## § 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. Pick some 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 subset like 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 be equal 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.