§ Nilradical is intersection of all prime ideals

§ Nilradical is contained in intersection of all prime ideals

Let x∈0x \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 0∈p0 \in p for all prime ideals, we have xn=0∈px^n = 0 \in p for xx. This means that xn=x⋅xn−1∈px^n = x \cdot x^{n-1} \in p, and hence x∈p∨xn−1∈px \in p \lor x^{n-1} \in p. If x∈px \in p we are done. If xn−1∈px^{n-1} \in p, recurse to get x∈px \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:
0→Rf→0 \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 f∈mf \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)m→0 \begin{aligned} 0 \rightarrow (R_f)_m \rightarrow 0 \end{aligned}
is exact. Thus, 0→Rf→00 \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. Pick some arbitrary element f≠0∈R/Nf \neq 0 \in R/N. We know ff is not nilpotent, so we naturally consider Sf≡{fi:i∈N}S_f \equiv \{ f^i : i \in \mathbb N \}.
  • The only thing one can do with a multiplicative subset like 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 S−1R/N=0S^{-1} R/N = 0, we have that 0∈S0 \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 S−1R=0S^{-1} R = 0

  • (Forward): Let SS contain zero. Then we must show that S−1R=0S^{-1} R = 0. Consider some element x/s∈S−1Rx/s \in S^{-1} R. We claim that x=0/1x = 0/1. To show this, we need to show that there exists an s′∈Ss' \in S such that xs′/s=0s′/1xs'/s = 0s'/1. That is, s′(x⋅1−0⋅s)=0s'(x \cdot 1 - 0 \cdot s) = 0. Choose s′=0s' = 0 and we are done. Thus every element is S−1RS^{-1}R is zero if SScontains zero.
  • (Backward): Let S−1R=0S^{-1} R = 0. We need to show that SS contains zero. Consider 1/1∈S−1R1/1 \in S^{-1} R. We have that 1/1=0/11/1 = 0/1. This means that there is an s′∈Ss' \in S such that s′1/1=s′0/1s'1/1 = s'0/1. Rearranging, this means that s′(1⋅1−1⋅0)=0s'(1 \cdot 1 - 1 \cdot 0) = 0. That is, s′1=0s'1 = 0, or s′=0s' = 0. Thus, the element s′s' must be zero for 11 to be equal to zero. Hence, for the ring to collapse, we must have 0=s′∈S0 = s' \in S. So, if S−1R=0S^{-1}R = 0, then SS contains zero.