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This is very much like Factoring polynomials to factors involving complex coefficients except that I'm concerned about bivariate polynomials, not univariate polynomials. Take for example the polynomial

$$3 - 6 x^2 + 3 x^4 - 10 y^2 + 6 x^2 y^2 + 3 y^4$$

I'd like to factor that over the complex numbers. If I know where this came from, I can do Factor[poly, Extension -> {Sqrt[3]}] to find that the above equals

$$\tfrac13 (3 - 3 x^2 + 2 \sqrt3 y - 3 y^2) (3 - 3 x^2 - 2 \sqrt3 y - 3 y^2)$$

but to find this I have to know, a priori, that $\sqrt3$ might be a useful extension.

Even more important, that I haven't forgotten any additional extensions which would have allowed me to factor this even more. (The latter is easy in the case above, but hard if the degrees get higher.) Edit: I found that this question can be answered using IrreduciblePolynomialQ[poly, Extension → All].

Is there any way I can compute a factorization of a polynomial from $\mathbb Z[x,y]$ into factors from $\mathbb C[x,y]$ (or equivalently $\bar{\mathbb Q}[x,y]$ i.e. with algebraic coefficients), without the use of any knowledge except for the polynomial itself?

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Here is an approach based on finding an approximate root, bumping to an approximate factor using GroebnerBasis, and resolving as an exact factor using RootApproximant.

poly = 3 - 6*x^2 + 3*x^4 - 10*y^2 + 6*x^2*y^2 + 3*y^4;

x0 = 11/7;
roots = y /. NSolve[poly /. x -> x0, WorkingPrecision -> 400];
root1 = First[roots];
fac = First[
   GroebnerBasis[{poly, (y - root1)^10, (x - x0)^17}, {y, x}, 
    MonomialOrder -> DegreeReverseLexicographic, 
    CoefficientDomain -> InexactNumbers]];
dtl = Chop[GroebnerBasis`DistributedTermsList[fac, {y, x}]];
newdtl = MapAt[RootApproximant, dtl, 
   Thread[{1, Range[Length[dtl[[1]]]], 2}]];
algfactor = GroebnerBasis`FromDistributedTermsList[newdtl]

(* Out[28]= -1 + x^2 + (2 y)/Sqrt[3] + y^2 *)

One can divide out by this factor, rinse, repeat. I'll skip that last and anyway I suspect it will not reduce further.

Factor[poly/algfactor, Extension -> Automatic]

(* Out[31]= -3 + 3 x^2 - 2 Sqrt[3] y + 3 y^2 *)

An explanation of the method, and the code from which I thoughtlessly cribbed, may be found here.

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  • $\begingroup$ You love that GroebnerBasis, don't you :D +1 $\endgroup$ – Sektor Mar 16 '15 at 19:10
  • $\begingroup$ That tl should be dtl. Unfortunately I lack rep to do a single-character edit here. $\endgroup$ – MvG Mar 16 '15 at 19:45
  • $\begingroup$ Thanks for pointing out the typo. GOt careless on the cut-and-paste (which I seem to like even more than Groebner bases). $\endgroup$ – Daniel Lichtblau Mar 16 '15 at 19:55
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    $\begingroup$ @Sektor, once you go Gröbner, it's hard to cut back... :) $\endgroup$ – J. M. is away May 3 '15 at 2:42
  • $\begingroup$ Tru dat, J.M. .. oops @Guess $\endgroup$ – Sektor May 3 '15 at 5:36

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