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Let $R'\in R$ be an integral extension of rings with $R'$ a $K$-algebra finitely generated . Consider a chain of different prime ideals in $R'$ , $P_{1}\subsetneq P_{2}\subsetneq P_{3}$

Such that there are prime ideals in $R$ , $ Q_{1},Q_{3} $ such that $ Q_{1}\bigcap R' = P_{1}$ and $ Q_{3}\bigcap R' = P_{3}$

Show there exits $Q_{2} \subset R$ such that $Q_{1}\subsetneq Q_{2}\subsetneq Q_{3}$

Show a counterexample that such $ Q_{2} $ might not satisfy $ Q_{2}\bigcap R' = P_{2}$

I have read this "Going between" property but I don't know how to apply going down theorem

I hava tryed to use the going-down theorem but I don't really know how to approach it .

Thank you .

Vegan Maths
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1 Answers1

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This is just spelling out MooS's linked answer and linking you to a counterexample.

First we reduce to the case that $R' \subseteq R$ is an extension of domains and $P_1 = Q_1 = 0$ by passing to the integral extension $R'/P_1 \subseteq R/Q_1$. Note that $R'$ is still an f.g. $K$-algebra.

Next we apply Noether Normalization: there exists a finite set of algebraically independent elements $\{y_i\}_{i=1}^n$ such that $R'$ is finitely generated as a module over $A = K[y_1, \ldots, y_n]$ (a fortiori an integral extension). The extension $A \subseteq R' \subseteq R$ is therefore integral, by transitivity. Moreover $A$ is a finite polynomial algebra over a field, so $A$ is a Unique Factorization Domain and hence is integrally closed.

We conclude that the extension $A \subseteq R$ has going down.

Recall that an algebraic extension of domains $B' \subseteq B$ is always essential, meaning that nonzero ideals contract to nonzero ideals. Indeed, for any $b \in B$ then there is a relation $b_0' + \sum_{i=1}^n b_i' b^i = 0$, with $b_i' \in B'$ and $b_0' \not= 0$, which shows that $b_0' \in bB \cap B'$.
From here you can deduce that for any integral extension $B' \subseteq B$ and proper chain of prime ideals $P_1 \subsetneq P_2$ of $B$, we get a proper chain of ideals $P_1 \cap B' \subsetneq P_2 \cap B'$.

Applying this to our situation, we get a proper chain of prime ideals $0 \subsetneq P_2 \cap A \subsetneq P_3 \cap A$ in $A$, such that $Q_3 \cap A = P_3 \cap A$. Now we can apply going down to deduce the existence of a prime $Q_2 \subseteq Q_3$ such that $Q_2 \cap A = P_2 \cap A$, and clearly $Q_2 \not= 0$.

Now for a counterexample, which shows that $Q_2$ doesn't necessarily contract to $P_2$, see here. As in user26857's answer there, you can use the extension $A = K[t^2 - 1, t^3 - t, z] \subseteq B = K[t,z]$, together with the primes $Q_3 = (t-1,z)$, $P_3 = Q_3 \cap A$, $P_2 = (z-t-1) \cap A$. The answer there shows that $P_2 \subsetneq P_3$ but that $Q_3$ does not contain a prime that contracts to $P_2$ (although $Q_3$ does contain a nonzero prime, as it rightly should by the above proof).

Badam Baplan
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