# The nilradical of the localisation is the localisation of the nilradical

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.

• "so that $x^k=0$" is not quite true. Recall when a fraction equals zero in a ring of fractions. – user26857 Feb 2 at 8:18
• 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! – nhmwhhxx Feb 2 at 10:02
• @nhmwhhxx: Yes, $x$ may not be in $N$, but $xtu$ is in $N$ and that's enough to conclude that $x/s=(xtu)/(stu)\in S^{-1}N$. – Shivering Soldier Feb 2 at 12:29

$$x^k/s^k = 0$$ 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.
• 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 $tx \in N$. – Trevor Gunn Mar 29 at 2:29