0
$\begingroup$

R is the real field. Every polynomial in R[x] has a nice factorization (of degree 1 and 2). What is the best result for arbitrary multivariable polynomials, say R[x,y],R[x,y,z]. Further more, if we allow n-th square root of a polynomial, like x^2+x*sqrt(y^2+1)+y^2, what is the best result for arbitrary polynomials. Thanks very much.

$\endgroup$
3
  • $\begingroup$ What do you mean by "best result"? Result of what kind, and what would best mean? $\endgroup$ Feb 19, 2012 at 5:26
  • 1
    $\begingroup$ In the polynomial ring cases $x^m+y$ is irreducible for all $m$. The second part of the question might not even make sense--the ring you are considering might not be a UFD. $\endgroup$ Feb 19, 2012 at 5:43
  • $\begingroup$ Many thanks. I apology for vague explanation concerning "best". In your example, we may choose x as a primary variable, and factor x^m+y viewing y as a constant. Take m=3 for example, that is (x^3+y)=(x+y^{1/3})(x^2-x*y^{1/3}+y^{2/3}). Which means that we will allow m-th root of y. So, in general, polynomials of 2 variable are "factorable", but the "factors" (which may contain m-th root) are kind of bad. I want to see whether such "bad factors" are controllable, say, they are polynomials of x and m-th root of y. Thanks a again for reading. $\endgroup$
    – Steve
    Feb 19, 2012 at 20:20

1 Answer 1

10
$\begingroup$

Even over the complex numbers, "most" polynomials in $k$ variables are irreducible once $k \geq 2$. You can see this by comparing dimensions. The space of polynomials $P$ of degree $n$ has dimension $D_k(n) = {n+k \choose k}$. If $n=n_1+n_2$ then pairs $(P_1,P_2)$ of polynomials of degree $n_1,n_2$ have dimension $D_k(n_1) + D_k(n_2)$, so products $P_1 P_2$ of such polynomials have dimension at most $D_k(n_1) + D_k(n_2) - 1$ (subtracting $1$ because $P_1 P_2 = (cP_1) (c^{-1} P_2)$ for all nonzero $c$). Except for $k=1$ (and $(n_1,n_2) = (n,0)$ or $(0,n)$), this is less than $D_k(n)$; therefore the space of reducible polynomials of degree $n$ must be smaller than the space of all polynomials of that degree.

$\endgroup$
2
  • 3
    $\begingroup$ This too little known result should be given in books (or lectures) on linear algebra as an application of the concept of dimension. $\endgroup$ Feb 19, 2012 at 9:00
  • $\begingroup$ Actually, it's a very nice trick that is folklore. I remember having this one in my linear algebra class. However, have never seen it in any literature. $\endgroup$ Feb 20, 2012 at 22:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.