# Spectrum of a ring (studied by Krull?) of rational functions

Let $$k$$ be an algebraically closed field and $$\mathbb A^2_k=\operatorname {Spec}k[x,y]$$ the affine plane over $$k$$.
Consider the ring $$R \subset k(x,y)$$ of the rational functions on the plane defined and constant on $$V(x)$$ (the $$y$$-axis $$x=0$$).
What is $$\operatorname {Spec}R$$ ?
(This is the geometric translation of an example due, I think, to Krull for which I unfortunately have no reference.)

Edit
Sorry,my definition of $$R$$ above is a bit ambiguous. What I mean is that $$R$$ consists of those fractions $$r(x,y)=\frac {p(x,y)}{q(x,y)}$$ which can be written as the quotient of two polynomials $$p(x,y),q(x,y)\in k[x,y]$$ such that $$q(0,y)\neq 0\in k[y]$$ and $$\frac {p(0,y)}{q(0,y)}\in k\subset k[y]$$.
For example the rational function $$\frac {y+x}{y-x}$$ mentioned by @YCor in the comments does belong to $$R$$ since $$y-0\neq0\in k[y]$$ and $$\frac {y+0}{y-0}=1\in k \subset k[y]$$

• I understand that $\frac{y+x}{y-x}\notin R$, right? (it's constant on this $x=0$ axis but not "defined" at $(0,0)$).
– YCor
Oct 16, 2020 at 12:46
• If my understanding is correct, it's a local ring, with two obvious prime ideals, 0 and the maximal one (zero on the vertical axis). Further prime ideals are obtained by choosing a non-vertical line $D$, intersecting the vertical axis at $p$ and mapping $f\in R$ to its germ at $p$ on $D$.
– YCor
Oct 16, 2020 at 12:55
• OK: so what remains from my previous comment is that this is a local ring with the two given prime ideals (0 and maximal), but I don't immediately see whether there are any others.
– YCor
Oct 16, 2020 at 15:06
• So just to clarify: Your ring consists of all $\Big(cp(Y)+Xf(X,Y)\Big)/\Big(p(Y)+Xg(X,Y)\Big)$ such that $p(Y)\neq 0$ and $c\in k$. Yes? Oct 16, 2020 at 18:15
• In other words, the localization $A$ of $k[x, y]$ at the prime ideal $(x)$ is a discrete valuation ring with residue field $k(y)$. We are looking at the preimage of $k\subseteq k(y)$ in $A$. So you can consider the abstract setting: consider a dvr $A$ with maximal ideal $\mathfrak{m}$ and a subfield $k \subseteq A/\mathfrak{m}$; what does the spectrum of the preimage of $k$ in $A$ look like? Oct 16, 2020 at 18:41

(Completing my comments above to an answer. Probably one can simplify this quite a bit.)

EDIT. The previous version mistakenly identified the ideal $$xA\cap R$$ with $$xR$$. In fact, the maximal ideal $$xA\cap R$$ of $$R$$ is not finitely generated: it is generated by $$\{xf\,:f\in k(y)\}$$ and a finite subset does not suffice. Similarly, $$xR$$ is not a prime ideal: we have $$xy^{-1}, xy\notin xR$$ but their product $$x^2\in xR$$.

The ring has exactly two prime ideals, $$(0)\subseteq \mathfrak{m}$$ where $$\mathfrak{m} = A\cap xk[x,y]$$.

Let $$A = k[x, y]_{(x)}$$ is the local ring at the generic point of the $$y$$-axis. This is a discrete valuation ring with maximal ideal $$\mathfrak{m}=(x)$$ and residue field $$A/\mathfrak{m} = k(y)$$.

The ring $$R$$ in question is the preimage of $$k\subseteq k(y)= A/\mathfrak{m}$$ in $$A$$. In other words, it is the fiber product $$R = A\times_{A/\mathfrak{m}} k$$.

EDIT (following Anton's comment below): a better reference for the following two paragraphs is stacks.math.columbia.edu/tag/0D2G Lemma 0B7J: the underlying space of the spectrum of the fiber product of the form $$A\times_{A/I} B$$ is the pushout of the corresponding underlying topological spaces of spectra.

