# Iitaka dimension is invariant under surjective morphism between smooth projective varieties

I would like to prove the following result (working on $$\mathbb{C}$$) but get trouble with the other direction.

Let $$f:Y'\rightarrow Y$$ be a surjective morphism between smooth projective varieties, then $$\kappa(Y',f^*D)=\kappa(Y,D)$$.

$$"\geq"$$: Note that we have for any divisor $$D$$ on $$Y$$ that $$H^0(Y',f^*\mathcal{O}_Y((mD))\cong H^0(Y,f_*f^*\mathcal{O}_Y((mD))\cong H^0(Y,\mathcal{O}_Y(mD)\otimes f_*\mathcal{O}_{Y'})$$ where for the second equality we applied the projection formula. But since $$f$$ is surjective, we have an inclusion $$\mathcal{O}_{Y}\hookrightarrow f_*\mathcal{O}_{Y'}$$ and hence $$H^0(Y,mD)\subseteq H^0(Y',f^*mD)$$ for all $$m$$. It follows that $$\kappa(Y',f^*D)\geq\kappa(Y,D)$$. But I do not find a nice way to show the other direction.

This is Proposition 1.5 in [1] below but I'll include the proof here for convenience. We'll prove something a bit stronger. Let $$k = \mathbb{C}$$ or any other algebraically closed field of characteristic zero.

Proposition: Suppose $$f : X \to Y$$ is a surjective morphism between normal projective varieties and $$L$$ is a line bundle on $$Y$$. Then $$\kappa(Y,L) = \kappa(X, f^*L)$$.

First we need an equivalent characterization of the Iitaka dimension. Let $$R(X,L) = \bigoplus_{m \geq 0} H^0(X, mL)$$ be the ring of sections of $$L$$. If $$R(X,L) \neq 0$$, then it is an integral domain with fraction field denoted $$Q(X,L)$$.

Lemma 1: If $$R(X,L) \neq 0$$, then $$\kappa(X,L) = \mathrm{tr.deg}_k Q(X,L) - 1$$.

Now consider the Stein factorization $$X \to Z \to Y$$ where $$g : X \to Z$$ has connected fibers and $$\pi : Z \to S$$ is finite. Then $$g_*\mathcal{O}_X = \mathcal{O}_Z$$. By the same computation as in the question, we have $$H^0(X, f^*(mL)) = H^0(Z, \pi^*(mL) \otimes g_*\mathcal{O}_X) = H^0(Z, \pi^*(mL))$$ so $$\kappa(X, f^*L) = \kappa(Z, \pi^*L)$$. Replacing $$f$$ with $$\pi$$, it suffices to prove the statement when $$f$$ is a finite surjection. Let $$F/k(Y)$$ be the Galois closure of $$k(Y)/k(X)$$ with Galois group $$G$$ and let $$X' \to X$$ be the normalization of $$X$$ in $$F$$. Then $$X' \to X$$ and $$X' \to Y$$ are finite Galois covers so it suffices to prove the statement when $$f$$ is finite Galois.

Lemma 2: Suppose $$f : X \to Y$$ is a finite Galois cover of normal varieties and let $$L$$ be a line bundle on $$Y$$. Then $$R(X, f^*L) \neq 0$$, then it is an integral ring extension of $$R(Y,L)$$.

Proof: $$f_*\mathcal{O}_X$$ is a finite $$\mathcal{O}_Y$$ algebra with a $$G$$-action such that $$(f_*\mathcal{O}_X)^G = \mathcal{O}_Y$$. Then by projection formula, $$f_*f^*(mL) = mL \otimes f_*\mathcal{O}_X$$ so $$R(X, f^*L)$$ inherits a $$G$$-action with invariant ring $$R(X,f^*L)^G = R(Y, L)$$. Now suppose $$r \in R(X,f^*L)$$. Then $$\varphi_r(T) := \prod_{g \in G} (T - g\cdot r)$$ is a $$G$$-invariant monic polynomial with $$r$$ as a root. Thus $$\varphi_r(T) \in R(Y,L)[T]$$ and exhibits $$r$$ as an integral element over $$R(Y,L)$$.

Now to finish the proof of the proposition, note that if $$R(X,f^*L) \neq 0$$, then $$R(X,f^*L)^G$$ contains $$k$$ and so $$R(X,f^*L) \neq 0$$ if and only if $$R(Y,L) \neq 0$$. In this case, $$R(X, f^*L)/ R(Y,L)$$ is an integral extension by Lemma 2, so $$Q(X, f^*L)/Q(Y,L)$$ is an algebraic field extension and so by Lemma 1, $$\kappa(X,f^*L) = \kappa(Y,L)$$.

[1] Mori, Shigefumi, Classification of higher-dimensional varieties, Algebraic geometry, Proc. Summer Res. Inst., Brunswick/Maine 1985, part 1, Proc. Symp. Pure Math. 46, 269-331 (1987). ZBL0656.14022.