Let $A\to B$ be a morphism of (unitary commutative) rings such that $B$ is module-finite over $A$ and there exists $f\in A$ which is a nonzerodivisor in $A$ and in $B$, with $A[1/f]\to B[1/f]$ an isomorphism.

My question is: is the diagram $A\to B \rightrightarrows B\otimes_A B$ exact ?

In geometric terms, this is more or less equivalent to the question whether descent of functions along a finite, scheme-theoretically birational morphism is possible.

Here are some comments:

  1. Of course, I'm mainly interested in the non-flat case (otherwise it's just faithfully flat descent).
  2. I computed several examples, including the normalization of the scheme formed by $n$ lines meeting in the origin in $\mathbf{A}^2$ (i.e. the scheme $V(x^n-y^n)\subset\mathbf{A}^2$); and the resolution $k[x^n,y^n,xy]\to k[x,y]$ of a plane quotient singularity. Any time, the answer is yes.
  3. I suspect that the result might be known and maybe present somewhere in the literature. The question is related also to descent of modules for universally injective morphisms, as exposed in the Stacks Project here but I could not find there something really close to the statement I'm after.

Thanks for any help!

EDIT (Jan 1st, 2016): the claim that the normalization of the scheme formed by $n$ lines meeting in the origin in $\mathbf{A}^2$ is an example is not true if $n>2$. Explanations are given in this MO answer.

  • $\begingroup$ You might check out Joe Ross's thesis. He had a theorem about descent for seminormalization. $\endgroup$ – Jason Starr Jun 23 '14 at 14:10
  • $\begingroup$ Ah, that's worth knowing indeed. Thanks. $\endgroup$ – Matthieu Romagny Jun 23 '14 at 14:19

This is not always true. For example, it fails for the inclusion $k[x^3,x^5]\to k[x]$, where $k$ is a field. This is because $x^7$ is in the equalizer: $x^7\otimes 1 = x^2\otimes x^5=x^5\otimes x^2 = 1\otimes x^7$.

Such examples are indeed in the literature. I saw this one in SGA3, exp V.

  • $\begingroup$ I'm probably being dense, but why is $x^2\otimes x^5=x^5\otimes x^2$? $\endgroup$ – Matthieu Romagny Jun 23 '14 at 9:33
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    $\begingroup$ OK, I get it: $x^2\otimes x^5=x^2\otimes x^3x^2=x^5\otimes x^2$. $\endgroup$ – Matthieu Romagny Jun 23 '14 at 9:42
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    $\begingroup$ Wow, that is surprising. I was also under the impression that the answer would be 'yes', having once established this in some examples where $A$ was a non-integrally closed order in the ring of integers $B$ of a number field. So now I wonder if there is a characterization of $A\rightarrow B$ for which the aforementioned complex is exact. $\endgroup$ – RP_ Jun 23 '14 at 10:40
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    $\begingroup$ @JBorger Not trying to claim anything but this was very similar to counterexample I was trying to think about!! I mean what's the first example of a finite morphism that's not an isomorphism it's gotta be the normalization $k[x^2,x^3] \hookrightarrow k[x]$ right? +1 by the way for this nice answer. $\endgroup$ – David Benjamin Lim Jun 23 '14 at 12:30

Apologies for a long comment masquerading as an answer.

I think you were on the right track when you mentioned universally injective ring homomorphisms, which are also often called pure ring homomorphisms.

You can find a well developed geometric theory of such homomorphisms in papers of Mesablishvili. The theory includes a generalization of the result that you and user52824 worked out in your answer. See, in particular, Theorem 6.5 of "More on descent theory for schemes," which states

"A quasi-compact morphism of schemes is stable schematically dominant if and only if it is pure if and only if it is a stable regular epimorphism."

Note that there is no "finite" hypothesis and no Noetherian hypothesis in the theorem.

The easiest way to get a pure ring homomorphism $R\to S$ is to have an $R$-module retraction $S\to R$. One often calls such a retraction a ``Reynolds operator'' because of the representation-theory situation (invariants of a linearly reductive group) that gives rise to such retractions. In fact, an elementary argument shows that if $S$ is finitely presented as an $R$-module (which is the situation you consider) and pure, then there is such a retraction.

For a proof that an $R$-algebra finitely presented as an $R$-module is pure if and only if it admits an $R$-module retraction onto $R$, one can see Corollary 5.3 of Hochster/Roberts "The purity of the Frobenius and local cohomology." The paper is now freely available online thanks to Elsevier's recent opening of access to older math articles. Although the corollary is in the middle of the paper, it belongs to a section about the basics of pure morphisms, and its proof is entirely contained in Section 5.

