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A variety is a separated reduced scheme of finite type over an algebraically closed field $k$ not necessarily affine.

Let $X$ be a normal variety and $X$ has only one isolated singularity at $x\in X$ and let $f:Y\rightarrow X$ be a canonical resolution of $X$ i.e $f$ is a resolution s.t $Y$ has at most canonical singularity and its canonical divisor $K_Y$ is $f$ ample.
I want to show that $f_*{K_X}-K_Y$ is effective. I know the outline of a proof but can't justify the detail i.e "the existence of a general hyperplane intersection".
Let me explain it. The outline is as follows.
Write $f_*{K_X}-K_Y=G_+-G_-$ the difference of effective divisor. I will show $G_-$ is zero.
Suppose $G_-> 0$.
Take codimension $\dim X-\dim f(G_-)$ "general hyperplane intersection" $X_0\subset X$.
Then
(i) $X_0\cap f(G_-)$ is non empty and zero dimensional.

Next, consider $f^{-1}(X_0)\subset Y$. Take codimension two dimensional "hyperplane intersection" $Y_0\subset f^{-1}(X_0)$ s.t
(i) $Y_0$ is smooth
(ii) $Y_0\cap G_-$ is a non zero divisor (codimension 1) of $Y_0$.

Put $C_+:=Y_0\cap G_+$ $C_-:=Y_0\cap G_-$look at $f|Y_0:Y_0\rightarrow X$ and a non zero divisor $C_-$. Since $C_-$ is contracted to a point, We can apply Hodge index theorem and we obtain self intersection number $(C^2)<0$. Hence $(f_*K_X-K_Y)*C_-=(G_+-G_-)*C_-=C_+*C_--(C_-^2)> 0$. On the other hand, $(f_*K_X-K_Y)*C_-=-K_Y*C_-<0$ since $K_Y$ is ample. contradiction.

Again my question is how can I construct $X_0$ and $Y_0$ which have desired property.

I know that Bertinis theorem is related but I can't apply the original form, for example in Hartshorne.

If possible, a detailed answer is appreciated.

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  • $\begingroup$ You should use the following two Bertini-type statements: (1) If $D \subset X$ is any smooth divisor, the general hyperplane section of $X$ intersects $D$ along a smooth subvariety. (2) More generally, given a linear system on a smooth projective variety, its general element is smooth outside the base locus of the system. $\endgroup$ Commented Apr 25 at 7:24
  • $\begingroup$ Thank you for your comments. I have to consider how to use them. $\endgroup$
    – George
    Commented Apr 25 at 9:10
  • $\begingroup$ I have some question. $\dim f(G_-) \subset X$ may not be a divisor . Can we apply the Bertinis theorem? After constructing $X_0$, what is a dimension of $f^{-1}(X_0)\cap G_0$? If the dimension is zero , I can't justify the proof. $\endgroup$
    – George
    Commented Apr 25 at 10:21

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