# Help with nilradicals

I know the definition of the nilradical of a ring, and I know that it is an ideal, but I don't know how to

"[...] determine the nilradicals of the rings $\mathbb{Z}/(12)$, $\mathbb{Z}/(n)$, and $\mathbb{Z}$."

Any help with how to approach this problem would be greatly appreciated!

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The nilradical contains all the nilpotent elements of a ring $R$, that is the $x\in R$ for which some power $x^n$ will be $0$. Take for example $6\in Z/12$. Is $6^n=0$ for some $n$, modulo $12$? You can just go through the elements of $Z/12$ to get an intuition. – Gregor Bruns Nov 15 '12 at 18:15
Dear @Carolus, Do you know that the nilradical is exactly the set of nilpotents? Or the intersection of the prime ideals? – Keenan Kidwell Nov 15 '12 at 18:18
@KeenanKidwell: The set of nilpotents. – Carolus Nov 15 '12 at 18:19

## 1 Answer

It is a standard theorem that the nilradical of a ring $R$ is the intersection of all the prime ideals in $R$.

So, for example, take $\mathbb{Z}/12$. Its prime ideals are (the images of) $(3)$ and $(2)$ (using the correspondence that prime ideals in a quotient ring $R/I$ are precisely the prime ideals in $R$ containing $I$). The intersection of $(3)$ and $(2)$ is the ideal $(6)$, which contains the two elements $0$ and $6$.

Edit: If you don't want to use the quoted theorem, then, at least in PID's, an element $x$ in $R/(r)$ is nilpotent if and only if $r|x^n$ for some $n$ (this is easy to prove!). So for the example $\mathbb{Z}/12$, we see that $12|6^2$, så $6^2=0$ is nilpotent.

Note that the condition $r|x^n$ for some $n$ is equivalent that all factors of $r$ must divide some factor of $x$.

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Ok, I think I get it. Is there some way of doing it without this standard theorem? The book I'm using (Artin) mentions the word nilradical only four times (once in the index) much less the theorem you are referring to. I think I am supposed to do the exercise without it (if this is possible?). – Carolus Nov 15 '12 at 18:36
@Carolus: I've edited my answer, adding another method. – Fredrik Meyer Nov 15 '12 at 18:58
It seems I have some more reading to do, I haven't covered PID's yet :) I will probably return to this question in a day or so. Thanks for the pointers though! – Carolus Nov 15 '12 at 19:03
A PID is just a commutative ring where every ideal is generated by one element (some author demand that it is a domain, i.e. has no zero divisors). So for example $\mathbb{Z}$ and all its quotiens are PID's. – Fredrik Meyer Nov 15 '12 at 19:33