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If we have a containment of prime ideals in a commutative ring with $1$ is the "larger" prime ideal necessarily of the higher height?

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up vote 5 down vote accepted

Remember, the height of a prime ideal $\mathfrak{p}$ is equal to the number of strict inclusions in a maximal chain of primes ending in $\mathfrak{p}$. So, if $\mathfrak{p}$ is strictly contained in $\mathfrak{q}$, then a maximal chain of primes ending in $\mathfrak{q}$ must necessarily have more inclusions - so, yes (provided that we are in a Noetherian ring, thus there are no infinite chains of prime ideals).

Let $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \ldots \subsetneq \mathfrak{p}_n = \mathfrak{p}$ be a maximal chain of primes with respect to inclusion (so that $ht(\mathfrak{p})=n$), then this chain can be extended to $\mathfrak{p}_0 \subsetneq \mathfrak{p}_1 \subsetneq \ldots \subsetneq \mathfrak{p}_n \subsetneq \mathfrak{q}$, which has one more inclusion than the original chain. Since $ht(\mathfrak{q})$ depends on the maximum length of a chain of primes, it may only increase from $n+1$ (since perhaps our extended chain is not of maximal length).

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Thanks for the response. I don't see why a maximal chain containing $q$ must have more inclusions. It's not necessary that such a chain will contain $p$. So it seems like it's possible we have two different chains one terminating at $p$ and another at $q$ such that the second one does not contain $p$ and have the same lengths. – Past Mustafa Feb 24 '12 at 0:51
But height depends on a maximal chain - so we certainly have a chain of primes that is longer when we end with $\mathfrak{q}$ (the chain that contains $\mathfrak{p}$) as opposed to ending with $\mathfrak{p}$, so creating a shorter chain is irrelevant. – Andrew Parker Feb 24 '12 at 1:09
I have edited my answer to clarify regarding my previous comment. – Andrew Parker Feb 24 '12 at 1:40
Thanks. I got confused between height of an ideal being the infimum of the heights of the primes containing it and height of a prime ideal. I was assuming it is the length of the smallest chain. – Past Mustafa Feb 24 '12 at 3:50
Although, we should say longest or maximum length chain and not maximal. – Past Mustafa Feb 24 '12 at 3:58

Without restricting to finite heights, the answer is no; we could have $\mathfrak{p}\subsetneq\mathfrak{q}$ but $\operatorname{ht}(\mathfrak{p})=\operatorname{ht}(\mathfrak{q})=\infty$. For example, in $R=\mathbb{C}[x_1,x_2,\ldots]$, the polynomial ring in infinitely many variables over $\mathbb{C}$, the prime ideal $\mathfrak{p}=(x_2,x_3,\ldots)$ is strictly contained inside $\mathfrak{q}=(x_1,x_2,\ldots)$, but both $\mathfrak{p}$ and $\mathfrak{q}$ contain infinitely long chains of strict inclusions of prime ideals, e.g.


so $\operatorname{ht}(\mathfrak{p})=\operatorname{ht}(\mathfrak{q})=\infty$.

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We always have $\mathrm{ht}(q)\ge \mathrm{ht}(p)+\mathrm{ht}(q/p)$. If $p\ne q$, then $\mathrm{ht}(q/p)>0.$ – user18119 Feb 24 '12 at 12:53
Agreed - though if $p$ and $q$ are both of infinite height, this is not a contradiction; or am I mistaken? – Zev Chonoles Feb 24 '12 at 16:04
Sorry for not being explicit. What I meant is even when $p, q$ have infinite height, one can say $q$ has bigger height than $p$ because $q/p$ has positive height. – user18119 Feb 24 '12 at 19:42

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