Derived categories of curves equivalent then the curves are isomorphic I am a beginner at derived categories and I'm looking for a proof of the following fact:
If $X$ and $Y$ are smooth projective curves such that $D^b(Coh\,X)$ is equivalent to $D^b(Coh\,Y)$ then $X$ and $Y$ are isomorphic.
Can anyone give me the explanation of this fact or provide a good reference for this?
As I understand, the crucial point here is that any object in $D^b(Coh \, X)$ splits as a sum of its cohomology sheaves but I don't know how to finish the proof.
Thank you.
 A: The non-elliptic case is easier. By a theorem of Bondal-Orlov, see [Huybrechts, Fourier-Mukai transforms in algebraic geometry, Theorem 4.11]: 
Theorem. Let $X$ and $Y$ be smooth projective varieties with equivalent derived category $D^b(X)$ and $D^b(Y)$. If the (anti)-canonical bundle of $X$ is ample, then  $X$ and $Y$ are isomorphic. 
For elliptic curve, we need a generalized theorem by Kawamata, See [Kawamata, D-equivalence and K-equivalence] or [Huybrechts, Fourier-Mukai transforms in algebraic geometry, Theorem 6.15]:
Theorem. Let $X$ and $Y$ be smooth projective varieties with equivalent derived category $D^b(X)$ and $D^b(Y)$. Then the (anti)-canonical bundle of $X$ is nef if and only if the (anti)-canonical bundle of $Y$ is nef.
This theorem implies if $X$ is elliptic curve, the so is $Y$.
By [Huybrechts, Fourier-Mukai transforms in algebraic geometry, Theorem 5.39], the hodge structuces of $X$ and $Y$ are the same, which yeilds that $X$ and $Y$ are isomorphic.
A: I don't know of a uniform way to prove it for all curves. For elliptic curves it follows from Hodge theory and for the rest it's a consequence of the Bondal-Orlov theorem. It's all explained in Huybrechts's excellent book on Fourier-Mukai transforms.
