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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.2012-11-15
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    Dear @Carolus, Do you know that the nilradical is exactly the set of nilpotents? Or the intersection of the prime ideals?2012-11-15
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    @KeenanKidwell: The set of nilpotents.2012-11-15

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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?).2012-11-15
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    @Carolus: I've edited my answer, adding another method.2012-11-15
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    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!2012-11-15
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    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.2012-11-15