Canonical bundle and canonical divisor in a $K3$ surface The algebraic definition of a $K3$ surface is this:

A smooth algebraic suface $X$ is called $K3$ if:
i) $X$ has trivial canonical bundle;
ii) $h^1(X,\mathcal{O}_X)=0$.

I know that i) means that the bundle $\Omega^n=\bigwedge_{i=1}^n\Omega$ (where $\Omega$ is the cotangent space) is trivial.
However, in many different sources I've found the first condition stated as $K_X=0$. That's the canonical divisor, right? How can that be zero? How is that equivalent to the canonical bundle being trivial?
 A: This is just the correspondence (on suitably nice schemes) between invertible sheaves up to isomorphism and divisors up to linear equivalence. Briefly, on a nice scheme, every divisor determines an invertible sheaf, and every invertible sheaf determines a divisor, and these two processes are mutually inverse (up to equivalence). On one side, invertible sheaves up to isomorphism form an abelian group under tensor product, and on the other, divisors up to linear equivalence form an abelian group under formal sum. 
Since this is an isomorphism of groups, we must have that the identity element on one side corresponds to the identity element on the other. On the line bundle/tensor product side, the identity element is $\mathcal{O}_X$, the structure sheaf: for any sheaf $\mathcal{F}$, we have that $\mathcal{F}\otimes_{\mathcal{O}_X} \mathcal{O}_X = \mathcal{F}$. On the divisor/formal sum side, the identity element is $0$: $D+0=D$ for any divisor $D$. So saying that $K_X=0$ is the same as saying the canonical bundle is isomorphic to the structure sheaf, which is just a rephrasing of saying that the canonical bundle is trivial.
