A question on fibered Calabi-Yau threefolds

Let $\phi:X\rightarrow \mathbb{P}^1$ be a fibered Calabi-Yau threefold with a general fiber $F$. The following are known

1. $\phi=\Phi_{mF}$ for some $m\in \mathbb{N}$, where $\Phi_D$ stands for the map associated to the complete linear system |D|.
2. The fiber $F$ is either abelian or K3 surface.

How can one prove the above facts?

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This should follow from the adjunction formula $K_{X} = K_{X/\mathbf P^1} + \phi^\ast K_{\mathbf P^1}$, the fact that $K_{\mathbf P^1} = \mathcal{O}_{\mathbf P^1}(-2)$ and the classification of surfaces with trivial canonical sheaf. – Ariyan Javanpeykar Feb 6 '13 at 23:59
Ariyan: I think perhaps it's clearer if one writes the more standard adjunction formula: $K_F=(K_X+F)_{|F}$. Since the fibres over 2 distinct points of $\mathbf{P}^1$ are linearly equivalent, we get $F_{|F}=O_F$, and then use classification of surfaces as you say. – user5117 Feb 7 '13 at 16:10
I agree that it's clearer. Forgive me if I'm wrong, but I think from your explanation it also follows that if $f:X\to \mathbf P^1$ is a flat projective morphism with $K_X$ trivial, then the canonical sheaf of the trivial fibre of $f$ is trivial. This is a slight generalization of the result Zheng is asking about. – Ariyan Javanpeykar Feb 7 '13 at 17:18
The other comments have already answered part 2. Part 1 is even easier. By taking Stein factorization and since $h^(O_X)=0$, we get a map $X\to\mathbb{P}^1\to\mathbb{P}^1$, where the composite is your original map. Thus, if $G$ is a general fiber of the first, then $G$ is irreducible and the map you started with is given by a sublinear system of $|mG|$. One has an exact sequence, $0\to O_X\to O_X(G)\to O_G(G)=O_G\to 0$. Since $H^1(O_X)=0$, we get that $h^0(O_X(G))=2$ and it is base point free giving the above map. Easy o see that $h^0(mG)=h^0(F)=2$. – Mohan Feb 7 '13 at 19:44
Ariyan: Right, all you need to use is that K_X is trivial. (I guess you meant to say "general fibre" instead of "trivial fibre".) By the way, notice that flatness of such a morphism is automatic in this case as long as X is irreducible Cohen-Macaulay (which it should be for the question to make sense anyway). Also, a bit more generally, if we had some smooth curve of genus g instead of P^1, the same argument would show that K_F is torsion (of order at most 2g). – user5117 Feb 7 '13 at 20:38