This is a result that seems pretty easy to prove, yet it is given as a corollary (3.12) in Atiyah Macdonald - I'm not sure if the previous result (that localisation commutes with taking radicals) is necessary to prove this, or if it is just given as a 'nice' application.
The statement is that
If $N$ is the nilradical of $A$ then $S^{-1}N$ is the nilradical of $S^{-1}A$.
If $x/s$ is some nilpotent in $S^{-1}A$, then $x^k/s^k = 0$ for some $k$, so that $x^k = 0$ and so $x \in N$, i.e. $x/s \in S^{-1}N$. Conversely if $x/s \in S^{-1}N$ then $x$ must be nilpotent in $A$, hence $x/s$ is nilpotent in $S^{-1}A$.
This proof seems incredibly straightforward so I'm not sure what I'm missing.
As per the comments, the line
is not correct. For $x^k/s^k =0 $ in $S^{-1}A$, all that this says is that there exist $t, u \in A$ so that $x^ktu = 0$ in $A$.
On the other hand, $xtu$ is in $N$, and so $x/s = (xtu)/(stu) \in S^{-1}N$, which is what we wanted.