EDIT: There seem to be some problems with theThe argument below, especially the case of K3 surfaces.that I will try to post a correct version later.
had is flawed and I wantdo not know how to argue that in the case in which X and Y are surfacesfix it. Here is a short description of one part that are not ruled, then you can decide if they are isomorphic or notI cannot address.
Computationally you can decide if the surfaces We are rational (P_2=q=0) or ruled (P_{12}=0). I will ignore these cases, even though it might be possible to decide them; in any case, you can easily determine the birational isomorphism class ofgiven two polarized K3 surfaces X and Y.
Compute the Kodaira dimension of the surfaces by computing the image under the linear system |12K|. If the surfaces are of general type, then we are more or less done since any isomorphism must come from a linear isomorphism of the ambient projective space under the "embedding" given by |5K|. If the Kodaira dimension is one, then you essentially reduce to computing the base curve of the elliptic fibration and the corresponding morphism to thedegrees jx-line: this information seems to be pretty computable!
If the Kodaira dimension is zero, then you look for exceptional curves on your surfaces (using Hilbert schemes) and contract them, until $K^2=0$. The final surfaces will be one of Enriques, bielliptic, Abelian and K3 surfaces; these classes are easily discernible by $p_g$ and qy respectively. NoteIt is clear that replacingwe can check if X and Y by their minimal models in these cases does not affect the computability of an isomorphism. (This reduction may not be needed, but it makes me feel safer; note also that you can probably do this more efficiently by looking for effective divisors in (multiples of) the anticanonical divisor.)
Enriquesare isomorphic as polarized surfaces seem to be decidable since their ample bundles do not vary, so you can get them all in the same projective space and reduce the isomorphism to an ambient linear isomorphism.
Bielliptic surfaces roughly reduce to the case of curves.
Abelian surfaces correspondbut we would like to genus two curves andcheck if they are therefore again recognizableisomorphic as surfaces.
Finally If we get to the interesting case: K3 surfaces! (Recall thoughcould prove that we left out rational surfaces.)
Determine very ample line bundles $A_X$there were finitely many polarizations on X and $A_Y$ onhaving degree Yy and this finite set were computable, then we would be done. Just toUnfortunately, there could be infinitely many such polarizations on the safe side, decide if the two embeddings for X and Y that you obtain are not the "same"! Assume they are not the same.
Here is where the real meat ofFor Abelian surfaces the argumentsituation is (and where I might have made my most serious mistake). Find models of X and Y overslightly better, since the same finite extensionnumber of Q and findpolarisations with a prime p thatgiven degree is of good non-supersingular reduction for both X and Y and reduce them modulo p. Using the known Tate-conjecture in this case we can compute the (geometric) Picard latticesfinite (up to tensoring with Q) of the reductions by looking for new divisors until we find enough whose intersection matrix has the appropriate rank. We can establish an isomorphism of the corresponding rational vector spaces with intersection form, find a vector in this vector space that is represented by an effective divisor in both surfaces and has positive square. The image under suchby a line bundletheorem of the two surfaces will allow us to find an isomorphism between the two sufaces in positive characteristicNarasimhan and Nori). The outcome of all this is that now we canTo find out what lattice do the two ample line bundles $A_X$ and $A_Y$ we started withpolarisations on X andan Abelian variety YA span in their respective Picard groups: it, one possibility is the lattice spanned by their images in the Picard group of the reduction!
So now we go back to our surfaces X and Y and their ample line bundles $A_X$ and $A_Y$. We know the degree andcompute the genusorbits of the ample line bundle $A_Y$automorphism group of YA underon the embeddingNeron-Severi group of XA given by. While this could be difficult over $A_Y$$\overline{\mathbb{Q}}$, and we look for suchit might be replaced by a divisor insimilar statement modulo a prime of good reduction, where at least Picard numbers can be computed by the Hilbert scheme:Tate conjecture. Also in this case, I do not know if there will be finitely many possibilitiesare further problems with this approach. We check all of them and we conclude!
I hope that the above is correct, or at leastbelieve that it is recoverable if it isn'tthis kills completely my previous post!