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Let $X$ be a projective K3 surface and $f: X \to Y$ a non etale, finite map, restricting to etale on non empty open $U \subset Y$ of degree prime to char of alg closed base field of $k$. Assume moreover $Y$ to be smooth. Non etaleness means that the ramification locus $R$ is non empty. Assume that $R$ is pure of dimension $1$, so union of curves.

(The concrete example I had in mind if that $f$ comes from quotient with resp to a non-symplectic automorphism $s: X \to X$ (ie non symplectic in sense that the induced action on $H(X,K_X)=H^0(X, \Omega_X^2)$ is not trivial) of finite order prime to char of $k$. Then $f: X \to Y:= X/ \langle s \rangle$, then $R$ is the fixed point locus by $s$, which we assume to be pure $1$-dimensional.

I'm interested in proving that then $Y$ is rational.

The approach I had in mind: The use Castelnovo criterion for rationality of surfaces. One step is to show that plurigenera $P_n(Y)$ vanish. By ramification formula we have for canonical divisors relation

$$ 0 \sim K_X \sim f^*K_Y +R $$

thus $f^*K_Y \sim -R$ so the pullback of $K_Y= \Omega_Y^2$ is anti effective as $R$ effective. But is it sufficient to conclude that plurigenera $P_n(Y) =H^0(Y, K_Y^n)$ vanish? In other words, to attain anti effectiveness for $K_Y$ from $f^*K_Y $?

Question: How can we deduce from anti effectiveness of $f^*K_Y $, anti effectiveness of $K_Y$?

#EDIT: One way to obtain this follows from Keerthi Madapusi's argument (...if I understood this correctly) by noticing that in our situation for every line bundle $L$ on $Y$ the adjunction map $L \to f_*f^* L$ is injective, so we can apply it to $K_Y$ and form global sections.

But what I'm wondering about if it is also possible to deduce anti effectiveness of $K_Y $ from anti effectiveness of $f^*K_Y $ by working direct of level of divisor classes modulo rational equivalence $\operatorname{Div}/ \sim $.

Can I for example deduce something like that if $ R =\sum_i^n C_i$ decomposition in irred curves $C_i$ and and $f$ restricts on $C_i$ to finite maps $f_i:C_i \to f(C_i)$ of degrees $r_i$, then $K_Y \sim -\sum_i^n r_if(C_i)$?

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  • $\begingroup$ For any line bundle L, there is an injective map from global sections over Y to those over X. Also, note that by the regularity of Y and purity of branch locus, R will always be an effective divisor. You don't need to assume this separately. $\endgroup$
    – Keerthi Madapusi
    Commented Nov 16 at 17:36
  • $\begingroup$ @KeerthiMadapusi: You mean the map induced from adjunction map $ L \to f_*f^*L$? If yes, but is it immediately clear that the induced map on global sections is injective? $\endgroup$
    – user267839
    Commented Nov 16 at 17:46
  • $\begingroup$ @KeerthiMadapusi: Or, on level of affine trivialization $U =\text{Spec}(A)$ of $Y$ + finiteness of $f$ this is really about ring map $A \to B$ whose kernel must be the radical of $A$ (as $f$ dominant), but as $Y$ smooth, $A$ is integral, so kernel is zero, so $L \to f_*f^*L$ is injective at each stalk and so on global sections. Not sure, if that's the most elegant way to see it. Can assumptions of $f$ weakened keeping this map still injective on line bundles? Eg, is it still injective if $f$ is not affine? $\endgroup$
    – user267839
    Commented Nov 16 at 17:57
  • $\begingroup$ @KeerthiMadapusi: This solves the initial problem. But, just from point of curiosity, do you know if it is also possible to reason purely on level of divisor /intersection calculus directly that $f^*K_Y$ effective implies $K_Y$ effective? $\endgroup$
    – user267839
    Commented Nov 16 at 18:11

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