To get endless examples of pure ring homomorphisms, you can thus use representation theory: let $k$ be a field, let $G$ be a finite group whose order is invertible in $k$, and let $V$ be a finite-dimensional representation of $G$ over $k$. The inclusion $k[V]^G \to k[V]$ is then a pure ring homomorphism, since we have a $k[V]^G$-module retraction (the Reyonlds operator mentioned earlier) $k[V]\to k[V]^G$ defined by averaging over $G$. This sort of example generalizes your quotient-singularity computation.

  • $\begingroup$ Dear anon, that's a great answer! In fact I thought a bit more and found out that result about descent along pure morphisms in the 1970 CRAS note 'Descente par morphismes purs' by J.-P. Olivier (no proof there). I read also in the The Stacks Project that Mesablishvili was the first to give full proofs of such statements. On the other hand I had made no connection with Reynolds operators. Thanks for these explanations and references! That's very useful to me. $\endgroup$ – Matthieu Romagny Jun 28 '14 at 11:55
  • $\begingroup$ ps: in Olivier's note the result is the Corollary to Thm. 2.6. $\endgroup$ – Matthieu Romagny Jun 28 '14 at 12:07

I now remembered a statement in Grothendieck's Bourbaki talk 190 on faithfully flat descent (Lemma 3 below) which allows to prove:

Proposition. Let $S$ be a locally noetherian scheme. Then any finite, universally scheme-theoretically dominant morphism $S'\to S$ is a strict epimorphism.

Remarks. 1) A finite morphism is an epi iff it is scheme-theoretically dominant. 2) There is a generalization due to Mesablishvili, see anon's answer below.

So the original question has a positive answer provided we strengthen the assumption so as to require it to hold universally i.e. after any base change $T\to S$. Moreover we may also weaken birational to dominant.

For the proof I shall use three lemmas:

Lemma 1. An epimorphism of schemes which admits a section is strict.

Lemma 2. Let $u:S'\to S$ be a finite, universally schematically dominant epimorphism of schemes. If $u$ becomes strict after base change by a strict finite epimorphism $v:T\to S$, then $u$ is strict.

Lemma 3 (Bourbaki talk 190). Let $S$ be a [EDIT] locally noetherian scheme. Then any finite epimorphism $u:S'\to S$ factors as a composition of a finite number of strict finite epimorphisms.

Let us first give:

Proof of the Proposition, given the lemmas. Let $u:S'\to S$ be a finite, universally scheme-theoretically dominant epimorphism with $S$ locally noetherian. By Lemma 3 we can write $u=u_1\dots u_n$ where each $u_i$ is a strict finite epimorphism. By Lemma 2 and induction, it is enough to prove that $u$ is strict after base change by itself. But after such a base change $u$ acquires a section, hence by Lemma 1 it is strict.

Here are now the proofs of the lemmas.

Proof of Lemma 1. This is well-known and easy.

Proof of Lemma 2. We easily reduce to the case where $S,S'$ are affine, say $S=\text{Spec}(A)$ and $S'=\text{Spec}(A')$. Then we get the diagram: $$ \begin{array}{ccccc} A & \stackrel{f}{\to} & A' & \rightrightarrows & A'\otimes_A A' & \\ \downarrow & & \downarrow & & \downarrow & \\ B & \to & B' & \rightrightarrows & B'\otimes_B B' & \simeq (A'\otimes_A A')\otimes_A B\\ \downarrow\downarrow & & \downarrow\downarrow & & & \\ B\otimes_A B & \to & B'\otimes_{A'} B' & & & \\ \end{array} $$ where $B':=B\otimes_A A'$ and the second row is assumed exact. Since $f$ is universally injective, then the map $B\otimes_A B\to B'\otimes_{A'} B'$ in the last row is injective. Then an easy diagram chase shows that the first row is exact.

[EDIT] Below I give the proof of Lemma 3 that was cooked up by myself and user52824 in his comments below. Incorporating the arguments suggested by user52824 required to edit the original proof sketch only in order for the reading to go smoothly. For these reasons, I reckoned that the historical record of the question (and answer) would be just as good if I edited directly in the original text rather than added a patch afterwards. The place where the arguments of user52824 appear is clearly indicated.

