For the following two ring extensions $R\subset S$:
1) $\mathbb{R}[y]\subset \mathbb{R}[x,y]/(x^2-y);$ 2) $\mathbb{Z}\subset\mathbb{Z}[x]/(x^2-3)$
find in each case,
i) a nonzero prime ideal $P$ in $R$ with a unique prime ideal $Q$ in $S$ lying over $P$, and $S/Q\simeq R/P$.
ii) a nonzero prime ideal $P$ in $R$ with a unique prime ideal $Q$ in $S$ lying over $P$, and $S/Q\not\simeq R/P$.
iii) a nonzero prime ideal $P$ in $R$ such that there are more than one prime ideal in $S$ lying over $P$.
I have been working on this problem for a while and I do not actually have clue about how to approach this one. I cannot seem to find a way to start looking for possible examples for i) and ii). But iii) seems a bit easier and I wonder if $(y)\subset(x,y),(y)$ and $(2)\subset(1+\sqrt{3}),(1-\sqrt{3})$ are prime ideals that will work.
Think geometric, work algebraic! You should think about the map on prime ideals induced by $\mathbb R[y] \subset \mathbb R[x,y]/(x^2-y)$ as the projection of the parabola onto the $y$-axis. For example if you pick the prime $(y-4)$, you get $(x-2,y-4)$ and $(x+2,y-4)$ lying over that prime. This is obtained by thinking of the point $y=4$ on the $y$-axis. The corresponding points on the parabola, that project onto that point, are $(x=2,y=4)$ and $(x=-2,y=4)$.
However, we have to deal with the fact that $\mathbb R$ is not algebraically closed, so the correspondence algebra <-> geometry fails a little bit. But we can deal with that: Given the prime $(y+1)$, there is no prime of the form $(x-a,y+1)$ lying over, since we would have to pick $a = i$. The trick is to replace $x-i$ by the minimal polynomial of $i$, so we obtain the prime $(x^2+1,y+1)$ lying over $y+1$. It is easily checked, that $(x^2+1,y+1)$ indeed is a prime containing $(x^2-y)$ and thus is a prime of $\mathbb R[x,y]/(x^2-y)$.
The extension $\mathbb Z \subset \mathbb Z[x]/(x^2-3)$ does not provide such an intuitive geometric interpretation, but you can deal with it the same way:
Instead of distinguishing between the cases $y>0, y < 0, y=0$, we should distinguish between the following three cases for a prime $p \in \mathbb Z$, which correspond to the cases above (and behave the same way):