Factorizing polynomials of several variables (in a different perespective) - MathOverflow

Factorizing polynomials of several variables (in a different perespective)

2010-06-10T07:28:14Z

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).

Please suggest a reading material or journal, if any.

[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.

Answer by coudy for Factorizing polynomials of several variables (in a different perespective)

2010-06-10T07:44:44Z

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".

Answer by Qiaochu Yuan for Factorizing polynomials of several variables (in a different perespective)

2010-06-10T10:45:36Z

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.

As David indicates, the truth is much more interesting: the zero set of polynomials in two or more variables generically describe interesting geometric structures called algebraic varieties, and the case in which the polynomials factor into linear factors corresponds to the most boring structure possible: a bunch of straight lines. What the fundamental theorem of algebra tells you that algebraic geometry in one dimension is boring; you have to go up to two dimensions or higher to see the interesting behavior. An introductory reference I highly recommend on this subject is Cox, Little, and O'Shea's Ideals, Varieties, and Algorithms.