Let me see, your question depends strongly on the variety your line bundle is over. The equivalence classes, algebraic or numeric, of line bundles are measuring some facts about your variety. the former is a cohomology group whereas the latter may have information on the rational curves that your variety may have. That is to say, we've got to say what is the base variety like. Supposing the variety $X$ over $k=\overline{k}$ is a projective scheme which is regular in codimension one, then two **effective** divisors algebraically equivalent give rise to the same line bundle $\mathcal{O}(D)$ (invertible sheaf) which eventually gives rise to morphism $\phi:X\rightarrow \mathbb{P}(H^0(X,mD))=\mathbb{P}^r$ (up to an automorphism of $\mathbb{P}^r$). Meaning we are getting a map $\phi$ out of the line bundle $\mathcal{O}(mD)$ which is going to be an embedding provided the linear system $|\mathcal{O}(mD)|$ separate points and separate tangent vectors. If the former property is not true, we may get only a rational map $\phi:X\rightarrow \mathbb{P}^r$, whereas the latter property is taking care of the singularities of the image $\phi(X)\subset \mathbb{P}^r$. The map $\phi$ though depends on the numerical class of the divisor we started with, then that's why I asked for an effective divisor at the beginning.