Talk:Étale morphism

I removed the reference to monic polynomials in the section concening the localization of k[x_1,...,x_n]/<f_1,...,f_n> since (a) I don't know what a monic polynomial in several variables is and (b) I don't believe such a condition is necessary.

RJChapman 18:57, 27 November 2006 (UTC)

Are the five defintions really equivalent without any hypothesis? J. Milne seems to imply in his lecture notes on etale cohomology (available from his website) on page 13 that the definitions involving the morphism to be locally of finite presentation are weaker than the definition flat+unramified.

-- According to Milne's book on etale cohomology, , p.30, for f : X --> Y they are equivalent: flat + unramified (which is by definition locally of finite type) is equivalent to locally of finite presentation and formally etale (i.e. the functor Hom_X(-, Y) does only see the topological structure of -). Jakob.scholbach 21:17, 17 April 2007 (UTC)

I changed the condition "flat+unramified" to "flat+G-unramified". The difference is that "unramified" requires locally of finite type whereas "G-unramified" requires locally of finite presentation. Milne works under the general hypothesis of all schemes being locally Noetherian where the notions coincide, but in general one should make the distinction. 84.59.181.40 (talk) 22:46, 16 August 2017 (UTC)

Pronunciation of étale

Could somebody add an IPA pronunciation of the term étale?

Baccala@freesoft.org 08:09, 29 September 2007 (UTC)

Added the correct French pronunciation of the word étale (/etal/), and reverted a recent change to /eɪtɑːl/. The latter likely reflects local usage (somewhere in the US?); however, also in all English-language discussions I've had where étale morphisms have been discussed in the UK and the US, people have quite invariably used essentially the French pronunciation. Thus, I'd stick to the French version. Stca74 (talk) 09:05, 10 August 2008 (UTC)

Then please link it as French rather than English, and make it obvious to the reader that it isn't English. kwami (talk) 11:22, 2 March 2010 (UTC)

Comparison of dimensions

It is not true that X -> Y étale implies dim X = dim Y. Juste take Y = Spec (DVR) and X = generic point. I will remove change this statement. Liu (talk) 19:50, 6 January 2011 (UTC)

Definition of standard étale

An anonymous user recently tried to correct the definition of standard étale. However, the definition formerly in the article and the definition given by the anonymous user are equivalent. I wanted to record the equivalence of the two for anyone else who came along:

We have φ : R → S a homomorphism of commutative rings, and we know f and g in R[x] such that f is monic and S = R[x]_g / f R[x]_g. Suppose first that the derivative f′ is a unit in R[x]_g. Then it's obvious that f′ is also a unit in S: If we choose h to be the inverse of f′ in R[x]_g, then h is still the inverse of f′ in S. For the other direction, assume that f′ is a unit in S. This means that there exists some h in R[x]_g such that f′h = 1 + fk, where k is some other element of R[x]_g. The trick is that we replace g by g₁ = (1 + fk)g. Of course, localizing R[x] at g isn't the same as localizing it at g₁; but that doesn't matter, because S localized at 1 + fk is isomorphic to S: The fk term vanishes in S, so all the localization does is make 1 a unit, which was already true. So we make the replacement, and then f′ is a unit in $R[x]_{g_{1}}$.

Geometrically, R[x] is a one-dimensional affine space over R (I'm leaving out the application of Spec here). The space R[x]_g is what you get by leaving out a hypersurface; you do this to remove all the bad points (where the morphism isn't étale). Localizing at g₁ simply leaves out more. The space S is defined in R[x] by looking at the solutions of f = 0 and leaving out the bad points using g. What we showed above is that leaving out bad points using g₁ produces the same effect; in other words, the hypersurface defined by 1 + fk doesn't meet S. Ozob (talk) 14:11, 4 December 2012 (UTC)

Definition of derivative

The article derivative doesn't provide a definition for polynomials over arbitrary rings. If there's a standard definition, I'm unaware of it, and the best abstract definition I know of is for derivations, which are many. Is there a unique derivation generated by setting its value on elements of the ground field to 0, and on generators of the polynomial ring to 1? Even then, it would only apply to algebras over a field, and that isn't required by the article. ᛭ LokiClock (talk) 00:31, 22 March 2013 (UTC)

It's what you get by formally applying the power rule and linearity:

$f(x)=a_{0}+a_{1}x+\cdots +a_{n}x^{n}\Rightarrow f'(x)=a_{1}+2a_{2}x+\cdots +na_{n}x^{n-1}.$

This is a derivation. It's canonical: The relative Kahler differentials $\Omega _{R[x]/R}$ are isomorphic to R with basis element dx, and the derivative is also the map $f\mapsto df$ followed by the isomorphism $R\cdot dx\cong R\cdot 1$. Ozob (talk) 00:57, 23 March 2013 (UTC)

Thanks. Using that information I was able to locate this derivative on Wikipedia. ᛭ LokiClock (talk) 05:09, 23 March 2013 (UTC)