This is confusing, there appear to be possibly two definitions of smoothness in algebraic geometry for a morphism $f: X \rightarrow Y$ of schemes of finite type over an arbitrary field $k$.

Definition 1 (Milne, Etale Cohomology pp.30-31): Let $f: X \rightarrow Y$ be locally of finite type. Then we say $f$ is smooth if it is flat and $\Omega_{X/Y}$ is locally free of rank the relative dimension. (For reference, this is the same definition as in the stacks project.)

Definition 2: (Hartshorne, III.10, p268). A morphism $f: X \rightarrow Y$ of schemes of finite type over a field $k$ is smooth of relative dimension $n$ if it is flat, the sheaf of relative differentials $\Omega_{X/Y}$ is locally free of rank $n$, and if: (*) If $X' \subset X$ and $Y' \subset Y$ are irreducible components such that $f(X') \subset Y'$ then $\dim X' = \dim Y' + n$.

In the situation of Hartshorne's definition, there is the extra hypothesis (*). Is this always satisfied, or are there just two definitions?

  • $\begingroup$ I think that Hartshorne is just including the definition of "relative dimension $n$" in his version. Milne says "of the relative dimension" so presumably he's assuming the reader knows what that means already. $\endgroup$ – Ramsey Sep 15 '12 at 17:58
  • $\begingroup$ There are several different definitions for smoothness, some of which are equivalent and some of which are not. Hartshorne's is slightly less general than the one given in EGA (see, e.g., user.math.uzh.ch/bruin/smooth.pdf pdf pp. 3/8), so (without trying to conjure up an example) I'd guess (*) is not always satisfied. $\endgroup$ – Benjamin Dickman Sep 15 '12 at 18:03
  • 2
    $\begingroup$ Hypothesis $(\ast)$ cannot be removed, as you can see via the special case when $Y={\rm{Spec}}(k)$ and $X={\rm{Spec}}(K)$ for a finite extension of fields. The $K$-vector space $\Omega^1_{X/Y}$ has some dimension $n≥0$, and $n>0$ if and only if $K/k$ is not separable (equivalently, not etale). So we need some condition to relate n to fiber dimensions. Hartshorne's $(\ast)$ is focusing exclusively on dimensions of generic fibers, so some work is needed to show that Definition 2 forces all fibers to have pure dimension $n$. Definition 1 is conceptually superior. $\endgroup$ – grp Sep 15 '12 at 21:31
  • 4
    $\begingroup$ The relative dimension of $f$ is a function on $Y$ (which is locally constant if $f$ is smooth in the EGA (or Milne) sense), but Hartshorne forces it to be globally constant on $Y$. In other words, a morphism of schemes of finite type over a field is smooth in the EGA sense if and only if it is locally smooth in the sense of Hartshorne. This means, for instance, that, according to Hartshorne's definition, a disjoint union of smooth schemes is not smooth, which is silly. $\endgroup$ – Denis-Charles Cisinski Sep 15 '12 at 21:41
  • 4
    $\begingroup$ @LMN: The main content in relating the two definitions is as follows. First, by flatness every irreducible component of $X$ dominates one of $Y$ and so under Def. 2 each generic fiber is of pure dimension $n$. To prove the same for all non-empty fibers, by base change to dvr's over $Y$ and the purity hypothesis for the generic fibers we can pass to irreducible components (after the dvr base change, using the easy nature of flatness over a dvr!) to reduce to the case of an irreducible scheme fppf over a dvr. Now apply results about flatness and dimension theory (or EGA IV$_3$, 14.3.10). $\endgroup$ – grp Sep 15 '12 at 23:59

Your Answer

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

Browse other questions tagged or ask your own question.