}(\breve{M},k) \cong \mathrm{Ext}^{d-n}_{B_0}(M_0,B_0)$.

Here and in what follows, I write $(-)_0 = (-) \otimes_A k$ for pullback to the closed fiber of $f$. By the local criterion of flatness, the theorem follows if we can show that $\mathrm{Tor}_1^{A}(\breve{M},k)=0$, i.e. that $\mathrm{Ext}^{d-1}_{B_0}(M_0,B_0) = 0$. Since $B_0$ is Gorenstein, it clearly suffices to prove that $M_0$ is a Cohen-Macaulay $B_0$-module of codimension $d$. This is the content of the next lemma, which must be well-known, but which I couldn’t find in ten minutes of googling. Do you know a reference for this?

Lemma 2. If $A \to B$ is a flat local map of Gorenstein local rings, and $M$ is a finitely generated $B$-module which is flat over $A$, then $M$ Cohen-Macaulay over $B$ implies $M_0$ Cohen-Macaulay over $B_0$ of the same codimension. Here $(-)_0 = (-) \otimes_A A/\mathfrak{m}_A$ as above.

To prove this, first note that $d:=\mathrm{codim}(M)=\mathrm{depth}(B) - \mathrm{depth}_B(M)$ since $B$ and $M$ are CM. Then $\mathrm{depth}_{B_0}(M_0) = \mathrm{depth}_B(M)-\mathrm{depth}(A)$ $= \mathrm{depth}(B)-d-\mathrm{depth}(A)=\mathrm{depth}(B_0)-d=\mathrm{dim}(B_0)-d,$ where the first equality holds e.g. by Theorem 23.3 in Matsumura’s Commutative Ring Theory. (One can also read off the equality $\mathrm{depth}_{B_0}(M_0) = \mathrm{dim}(B_0)-d$ from the spectral sequence argument above.)

So now we just need to show that $d=\mathrm{dim}(B_0)-\mathrm{dim}(M_0)$. But we can check this last equality with $(-)_0$ replaced by $-\otimes_A A'$ where $A\twoheadrightarrow A'$ is any Artinian quotient. In particular we can assume that $A'=A/I$ where $I \subset A$ is an $\mathfrak{m}_A$-primary ideal generated by a regular sequence. Then

$d = \mathrm{dim}(B)-\mathrm{dim}(M) = {least\; n\; with\; \mathrm{Ext}^n_B(M,B) \neq 0}$
$= {least\; n\; with\; \mathrm{Ext}^n_{B/I}(M/I,B/I) \neq 0} = \mathrm{dim}(B/I)-\mathrm{dim}(M/I)$

where the first equality holds by definition, the second and fourth equalities are a well-known property of f.g. modules over CM local rings, and the third equality follows from the facts that $B$ is Gorenstein and that (thanks to our flatness assumptions) $I$ is generated by an $M$-regular sequence which is also a $B$-regular sequence. (To actually prove the third equality, use induction on the number of generators of I.)  Thus $d = \mathrm{dim}(B/I)-\mathrm{dim}(M/I) = \mathrm{dim}(B_0)-\mathrm{dim}(M_0)$ as desired.

Q1. Does the theorem hold more generally?

Q2. Is there a reference for Lemma 2 in the literature? This can’t be a new observation.

Q3. Can the Gorenstein condition in Lemma 2 be weakened? This seems unlikely to me.

## 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:

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

1. ($A[[x_1,\dots,x_n]]$ is quasi-excellent and) $A[[x_1,\dots,x_n]] \to B[[x_1,\dots,x_n]]$ is a regular ring map.
2. 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. $\square$

## Tales from the dropbox

As today’s service to the number theory entertainment complex, I could write an ill-advised rant about the abc conjecture situation. Instead, here’s another dropbox link. This time it’s some notes I wrote while trying to understand Kohlhaase’s paper Smooth duality in natural characteristicWarning: These are only about 40% done. The only original content (so far) is Proposition 1.6 and Theorem 1.12.

## Vanishing for O+/p-cohomology of Stein spaces

OK I’m too lazy to convert this into a proper blog post, but Proposition 0.4 here is something I needed recently, which others might find useful.  Lemma 0.2 might also come in handy.  Please ask in the comments if you’d like more details!

Update (Sept. 29). James Newton pointed out that it’s not actually clear how to fill in the details in the “proof” of Lemma 0.1.i. Therefore, the results in Lemmas 0.1 and 0.2 of the document linked above are currently unproved in the generality stated there. Fortunately, I was able to recover some weaker versions of Lemmas 0.1 and 0.2 which still suffice for the intended application to cohomology of Stein spaces. The corrected writeup is available here.