Posted by Jason Polak on 05. January 2018 · Write a comment · Categories: commutative-algebra · Tags:

Let $R$ be a commutative ring and $(p)$ be a principal prime ideal. What can be said about the intersection $\cap_{k=1}^\infty (p)^k$? Let’s abbreviate this $\cap (p)^k$ (I like to use the convention that when limits are not specified, then the operation like intersection is taken over all possible indices).

Let’s try an example. For the integers, every principal prime is of the form $(p)$ where $p$ is a prime number or zero. And $(p)^k = (p^k)$ so $\cap (p)^k = (0)$. In fact if $R$ is any Noetherian integral domain then $\cap (p)^k = 0$.

If $R$ is not an integral domain then $\cap (p)^k$ is not necessarily zero. For example, let $S$ be an integral domain and let $R = S\times S$. In $S\times S$, the prime ideal generated by the single element $p = (1,0)$ is its own $k$-th power for all $k$. So $\cap (p)^k = p$.

Of course, it is impossible that in an integral domain to have $(p) = (p)^2$ for some principal prime $p$ unless $p = 0$. Of course, it is possible in an integral domain to have $P = P^2$ for a nonzero prime ideal $P$ that is necessarily not principal. Just take a “polynomial” ring over a field where the powers are allowed to be all nonnegative rationals; that is, a ring of the form $k[\Q^+]$ where $\Q^+$ is the monoid of all nonnegative rational numbers under multiplication. In the case of $k[\Q^+]$, a prime such that $P^2 = P$ would be the prime $P$ generated by all elements of the form $x^q$ where $q \gt 0$ is a rational number.

I will leave the reader with the following question:

Does there exist an integral domain, necessarily non-Noetherian, that contains a principal prime $(p)$ with $\cap (p)^k\not= 0$?
Posted by Jason Polak on 30. December 2017 · Write a comment · Categories: commutative-algebra · Tags: ,

Imposing structure on the poset of prime ideals of a ring $R$ is one way to gain a hold onto its structure. The poset of prime ideals of $R$ is simply a fancy term for the set of prime ideals of $R$, partially ordered by inclusion. Usually this set is not totally ordered: in the ring of integers $\Z$ for instance, the prime ideals $(2)$ and $(3)$ cannot be compared by inclusion. It seems to me that requiring the poset of primes to be totally ordered is a strict condition indeed.

Here is one type of domain in which the prime ideals are totally ordered: the valuation domain.
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