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We know if a normal singular $\mathbb{Q}$-factorial variety $X$ has a crepant resolution, then it is canonical but not terminal. What can we say about the converse? To be more precise:

Is there a normal singular $\mathbb{Q}$-factorial $X$ which is canonical but not terminal, yet admitting no crepant resolution?

Edit: In light of the first comment, I guess what I really want to ask is if the above situation occurs if $X$ only has one canonical point. (I believe then discrep=0 is achieved by considering a sequence of resolutions with positive discrepancy?)

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    $\begingroup$ Does "resolution" here really mean "resolution" or just "partial resolution"? If the former the answer seems to be "yes" for a fairly trivial reason: let $X$ be a variety with one terminal point and one canonical point. $\endgroup$
    – Lazzaro Campeotti
    Commented Nov 10, 2016 at 9:52
  • $\begingroup$ That's a good point. I will edit my question accordingly. $\endgroup$
    – Xuqiang QIN
    Commented Nov 10, 2016 at 16:32
  • $\begingroup$ If a canonical singularity has a crepant resolution, then it should be Gorenstein. So find a non-Gorenstein canonical singularity is enough for your question. For example with Gorenstein canonical singularities, I think one can produce such example by taking canonical model of a Gorenstein terminal singularity. $\endgroup$
    – Chen Jiang
    Commented Nov 13, 2016 at 2:49
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    $\begingroup$ @ChenJiang: why is it that a canonical singularity with a crepant resolution would have to be Gorenstein? $\endgroup$
    – Sándor Kovács
    Commented Nov 16, 2016 at 6:16
  • $\begingroup$ @SándorKovács: say $Y\to X$ is the crepant resolution such that $K_Y=f^*K_X$, then $K_Y$ is Cartier implies $K_X$ is also Cartier. Did I miss something? $\endgroup$
    – Chen Jiang
    Commented Nov 16, 2016 at 8:04

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