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Looking at almost 10 year old notes from a course on commutative algebra, I found the following exercise which I had apparantly not done back then:

Task: Compute the spectrum of the ring $R = \mathbb{Z}[x,y]/(x(2+xy),y(2+xy))$, where $\deg(x) = 1$ and $\deg(y) = -1$.

Out of curiosity I started doing it, but cannot seem to get a full answer (probably I am also a bit rusty...). Let me show you what I am able to do:

Let $P$ be a prime ideal of $R$. Since $$y(2+xy) = 0 \in P,$$ we have $y \in P$ or $2+xy \in P$. Let us for now assume that $y \in P$. Then $R / P$ is a quotient of $R / (y) \cong \mathbb{Z}[x]/(2x)$. It is not difficult to understand the spectrum of the $\mathbb{Z}[x]/(2x)$ as these correspond to the prime ideals of $\mathbb{Z}[x]$ which contain $2x$. Therefore I am able to determine these (up to making a silly mistake):

$\bullet$ $(p,x)$ where $p$ is a prime or $p = 0$

$\bullet$ A maximal ideal of $\mathbb{F}_2[x]$ (lying over the homogeneous prime ideal $(2)$)

$\bullet$ $(2)$

From this we can then also find the prime ideals $P \subset R$ which contain $y$. So let us now assume that $y \not\in P$ and $2+xy \in P$. In this case I don't really seem to get anywhere. Maybe one of you can come up with a good way of handling this case.

Con
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1 Answers1

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I assume you are familiar with the classification of prime ideals in the polynomial ring $R[x]$ for a PID $R$; see, for example, MSE/174595. Additionally, I assume you know the classification of prime ideals in localizations; see here.

Since $R/(x) = \mathbb{Z}[y]/(2y)$ and $R/(y) = \mathbb{Z}[x]/(2x)$, the prime ideals of $R$ that contain $x$ or $y$ are well understood.

Thus, it remains to consider the prime ideals that contain neither $x$ nor $y$. These correspond to the prime ideals in the localization $R[x^{-1}, y^{-1}]$. In this ring, the equation
$$0 = x(2 + xy)$$ implies $2 + xy = 0$. Since $xy$ is invertible in this localization, it follows that $2$ must also be invertible. Moreover, $y$ is uniquely determined by $x$, specifically,
$$ y = -\frac{2}{x}.$$ This shows that $$ R[x^{-1}, y^{-1}] = \mathbb{Z}[\tfrac{1}{2}]\,[x, x^{-1}]. $$ The prime ideals of this ring correspond to the prime ideals of $\mathbb{Z}[x]$ that do not contain $x$ or $2$, which are already well understood.

Martin Brandenburg
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  • Thanks a lot! I did not consider $R[x^{-1},y^{-1}]$ and this was exactly what I was missing. Once I am allowed to, I will also award you the bounty. – Con Feb 18 '25 at 22:50