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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.

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    $\begingroup$ "so that $x^k=0$" is not quite true. Recall when a fraction equals zero in a ring of fractions. $\endgroup$
    – user26857
    Feb 2 '20 at 8:18
  • $\begingroup$ Perfect, I knew I was missing something simple. If $x^k = 0$ in $S^{-1}A$ then $x^k t u = 0$, so we can't say $x \in N$. Thanks! $\endgroup$
    – mi.f.zh
    Feb 2 '20 at 10:02
  • $\begingroup$ @nhmwhhxx: Yes, $x$ may not be in $N $, but $xtu$ is in $N$ and that's enough to conclude that $x/s=(xtu)/(stu)\in S^{-1}N $. $\endgroup$ Feb 2 '20 at 12:29
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As per the comments, the line

$x^k/s^k = 0$ so that $x^k = 0$

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.

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  • $\begingroup$ If $a/b = 0$ in $S^{-1}A$ then $ta = 0$ for some $t \in S$. You just need one element, not two and that element is in $S$ not just in $A$. So we have $tx^k = 0$ which means $(tx)^k = 0$ which means $tx \in N$. $\endgroup$ Mar 29 '20 at 2:29

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