# Irreducible polynomial in $\mathbb C[x_1,x_2]$ also irreducible in $\mathbb C[x_1,x_2,…x_r]$? [duplicate]

Let $$f_1(x_1), f_2(x_2)$$ be polynomials in a single variable, of relatively prime degree, with complex coefficients. If $$f_1(x_1)+f_2(x_2)$$ is irreducible in $$\mathbb C[x_1,x_2]$$, then is it irreducible in $$\mathbb C[x_1,x_2,...x_r]$$ for every $$r\ge 3$$ ?

## marked as duplicate by Bill Dubuque, Community♦Feb 11 at 21:01

Yes; if $$f:=f_1(x_1)+f_2(x_2)$$ is irreducible in $$\Bbb{C}[x_1,x_2]$$ then it is prime, so the quotient $$\Bbb{C}[x_1,x_2]/(f)$$ is an integral domain. For every $$r\geq3$$ you have $$\Bbb{C}[x_1,\ldots,x_r]/(f)\cong(\Bbb{C}[x_1,x_2]/(f))[x_3,\ldots,x_r],$$ which is again an integral domain, so $$f$$ is also prime and hence irreducible in $$\Bbb{C}[x_1,\ldots,x_r]$$.
Alternatively, suppose $$f$$ is reducible in $$\Bbb{C}[x_1,\ldots,x_r]$$ for some $$r\geq3$$, and write $$f=gh$$ with $$g,h\in\Bbb{C}[x_1,\ldots,x_r]$$ nonconstant. Because $$f$$ is irreducible in $$\Bbb{C}[x_1,x_2]$$ either $$g$$ or $$h$$ must have a monomial term that is a multiple of some $$x_k$$ for some $$k\geq3$$. That is to say, there exists some $$k\geq3$$ such that either $$\deg_{x_k}g>0$$ or $$\deg_{x_k}h>0$$. Then $$\deg_{x_k}f=\deg_{x_k}gh=\deg_{x_k}g+\deg_{x_k}h>0,$$ a contradiction. So $$f$$ is also irreducible in $$\Bbb{C}[x_1,\ldots,x_r]$$.
If $$D$$ is any domain and if for $$a\in D$$, $$a\neq 0$$, we have $$a=bc$$ for $$b,c\in D[x]$$, then $$b,c\in D$$. To see this, note that $$D[x]$$ is also a domain, so in $$D[x]$$ we have: $$0=\deg(a)=\deg(bc)=\deg (b)+\deg(c)$$, where from $$\deg(b)=\deg(c)=0$$, i.e. $$b,c\in D$$. (We use here that for $$a\in D[x]$$ such that $$a\neq 0$$: $$a\in D$$ iff $$\deg(a)=0$$.)
Now by induction we see that any factorization in $$D[x_1,\ldots,x_n]$$ of an element from $$D$$ is already a factorization in $$D$$. Here, if you have an irreducible element in $$D=\mathbb C[x_1,x_2]$$, it stays irreducible in $$D[x_3,\ldots,x_n]=\mathbb C[x_1,\ldots,x_n]$$.