Question: Here's a question that I've wondered about somewhat idly from time to time.
I think it would be hard, but even a partial result would be interesting.
Hopefully someone here can make some headway.

Fix an algebraically closed field k (one can also relax this, e.g. k = $\mathbb{Q}$
is very interesting). Suppose that X and Y are two closed subschemes of affine or projective spaces over k given by explicit equations.

*Is there an algorithm to decide whether X and Y are isomorphic?*

A few comments.

1. I'm being a bit sloppy about the use of the word "algorithm". One can assume that the operations of k are Turing computable, or else choose a different model of computation.
2. For affine schemes, this can be stated as the isomorphism problem for finitely generated
commutative k-algebras. I expect that the isomorphism problem for noncommutative finitely presented associative algebras would be false, since it seems close to the corresponding problem for groups which is undecidable. But the reduction is not quite clear to me. Is this known?
3. To give a positive solution, one would need an effective threshold d, so that either
there is no isomorphism at all or there exists an isomorphism defined by a rational expression with degree (of the numerator and denominator) bounded by $d$. I have no
feeling for whether this reasonable or not.
4. For the simplest case, where X is a smooth projective plane curve, a positive
solution would amount showing that the corresponding point in the moduli space $M\_g$, is
computable ($g= (deg X-1)(deg X-2)/2$ ). This looks challenging, but somehow it feels tractable to me.

Added Note: While I think spontaneous discussion is good, I would like to encourage
anyone who has a substantial result about this to write it down carefully and perhaps post it on the arxiv.

Answer: 
The argument that I had is flawed and I do not know how to fix it. Here is a short description of one part that I cannot address. We are given two polarized K3 surfaces *X* and *Y* of degrees *x* and *y* respectively. It is clear that we can check if *X* and *Y* are isomorphic as polarized surfaces, but we would like to check if they are isomorphic as surfaces. If we could prove that there were finitely many polarizations on *X* having degree *y* and this finite set were computable, then we would be done. Unfortunately, there could be infinitely many such polarizations on *X*.

For Abelian surfaces the situation is slightly better, since the number of polarisations with a given degree is finite (by a theorem of Narasimhan and Nori). To find the polarisations on an Abelian variety *A*, one possibility is to compute the orbits of the automorphism group of *A* on the Neron-Severi group of *A*. While this could be difficult over $\overline{\mathbb{Q}}$, it might be replaced by a similar statement modulo a prime of good reduction, where at least Picard numbers can be computed by the Tate conjecture. Also in this case, I do not know if there are further problems with this approach.

I believe that this kills completely my previous post!
It seems that one can embed 10th Hilbert problem over $\mathbb Z$ into this question, and for that the answer is not known.