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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.

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Use the Correspondence Theorem. What are the ideals that contain $(xy-2)$? – Matt Dec 7 '12 at 4:57
See my answer here. If you want to have $\mathbb R[X,Y]/(XY-1)$ instead of $\mathbb R[X,Y]/(XY-2)$ (in order to use what I said in that topic), then use the automorphism of $\mathbb R[X,Y]$ that sends $X$ to $X/2$ and $Y$ to itself. – user26857 Dec 7 '12 at 20:15
up vote 3 down vote accepted

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).

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What you say is very explicit "geometrically." Moding out by $xy-2$ is saying set it equal to zero which is the hyperbola $xy-2=0.$ If you graph this on the xy-axis, then think about the setting $(x-a, y-b)$ to 0, these correspond to points (a,b). Thus the maximal ideals are just the $(x-a, y-b)$ for which you get a point $(a,b)$ on the hyperbola $xy-2=0$. – Matt Dec 7 '12 at 5:44
@Matt Right, I almost mentioned it, but then I worried it would obscure the whole thing if Anri has never heard this algebraic-geometric correspondence. – jathd Dec 7 '12 at 5:49
Dear jathd, this is not correct: there are other maximal ideals in $\mathbb{R}[x,y]/(xy-2)=\mathbb R[\bar x, \bar y]$, like for example $(\bar x^2+1)$. – Georges Elencwajg Dec 7 '12 at 7:01
Thanks a lot. I realize I was not approaching this the right way – Anri Rembeci Dec 7 '12 at 7:26
I still think that your answer is far from complete. – user26857 Jan 6 '13 at 1:29

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