# Integral Domain but not a UFD

Let $R = \mathbb{Z}[x,y]$. Find ideals $I$ such that.

1. $R/I$ is an integral domain but not a UFD
2. 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.

• What does $z=1,x,$ and $z^2-1$ in $R/I$ mean? You want to find three elements of $R/I$ which are roots of $z^2-1$. $z$ is a variable, not an element of $R/I$. – Thomas Andrews Jan 1 '15 at 6:33
• @ThomasAndrews My bad, I meant if $z = -1, 1$ or $x$. Then $z^2 -1 = 0$ in $R/I$ – algor207 Jan 1 '15 at 6:37
• So, $x^2-1$ is divisible by $x^2+1$? – Thomas Andrews Jan 1 '15 at 6:38
• @ThomasAndrews Ah!, what I wrote is complete garbage. As of now, I still can't figure it out. PS. I've also deleted the garbage I wrote. – algor207 Jan 1 '15 at 6:47
• I thought you wanted it such that $x+I$ satisfied the equation $z^2-1$ as well as $1+I$ and $-1+I$? I think that works if you just choose the ideal so that this happens. – user1537366 Jan 1 '15 at 6:56

1. Let $I=(3,x^2+y^2-1)$. Then $R/I\simeq(\mathbb Z/3\mathbb Z)[x,y]/(x^2+y^2-1)$.
2. Let $I=(3,x^2-1)$. There are at least three roots of $z^2-1$ in $R/I$: $1,2$, and $x\bmod I$.
• For 1, Is it true because $2 = 2 \cdot 1 = 2 \cdot (x^2 + y^2)$? Thus also $I = (3, (x + y)^2 - 1)$ could work. i.e $2 = 2 \cdot 2 = (x + y) \cdot (x+y)$? – algor207 Jan 1 '15 at 16:24
• @algor207 No: $2$ is invertible in $R/I$. It works for $y^2=(x-1)(x+1)$. – user26857 Jan 1 '15 at 17:03
• @user1537366 For this: $x(x-y)=1$. – user26857 Jan 5 '15 at 15:08