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.
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$\begingroup$ What do you mean by "best result"? Result of what kind, and what would best mean? $\endgroup$– Ryan BudneyFeb 19, 2012 at 5:26
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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. $\endgroup$– Guillermo MantillaFeb 19, 2012 at 5:43
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$\begingroup$ 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. $\endgroup$– SteveFeb 19, 2012 at 20:20
1 Answer
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.
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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
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$\begingroup$ 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