Let X be a smooth variety, take the proj of canonical ring of X and denote it by Y. Is Y always a canonical variety? I know it's true for general type variety. Thank in advance.


1 Answer 1


I believe that $(Y,B)$ is always klt for some boundary $B$. In fact by Theorem 5.2 of https://projecteuclid.org/download/pdf_1/euclid.jdg/1090347529, after passing to a truncation of the pluricanonical rings, there is a klt pair of log general type $(Y',B')$ with isomorphic pluricanonical ring (more precisely $R(K_X)^{(a)}\cong R(K_{X'}+B')^{(b)}$ for appropriate integers $a,b>0$). But then $Y={\rm Proj} R(K_X)\cong {\rm Proj} R(K_{Y'}+B')$ and $Y'\to Y$ is the log canonical model of $(Y',B')$ which has klt singularities as $(Y',B')$ has klt singularities.

I also believe that $Y$ may not be canonical. I think one can construct examples where $Y$ is not canonical by considering a surface $S$ of general type with a cyclic group $G$ of order $n$ acting on it such that $S/G$ has non-canonical singularities (eg. $n=3$ and locally $g(x,y)=(\xi x, \xi ^2 y)$ where $\xi$ is a primitive third root of 1) and $E$ an elliptic curve where $G$ acts via a translation of order $n$. Then $X=(S\times E)/G$ should be smooth but $Y=S/G$ is klt but not canonical (I did not check the details!).


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