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I've been stuck on this problem for a while. If it is reducible, then $f=gh$, where $g,h$ are of degree $1$ (they cannot be of degree $0$, since constant terms are units). Then when considered as elements of $\mathbb{C}[x][y]$, $$g(y)=a_0(x)+a_1(x)y$$ $$h(y)=b_0(x)+b_1(x)y$$ where $a_0(x),a_1(x),b_0(x),b_1(x)$ are only dependent on $x$. Thus, $$x^2+y^2+1=gh=(a_0(x)+a_1(x)y)(b_0(x)+b_1(x)y)$$ This implies that $a_1(x)$ and $b_1(x)$ cannot involve $x$, since their product is $1$. Thus, $$x^2+y^2+1=gh=(a_0(x)+a_1y)(b_0(x)+b_1y)$$ for some $a_1, b_1 \in \mathbb{C}$. But I do not know what to do next.

Pawel
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    Maybe somehow there is a way to show that $a_1, b_1$ have to be $1$? – Pawel Aug 07 '16 at 14:14
  • For any integral domain $R$, the degree function from $R[x]\to \Bbb N$ is additive. in the sense that $deg(fg)=deg(f)+deg(g)$. Now use this when $R=\Bbb C[x]$ and similarly with $R=\Bbb C[y]$ – PVAL-inactive Aug 07 '16 at 14:16
  • @PVAL , would you mind expanding on that? I know this property but I have a hard time applying it... – Pawel Aug 07 '16 at 14:19
  • A generalization is discussed here: https://math.stackexchange.com/questions/486668/x2-y2-z2-is-irreducible-in-mathbb-c-x-y-z – Watson May 19 '18 at 14:54

3 Answers3

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Geometric Solution

Consider the homogeneous polynomial $F(x,y,z):=x^2+y^2+z^2$ in $\mathbb{C}[x,y,z]$. Then, $F$ is irreducible over $\mathbb{C}$ if and only if $F$, $\frac{\partial F}{\partial x}$, $\frac{\partial F}{\partial y}$, and $\frac{\partial F}{\partial z}$ have no nonzero common root (i.e., they have no common root in $\mathbb{CP}^2$). As the only common root of $x^2+y^2+z^2$, $2x$, $2y$, and $2z$ is $(x,y,z)=(0,0,0)$, we conclude that $F$ is indeed irreducible. Hence, $f(x,y):=F(x,y,1)$ is also irreducible.

Solution with Eisenstein's Criterion

Prove that $y^2+1$ has no repeated prime factor in $\mathbb{C}[y]$. Then, consider $f(x,y)=x^2+y^2+1$ as an element of $\big(\mathbb{C}[y]\big)[x]$. Apply Eisenstein's Criterion with respect to one of the irreducible factors of $y^2+1$.

Batominovski
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Continuing from where you stopped: $$ a_0(x)b_0(x)+(a_0(x)b_1(x)+a_1(x)b_0(x))+a_1(x)b_1(x)y^2=x^2+y^2+1 $$ tells you that $$ a_1(x)b_1(x)=1 $$ so they're constant and inverse of one another. Hence it's not restrictive to assume they're $1$. Therefore $b_0(x)=-a_0(x)$ and finally $$ -a_0(x)^2=x^2+1 $$ which is impossible, because $x^2+1=(x+i)(x-i)$ has no repeated factor in $\mathbb{C}[x]$.

egreg
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$a_0(x)b_0(x)=x^2+1$ so, up to bringing a constant on one side and switching names, either of these holds:

  1. $a_0(x)=x^2+1$ and $b_0(x)=1$. But then you'd have $0=a_0(x)b_1+b_0(x)a_1=b_1(x^2+1)+a_1$, which is impossible for $b_1\ne0$

  2. $a_0(x)=x+i$ and $b_0(x)=x-i$. But then you'd have $b_1(x+i)+a_1(x-i)=0$ which does not leave much choice for $a_1$ and $b_1$...

  • Thus, $b_1+a_1=0 \implies b_1=-a_1$, so they have to be additive inverses of each other. But we also know that $b_1a_1=1$, so $$b_1(-b_1)=1$$ Thus, $b_1^2=-1$, implying that $b_1=\pm i$. Similarly, $a_1= \pm i$. Thus, I just need to check that none of these options will yield a correct factorization? – Pawel Aug 07 '16 at 14:34
  • I would have gone $b_1i-a_1i=0$ instead of the product. –  Aug 07 '16 at 14:36
  • Thus, $b_1+b_1=0$ implying that $b_1=0$, a contradiction. Thanks! – Pawel Aug 07 '16 at 14:42