1
$\begingroup$

Let $X$ be a scheme of finite type over a field $k$. This reference page (https://ncatlab.org/nlab/show/smooth+scheme) defines $X$ to be a smooth scheme if for each $x \in \tilde{X} := X \times_k \overline{k}$, the stalk $\mathcal O_{\tilde{X},x}$ at $x$ is a regular local ring.

I have a few questions:

(i): Why is this defined in terms of an algebraic closure of $k$? This definition seems at odds with the relative perspective which is common in modern algebraic geometry. Is there a more intrinsic definition?

(ii): There are nonclosed points in $\tilde{X}$. For classical varieties over an algebraically closed field, all points are closed, and such a variety $Y$ is smooth if each stalk $\mathcal O_{Y,y}$ is a regular local ring. So this new definition seems to be saying something stronger. Take for example the case of a classical affine variety: if $A$ is a reduced, finitely generated $\overline{k}$-algebra, then in the classical definition, being smooth means $A_{\mathfrak m}$ is a regular local ring for each maximal ideal $\mathfrak m$ of $A$, while the definition here seems to require more, namely that $A_{\mathfrak p}$ be regular local for each prime ideal $\mathfrak p$ of $A$. What is the connection between the classical definition and this one?

(iii): The page I linked to says that there are other notions of smoothness for schemes. What are some of these other notions?

$\endgroup$
  • $\begingroup$ Just a comment on (i): since the algebraic closure of a field is unique (up to isomorphism) a choice of algebraic closure of $k$ is, pretty much, intrinsic. Is that fair? If the scheme were over some non-field, say $\mathbb{Z}$, then there would be a lot of choice for the algebraically closed field, but since the scheme is over a field there is, really, no choice. $\endgroup$ – User0112358 Feb 27 '17 at 1:44
  • $\begingroup$ That's fine. I wasn't worried about the dependence on the choice of algebraic closure, I was just noticing the contrast between the Grothendieck method and the Weil method and the apparent exception here: for Weil, one works with the points of a variety over an algebraically closed field to answer rationality questions. $\endgroup$ – D_S Feb 27 '17 at 1:48
2
$\begingroup$

There is essentially only one definition of smoothness for a morphism locally of finite type of locally noetherian schemes, namely the one introduced by Grothendieck in SGA 1 and EGA IV. Usually, the definition of smoothness includes these finiteness hypothesis or even finite type instead of locally of finite type (and this is the only source of ambiguity in the definition: what finiteness hypotheses are assumed). Without finiteness assumptions, there is also a unique definition, but it depends on the topology (e.g., for a local homomorphism of local rings, both the topology of the maximal ideal and the discrete topology are interesting, and the associated notions of smoothness are different).

However there are several equivalent ways to define smoothness. You can see all this in EGA IV, mainly parts 1 and 4.

When considering locally noetherian schemes over a field, Grothendieck proved that smoothness is equivalent to geometric regularity. The definition in the first lines of your question is not the definition of smoothness, but the definition of geometric regularity, so it works (as a definition for smoothness) only over a field and with noetherian hypotheses. However, even the definition of geometric regularity can be made without using the algebraic closure of the field (in fact the usual way of defining it is without mentioning the algebraic closure; see e.g. EGA IV again).

Finally, $A_{\mathfrak m}$ is regular for all maximal $\mathfrak m$ if and only if $A_{\mathfrak p}$ is regular for all maximal $\mathfrak p$ by a classical result of Serre.

If you are willing to assume in some situations finiteness hypotheses in order to simplify, you can see a shorter exposition with shorter proofs of all these things in Bourbaki, Algèbre Commutative, chap. X (in terms of commutative rings instead of schemes, but all these questions are local).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.