$\begin{aligned}
0 \rightarrow R_f \rightarrow 0
\end{aligned}$

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.
- 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
contained in*not*prime ideal. Picksome arbitrary element $f \neq 0 \in R/N$. We know $f$ is not nilpotent, so we naturally consider$S_f \equiv \{ f^i : i \in \mathbb N \}$.*some* - 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.

- (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.