TITLE: Spectrum of a ring (studied by Krull?) of rational functions
QUESTION [6 upvotes]: 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 $\frac {y+0}{y-0}=1\in k \subset k[y]$

REPLY [3 votes]: 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.