Maximal ideals in $\mathbb{R}[x,y]/(xy-2)$? I'm working on a practice exam and one of the questions asks if there are any maximal ideals in $\mathbb{R}[x,y]/(xy-2)$ and, if so, to find one of them. 
Initially, I thought the quotient ring was a field because xy-2 is clearly irreducible; however, I forgot that this only applies to PIDs. My suspicion is that $(\bar{x})$ is a maximal ideal in the quotient ring, but I don't know how to demonstrate this. 
I was also thinking about the possibility of the quotient ring being local, although I'm not sure if this if the right direction. 
Thanks a lot. 
 A: The ideal $(\bar x)$ can't be maximal, because $\bar x$ is invertible (its inverse is $\bar y/2$), so $(\bar x)$ is the whole quotient.
Let's follow Matt's advice. First, the maximal ideals in $\mathbf R[x,y]$ are the ideals of the form $(x-a,y-b)$, for $a,b\in\mathbf R$. We need to find $a$ and $b$ such that $(xy-2)\subset(x-a,y-b)$. Let's remember what it means for $f\in\mathbf R[x,y]$ to be in $(xy-2)$: it means that $f$ is a multiple of $xy-2$. And what it means for $f$ to be in $(x-a,y-b)$: it means that $f(a,b)=0$. So you just need to find $a$ and $b$ such that if $f$ is a multiple of $xy-2$, then $f(a,b)=0$.
Then, you can check that $(\bar x-a,\bar y-b)$ is indeed a maximal ideal in $\mathbf R[x,y]/(xy-2)$.
Edit: As Georges points out in the comments, the maximal ideals of $\mathbf R[x,y]/(xy-2)$ are not all of the form $(\bar x-a, \bar y-b)$. They would be if $\mathbf R$ was algebraically closed, but it's not. Yet it is still true that all ideals of the form $(\bar x-a,\bar y-b)$ are maximal, and you can find suitable real numbers $a$ and $b$ (because $xy-2$ has a root in $\mathbf R^2$), so the above happens to work (but mostly by chance).
