Prove that the set of nilpotent elements of a ring is the intersection of its prime ideals.
I know these two useful facts:
{nilpotent elements}$=\sqrt{0}$
$\sqrt{I}= \bigcap$ of prime ideals containing I
First, I need to show that the nilpotent elements are in fact an ideal and then I need to show it's prime.
Any specific suggestions would be appreciated. I just don't know how to make this proof flow and exactly what leads to what.
I read through this question and comments as well:
Prime ideal and nilpotent elements
Best Answer
(Assuming the rings are commutative with identity, else the statement is false.)
As Norbert says, your two useful facts immediately combine to say that the set of nilpotent elements is equal to the intersection of prime ideals containing the ideal $\{0\}$, and that means all prime ideals.
Well sure, this is possible, although it is not necessary. You can find how to do this in a few other questions on the site, starting here The set of all nilpotent elements is an ideal
That... is not in the question at all. Showing something is equal to an intersection of prime ideals does not mean it itself is prime. Most of the time, the ideal of nilpotent elements is not prime.
Ok, fair enough. The outline of the proof is: show the two sets are equal by showing each one is a subset of the other.
On one hand, each nilpotent element is contained in each prime ideal. Thus the nilpotent elements are in the intersection of primes.
The other direction is a little harder, and usually we use a lemma: every ideal maximal with respect to being disjoint from a multiplicative subset of R is a prime ideal of R. Using this, we can show that for each nonnilpotent element, there is a prime ideal disjoint from the powers of that element, and hence the prime does not contain the element. This shows how nonnilpotents are excluded from the intersection, so that everything there is nilpotent.
As a corollary, you get that the nilpotents form an ideal, and it is not really necessary to prove this ahead of time.