Beauville Exercise VII.7 (3)-A proof that $\kappa(X)\geq \kappa(Y)$ for $f\colon X\to Y$ surjective morphism of smooth projective varieties

Question

Here $\kappa(X)$ denotes the Kodaira dimension of a smooth projective variety $X$.

Question 1:

I would like to solve Exercise VII.7 (3) from the Beauville book "Complex Algebraic Surfaces":

Let $\pi \colon X\to Y$ be a surjective morphism, then $\kappa(X)\geq \kappa(Y)$. Moreover, if $\pi$ is étale, then the equality holds.

I have already looked around and I have found these two questions that could be related to my problem

Iitaka dimension is invariant under surjective morphism between smooth projective varieties

Rational maps and Kodaira dimension

However, I don't understand so much what they mean, so I decided to ask once more here.

My idea is simply the following:

By the Hurwitz formula, we have $dK_X=\pi^*(dK_Y)+D$ for any $d\geq 1$, where $D$ is an effective divisor (in particular, $D$ is $d$-times the ramification divisor of $\pi$).

Therefore, the pullback $\pi^*\colon H^0(Y, dK_Y)\hookrightarrow H^0(X, \pi_1^*(dK_Y))\subseteq H^0(X, dK_X)$ is injective. In other words, $h^0(dK_X)\geq h^0(dK_Y)$. Define $k:=\kappa(S)$; if $k\leq 0$, then the thesis follows immediately by the definition of the Kodaira dimension. Otherwise, we would get $$ \limsup_{d\to \infty}\frac{h^0(dK_X)}{d^k}\geq \limsup_{d\to \infty}\frac{h^0(dK_Y)}{d^k}, $$ hence $k(Y)\leq k=k(X)$. Suppose now that $\pi$ is étale. This means for me just that $f$ is unramified, and so basically $D=0$. How can I conclude from here that $\kappa(X)=\kappa(Y)$?

Is maybe true that $h^0(dK_X)=h^0(dK_Y)$ for each $d\geq 1$?

What I have thought is that if $D=0$, then there exists at each point an open neighborhood $U$ (not necessarily Zariski open) such that $f\colon U\to f(U)=:V $ is an isomorphism, and therefore $\pi^*\colon H^0(V, dK_V)\to H^0(U, dK_U)$ is an isomorphism too. How could I conclude from here that then $\pi^*\colon H^0(Y, dK_Y)\to H^0(X, dK_X)$ is an isomorphism?

Question 2:

The second question is to generalize the previous result as follows:

Let $\pi\colon X\to Y$ be a surjective morphism of algebraic varieties, with $Y$ and $X$ not necessarily smooth. Then $\kappa(X)\geq \kappa(Y)$.

I think I have found a solution in the case of surfaces, but I don't know which should be the strategy in the higher dimension:

Let $\rho_X\colon \widehat{X}\to X$ and $\rho_Y\colon \widehat{Y}\to Y$ be two resolutions of the singularities of $X$ and $Y$. Consider the natural rational map $\widehat{X}\dashrightarrow \widehat{Y}$ and resolve its indeterminacy by a finite number of blow-ups $b\colon \smash{\widehat{X}}'\to \widehat{X}$. We have therefore a morphism $\pi\colon \smash{\widehat{X}}'\to \widehat{Y}$, which is surjective since $f$ is surjective too. Apply now the previous result to the map $\pi$ to get $\kappa(\smash{\widehat{X}}')\geq \kappa(\widehat{Y})$. However, the Kodaira dimension is a birational invariant, so

$$\kappa(X)=\kappa(\widehat{X})=\kappa(\smash{\widehat{X}}')\geq \kappa(\widehat{Y})=\kappa(Y).$$

Is it possible to generalize this? If not, which is the idea to prove it in a higher dimension?

Answer 1

It is necessary to assume that $k$ has characteristic zero (the book is probably only considering $k = \mathbb{C}$) since otherwise there are surjective inseparable maps $\mathbb{P}^2 \to X$ where $X$ is general type.

