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All rings are assumed to be commutative and unital, with all homomorphisms unital as well.

On last week's homework, there was a mistake in one of the questions:

(2.5) Let $R\to S$ be a morphism of commutative rings giving $S$ an $R$-algebra structure. Suppose that the induced maps $R\to S_{\mathfrak{p}}$ are formally étale for all prime ideals $\mathfrak{p}\subset S$. Then $R\to S$ is formally étale.

According to our professor, the exercise should have stated additionally that $S$ was finitely presented over $R$ (which allows us to prove that $S$ is in fact étale over $R$ rather than just formally étale). This is not too hard to do and is left as an exercise. (You can also find it in EGA).

However, I'm interested in seeing either a counterexample or a proof for the stronger claim in the grey box.

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Why not post this as an answer to Giant Laser Cannons question? Just to get reputation? –  Steven Gubkin Mar 10 '10 at 22:02
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I'm sure Bjorn Poonen would have been amenable to simply adding a disclaimer along the lines of "[Note: this is not a complete argument; see below answer posted by fpqc for a full solution]" to his answer to the other question. –  Reid Barton Mar 10 '10 at 23:31
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I feel that the reposting is drawing an undeserved level of ire. –  Tyler Lawson Mar 10 '10 at 23:40
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I'm voting to close. Reason: exact duplicate. –  S. Carnahan Mar 10 '10 at 23:41
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It just seems unnecessary. What actual problem does it solve that simply posting an answer to the original question does not solve? ("The answer will not be on a green background" is not an actual problem.) –  Reid Barton Mar 10 '10 at 23:43

3 Answers 3

up vote 5 down vote accepted

EDIT: Don't bother reading my partial solution. Brian Conrad pointed out that an easier way to do what I did is to use the equivalent definition of formally unramified in terms of Kähler differentials. And later on, fpqc posted below a complete solution passed on by Mel Hochster, who got it from Luc Illusie, who got it from ???.

OLD ANSWER: Here is a half-answer. I'll prove half the conclusion, but on the plus side I'll use only half the hypothesis! Namely, I will prove that if $R \to S_{\mathfrak{p}}$ is formally unramified for all primes $\mathfrak{p} \subset S$, then $R \to S$ is formally unramified.

Let $A$ be an $R$-algebra, and let $I \subseteq A$ be a nilpotent ideal. Given $R$-algebra homomorphisms $f,g \colon S \to A$ that become equal when composed with $A \to A/I$, we must prove that $f=g$. Fix $\mathfrak{p}\subset S$. Then the localizations $A_{\mathfrak{p}} := S_{\mathfrak{p}} \otimes_{S,f} A$ and $S_{\mathfrak{p}} \otimes_{S,g} A$ of $A$ (defined viewing $A$ as an $S$-algebra in the two different ways) are naturally isomorphic, since adjoining the inverse of an $a \in A$ to $A$ automatically makes $a+\epsilon$ invertible for any nilpotent $\epsilon$ (use the geometric series). Now $f$ and $g$ induce $R$-algebra homomorphisms $f_{\mathfrak{p}},g_{\mathfrak{p}} \colon S_{\mathfrak{p}} \to A_{\mathfrak{p}}$ that become equal when we compose with $A_{\mathfrak{p}} \to A_{\mathfrak{p}}/I A_{\mathfrak{p}}$. Since $R \to S_{\mathfrak{p}}$ is formally unramified, this means that $f_{\mathfrak{p}} = g_{\mathfrak{p}}$. In other words, for every $s \in S$, the difference $f(s)-g(s)$ maps to zero in $A_{\mathfrak{p}}$ for every $\mathfrak{p}$. An element in an $S$-module that becomes $0$ after localizing at each prime ideal of $S$ is $0$, so $f(s)=g(s)$ for all $s$. So $f=g$.

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And here's my plan for improving the environment: start by dismantling all light-beam weapons of any size... :) –  Bjorn Poonen Mar 2 '10 at 15:15
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Bjorn, a shorter way to say this "half" is that formally unramified is precisely Omega^1_{S/R} = 0, and for any prime P of S we have (Omega^1_{S/R})_P = Omega^1_{S_P/R} = 0, so we're done. Alas, that is where I got stuck; it is the existence which is the real problem. The only idea I had was to try to prove that a commutative ring C is the inverse limit of the natural diagram among all of its localizations. That would solve the problem, but I had no idea how to do it (yet it also seems like an "interesting" question in its own way). Any ideas? –  BCnrd Mar 2 '10 at 15:58
    
