§ Nilradical is intersection of all prime ideals

§ Nilradical is contained in intersection of all prime ideals

Let x0x \in \sqrt 0. We must show that it is contained in all prime ideals. Since xx is in the nilradical, xx is nilpotent, hence xn=0x^n = 0 for some nn. Let pp be an arbitrary prime ideal. Since 0p0 \in p for all prime ideals, we have xn=0px^n = 0 \in p for xx. This means that xn=xxn1px^n = x \cdot x^{n-1} \in p, and hence xpxn1px \in p \lor x^{n-1} \in p. If xpx \in p we are done. If xn1px^{n-1} \in p, recurse to get xpx \in p eventually.

§ Proof 1: Intersection of all prime ideals is contained in the Nilradical

Let ff be in the intersection of all prime ideals. We wish to show that ff is contained in the nilradical (that is, ff is nilpotent). We know that RfR_f (RR localized at ff) collapses to the zero ring iff ff is nilpotent. So we wish to show that the sequence:
0Rf0 \begin{aligned} 0 \rightarrow R_f \rightarrow 0 \end{aligned}
is exact. But exactness is a local property, so it suffices to check against each (Rf)m(R_f)_m for all maximal ideals mm. Since (Rf)m=(Rm)f(R_f)_m = (R_m)_f (localizations commute), let's reason about (Rm)f(R_m)_f. We know that RmR_m is a local ring as mm is prime (it is maximal), and thus RmR_m has only a single ideal mm. Since fmf \in m for all maximal ideal mm (since ff lives in the intersection of all prime ideals), localizing at ff in RmR_m blows up the only remaining ideal, collapsing us the ring to give us the zero ring. Thus, for each maximal ideal mm, we have that:
0(Rf)m0 \begin{aligned} 0 \rightarrow (R_f)_m \rightarrow 0 \end{aligned}
is exact. Thus, 0Rf00 \rightarrow R_f \rightarrow 0 is exact. Hence, ff is nilpotent, or ff belongs to the nilradical.

§ Proof 2: Intersection of all prime ideals is contained in the Nilradical

  • Quotient the ring RR by the nilradical NN.
  • The statement in R/NR/N 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. Picksome arbitrary element f0R/Nf \neq 0 \in R/N. We know ff is not nilpotent, so we naturally considerSf{fi:iN}S_f \equiv \{ f^i : i \in \mathbb N \}.
  • The only thing one can do with a multiplicative subsetlike that is to localize. So we localize the ring R/NR/N at SS.
  • If all prime ideals contain the function ff,then localizing at ff 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 S1R/N=0S^{-1} R/N = 0, we have that 0S0 \in S. So some fi=0f^i = 0. This contradicts the assumption that no element of R/NR/Nis nilpotent. Thus we are done.

§ Lemma: SS contains zero iff S1R=0S^{-1} R = 0

  • (Forward): Let SS contain zero. Then we must show that S1R=0S^{-1} R = 0. Consider some element x/sS1Rx/s \in S^{-1} R.We claim that x=0/1x = 0/1. To show this, we need to show that there exists an sSs' \in S such that xs/s=0s/1xs'/s = 0s'/1.That is, s(x10s)=0s'(x \cdot 1 - 0 \cdot s) = 0. Choose s=0s' = 0 and we are done. Thus every element is S1RS^{-1}R is zero if SScontains zero.
  • (Backward): Let S1R=0S^{-1} R = 0. We need to show that SS contains zero. Consider 1/1S1R1/1 \in S^{-1} R. We have that 1/1=0/11/1 = 0/1.This means that there is an sSs' \in S such that s1/1=s0/1s'1/1 = s'0/1. Rearranging, this means thats(1110)=0s'(1 \cdot 1 - 1 \cdot 0) = 0. That is, s1=0s'1 = 0, or s=0s' = 0. Thus, the element ss' must be zero for 11 to beequal to zero. Hence, for the ring to collapse, we must have 0=sS0 = s' \in S. So, if S1R=0S^{-1}R = 0, then SS contains zero.