Q: $f(x)=x^2+y^2+1$ is irreducible in $\mathbb{C}[x,y]$ 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.

A: 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 $y^2+1$.

A: 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 $\mathbb{C}[x]$.

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


*

*$a_0(x)=x^2+1$ 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 $b_1\ne0$

*$a_0(x)=x+i$ 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 $b_1$...
