Date: Fri, 29 May 2020 15:00:22 +0700
From: Nicolas Addington <adding@uoregon.edu>
Subject: supplemental

Dear Students,

The following commutative algebra question came up today, and I skipped 
over it but promised an email for those who were interested.  I see you 
were also discussing it in the chat.

Proposition: Let A be a Noetherian commutative ring and p a minimal 
prime ideal.  Then every s in p is a zero-divisor.

I'll follow a nice proof given in a MathOverflow comment by John 
Christian Ottem, https://mathoverflow.net/q/67382/16914 .

Consider the localization A_p.  Primes in A_p correspond to primes in A 
contained in p; thus A_p has only one prime, so every element of A_p is 
either a unit or nilpotent.  (Because the nilradical is the intersection 
of all primes.)

In particular s/1 is nilpotent, so s^n/1 = 0/1 in A_p for some n, so 
there is a t \notin p such that ts^n = 0 in A, so s is a zero-divisor.

----

We used the fact above to prove that if r is not a zero-divisor then

         dim(R/r) <= dim R - 1.

I also skipped over the reverse inequality, which holds whether r is a 
zero-divisor or not:

         dim(R/r) >= dim R - 1.

This is essentially Krull's principal ideal theorem, and you can read it 
in any commutative algebra book but the proof is not short.


Best,
Nick