Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Could any tell me if a multivariate polynomial generated from the sum of irreducible single variable polynomial is irreducible? For example, f(x)=x^2+2x+2, g(x)=x^2+3x+3, h(x)=x^3+2x^2+2x+2 all of them are irreducible, then what about f(x,y,z) = f(x)+g(y)+h(z)?

share|cite|improve this question

2 Answers 2

up vote 6 down vote accepted


share|cite|improve this answer

For two variables, see W. Feit, Some consequences of the classification of finite simple groups, in The Santa Cruz Conference on Finite Groups, Proc. Sympos. Pure Math. 37, American Mathematical Society, Providence, RI, 1980, pp. 175-181. The result is the following. A polynomial $f(x)\in\mathbb{C}[x]$ is indecomposable if whenever $f(x)=r(s(x))$ for polynomials $r(x),s(x)$, then either $\deg r(x)=1$ or $\deg s(x)=1$. Suppose that $f(x)$ and $g(x)$ are nonconstant indecomposable polynomials in $\mathbb{C}[x]$ such that $f(x)-g(y)$ factors in $\mathbb{C}[x,y]$. Then either $g(x)=f(ax+b)$ for some $a,b\in \mathbb{C}$, or else $$ \deg f(x) = \deg g(x) = 7,\ 11,\ 13,\ 15,\ 21,\ \mathrm{or}\ 31. $$ Moreover, this result is best possible in the sense that for $n= 7,11,13, 15, 21$, or $31$, there exist indecomposable polynomials $f(x)$, $g(x)$ of degree $n$ such that $f(x)\neq g(ax+b)$ and $f(x)-g(y)$ factors. The proof uses the classification of finite simple groups!

share|cite|improve this answer
The proof uses the classification of finite simple groups! Simply amazing ! – Chandan Singh Dalawat Nov 27 '11 at 15:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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