Does smoothness descend along flat morphisms? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T01:13:00Z http://mathoverflow.net/feeds/question/89152 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89152/does-smoothness-descend-along-flat-morphisms Does smoothness descend along flat morphisms? Anton Geraschenko 2012-02-22T01:06:16Z 2012-08-31T21:08:20Z <blockquote> <p>Suppose $f:X\to Y$ is a flat morphism of schemes. If $X$ is smooth at $x$, must $Y$ be smooth at $f(x)$?</p> </blockquote> <p>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:</p> <blockquote> <p>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?</p> </blockquote> <p>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.</p> http://mathoverflow.net/questions/89152/does-smoothness-descend-along-flat-morphisms/89155#89155 Answer by Karl Schwede for Does smoothness descend along flat morphisms? Karl Schwede 2012-02-22T01:31:33Z 2012-02-22T01:38:02Z <p>Here's a direct reference:</p> <p><a href="http://www.math.lsa.umich.edu/~hochster/615W04/L02_18.ps" rel="nofollow">Hochster Math 615 - 2004, February 18th</a></p> <p>It's a corollary of the characterization of regularity by the projective dimension of $R/m$.</p> <p>I found this by tracking references from page 35 of Mel Hochster's notes here:</p> <p><a href="http://www.math.lsa.umich.edu/~hochster/615W10/615.pdf" rel="nofollow">Hochster Math 615 - 2010</a></p> http://mathoverflow.net/questions/89152/does-smoothness-descend-along-flat-morphisms/89156#89156 Answer by Akhil Mathew for Does smoothness descend along flat morphisms? Akhil Mathew 2012-02-22T02:01:46Z 2012-02-22T02:01:46Z <p>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</p> <p>$$0 \to M_0 \to M_1 \to \dots \to M_n \to M \to 0$$ 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 $A$. </p> <p>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. </p> http://mathoverflow.net/questions/89152/does-smoothness-descend-along-flat-morphisms/91093#91093 Answer by Andrew Stout for Does smoothness descend along flat morphisms? Andrew Stout 2012-03-13T16:24:30Z 2012-03-14T16:22:22Z <p>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. </p> <p>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. </p> <p>This last point follows because if it were smooth, we should have an exact sequence</p> <p>$$0\rightarrow f^* \Omega_{k'/k} \rightarrow \Omega_{\mathbb{P}_{k'}^1/k} \rightarrow \Omega_{\mathbb{P}_{k'}^{1}/k'}\rightarrow 0$$</p> <p>which is actually</p> <p>$$0 \rightarrow \mathcal{O}_{\mathbb{P}_{k'}^1} \rightarrow \Omega_{\mathbb{P}_{k'}^1/k} \rightarrow \mathcal{O}_{\mathbb{P}_{k'}^1}(-2) \rightarrow 0$$</p> <p>which is absurd as smoothness means that, after taking stalks, the middle module is free of rank 1. </p> <p>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. </p> http://mathoverflow.net/questions/89152/does-smoothness-descend-along-flat-morphisms/106071#106071 Answer by Qing Liu for Does smoothness descend along flat morphisms? Qing Liu 2012-08-31T21:08:20Z 2012-08-31T21:08:20Z <p>The answer to the first question is yes and (probably you known) is a consequence of the second question. </p> <p>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 $y$. </p> <p>Consider the homomorphisms of local rings $$O_{S,s} \to O_{Y, y}\to O_{X,x}$$<br> 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). </p>