6
$\begingroup$

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.

$\endgroup$

1 Answer 1

5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.