# A property of canonical singularity

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($$\oplus_{m\geqq 0}{g_1}_*{f_1}_*\mathcal{O}(mK_{Y_i})$$)

• Probably, you intended to write $K_Y=f^*K_X+\sum_{i=1}^r m_iE_i$ and the $a_i$ are $m_i$? Apr 23 at 22:00
• 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) Apr 23 at 22:11
• @user267839 Thanks. You are right. Apr 23 at 22:52
• @user267839 I mean for all non negative m. Could you see my motivation added in the question. Apr 23 at 22:53
• @user267839 Hartog's argument might be used for my porpose too. Thanks. Apr 23 at 22:55

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.