Let $R = \mathbb{Z}[x,y]$. Find ideals $I$ such that.
- $R/I$ is an integral domain but not a UFD
- The polynomial $z^2 – 1$ has more than two roots in $R/I$.
For 1, I have $I = (x^2 – xy -1)$ which I think is irreducible and so far can't show why. Then in $R/I$, $xy = (x-1)(x+1)$.
These solutions don't 'feel' right.
Best Answer
Let $I=(3,x^2+y^2-1)$. Then $R/I\simeq(\mathbb Z/3\mathbb Z)[x,y]/(x^2+y^2-1)$.
Let $I=(3,x^2-1)$. There are at least three roots of $z^2-1$ in $R/I$: $1,2$, and $x\bmod I$.