Show that this map has not the going-down property. Let $A= k[x_1,x_2,y] / (x_2^2-x_1^2(x_1+1))$ and $Spec(A) \to Spec(k[x_1,x_2,y])$ the natural inclusion induced by the projection $k[x_1,x_2,y] \to A$. Consider the map $f : Spec(k[x,y]) \to Spec(A)$ induced by the ring homomorphism $A \to k[x,y]$, $x_1 \mapsto x^2 - 1$, $x_2 \mapsto x(x^2-1)$, $y \mapsto y$.

Show that $f$ does not have the going-down property.

Hint: consider the prime ideals $(x-1,y)$ and $(y-(x+1))$.
Unfortunately, despite the presence of the hint I don't understand how exactly I have to proceed...please, is there someone who could help me in same way?
Thank you very much.
Cheers
 A: $\newcommand{\ideal}[1]{{\mathfrak{#1}}}$
$\newcommand{\spec}[1]{{\mathrm{Spec}}\left({#1}\right)}$
Call $\ideal{p} = (y - (x+1))$ and $\ideal{m} = (x-1,y)$, ideals of $k[x,y]$. Call further $\ideal{n} = \ideal{m} \cap A = (x_1, x_2, y)$. 
Now $\ideal{p} \subseteq \ideal{m}' = (y,x+1)$ and $\ideal{m}' \cap A = \ideal{n}$. So 
$$\ideal{p} \cap A \subseteq \ideal{m}' \cap A = \ideal{n} = \ideal{m} \cap A$$
Now I contend that $\ideal{p}$ is the only ideal of $k[x,y]$ that lies over $\ideal{p} \cap A$. So, as $\ideal{p} \not\subseteq \ideal{m}$ going-down is not fulfilled.
It remains to prove that $\ideal{p}$ is the only ideal of $k[x,y]$ which lies over $\ideal{p} \cap A$. This follows from the isomorphism
between $D(x_1) \subseteq \spec{A}$ and $D((x-1)(x+1)) \subseteq \spec{k[x,y]}$. This corresponds to the isomorphism
$$A_{x_1} \cong k[x,y]_{x^2-1}$$
which I have not checked in full detail, but which must exist for geometric reasons: The lines $V(x-1)$ and $V(x+1)$ in $\spec{k[x,y]}$ correspond
to the line $l$ which is the extrusion of the nodal point $(x_1=x_2 =0)$
of the nodal curve $C = V(x_2^2-x_1^2\,(x_1+1))$ along the $y$-axis. All other points get mapped 1-1. As $\ideal{p} \in D(x^2-1)$ and $\ideal{p} \cap A \in D(x_1)$, only $\ideal{p}$ can lie over $\ideal{p} \cap A$.
