A ring extension of $\mathbb Z$ in which $x^4+y^4+1$ is reducible My first instinct was the ring of two variable polynomials modulo $a^4+b^4+1$.($\mathbb Z[a,b]/(a^4+b^4+1)$)
But I haven’t managed to reduce it to linear factors.
Can any of you help?
 A: Your question is very unclear. From the title, it looks as if you’re looking for a ring $A$ containing $\Bbb Z$ such that the polynomial $x^4+y^4+1\in\Bbb Z[x,y]$ becomes reducible when viewed as a polynomial in $A[x,y]$.
Perhaps you were thinking of your question as an analog to the one-variable situation, where the $\Bbb Z$-polynomial $x^4+1$ becomes reducible, first over the Gaussian numbers, where it’s $(x^2-i)(x^2+i)$, and better over the ring gotten by adjoining a primitive eighth root of unity $\zeta=\zeta_8=\frac{1+i}{\sqrt2}$, as $x^4+1=(x-\zeta)(x-\zeta^7)(x-\zeta^3)(x-\zeta^5)$.
Now, lamentably, if you’re restricting your rings $A$ to subrings of the complex numbers, $x^4+y^4+1$ remains irreducible no matter what $A$ is. The reason is that as soon as you get into multivariable polynomials, you’re suddenly no longer doing basic number theory nor abstract algebra, but Algebraic Geometry.
Geometrically, one thinks of $x^4+y^4+1=0$ as describing a “curve” in the (affine) plane over the complex numbers; topologically it’s a surface. And if you look at it in the way you should, as a curve in the projective plane over $\Bbb C$, it becomes a compact Riemann surface, of genus three. It’s only when, as a curve, it consists of two (possibly intersecting) curves in the plane that the original polynomial would have been reducible.
Since the algebraic geometers’ curve, topologists’ Riemann surface, defined by your polynomial equation $x^4+y^4+1=0$, does not break up into two separate, perhaps intersecting, curves/surfaces, your polynomial is “absolutely irreducible”, never becomes reducible no matter what field you consider its coefficients to lie in.
A: Homogenize and reduce to: is $P(x,y,z)=x^4 + y^4 + z^4$ reducible in $K[x,y,z]$, for some ring $K\supset \mathbb{Z}$?  If it is, then some (easy to write) system of equations with coefficients in $\mathbb{Z}$ has a solution in $K$. But that means ( Hilbert nullstellensatz) that the system has a solution in $\mathbb{C}$ ( or even in
$\bar{\mathbb{Q}}$, the algebraic closure of $\mathbb{Q}$).
So assume that $P(x,y,z) = P_1(x,y,z)\cdot P_2(x,y,z)$, with $P_1$, $P_2 \in \mathbb{C}[x,y,z]$, homogenous, of positive degree, and $\deg P_1 + \deg P_2 = 4$.
The complex projective curves $P_1=0$, and $P_2=0$ have a non-void intersection ( Bezout's theorem implies this). Let $(x_0\colon y_0\colon z_0)$ a common point of these curves, that is $P_i(x_0, y_0, z_0)=0$.
But this implies with the product rule
$$\frac{\partial P}{\partial x}(x_0, y_0, z_0) = \frac{\partial P}{\partial y}(x_0, y_0, z_0)=\frac{\partial P}{\partial z}(x_0, y_0, z_0)=0$$
(a singular point of the curve $P(x,y,z) = 0$).
But this implies $x_0 = y_0 = z_0 = 0$, contradiction. ( a point in the projective plane has at least one of the projective coordinates $\ne 0$).
Note: This works for any $P$ homogenous of $3$ variables for which the curve $P=0$ is non-singular.
For comparison, the polynomial $x^4 + 2 x^2 y^2 + y^4 - z^4$ is reducible. It also has two singular points: (at infinity) $(1\colon \pm i\colon 0)$.
