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Let $X$ be a normal variety with only one singularity at $x$ and $(X,x)$ is a canonical singularity i.e. $(X.x)$ satisfies $(i)$ and $(ii)$.
$(i)$ $(X,x)$ is a $\mathbb{Q}$ Goreinstein singularity.
$(ii)$ For any resolution $f:Y \rightarrow X$, we can write $K_Y=f^*K_X+\sum_{i=1}^r m_iE_i$ with $m_i\geqq 0$

How can I prove that for any resolution $f:Y\rightarrow X$ and $m\in > \mathbb{Z}_{\geqq 0}$, $f_*\mathcal{O}(mK_Y)\cong \mathcal{O}(mK_X)$?

I don't know even what is a morphism.

Motivation. I want to show the unique ness of "canonical resolution"
A resolution of $f:Y\rightarrow X$ is called canonical resolution if
(i) $Y$ has at worst canonical singularity.
(ii) $K_Y$ is $f$ ample.
My strategy is this.
Let $f_i:Y_i\rightarrow X$ be two canonical resolution.
Make two common resolution $g_1:X\rightarrow Y_1$ and $g_2:X\rightarrow Y_2$.
From (ii), we get $Y_i\cong$ Proj($\oplus_{m\geqq 0}{f_i}*\mathcal{O}(mK_{Y_i})$).
So if we can show ${g_i}_*\mathcal{O}(mK_{X})\cong \mathcal{O}(mK_{Y_i})$, we can show $Y_1\cong Y_2\cong $Proj($\oplus_{m\geqq 0}{g_1}_*{f_1}_*\mathcal{O}(mK_{Y_i})$)

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  • $\begingroup$ Probably, you intended to write $K_Y=f^*K_X+\sum_{i=1}^r m_iE_i$ and the $a_i$ are $m_i$? $\endgroup$
    – user267839
    Commented Apr 23 at 22:00
  • $\begingroup$ Do you really mean for all $m $ or big enough? Since $K_X$ is $\mathbb{Q}$-Cartier it's multiple for big enough $m$ becomes Cartier, so invertible. But as $X$ is normal, you can use Hartog's argument on the base as $x$ has codim $\ge 2$ (if your $X$ is not a curve) $\endgroup$
    – user267839
    Commented Apr 23 at 22:11
  • $\begingroup$ @user267839 Thanks. You are right. $\endgroup$
    – George
    Commented Apr 23 at 22:52
  • $\begingroup$ @user267839 I mean for all non negative m. Could you see my motivation added in the question. $\endgroup$
    – George
    Commented Apr 23 at 22:53
  • $\begingroup$ @user267839 Hartog's argument might be used for my porpose too. Thanks. $\endgroup$
    – George
    Commented Apr 23 at 22:55

1 Answer 1

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This essentially follows from Claim 2.3.1, p. 39 of

J. Kollár, S. Kovács: Singularities of the Minimal Model Program, Cambridge Tracts in Mathematics 200 (2013).

The Authors prove the result for $m=1$, but the same proof works without modifications for every $m \geq 1$. Compare also with Claim 2.3.2 (same page), Proposition 2.17 (p. 48) and Proposition 2.18 (p. 49).

The last result, applied with $S=X, \, p_X=\text{id}_X$, $\Delta=\Delta_Y=\emptyset$, gives precisely the statement you need.

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