## § Nilradical is intersection of all prime ideals

#### § Nilradical is contained in intersection of all prime ideals

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

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

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

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

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

#### § Lemma: $S$ contains zero iff $S^{-1} R = 0$

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