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Let $X$ and $Y$ be varieties with ample canonical sheaf. (We work over the complex numbers and varieties are smooth projective and connected.)

Suppose that $X$ and $Y$ are birationally equivalent. Are $X$ and $Y$ isomorphic?

Let $d$ be the dimension of $X$.

If $d=1$, the answer is yes. (Curves are determined by their function field.)

If $d=2$, the answer is yes. (The minimal model of a surface exists and is unique.)

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They are isomorphic for any d. Indeed, they are isomorphic to Proj of the canonical ring, which is a birational invariant. It's not a research level question... –  Lev Borisov Aug 23 at 12:25
If $X$ is isomorphic to the Proj of the canonical ring (defined by the canonical sheaf), then this means (in particular) that the canonical ring is finitely generated. I believe the latter has only been established recently by Birkar, Hacon and McKernan. Moreover, I don't see why the Proj of the canonical ring is isomorphic to the variety because it is not smooth in general. –  Pieter Aug 23 at 12:41
You have assumed that K is ample. In this case the finite generation is trivial (it's true for any ample line bundle). –  Lev Borisov Aug 23 at 12:56
@Pieter: don't forget Cascini! –  Artie Prendergast-Smith Aug 23 at 21:51
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