Proof of Lemma 3. I don't know any place where a proof appears. Here is one. Define $\mathcal{O}_S$-algebras $A_n$ as follows: set $A_0=u_*\mathcal{O}_{S'}$ and $$ A_{n+1}:=\{a\in A_n\,;\, a\otimes 1=1\otimes a \mbox{ in } A_n\otimes_{\mathcal{O}_S} A_n\}. $$ By the very definition, we see that the kernel of the surjective morphism $A_n\otimes_{\mathcal{O}_S} A_n\to A_n\otimes_{A_{n+1}} A_n$ is equal to $0$, since it is generated by the elements of the form $a\otimes 1-1\otimes a$ for $a\in A_{n+1}$. Hence the diagram $$ A_{n+1} \to A_n \rightrightarrows A_n\otimes_{\mathcal{O}_S} A_n \simeq A_n\otimes_{A_{n+1}} A_n $$ is exact. That is, the morphism of schemes $\text{Spec}(A_n)\to\text{Spec}(A_{n+1})$ is a strict finite epimorphism. Assume that the decreasing sequence $A_1\supset A_2\supset\dots$ stabilizes. Then there is $N\ge 0$ such that $A_N=A_{N+1}$. Thus $a\otimes 1=1\otimes a$ in $A_N\otimes_{\mathcal{O}_S} A_N$, for all $a\in A_N$. Hence the multiplication map $A_N\otimes_{\mathcal{O}_S} A_N\to A_N$ is an isomorphism, hence $\mathcal{O}_S\to A_N$ is a finite epimorphism of sheaves of rings, hence an isomorphism and we are done.

It remains to prove that the sequence stabilizes. We follow the hints given by user52824 in the comment below. First note that the formation of $A_n$ commutes with localization, and that the question of stabilization of the sequence of algebras is Zariski-local on $S$. Moreover if for some point $s\in S$ the sequence of germs $A_{n,s}$ stabilizes, then the isomorphism $\mathcal{O}_{S,s}\to A_{n,s}$ extends in a neighbourhood of $s$ and hence we may assume that $S$ is a local scheme with closed point $s$. In particular, we may assume that $S$ (local or not) has finite Krull dimension $d$, and we shall argue by induction on $d$. If $d=0$ then the rings $A_n$ have finite length and hence the sequence must stabilize. If $d\ge 1$ then since $S\setminus \{s\}$ has dimension $<d$ we know that away from $s$, the sequence $A_n$ stabilizes at some $N$. By the same argument as before, we know moreover that away from $s$, the map $\mathcal{O}_S\to A_N$ is an isomorphism. It follows that the $\mathcal{O}_S$-module $A_n/\mathcal{O}_S$ has finite length, for all $n\ge N$. Thus the sequence $A_n/\mathcal{O}_S$ stabilizes, hence also $A_n$ and we are done.

To finish with, I'll make a comment on the nice counterexample given by James. It shows that dominant instead of universally dominant is not enough. It turns out to be not so easy to prove that the extension $k[x^3,x^5]\to k[x]$ is not universally injective (here is a hint: write $u=x^3$ and $v=x^5$ so $A:=k[x^3,x^5]\simeq k[u,v]/(u^5-v^3)$, and show that after the base change $A\to A/(u-v)$ the nonzero element $c=u^3-u$ maps to $0$). This leads to the question to find nice, easily recognizable classes of (non faithfully flat) universally dominant scheme morphisms. Apart from isolated examples, I must say that I don't have much in stock.

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    $\begingroup$ Lemma 3 is too global ($S$ could be highly disconnected, requiring ever more steps of factorization over different connected components), so use noetherian $S$. You also need a dominance hypothesis in Lemma 3 (so $A$ is a subring of $A$'); you didn't define "strict", but I guess the meaning. Then everything is inside the $A$-finite $A'$, so by "spreading out" we can assume $A$ is local. By dimension induction, all OK over the (usually non-affine!) punctured Spec -- the trick from EGA proof of ZMT -- and then by length consideration of special fiber we get stabilization for Lemma 3. QED $\endgroup$ – user27920 Jun 25 '14 at 14:36
  • $\begingroup$ Whoops, thanks for pointing out the parasitic 'locally' in the statement of Lemma 3. As for the dominance assumption, it is in fact equivalent to S'->S being epic. As for 'strict epi', I didn't define it but it is standard, see e.g. Grothendieck's Bourbaki 190. Finally thank you for the ideas to fill in the hole in the proof, I'll edit so as to incorporate this to the answer above. $\endgroup$ – Matthieu Romagny Jun 26 '14 at 7:27
  • $\begingroup$ Dear user52824, I completed the proof with your insightful suggestions. Thanks! $\endgroup$ – Matthieu Romagny Jun 26 '14 at 12:40
  • $\begingroup$ Dear Matthieu: You're welcome. When I first saw the question (before counterexamples were given) I had tried and failed to make a counterexample with $k[x^2,x^3]$ and then tried to make an affirmative proof, eventually reducing to the local case with everything known away from the special fiber, and then got stuck trying to push through a length argument there. So I was glad to see that aspects of the same argument could be applied to your salvage based on Lemma 3. $\endgroup$ – user27920 Jun 26 '14 at 21:56

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