Riemann-Roch for Zero Cycles on a Surface

As far as I understand, it is not difficult to formulate a the theorem of Riemann-Roch for divisors on any variety. I have been given the impression, that one can still formulate such theorems for cycles of arbitrary codimension. What I'd really be interested to know is:

What is meant by the Riemann-Roch theorem for zero cycles on a surface?

At this point, I will truly show my ignorance, I am aware that there is such a thing as the Grothendieck-Riemann-Roch theorem which, I do not understand, but know makes some statement involving some objects I do have vague familiarity with: vector bundles, genus, chern classes,... I am also aware that there are well known procedures for going between line bundles and divisors. So, in my naive mind, this makes me suspect there is some route between vector bundles and cycles of higher codimension and that the Grothendieck-Riemann-Roch theorem in the context of codimension 2 on a surface is what I am looking for. How far from the actual situation is this? I would truly appreciate if anyone could clarify any relationships between vector bundles of rank r and cycles of codimension n-r, if such a thing exists!

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The Hirzebruch-Riemann-Roch theorem gives a formula for the Euler characteristic of a vector bundle $E$ on a projective variety $X$ in terms of the Chern classes of $E$ and the Chern classes of the tangent bundle $T_X$. The Grothendieck-Riemann-Roch theorem is a stronger theorem which generalizes this to the relative case of a family of varieties $\mathcal X/S$; the Hirzebruch-Riemann-Roch theorem is the special case where $S$ is a point.
Thus, one can compute the correct Riemann-Roch formula for $0$-cycles on surfaces the same way one computes the Riemann-Roch formula for $0$-cycles on curves. If $C$ is an effective zero−cycle, $\chi_a(\mathcal O_c)=\operatorname{deg} C$, so $\chi_a(I(C))=\chi_a(\mathcal O_X)-\operatorname{deg} C$. It is not clear to me whether ideal sheaves are well-defined for ineffective zero-cycles. –  Will Sawin Oct 22 '12 at 17:03