7

Let $X$ be a smooth projective integral variety over an algebraically closed field $k$. Let $Y$ be a (not necessarily smooth) projective integral variety over $k$.

Assume that $D^b(X) \cong D^b(Y)$.

Does it follow that $Y$ is smooth?

Edit: Here $D^b(X)$ (resp. $D^b(Y)$) is $D^b(Coh(X))$ (resp. $D^b(Coh(Y))$).

Marco
  • 71
  • 2
  • 7
    Isn't Serre's homological criterion that $Y$ is smooth if and only if, for every pair of objects $A$, $B$ of $D^b(Y)$, $\text{Hom}_{D^b(Y)}(A,B[n])$ equals ${0}$ for all but finitely many choices of $n$? – Jason Starr Jan 19 '17 at 11:53
  • 5
    What Jason mentioned sometimes is called Ext-finiteness and it is, indeed, a criterion of smoothness. – Sasha Jan 19 '17 at 11:57
  • 1
    Thank you for your comments. Is there a reference where I can find the statement with proof? I looked around quickly but couldn't find it. – Marco Jan 19 '17 at 12:00
  • 2
    I just read in another MO post that the correct attribution is due to Auslander-Buchsbaum-Serre, cf. http://mathoverflow.net/questions/103530/serres-theorem-about-regularity-and-homological-dimension – Jason Starr Jan 19 '17 at 12:46
  • 7
    Somewhere Count Dracula is smiling that I misattributed a result: http://mathoverflow.net/questions/259808/hartshorne-conjecture-on-vector-bundles-on-projective-spaces#comment641171_259808 – Jason Starr Jan 19 '17 at 12:48

1 Answers1

5

I shall elaborate on the comments and provide a sketch of the proof (for which I unfortunately don't know a reference). As $k$ is a perfect field, $X$ is smooth if and only if $X$ is regular. By a well-known theorem of Serre $X$ is regular if and only if for every closed point $x\in X$ the local ring $\mathscr{O}_{X,x}$ is of finite homological dimension,

Theorem. The following are equivalent:

(i) $X$ is regular;

(ii) $Coh(X)$ has finite homological dimension;

(iii) $D^b(X)$ is Ext-finite.

Proof. (i)$ \Rightarrow $(ii): Serre duality.

(ii)$ \Rightarrow $(iii): By assumption the condition required by Ext-finiteness holds for all complexes concentrated in one degree, and it follows for all bounded complexes for instance by use of standard spectral sequences (see remark 3.7 in Huybrecht's `Fourier-Mukai transforms').

(iii)$ \Rightarrow $(i): Suppose that $X$ is not regular, let $x\in X$ be a closed singular point. Let $\mathscr{O}_x$ be the skyscraper sheaf with stalk $\kappa(x)$ at $x$. Then the local-to-global spectral sequence yields isomorphisms $\mathrm{Ext}^n_X(\mathscr{O}_x,\mathscr{O}_x)\simeq\mathrm{Ext}^n_{\mathscr{O}_{X,x}}(\kappa(x),\kappa(x))$. The latter is nonzero for infinitely many $n$ because $\mathscr{O}_{X,x}$ has infinite homological dimension. Hence $D^b(X)$ is not Ext-finite.

ssx
  • 2,729