0

The question comes from Liu's book :

2.9. Let $X$, $Y$ be locally Noetherian integral schemes. Let $f : X \to Y$ be a dominant morphism of finite type. We say that $f$ is generically separable (resp. generically étale) if $k(X)/k(Y)$ is a separable (resp.finite separable) extension.

(a) Show that if $f$ is generically separable (resp. generically étale), then there exists a non-empty open subset $U$ of $X$ such that $f|_{U} : U \to Y$ is smooth (resp. étale).

First we say that a finitely generated extension $K$ of $k$ (i.e. $K = k(f_{1}, ..., f_{n})$ for some $f_{i} \in K$) is separable if K is finite separable over a purely transcendental extension of k. This is equivalent of saying that Then $\Omega_{K/k}^{1}$ is a K-vector space of dimension $\operatorname{trdeg}_{k}K$.

Now what I've done .

Let $\xi, \xi'$ the respective generic points of de $X, Y$. Let $n := \operatorname{trdeg}_{K(Y)}K(X)$. We can suppose $X = Spec(A)$ and $Y = Spec(B)$. Being generically separable (respectively generically étale) is equivalent to $\dim_{K(X)}\Omega^{1}_{K(X)/K(Y)} = n$ (respectively $\dim_{K(X)}\Omega^{1}_{K(X)/K(Y)} = 0$ ). If I show $f$ is smooth at $\xi$ (respectively étale at $\xi$) it would be ok since the set of points where $f$ is smooth (respectively étale) is open. But we have, $\Omega^{1}_{K(X)/K(Y)} = \Omega^{1}_{\mathcal{O}_{X, \xi}/\mathcal{O}_{Y, \xi'}}$. So,being generically separable (respectively generically étale) implies $\Omega^{1}_{X/Y}$ is a locally free in a neighborhood of $\xi$ of rank $n$ (respectively is a locally zero in a neighborhood $\xi$). This comes from exercise $1.12.$ $(a)$ page $173$ of the book and from the coherence of $\Omega_{X/Y}^{1}$ (since $f$ is of finite type). Hence for all $x \in X_{\xi'}$, $\Omega^{1}_{\mathcal{O}_{X, x}/\mathcal{O}_{Y, \xi'}} \simeq \Omega^{1}_{X_{\xi'}/k(\xi'), x} \otimes k(x)$ by the exercice $2.5.$ $(b)$ page $226$. Now I don't know what to do?

KReiser
  • 65,137
Analyse300
  • 89
  • 3
    What is your definition of smoothness? – Kenta S Aug 23 '23 at 02:06
  • Let $X, Y$ be two locally noetherian schemes. We say $f : X \mapsto Y$ is smooth at $x$ if the fiber $X_{f(x)}$ (which is a $k(f(x))$ scheme) is smooth near $x$ over $k(f(x))$. – Analyse300 Aug 23 '23 at 07:59
  • I want to show $f$ is smooth at $\xi$ (respectively étale at $\xi$). – Analyse300 Aug 23 '23 at 08:07
  • It's enough to show $\Omega^{1}{X{\xi'}/k(\xi'), \xi}$ is free of rank $dim_{\xi}X_{\xi'}$ (respectively $\Omega^{1}{X/Y, \xi } = 0$ since smooth implies flat and since $\Omega^{1}{X/Y, \xi } = 0$ will implie unramified at $\xi$ hence étale at $\xi$). – Analyse300 Aug 23 '23 at 08:14
  • The étale case is then easy and it's just the smoothness I still searching for. – Analyse300 Aug 23 '23 at 08:18
  • Note that $\Omega^{1}{\mathcal{O}{X, \xi}/\mathcal{O}{Y, \xi'}} \simeq \Omega^{1}{X_{\xi'}/k(\xi'), \xi} \otimes k(\xi)$ is a $k(\xi)$ vector space of dimension $trdeg_{K(Y)}K(X)$. If the fiber $X_{\xi'}$ were integral, I could show this implies $\Omega^{1}{X{\xi'}/k(\xi'), \xi}$ is free of rank $trdeg_{K(Y)}K(X) = dim(X_{\xi'}) = dim_{\xi}X_{\xi'}$. – Analyse300 Aug 23 '23 at 16:51
  • Your definition of smoothness does not look sufficient. You need to assume that $f$ is flat of finite type near (or at – this is equivalent [not obviously so, though]) $x$! – Aphelli Aug 31 '23 at 15:01
  • Right I forget of finite type and flat! – Analyse300 Aug 31 '23 at 16:50
  • thank you very much. – Analyse300 Aug 31 '23 at 16:51
  • 1
    I’m not sure how you can show such a statement without some sort of openness property (eg “the set of points at which $f$ is étale is open”, or maybe replace “étale” with “flat”). Anyway, I think the smooth case follows from the étale case, because after shrinking $X,Y$, you can assume that $f$ comes from a generically étale $X \rightarrow \mathbb{A}^n_Y$. – Aphelli Sep 01 '23 at 08:49

1 Answers1

1

Both statements follow from standard openness properties and an alternate characterization of smoothness in terms of the rank of $\Omega_{X/Y}$.

Theorem. (Openness of the flat locus, Stacks 0938). Suppose $f:X\to S$ is a morphism of schemes which is locally of finite presentation. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_X$-module which is locally of finite presentation. Then the set $$U=\{x\in X \mid \mathcal{F} \text{ is flat over } S \text{ at } x\}$$ is open in $X$.

This theorem is a bit of a bear to prove yourself; hopefully Liu has spent some time on this breaking it up or proving it himself.

Lemma. The rank of a coherent sheaf on a noetherian scheme is upper semi-continuous, and if such a sheaf has a free stalk at a point, it is locally free in a neighborhood. (Refer to here on MSE and here on MSE for instance.)

Lemma. (Stacks 01V9) A map of schemes $f:X\to S$ is smooth at $x\in X$ iff the map $\mathcal{O}_{S,f(x)}\to\mathcal{O}_{X,x}$ is flat and $\Omega_{X/S,x}$ can be generated by at most $\dim_x X_{f(x)}$ elements.

Now we can look at the situation for the maps of generic points, apply the results above, and win.

Let $\eta$ be the generic point of $X$ and $\theta$ be the generic point of $Y$. Then the map $\mathcal{O}_{Y,\theta}\to\mathcal{O}_{X,\eta}$ is just $k(Y)\to k(X)$, which is a field extension and therefore flat. The stalk of $\Omega_{X/Y}$ at $\eta$ is the module of differentials associated to the field extension $k(Y)\subset k(X)$, and so by generic separability, it is a $k(X)$-vector space of dimension $\operatorname{trdeg} k(X) -\operatorname{trdeg} k(Y)$ and hence $\Omega_{X/Y}$ is free of rank $\operatorname{trdeg} k(X) -\operatorname{trdeg} k(Y)$ near $\eta$. Now $f$ is flat in a neighborhood of $\eta$ by the theorem, $\Omega_{X/Y}$ is locally free of rank $\operatorname{trdeg} k(X) -\operatorname{trdeg} k(Y)$ in a neighborhood of $\eta$ by the first lemma, and so on the common open neighborhood $U$ of $\eta$ where both these conditions hold we may conclude that $f$ is smooth and we're done. When $k(Y)\to k(X)$ is finite, $\Omega_{X/Y}$ is zero in a neighborhood of $\eta$, and we may conclude that $f$ is in fact smooth of relative dimension zero, or etale.

KReiser
  • 65,137