Question 1

As stated, if $\pi : X \to Y$ is a surjective map of smooth proper varieties of the same dimension then $H^0(Y, \omega_Y^{\otimes k}) \to H^0(X, \omega_X^{\otimes k})$ is injective for any $k \ge 0$. If $\xi \in X$ and $\eta \in Y$ are the generic points then these spaces of sections embed into $\omega^{\otimes k}_{Y, \eta}$ and $\omega^{\otimes k}_{X, \xi}$ (using integrality and that these sheaves are vector bundles) and the map $\omega^{\otimes k}_{Y, \eta} \to \omega^{\otimes k}_{X, \xi}$ is injective.

Now we need to show equality in the case $\pi$ is etale. As pointed out in the comments, the claim that $h^0(X, \omega_X^{\otimes k}) = h^0(Y, \omega_Y^{\otimes k})$ is false, only the Kodaira dimensions are equal.

Since $\pi$ is proper and etale, it is finite. Therefore, $\pi_* \mathcal{O}_X$ is a vector bundle. Since $\pi$ is etale $\pi^* \omega_Y^{\otimes k} = \omega_X^{\otimes k}$. Therefore, using the projection formula,

$$ H^0(X, \omega_X^{\otimes k}) = H^0(Y, \omega_Y^{\otimes k} \otimes \pi_* \mathcal{O}_X) $$

Therefore, it suffices to show that these groups grow in $k$ as a polynomial of degree at most $\kappa(Y)$. By taking the Galois closure $f : \tilde{X} \to X \to Y$ we may assume that $X \to Y$ is a $G$-cover for some finite group $G$. Therefore $(f_* \mathcal{O}_X)^G = \mathcal{O}_Y$ Write,

$$ R(Y, \omega_Y) = \bigoplus_{k \ge 0} H^0(Y, \omega_Y^{\otimes k}) \\ R(X, \omega_X) = \bigoplus_{k \ge 0} H^0(X, \omega_X^{\otimes k}) = \bigoplus_{k \ge 0} H^0(Y, \omega_Y^{\otimes k} \otimes f_* \mathcal{O}_X) $$

The injection $R(Y, \omega_Y) \to R(X, \omega_X)$ is a ring map and $R(X, \omega_X)^G = R(Y, \omega_Y)$. It is a general fact that a ring is integral over its $G$-invariants under a finite group action. This is because any $r \in R$ satisfies,

$$ f(x) = \prod_{g \in G}(x - g(r)) $$

and the coefficients are in $R^G$ so $r$ is integral over $r$. In particular, if $R$ is finitely generated over $R^G$ then it is a finite extension. It is a (hard) fact that $R(Y, \omega_Y)$ and $R(X, \omega_X)$ are finitely generated over $k$. Therefore, $R(Y, \omega_Y) \to R(X, \omega_X)$ is a finite extension and hence if $N$ is the number of generators then,

$$ h^0(X, \omega_X^{\otimes k}) \le N h^0(Y, \omega_Y^{\otimes k}) $$

giving the required bound.

Question 2

By the first part, $\kappa(X)$ is a birational invariant for smooth varieties. Therefore, a good definition of $\kappa(X)$ for $X$ singular is to choose a resolution of singularities $\tilde{X} \to X$ and define $\kappa(X) := \kappa(\tilde{X})$ which is independent of the choice of resolution because any two are birational. Given a surjective map $f : X \to Y$ of proper varieties of the same dimension we choose resolutions of singularities $\pi_X : \tilde{X} \to X$ and $\pi_Y : \tilde{Y} \to Y$ and there is only a rational map between $\tilde{X}$ and $\tilde{Y}$. The closure of its graph gives $\Gamma \subset \tilde{X} \times \tilde{Y}$ which maps $\Gamma \to \tilde{X}$ birationally. Choose a resolution of singularities $\tilde{\Gamma} \to \Gamma$. Therefore we get,

$$ \tilde{Y} \leftarrow \tilde{\Gamma} \rightarrow \tilde{X} $$

of surjective maps of smooth varieties of the same dimension. Since $\tilde{\Gamma} \to \tilde{X}$ is birational $\kappa(\tilde{\Gamma}) = \kappa(\tilde{X}) = \kappa(X)$ and $\kappa(\tilde{\Gamma}) \ge \kappa(\tilde{Y}) = \kappa(Y)$ so we conclude.