When $R$ is not Noetherian, what happens to exercise 15.16 (p.296) of Sharp's book "Steps in Commutative Algebra"

Question

Krull's height theorem says that if $R$ is a Noetherian ring and $I$ is a proper ideal generated by $n$ elements of $R$, then $\operatorname{ht} I\le n$.
In the absence of the Noetherian hypothesis, the conclusion fails.

When $R$ is not Noetherian, what happens to exercise 15.16 (p.296) of Sharp's Steps in Commutative Algebra which says:
Let $R$ be a commutative Noetherian ring, and let $a\in R$ be a non-unit and a non-zerodivisor. Let $P \in \operatorname{Spec}(R)$ be such that $a\in P$. Prove that $\operatorname{ht}_{R/a}P/Ra=\operatorname{ht}_RP-1.$

I don't understand what's good for that preamble on Krull's principal ideal theorem. — user26857, Jan 01 '16 at 21:21

score 0 · Accepted Answer · edited Apr 13 '17 at 12:20

0

Use this example. Set $a=X$ and notice that $\operatorname{ht} P/(a)=0$ while $\operatorname{ht} P-1=1$.

edited Apr 13 '17 at 12:20

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answered Jan 01 '16 at 21:18

user26857

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When $R$ is not Noetherian, what happens to exercise 15.16 (p.296) of Sharp's book "Steps in Commutative Algebra"

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