Factorizing polynomials of several variables (in a different perespective) - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-19T06:45:04Z http://mathoverflow.net/feeds/question/27657 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/27657/factorizing-polynomials-of-several-variables-in-a-different-perespective Factorizing polynomials of several variables (in a different perespective) To be cont'd 2010-06-10T07:28:14Z 2010-06-10T15:05:35Z <p>I am looking for factorization of polynomials of several variables in the way outlined below.</p> <p>Consider a second degree polynomial of two variables over the complex numbers.</p> <p>"P(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey + F" (see the edit below)</p> <p>Experimenting with some polynomials of this sort showed me that factorization is possible in the following way.</p> <pre><code> "P(x,y) = (ax + by + c)(dx + ey + f)" (see the edit below) , </code></pre> <p>the coefficients being over the complex numbers. </p> <p>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). </p> <p>Please suggest a reading material or journal, if any.</p> <p>[EDIT: I am sorry, I erred. I have edited my question. The edit is that the polynomial has no constant term:<br> P(x,y) = Ax^2 + Bxy + Cy^2 + Dx + Ey<br> and in the expected factorization, the last linear factor does not have a constant term, too:<br> P(x,y) = (ax + by + c)(dx + ey + f) ]</p> <p>I kept the original question as it is for documentation purposes. </p> http://mathoverflow.net/questions/27657/factorizing-polynomials-of-several-variables-in-a-different-perespective/27659#27659 Answer by coudy for Factorizing polynomials of several variables (in a different perespective) coudy 2010-06-10T07:44:44Z 2010-06-10T07:44:44Z <p>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 $S^1\times S^1$. </p> <p>As a reference, I can point to Lang "Algebra", or Jacobson "Basic algebra".</p> http://mathoverflow.net/questions/27657/factorizing-polynomials-of-several-variables-in-a-different-perespective/27676#27676 Answer by Qiaochu Yuan for Factorizing polynomials of several variables (in a different perespective) Qiaochu Yuan 2010-06-10T10:45:36Z 2010-06-10T10:45:36Z <p>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.</p> <p>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 <a href="http://en.wikipedia.org/wiki/Algebraic_variety" rel="nofollow">algebraic varieties</a>, 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 <a href="http://www.amazon.com/Ideals-Varieties-Algorithms-Computational-Undergraduate/dp/0387946802" rel="nofollow">Ideals, Varieties, and Algorithms</a>.</p>