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.

  • 6
    $\begingroup$ You should be careful to distinguish "smooth" from "regular". Here you obviously mean "regular". $\endgroup$ Feb 23, 2012 at 10:32

4 Answers 4


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.


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


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.


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).


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.