Does smoothness descend along flat morphisms? - MathOverflow

Does smoothness descend along flat morphisms?

2012-02-22T01:06:16Z

Suppose $f:X\to Y$ is a flat morphism of schemes. If $X$ is smooth at $x$, must $Y$ be smooth at $f(x)$?

If $f$ is locally finitely presented, then it is open (using EGA IV 1.10.4), so after replacing $Y$ by $f(X)$, we can assume $f$ is faithfully flat. I'd be happy to understand even the case where $X$ and $Y$ are local:

Suppose $R$ and $S$ are local rings and $R\to S$ is a local homomorphism with $S$ (faithfully) flat over $R$. If $S$ is regular, must $R$ be regular?

Note that I'm not asking if smoothness is "flat local"; there are certainly flat morphisms from singular things to smooth things (e.g. $k[x,y]/(x^2-y^2)$ is flat over $k[x]$). The question is whether there are flat morphisms from smooth schemes which hit singular points.

Answer by Karl Schwede for Does smoothness descend along flat morphisms?

2012-02-22T01:31:33Z

Here's a direct reference:

Hochster Math 615 - 2004, February 18th

It's a corollary of the characterization of regularity by the projective dimension of $R/m$.

I found this by tracking references from page 35 of Mel Hochster's notes here:

Hochster Math 615 - 2010

Answer by Akhil Mathew for Does smoothness descend along flat morphisms?

2012-02-22T02:01:46Z

EGA 0-IV, 17.3.3 has the second claim: if $A \to B$ is a local homomorphism of local noetherian rings, and $B$ is regular and $A$-flat, then $A$ is regular. The strategy is to use the fact that if $B$ is faithfully flat over $A$, then the (global) projective dimension of $A$ is at most the projective dimension of $B$, and Serre's characterization of regular local rings as those with finite global dimension. For instance, suppose that $\mathrm{proj} \dim B = n$, and consider a resolution

$$0 \to M_0 \to M_1 \to \dots \to M_n \to M \to 0$$ of $A$-modules, where all the $M_i$ except possibly $M_0$ are projective. We can assume without loss of generality that everything is finitely generated. Tensoring with $B$ gives a resolution: $$0 \to M_0 \otimes_A B \to M_1 \otimes_A B \to \dots \to M_n \otimes_A B\to M\otimes_A B \to 0$$ where, by the condition on $B$, we find that $M_0 \otimes_A B$ is projective. Thus $M_0$ is projective over $A$.

For the last step, I used the fact that projectivity descends under faithfully flat extensions; in general, this is a theorem of Raynaud-Gruson, but it follows directly if everything is noetherian and finitely generated, since then projectivity is equivalent to flatness.

Answer by Andrew Stout for Does smoothness descend along flat morphisms?

2012-03-13T16:24:30Z

Your second question is answered affirmatively by EGA IV_2 Corollary 6.5.2, which references EGA IV_1 $\S0$ 17.3.3(i). Here you only need to assume that $f: X \rightarrow Y$ is a flat morphism of locally Noetherian schemes.

In general, smoothness is a stronger condition than regularity. For example, if $k'$ is a non-separable field extension of a field $k$, then the structure morphism $f:\mathbb{P}_{k'}^{1} \rightarrow \mbox{Spec}(k)$ is regular, but it is not smooth.

This last point follows because if it were smooth, we should have an exact sequence

$$0\rightarrow f^* \Omega_{k'/k} \rightarrow \Omega_{\mathbb{P}_{k'}^1/k} \rightarrow \Omega_{\mathbb{P}_{k'}^{1}/k'}\rightarrow 0$$

which is actually

$$0 \rightarrow \mathcal{O}_{\mathbb{P}_{k'}^1} \rightarrow \Omega_{\mathbb{P}_{k'}^1/k} \rightarrow \mathcal{O}_{\mathbb{P}_{k'}^1}(-2) \rightarrow 0$$

which is absurd as smoothness means that, after taking stalks, the middle module is free of rank 1.

If you work in the category of schemes of finite type over a perfect field then they are equivalent (cf., Liu's Algebraic Geometry and Arithmetic Curves, Chapter 4 Corollary 3.33), which answers your first question.

Answer by Qing Liu for Does smoothness descend along flat morphisms?

2012-08-31T21:08:20Z

The answer to the first question is yes and (probably you known) is a consequence of the second question.

Suppose $f : X\to Y$ is a flat morphism of schemes over $S$ with $X$ smooth. Let $y=f(x)$ and let $s$ be the image of $x$ in $S$. By the positive answer to the second question, the geometric fiber $Y_{\bar{s}}$ is regular. So it only remains to see that$Y$ is flat over $S$ at $y$.

Consider the homomorphisms of local rings $$ O_{S,s} \to O_{Y, y}\to O_{X,x}$$
The second is flat by hypothesis, hence faithfully flat (because we deal with local rings), and the composition is flat by the smoothness of $X\to S$. So the first one is flat (easily seen using definition of flatness).