# Factorizing polynomials of several variables (in a different perespective)

I am looking for factorization of polynomials of several variables in the way outlined below.

Consider a second degree polynomial of two variables over the complex numbers.

"P(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F" (see the edit below)

Experimenting with some polynomials of this sort showed me that factorization is possible in the following way.

     "P(x,y) = (ax + by + c)(dx + ey + f)"  (see the edit below) ,


the coefficients being over the complex numbers.

So, given an nth degree polynomial in n variables without a constant term, is it always possible to factorize it into n linear factors each having n variables in the above way? (This rings bells about the fundamental theorem of algebra).

[EDIT: I am sorry, I erred. I have edited my question. The edit is that the polynomial has no constant term:
P(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey
and in the expected factorization, the last linear factor does not have a constant term, too:
P(x,y) = (ax + by + c)(dx + ey + f) ]

I kept the original question as it is for documentation purposes.

• This is not at all true. An equation of the type you consider is a conic; it will factor in the proposed way if and only if that it is a union of two lines (or the same line twice). For most values of (A,B,C,D,E,F), you will get a smooth conic and the equation will not factor. Jun 10, 2010 at 7:36
• How about $x^2+y^2+1$? Jun 10, 2010 at 7:37
• And how about x^2+y^2+y? Jun 10, 2010 at 15:17
• @ Elohemahab - Try graphing the zero set of x^2 + y^2 + y and understand that this shape could not be the shape you would get by plotting the zero set of something which is factored linearly. We are trying to get across that you are saying that every polynomial zero set should look like a collection of lines and points, but this is clearly not so. Jun 10, 2010 at 15:29
• Maybe even easier: try y-x^2 Jun 10, 2010 at 15:29

Let K be a field. The ring $K[X_1,...,X_n]$ is factorial, which means that any polynomial in n variables can be factored into a product of irreducible polynomials. But of course, these polynomials are not of degree 1 in general. If a polynomial can be factored as a product of terms of degree one, then its zero set is a finite union of hyperplanes. An interesting family of irreducible polynomials in C[X,Y] is given by Y^2=X(X-1)(X-L), L different from 0 and 1. The zero set in $P^2(C)$ is called an elliptic curve and it is diffeomorphic to a torus $S^1\times S^1$.

As a reference, I can point to Lang "Algebra", or Jacobson "Basic algebra".

• But it seems that actually factoring polynomial in n variables is not very easy? GAP 4 still doesn't have a proper algorithm? Jun 10, 2010 at 7:52
• @Vrem. There are algorithms for factoring polynomials. But even for integer coefficients polynomials in one variable, a polynomial time algorithm is not so old (LLL algorithm, 1982). This algorithm can be used to factorize multivariate polynomials with coefficiens in a number field, in polynomial time (see e.g. Lenstra, Siam J. Comput. 1987). Arguably, factoring polynomials by hand is pretty tedious. Jun 10, 2010 at 8:31
No. For example, the polynomial $x^2 + (y^2 + 1)$ factors into linear factors if and only if $- y^2 - 1$ is the square of an element in $\mathbb{C}[y]$, which it's not.