# 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

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?

• It is not true that $h^0(dK_X) = h^0(dK_Y)$; think about the case where $Y$ is a smooth projective curve of genus $g \geq 2$. Then $X$ has genus $(\deg f)(g-1)+1 > g$, so $h^0(K_X) > h^0(K_Y)$. Commented Feb 14, 2023 at 16:44

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.

• You should assume that $\pi$ is generically finite. Commented Feb 14, 2023 at 22:00
• @JasonStarr is this necessary? I think generic smoothness is enough since pullbacks of forms along smooth maps is injective (you can factor a smooth map locally as etale maps and affine projects each of which is injective on forms). Commented Feb 14, 2023 at 23:34
• For $X$ equal to $Y\times \mathbb{P}^1$ with $\pi$ equal to the first projection, the Kodaira dimension of $X$ is negative independent of the Kodaira dimension of $Y$. Commented Feb 15, 2023 at 0:02
• Ah of course. k-forms pullback to k-forms injectivlely but these are not top forms unless dimensions are equal. Commented Feb 15, 2023 at 2:17
• Thank you for your answer Ben C, I have some comments: Question 1: I don't know how to show this, can you give me some hints? Question 2: I think it is not so easy, because you are starting from a surjective map of singular varieties. You can use the fact of Question 1 only if you construct a surjective map between some desingularization of them. How can I construct such a map? I had that idea in the case of surfaces, but l I don't think I can use the resolution of indeterminacy in the higher dimension. Commented Feb 15, 2023 at 11:11