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Let $A \subset B$ be an integral ring extension and assume that $A$ is a finitely generated $K$-algebra over some field $K$. Let $P_1\subsetneq P_3$ be prime ideals of $A$ and let $Q_1\subsetneq Q_3$ be prime ideals in $B$ lying over $P_1$ and $P_3$, respectively.

Show that if there is a prime ideal $P_2$ of $A$ such that $P_1\subsetneq P_2\subsetneq P_3$ then there is also a prime ideal $Q_2$ of $B$ such that $Q_1\subsetneq Q_2\subsetneq Q_3$.

By Noether Normalization, $A$ is finite over some $K[x_1,\dotsc,x_n]$ and thus integral. But I have no ideal how to use this. I am also thinking about height of the prime ideals, but most theorems I know require extra assumptions.

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First of all, you should consider the integral extension $A/P_1 \subset B/Q_1$. Then you are down to this statement:

We have an integral extension of integral domains $A \subset B$. Given $0 \subset P_2 \subset P_3$ in $A$ and $Q_3 \subset B$ over $P_3$, there exsists a prime $Q_2$ between $0$ and $Q_3$.

The good thing about that: The inclusion $0 \subset Q_2$ is automatically fulfilled, we do not have to worry about this.

Now it is an easy one: After Noether Normalization of $A$ (! $A$ is not finite over $K$ !), $B$ is integral over a normal integral domain and then you can use Going Down.

MooS
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  • If $A$ is not finite over $K$, what is the normal integral domain that $B$ is integral over?? –  Mar 02 '15 at 08:33
  • $K[x_1, \dotsc, x_n]$. $A$ is finite over that polynomial ring, so $B$ is integral over it by the transitivity of integral extension. But $A$ is definitely not finite over $K$. I mean $A$ has dimension $\geq 2$. Any ring finite over $K$ is zero-dimensional. – MooS Mar 02 '15 at 08:34
  • But does this mean we are applying Going Down to $B$ integral over $K[x_1,\dotsc,x_n]$? –  Mar 02 '15 at 08:44
  • Yes, exactly this. – MooS Mar 02 '15 at 08:44
  • This might be a still stupid question. But to apply Going Down, we need a $P_2\subset P_3$ in $K[x_1,\dotsc,x_3]$ to start with. How do we get them from the original ones in $A$? –  Mar 02 '15 at 08:46
  • Well, this is indeed not a clever question :D Just intersect the given chain in $A$ with $K[x_1, \dotsc, x_n]$. – MooS Mar 02 '15 at 08:49
  • We do not require this, since we cannot deduce it. We only can deduce that $Q_2$ lies over $P_2 \cap k[x_1, \dotsc, x_n]$. You might accept the answer btw. – MooS Mar 02 '15 at 09:03