§ Nilradical is intersection of all prime ideals
§ Nilradical is contained in intersection of all prime ideals
Let . We must show that it is contained in all prime ideals.
Since is in the nilradical, is nilpotent, hence for some . Let be an
arbitrary prime ideal. Since for all prime ideals, we have
for . This means that ,
and hence . If we are done.
If , recurse to get eventually.
§ Proof 1: Intersection of all prime ideals is contained in the Nilradical
Let be in the intersection of all prime ideals. We wish to show that
is contained in the nilradical (that is, is nilpotent). We know that (
localized at ) collapses to the zero ring iff is nilpotent. So we wish to
show that the sequence:
is exact. But exactness is a local property, so it suffices to check against each for
all maximal ideals . Since (localizations commute), let's reason about .
We know that is a local ring as is prime (it is maximal), and thus has only a single
ideal . Since for all maximal ideal (since lives in the intersection of all prime
ideals), localizing at in blows up the only remaining ideal, collapsing us the ring to give
us the zero ring. Thus, for each maximal ideal , we have that:
is exact. Thus, is exact. Hence, is nilpotent, or belongs to the
§ Proof 2: Intersection of all prime ideals is contained in the Nilradical
- Quotient the ring by the nilradical .
- The statement in becomes "in a ring with no ninpotents, intersection of all primes is zero".
- This means that every non-zero element is not contained in some prime ideal. Pick some arbitrary element . We know is not nilpotent, so we naturally consider .
- The only thing one can do with a multiplicative subset like that is to localize. So we localize the ring at .
- If all prime ideals contain the function , then localizing at destroys all prime ideals, thus blows up all maximal ideals, thereby collapsing the ring into the zero ring (the ring has no maximal ideals, so the ring is the zero ring).
- Since , we have that . So some . This contradicts the assumption that no element of is nilpotent. Thus we are done.
§ Lemma: contains zero iff
- (Forward): Let contain zero. Then we must show that . Consider some element . We claim that . To show this, we need to show that there exists an such that . That is, . Choose and we are done. Thus every element is is zero if contains zero.
- (Backward): Let . We need to show that contains zero. Consider . We have that . This means that there is an such that . Rearranging, this means that . That is, , or . Thus, the element must be zero for to be equal to zero. Hence, for the ring to collapse, we must have . So, if , then contains zero.