Formally étale at all primes does not imply formally étale. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T11:48:37Z http://mathoverflow.net/feeds/question/16132 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/16132/formally-etale-at-all-primes-does-not-imply-formally-etale Formally étale at all primes does not imply formally étale. Harry Gindi 2010-02-23T06:04:48Z 2010-03-14T02:08:02Z <p>All rings are assumed to be commutative and unital, with all homomorphisms unital as well.</p> <p>On last week's homework, there was a mistake in one of the questions:</p> <blockquote> <blockquote> <p><strong>(2.5)</strong> Let $R\to S$ be a morphism of commutative rings giving $S$ an $R$-algebra structure. Suppose that the induced maps <code>$R\to S_{\mathfrak{p}}$</code> are formally étale for all prime ideals $\mathfrak{p}\subset S$. Then $R\to S$ is formally étale. </p> </blockquote> </blockquote> <p>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). </p> <p>However, I'm interested in seeing either a counterexample or a proof for the stronger claim in the grey box. </p> http://mathoverflow.net/questions/16132/formally-etale-at-all-primes-does-not-imply-formally-etale/16144#16144 Answer by James Borger for Formally étale at all primes does not imply formally étale. James Borger 2010-02-23T11:27:37Z 2010-02-23T11:27:37Z <p>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.</p> http://mathoverflow.net/questions/16132/formally-etale-at-all-primes-does-not-imply-formally-etale/16862#16862 Answer by Bjorn Poonen for Formally étale at all primes does not imply formally étale. Bjorn Poonen 2010-03-02T15:14:48Z 2010-03-14T02:08:02Z <p><strong>EDIT:</strong> 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 <em>complete</em> solution passed on by Mel Hochster, who got it from Luc Illusie, who got it from ???.</p> <p><strong>OLD ANSWER:</strong> 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 <code>$R \to S_{\mathfrak{p}}$</code> is formally unramified for all primes $\mathfrak{p} \subset S$, then $R \to S$ is formally unramified.</p> <p>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 <code>$f_{\mathfrak{p}},g_{\mathfrak{p}} \colon S_{\mathfrak{p}} \to A_{\mathfrak{p}}$</code> that become equal when we compose with $A_{\mathfrak{p}} \to A_{\mathfrak{p}}/I A_{\mathfrak{p}}$. Since <code>$R \to S_{\mathfrak{p}}$</code> 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 $f=g$.</p> http://mathoverflow.net/questions/16132/formally-etale-at-all-primes-does-not-imply-formally-etale/17775#17775 Answer by Harry Gindi for Formally étale at all primes does not imply formally étale. Harry Gindi 2010-03-10T23:09:05Z 2010-03-10T23:48:57Z <p>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 $\mathfrak{p}\subset S$.</p> <p>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.</p> <p>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. </p> <p>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 <code>$$\bar{F}_j+ \sum_{m\in M}\overline{\frac{\partial F_j}{\partial X_m}}\delta_m=0.$$</code></p> <p>Rearranging, we get a system of equations indexed by $J$<code>$$(*)_{j\in J}\sum_{m\in M}\overline{\frac{\partial F_j}{\partial X_m}}\delta_m=-\overline{F}_j.$$</code></p> <p>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 <code>$(*)_{j_k}$</code>, for each $m$, we get an equation of the form <code>$$(**)_m \quad \delta_m=-(s_{m,1}\overline{F}_{j_{m,1}}+\cdots + s_{m,h_m}\overline{F}_{j_{m,h_m}}).$$</code></p> <p>Showing that these define solutions for all of the equations <code>$(*)_j$</code> 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.</p> <p>(Note: This is not my proof. I've paraphrased the proof communicated to me by Mel Hochster.)</p> <p>Edit: Fixed LaTeX using Scott's suggestion.</p>