# Smooth scheme of finite type over a field, some questions

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?

• 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. – User0112358 Feb 27 '17 at 1:44
• 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. – D_S Feb 27 '17 at 1:48

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.