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

Question

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 $\frac {y+0}{y-0}=1\in k \subset k[y]$

Answer 1

(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 $\operatorname{Spec} A\to \operatorname{Spec} R$.

Answer 2

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.

Answer 3

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.