OK, thank you, Brian; that is much simpler. –  Bjorn Poonen Mar 2 '10 at 18:26
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@Brian: Regarding your question about a commutative ring C is necessarily the inverse limit of its localizations at primes: The answer is no. If C is an infinite product of fields k_i, the primes form an antichain, so the inverse limit is the product of the residue fields, which include not only the k_i but also all the nonprincipal ultraproducts of the k_i. –  Bjorn Poonen Mar 3 '10 at 3:12
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Bjorn, I agree that probably it isn't true. But this has the feeling of the kinds of point-set topology questions on which one can waste tremendous amounts of time as an undergraduate: either find a plausible counterexample or proof with little effort in little time, or best not to dwell much on it. Now that you killed the only idea I had on it, I won't try to find another idea for it. –  BCnrd Mar 3 '10 at 8:32

Using the module of Kähler differentials, it is easy to show that $R\to S$ is formally unramified if and only if the induced maps $R\to S_{\mathfrak{p}}$ are formally unramified for all primes $\mathfrak{p}\subset S$.

Consider a presentation of $S$ over $R$ as $R[X]/I$ in generators and relations, where $R[X]:=R[X_m]_{m\in M}$ is a polynomial ring in a possibly infinite family of indeterminates indexed by $M$, and $I\subset R[X]$ is an ideal. Fix a family of generators of $I=(F_j)_{j\in J}$ indexed by $J$, again not necessarily finite.

It is enough to show that $R\to S$ is formally smooth. This is equivalent to showing that there exists a morphism of $R$-algebras that is a splitting for the canonical projection $\pi:R[X]/I^2 \to R[X]/I=S$, which will necessarily be unique because $R\to S$ is formally unramified.

Let $\overline{X}_m$ denote the image of $X_m$ in $R[X]/I^2$. We must find elements $\delta_m\in I/I^2$ such that $(\forall j\in J)F_j(X_m + \delta_m)=0$. We rewrite this using Taylor's formula as $$\bar{F}_j+ \sum_{m\in M}\overline{\frac{\partial F_j}{\partial X_m}}\delta_m=0.$$

Rearranging, we get a system of equations indexed by $J$$$(*){j\in J}\sum{m\in M}\overline{\frac{\partial F_j}{\partial X_m}}\delta_m=-\overline{F}_j.$$

We wish to find a unique solution for this system in the $\delta_m$. Since $\Omega_{S/R}=0$, each $dX_m\in \Omega_{R[X]/R}$ is an $S$-linear combination $dX_m=s_{m,1}dF_{j_{m,1}}+\cdots + s_{m,h_m}dF_{j_{m,h_m}}$. If we use the $s_{m,k}$ as coefficients to form $S$-linear combinations of the equations $(*)_{j_k}$, for each $m$, we get an equation of the form $$(**)_m \quad \delta_m=-(s_{m,1}\overline{F}_{j_{m,1}}+\cdots + s_{m,h_m}\overline{F}_{j_{m,h_m}}).$$

Showing that these define solutions for all of the equations $(*)_j$ is not immediate, but it is a local question on $S$. However, our local rings $S_{\mathfrak{p}}$ are all formally étale, so the local conditions are satisfied. Then this proves the global claim.

(Note: This is not my proof. I've paraphrased the proof communicated to me by Mel Hochster.)

Edit: Fixed LaTeX using Scott's suggestion.

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I recommend using \overline instead of \bar. –  S. Carnahan Mar 10 '10 at 23:43
    
My edit is in reference to this comment. I replaced the bars with overlines a while ago. –  Harry Gindi Mar 11 '10 at 1:07
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This is very nice, though since at the end you pass to a localization (so no longer quite working with a "presentation" as an $R$-algebra anymore), I think it is better if before you launch into the calculation you generalize the setup to allow $S$ to more generally be a localization of $R[X]/I$ at some unspecified multiplicative set (so can then ramp it up to the complement of a prime later in the argument without departing from the initial setup of the argument). The usual business of stating hypotheses in sufficient generality for what comes later in the proof. Just stylistic... –  BCnrd Mar 11 '10 at 2:49

For just about any ring $R$ and any prime ideal $Q$, taking $S=R_Q$ will give you a counterexample, simply because $R_Q$ usually won't be finitely generated as an $R$-algebra. For example $R=k[x]$ and $Q=(x)$ works.

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We're not looking for étale, we're looking for formally étale. So it just has to satisfy the nilpotent lifting properties. –  Harry Gindi Feb 23 '10 at 11:42
    
In particular, if $S=R_Q$ and for every prime $\mathfrak{p}$ of $S$, $S_{\mathfrak{p}}$ is formally étale over $R$, but $S$ only has one maximal ideal, and localizing at this maximal ideal gives us an isomorphic ring since we're localizing away units, so in particular $S_Q$=$R_Q$, which is then by assumption formally étale. –  Harry Gindi Feb 23 '10 at 11:46
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Whoops, my mistake. I misread the question. Thanks for the correction. –  JBorger Feb 23 '10 at 12:20

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