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 $r\ge 3$ ?
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Duplicate of If $p\in R[X_1,\dots,X_n]$ is irreducible, is it still irreducible in $R[X_1,\dots,X_n,\dots,X_N]$? – Bill Dubuque Feb 11 '19 at 15:06
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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 $\Bbb{C}[x_1,\ldots,x_r]$.
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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 $D[x_3,\ldots,x_n]=\mathbb C[x_1,\ldots,x_n]$.
