Counterexample to going down Theorem 
Let $k$ be a field. Let $R=k[X,Y],\ A=k[X,Y,\frac{X}{Y}]$. Show that going down doesn't hold for $A/R$.

My approach: I took the map $\phi: \operatorname{Spec} A\to \operatorname{Spec} R,\ q\mapsto q\cap R$ to verify the preimages of $q$ in $R$. Sadly, I have not been able to find a counterexample to my problem. I suspect that one could exploit the fact that $k[Y]\in \operatorname{Spec} R, \ k[Y]\not\in \operatorname{Spec} A$. However, to make this work as a counterexample, I would have to find a prime ideal in $R$ containing $k[Y]$ and a prime ideal in $A$ lying above it. This, however, is impossible due to $k[Y]$ being a maximal ideal. Any help is greatly appreciated!
 A: You should note that this extension is not integral as $X/Y$ is not integral over $k[X,Y]$. Indeed this element is going to be important in constructing your counterexample. The key observation is $X=(X/Y)\cdot Y$.
Consider the chain of ideals $\mathfrak{p}_1\supset\mathfrak{p}_2$ where $\mathfrak{p}_1=(X,Y), \mathfrak{p}_2=(X)$ in $k[X,Y]$.
This is a chain of prime ideals(if you don't know why these are prime ideals, you should check they are by looking at the quotients).
Now, $\mathfrak{q}_1=(X,Y)\subset k[X,Y,X/Y]$ is a prime ideal. Indeed the quotient is $k[X/Y]\cong k[Z]$ which is an integral domain. Note that $\mathfrak{q}_1$ lies over $\mathfrak{p}_1$.
Now I want to find a prime ideal $\frak{q}_2$ lying over $\mathfrak{p}_2=(X)$. Is $(X)\subset k[X,Y,X/Y]$ a prime ideal? Unfortunately it isn't as $X=(X/Y)\cdot Y$.
So what is a prime ideal lying over $\mathfrak{p}_1$? It turns out $(X/Y)$ is the only choice. But if we let $\mathfrak{q}_2=(X/Y)$ then $\mathfrak{q}_1\not\supset \mathfrak{q}_2$.
