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Let $\phi:X\dashrightarrow Y$ be a generically finite, dominant rational map between smooth projective varieties over $\mathbb{C}$.

Assume that $Y$ is of general type. May we conclude then that $X$ is of general type as well?

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    $\begingroup$ Yes that's true. $\endgroup$ – Donu Arapura Apr 10 '17 at 23:07
  • $\begingroup$ Thanks. Could you briefly explain to me why? It is clear for curves. $\endgroup$ – TopGatLu Apr 10 '17 at 23:10
  • $\begingroup$ Look at the transcendence degrees of the canonical rings. $\endgroup$ – Francesco Polizzi Apr 11 '17 at 5:43
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    $\begingroup$ In more details: for each $n$ there is an injective map $\phi^*:H^0(Y,K_Y^{{\scriptscriptstyle\otimes} n})\rightarrow H^0(X,K_X^{{\scriptscriptstyle\otimes} n})$ (note that $\phi$ is well-defined outside a set of codimension $\geq 2$). $\endgroup$ – abx Apr 11 '17 at 6:30
  • $\begingroup$ Right. I expanded the comment in an answer. $\endgroup$ – Francesco Polizzi Apr 11 '17 at 9:38
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As abx remarked in his comment above, for any $n \geq 1$ there is an injective pullback map $\phi^* \colon H^0(Y, \, nK_Y) \to H^0(X, \, nK_X)$, hence we have an injective map of canonical rings $$\phi^* \colon \mathcal R(Y, \, K_Y) \to \mathcal R(X, \, K_X).$$ In particular, the transcendence degree of $\mathcal R(Y, \, K_Y)$ is less than or equal to the transcendence degree of $\mathcal R(X, \, K_X)$.

On the other hand, the Kodaira dimension equals the trancendence degree of the canonical ring minus $1$, so if $Y$ is of general type and $\dim Y = \dim X =n$ we have $$ n+1 = \mathrm{tr\,deg} \,\mathcal R(Y, \, K_Y) \leq \mathrm{tr\,deg} \, \mathcal R(X, \, K_X) \leq n+1,$$ and this shows that $X$ is of general type as well.

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  • $\begingroup$ Thanks a lot. It seems to me that for your argument you just need that $\phi$ is dominant. Is this correct? $\endgroup$ – TopGatLu Apr 12 '17 at 2:53
  • $\begingroup$ You need both the assumption dominant and generically finite, since they guarantee that the pullback $\phi^*$ is injective on canonical rings. Only dominant is not sufficient: a curve $C$ of genus $\geq 2$ is of general type, we have a dominant (actually surjective) map $$C \times \mathbb{P}^1 \to C,$$ but $C \times \mathbb{P}^1$ is not of general type. For fiber spaces, the main result about Kodaira dimensions is the Iitaka conjecture, see en.wikipedia.org/wiki/Iitaka_dimension#Iitaka_conjecture $\endgroup$ – Francesco Polizzi Apr 12 '17 at 5:49
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This is a point I've been confused about proving several times recently, so I'm going to write out the argument in (potentially excessive) detail.

As abx points out, $\phi^*$ gives an injection $H^0(Y, K_Y^{\otimes n}) \hookrightarrow H^0(X, K_X^{\otimes n})$, which implies the stronger statement that $P_n(Y) \leq P_n(X)$ for all $n$, so in particular the Kodaira dimension of $Y$ is at most that of $X$ (the inequality is often strict, e.g. curves of any genus have finite morphisms to $\mathbb{P}^1$). Let's show $n = 1$ for ease of notation, but the argument is identical for all $n$.

EDIT: My definition of Kodaira dimension is that it is the order of growth of the plurigenera, so if $P_n(Y) \leq P_n(X)$ for all $n$, the fact that $\kappa(Y) \leq \kappa(X)$ is immediate. The fact that this is equivalent to the statement about the transcendence degree of the pluricanonical ring, used in Franscesco Polizzi's answer, is not trivial and uses the asymptotic Riemann-Roch theorem.

There is an open set $U \subseteq X$ such that $\phi|_U: U \rightarrow Y$ is an actual morphism (of course, $\phi$ is literally equal to $\phi|_U$ by the definition of a rational morphism, but I want to emphasize that the domain is $U$). Because your varieties are normal and proper, you can ensure that $U$ has codimension at least $2$ (by the valuative criterion and the fact that the local rings of codimension $1$ points are DVR's). By the $S_2$ property of normality, this implies that $K_X \rightarrow \iota_* ((K_X)|_U) = \iota_* K_U$ is an isomorphism, with $\iota: U \hookrightarrow Y$ the inclusion. (I'm using all over the place that $K_S|_V = K_V$ for $V \subseteq S$ with $S$ smooth, which is obvious from the construction at least in the smooth case).

Now, the natural morphism $K_Y \rightarrow (\phi|_U)_* \phi|_U^* K_Y$ is injective, because $K_Y$ is a locally free sheaf on the smooth variety $Y$ and $\phi|_U$ is dominant (see this question for a discussion of what could go wrong if you drop the locally free hypothesis). Then, we can compose this with the pushforward of the natural morphism of sheaves $\phi|_U^* K_Y \rightarrow K_U$ to get a morphism of sheaves $K_Y \rightarrow (\phi|_U)_* K_U$. On sections, we get a morphism $H^0(V, K_V) \rightarrow H^0(W, K_W)$ for any dense open $V \subseteq Y$, with $W = (\phi_U)^{-1}(V)$, compatible with restrictions. In particular, this defines the pullback $H^0(Y, K_Y) \rightarrow H^0(U, K_U) = H^0(X, K_X)$. For a dense open $V \subseteq Y$, we have a diagram:

$ \require{AMScd} \begin{CD} H^0(Y, K_Y) @>>> H^0(X, K_X)\\ @VVV @VVV \\ H^0(V, K_V) @>>> H(W, K_W) \end{CD} $

The vertical arrows are injections because the canonical sheaves are locally free and everything in sight is an integral variety. Thus, if we can choose some $V$ such that the bottom arrow is injective, the top arrow is automatically injective.

Now, because you are working over a characteristic $0$ field, a generically finite dominant morphism is generically étale. This means that there is a dense open set $V \subseteq Y$ such that $W:= \phi|_U^{-1}(V) \rightarrow V$ is an étale morphism.

Aside: this fact is false in positive characteristic (i.e. Frobenius morphisms are everywhere non-étale). On the other hand, I imagine your statement is still true in positive characteristic since inseparable morphisms "don't change anything", but I don't know how to prove this.

EDIT: The desired statement is not true in positive characteristic, as the example in the comments shows. So generic smoothness is really essential here.

Now, the relative canonical sheaf is zero for étale morphism, so the canonical exact sequence $0 \rightarrow f^* K_T \rightarrow K_S \rightarrow K_{S/T} \rightarrow 0$ for a smooth morphism $f: S \rightarrow T$ shows that the natural morphism $\phi|_U^* K_V \rightarrow K_W$ is an isomorphism. Then, because we already know $K_V \rightarrow (\phi|_U)_* (\phi_U)^* K_V$ is injective, we're done.

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    $\begingroup$ Actually, the statement is false in characteristic $p$. In "An Example of Unirational Surfaces in Characteristic $p$", Shioda shows that the Fermat surface of degree $n$ is unirational if there is a power $q$ of $p$ such that $q \equiv -1 \pmod n$. This implies that there are Fermat surfaces of general type which admit generically finite dominant rational maps from $\mathbb{P}^2$. $\endgroup$ – user36254 Apr 11 '17 at 17:04

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