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I'm trying to prove that the set $\mathrm{nilrad}(A)$ of nilpotent elements of $A$ is an ideal

Pf/ if $g\in\mathrm{nilrad}(A)$, then $g^n = 0$, for some $n>0$. Let $h$ be an element of $\mathrm{nilrad}(A)$. then $(gh)^n = (hg)^n = g^nh^n = 0\cdot0 = 0.$

also, $(-g)^n = (-1)^ng^n = (-1)^n\cdot 0 = 0.$

Now we show that $\mathrm{nilrad}(A)$ is closed under elements of $A$. Let $i$ be an element of $A$; then, $(gi)^n = g^ni^n = 0\cdot i^n = 0.$

$(g+i)^n = \sum_{i=0}^n \binom{n}{k}g^ni^{n-k}$

Also, how can we show that the set radical $\mathrm{rad}(I)$ for an ideal $I$ is an ideal? Pretty much trying to organize the proof. Thanks.

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    I am, but have been somewhat careful/judicious2012-06-01

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Yet another way: if we put $\,\,\mathcal{P}:=\{P\leq A\,;\,P\,\text{ is prime}\}\,$ , then in fact $\,\,\displaystyle{nilrad(A)=\bigcap_{P\in\mathcal{P}}P}$ :

Hints for the proof:

(1) $\,\displaystyle{x\in nilrad(A)\Longrightarrow \exists\,n\in\mathbb{N}\,\,s.t.\,\,x^n=0\Longrightarrow x^n\in P\,,\,\forall P\in\mathcal{P}}$ ...

(2) Suppose $\, \displaystyle{x\in\bigcap_{P\in\mathcal{P}}P-nilrad(A)\Longrightarrow \mathcal{X}:=\{x^n\}}\,\,$ is a closed set of $\,A\,$ not containing zero $\,\Longrightarrow \text{the set} \,\,T_{\mathcal{X}}:=\{I\leq A\,\,;\,\,I\cap\mathcal{X}=\emptyset\}\,$ has a maximal element which is a prime ideal of $\,A$

Sketch of proof for the last part: partial order $\,T_{\mathcal{X}}\,$ in the "usual way", apply Zorn's lemma and get a maximal element which is proved to be a prime ideal by assuming there's a couple of elements not belonging to this ideal but whose product does belong...

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I believe the for the first part you should be able to finish it yourself. For the second, suppose you have an ideal $I$ and as always the canonical projection $\phi: A \longrightarrow A/I$. Now the nilradical in here is an ideal as you will have proved. Since the preimage of an ideal is an ideal, consider $\phi^{-1}(\textrm{nilrad}(A/I))$.

I claim that this is the radical of $I$. Can you prove this bit?