Restriction of the Canonical Divisor $K_X$ to a general fiber

Let $\ f:X\to Z$ be a surjective morphism between two smooth projective varieties with connected fibers $(f_*\mathcal{O}_X=\mathcal{O}_X)$. Let $F$ be a general fiber of $f$ and $\mbox{dim } F<(\mbox{dim } X -1)$, i.e., $F$ is not a divisor of $X$. Then how do you prove that $(K_X)|_F\sim K_F$, where $K_X$ and $K_F$ are the corresponding canonical divisors?

By the general adjunction formula of Hartshorne, $\mathcal{O}_F(K_F)\cong \mathcal{O}_X(K_X)\otimes \wedge^r \mathcal{N}_F$, where $\mathcal{N}_F$ is the normal bundle of $F$ and $r$ is the codimension of $F$ in $X$. That means for a general fiber $F$ we must have $\wedge^r \mathcal{N}_F\cong \mathcal{O}_F$. How do you prove this? I am unable to do it.

I see that in the case of ruled surfaces this follows very easily since the fibers are $\mathbb{P}^1$, thus normal bundle is determined by it's degree.

Also, what about an arbitrary smooth closed fiber $F_0$ of $f$? Is it still true? A paper of Mori [Classification of higher dimensional varieties, 1987, Proposition 2.1(iii)] claims this should still be true for such $F_0$.

$\star$ Everything here is in characteristic $0$.

• If a fiber F is smooth (as a scheme, i.e., no reduced structure), then the differential of f gives an isomorphism of the normal bundle to F and the tangent space to Z at the corresponding point. In particular, the normal bundle (and hence its top exterior power) is trivial. – t3suji Sep 19 '15 at 23:51