Algebraic Geometry – Rational Contraction and Proj of Section Ring

Question

I am reading the paper "Mori Dream Spaces and GIT" by Hu and Keel.

https://arxiv.org/abs/math/0004017

I cannot understand the proof of Lemma 1.6 in it.

Let $X$ be a normal projective variety.
Assume that $D$ is a divisor on $X$ such that $R(X,D) = \bigoplus_{m \in \mathbb{Z}_{\ge0} } H^0(X,\mathcal{O}_X(mD))$ is finitely generated.

Lemma 1.6 states that the natural rational map $\varphi_D \colon X \dashrightarrow Y := Proj(R(X,D))$ is a rational contraction, i.e, there exits a resolution $\Phi \colon \tilde{X} \to Y$ of $\varphi_D$ such that $\Phi_*(\mathcal{O}_{\tilde{X}} (E)) \simeq \mathcal{O}_Y$ for any effective exceptional divisor $E$.
Moreover, $D$ can be written as $ D = \varphi_D^*(A) + F $ for an ample divisor $A$ on $Y$ and a $\varphi_D$-fixed divisor $F$.

Outline of the proof of the paper is as follows:

(1) The finite generation of $R(X,D)$ implies that $D$ can be written as $D = M+F$ for a movable divisor $M$ and an effective divisor $F$ such that
$Sym_k(H^0(X,M)) \to H^0(X,kM)$ is surjective and $H^0(X,kM) \to H^0(X,kM +rF)$ is an isomorphism for any $k>0$ and $r>0$ after replacing $D$ by its multiple.

(2) After taking appropriate resolution, we may assume that $\varphi_D$ is a regular
and $M = \varphi_D^* A $. Now we can see $F$ is $\varphi_D$-fixed by the isoromorphism $H^0(X,kM) \simeq H^0(X,kM +rF)$.

Here is my question :

[1] How can wee see that $\varphi_D$ is a rational contraction? Is it a well-known result?

[2] Why does the isomorphism $H^0(X,kM) \simeq H^0(X,kM +rF)$ implies that
$F$ is $\varphi_D$-fixed?

Can anyone who understands these results answer my questions?

Answer 1

Let's assume that $|kD|=|kM|+kF$ where $|M|$ is base point free and moreover $Sym ^kH^0(M)\to H^0(kM)$ is surjective for any $k>0$. This can be achieved replacing $X$ by an appropriate resolution and $k$ by a multiple. Let $f:X\to Y$ be the morphism induced by $|M|$ so that $f^*A=M$.

For [1], we wish to show that $f_* \mathcal O _X=\mathcal O _Y$. The Stein factorization $X\to Z\to Y$ is defined by ${\rm Spec} f_* \mathcal O _X$. We have $H^0(X,kM)=H^0(f_*(kM))=H^0(f_* \mathcal O _X\otimes \mathcal O _Y(kA))$ and so $|kM|$ in fact defines the morphism $X\to Y$. Since $Sym ^kH^0(M)\to H^0(kM)$ is surjective, then $|M|$ defines the same morphism.

For [2], Since $H^0(X,kM)=H^0(X,kM+rF)$, then $H^0(\mathcal O _Y(kA))\cong H^0(\mathcal O _Y(kA)\otimes f_* \mathcal O _X(rF))$. For $k\gg r$, the sheaf $\mathcal O _Y(kA)\otimes f_* \mathcal O _X(rF)$ is globally generated and so $f_* \mathcal O _X(rF)\cong \mathcal O _Y$. (See also Lemma 3.2 https://arxiv.org/pdf/1104.4981.pdf).

Hope this helps.