Is $\frac{\mathbb{Z}[X,Y]}{(2, X)}$ a Euclidean Domain?
Definition for a Euclidean Domain (ED) I tend to use is the following: Let $R$ be an integral domain and $R$ supports at least one Euclidean function, that is there at least exits one $\phi$ such that $\phi: R^{*}\to \mathbb{N}_0$ so that for any $a,b\in R^{*}$, either $a|b$ or $\exists r,s\in R$ such that $b=ar+s$ and $\phi(s)<\phi(a).$ (This definition is not very important to my question but I put it out here so that we don't have any potential confusions.)
Now I recognise all elements in $\frac{\mathbb{Z}[X,Y]}{(2, X)}$ are in the form $f(y)+(2,X)$ such that $f(y)\in \mathbb{Z}[Y]$ and the constant term of $f(y)$ has to be odd.
I had a couple ideas of proving or disproving if an integral domain is an ED. One would be if I could prove it is not a PID. Also I thought about the possibility of having an isomorphism to other, more familiar, rings. However, I did not manage to get far with either, could someone please help me to go a little further?
Best Answer
Hint: Consider the map $\mathbb{Z}[X,Y] \to \mathbb{F}_2[Y]$ given by $f(X,Y) \mapsto f(0,Y) \bmod 2$. Prove that this map is a surjective ring homomorphism and that its kernel is $(2,X)$.