From Dummit & Foote:
Proof: Passing to $R/I$. Proposition 11 shows that it suffices to prove this result for $I = 0$ ...
Proposition 11:
... $(\text{rad}\ I)/I$ is the nilradical of $R/I$. In particular $R/I$ has no nilpotent elements if and only if $I =\text{rad} \ I$.
How does it suffice to show the special case?
On
Thanks @rschwieb.
We want to show $\text{ rad}\ I = \bigcap\limits_{J \supset I; J \text{ prime}} J$,
given that in $R/I$, $\text{ rad } (0) = \bigcap\limits_{J \text{ prime}} J$.
There is a bijective correspondence between prime ideals $\supset I$ and prime ideals of $R/I$, $\phi : J \mapsto J/I$, and $\phi^{-1} : J/I \to J$. Rewriting the $\bigcap$ formula in terms of $R/I$ gives:
$\text{rad } (I/I) = \bigcap\limits_{J/I \text{ prime}} J/I$.
Applying $\phi^{-1} = \psi$ to this gives the $\text{rad} \ I$ formula as long as $\psi$ commutes with $\text{rad } $, ie. $\psi \circ \text{rad } \ I$ = $\text{rad} \circ \psi (I)$.
$\text{rad} \circ \psi(I) = \{ \bar{x} \in R/I : \bar{x}^n \in \psi(I)\} = \{\bar{x} \in R/I : x^n \in I\} = \psi \circ \text{rad } (I)$.
If you prove it for the zero ideal in any ring, then you have proven it for the zero ideal in $R/I$, but by correspondence, the primes in $R/I$ correspond to primes of $R$ containing $I$.