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 $\operatorname{ht}_{R/a}P/Ra=\operatorname{ht}_RP-1.$

  • $\begingroup$ I don't understand what's good for that preamble on Krull's principal ideal theorem. $\endgroup$ – user26857 Jan 1 '16 at 21:21

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.