# 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.

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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. – David Speyer Jun 10 '10 at 7:36
How about $x^2+y^2+1$? – Hailong Dao Jun 10 '10 at 7:37
And how about x^2+y^2+y? – Gjergji Zaimi Jun 10 '10 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. – Steven Gubkin Jun 10 '10 at 15:29
Maybe even easier: try y-x^2 – Steven Gubkin Jun 10 '10 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".

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But it seems that actually factoring polynomial in n variables is not very easy? GAP 4 still doesn't have a proper algorithm? – John Vrem Jun 10 '10 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. – coudy Jun 10 '10 at 8:31
See also mathoverflow.net/questions/14076/… . – Qiaochu Yuan Jun 10 '10 at 10:49

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.

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Or, if you are afraid of computers, Miles Ried's book is a very nice intro. – Steven Gubkin Jun 10 '10 at 13:13
"afraid of computers"...is it because I have not latex-ed up my formulas up there, thanks anyway. – Unknown Jun 12 '10 at 7:12
No, it is because I am afraid of computers (well, of programming). Coincidentally I am trying to overcome my fear this summer by working through Ideals, Varieties, and Algorithms. – Steven Gubkin Jun 12 '10 at 20:36
It turns out Miles is an intriguing introduction, thanks very much. It is in our library, too. – Unknown Jun 26 '10 at 18:02