By Stacks Project, Tag 07RS https://stacks.math.columbia.edu/tag/07RS, $$\operatorname{Spec} R$$ is the pushout of $$\operatorname{Spec}k \leftarrow \operatorname{Spec} A/\mathfrak{m} \to \operatorname{Spec} A$$.

By Theorem 3.4 (and its proof) in Schwede's paper http://www-personal.umich.edu/~kschwede/SchemeWithoutPoints.pdf , we get that the underlying space of $$\operatorname{Spec} R$$ is the corresponding pushout in spaces. But $$\operatorname{Spec} A/\mathfrak{m}\to \operatorname{Spec} k$$ is a homeomorphism, and hence so is $$\operatorname{Spec} A\to \operatorname{Spec} R$$.

• Thanks a lot for this fine answer. Oct 16, 2020 at 19:06
• Thanks again, Piotr: your quite subtle Edit perfectly explains why I was uneasy with my calculations, which prompted me to post the question in the first place. (By the way, are you of Russian origin as your first name seems to indicate?) Oct 16, 2020 at 20:09
• There is no mistake if you don't forget to take the radical: $rad(xR)=xA$, because $(Ax)^2\subset x(Ax)\subset xR$. By the way, $Ax\subset R$, so you don't need the intersection in $Ax\cap R$. Oct 17, 2020 at 8:28
• I think a better reference is stacks.math.columbia.edu/tag/0D2G Lemma 15.6.2 Oct 17, 2020 at 10:30
• @AntonMellit perfect, thanks! Oct 17, 2020 at 10:48

Your ring $$A$$ is a pullback

$$\matrix{ A&\rightarrow & k\cr \downarrow &&\downarrow\cr k[X,Y]_{(X)}&\rightarrow& k(Y)\cr}$$

where the right arrows send $$X$$ to zero.

If you invert $$X$$, the fields on the right become $$0$$, so the downarrow on the left becomes an isomorphism $$A[X^{-1}]=k(X,Y)$$. Thus all nonzero primes in $$A$$ contain $$X$$.

If you go mod $$X$$, $$A$$ becomes the field $$k(Y)$$. Thus $$(X)$$ is maximal. Because it is contained in all nonzero primes, it is the only nonzero prime.

• Thanks a lot for this fine answer. Oct 16, 2020 at 19:04
• Very similar to my answer, and with the same mistake: $(X)$ is not a prime ideal, since $XY$ and $XY^{-1}$ are not in $(X)$ but their product is. The correct ideal is $Xk[X,Y]_{(X)}\cap A$. Oct 16, 2020 at 19:42
• @PiotrAchinger: Yes, I think our answers were about 12 seconds apart. (Actually you beat me by a little more than that with your comments, which were posted while I was composing my answer.) Thanks for the correction; I'll leave the answer and your comment in place. Oct 16, 2020 at 20:45
• Are you implicitly using some property of fiber products and localisations? Oct 17, 2020 at 8:45
• @AntonMellit : I am using only that localization is exact, hence preserves pullbacks. Oct 17, 2020 at 14:07

I will join the party if you don't mind.

Let $$A=k[x,y]_{(x)}$$. This is the ring of fractions $$p/q$$ such that $$q$$ is not divisible by $$x$$. Our ring $$R$$ can be written as $$R=k+x A$$.

Inverting $$x$$ produces $$R[x^{-1}] = A[x^{-1}]=k(x,y)$$. The latter is a field with a unique prime ideal $$(0)$$. Thus the only prime ideal not containing $$x$$ is $$(0)$$.

The radical of $$(x)$$ is $$x A$$ because $$(x A)^2 \subset x ( x A) \subset x R$$ implies $$xA\subset \sqrt{(x)}$$ and $$x A$$ is maximal. So the only prime ideal containing $$x$$ is $$x A$$.

Any prime ideal either contains $$x$$ or it doesn't, so $$(0), xA$$ is the complete list.