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The set of all nilpotent element is an ideal of R

An element $a$ of a ring $R$ is nilpotent if $a^n = 0$ for some positive integer $n$. Let $R$ be a commutative ring, and let $N$ be the set of all nilpotent elements of $R$.

(a) I'm trying to show that $N$ is an ideal, and that the only nilpotent element of $R/N$ is the zero element.

(c) What are the nilpotent elements of $R = \mathbb{Z}_{24}$? And what is the quotient ring $R/N$ in that case? (what known ring is it isomorphic to?)

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marked as duplicate by azarel, Benjamin Lim, Kannappan Sampath, Gerry Myerson, Asaf Karagila Apr 16 '12 at 6:20

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

I just finished giving hints on why the set is an ideal.

If $a+N$ is nilpotent in $R/N$, then $(a+N)^n = a^n+N = 0+N$ for some $n\gt 0$. Therefore, $a^n\in N$, hence $a^n$ is nilpotent, so there exists $k\gt 0$ such that $(a^n)^k = a^{nk}=0$. Therefore, $a$ is nilpotent, so $a\in N$, hence $a+N = 0+N$.

(The ideal generated by all nilpotent elements in a commutative ring is a radical ideal: if $x^n\in N$ for some $n\gt 0$, then $x\in N$).

An element $a + 24\mathbb{Z}$ is nilpotent in $\mathbb{Z}_{24}$ if and only if there exists $n\gt 0$ such that $a^n+24\mathbb{Z}=0+24\mathbb{Z}$; that is, if and only if there exists $n\gt 0$ such that $24|a^n$. Since $24 = 2^3\times 3$, what do you need to be true of $n$ in order for $a^n$ to be a multiple of $24$?

Once you know what the set of nilpotent elements is, remember that any quotient of $\mathbb{Z}_{24}$ is a quotient of $\mathbb{Z}$ (by the Isomorphism Theorems), so it should be pretty easy to figure out what $R/N$ is in this case.

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The first question on why the set of all nilpotent elements in a commutative ring $R$ is an ideal (also called the nilradical of $R$, denoted by $\mathfrak{R}$) has already been answered numerous times on this site. I will tell you why $R/\mathfrak{R}$ has no nilpotent elements.

Suppose in the quotient we have an element $\overline{a}$ such that $\overline{a}^n = 0$. But then multiplication in the quotient is well defined so that $\overline{a}^n = \overline{a^n}$. This means that $a^n$ must be in $\mathfrak{R}$, so that there exists a $k$ such that $(a^n)^k =0$. But then this means that $a^{nk} = 0$ so that $a \in \mathfrak{R}$. In otherwords, in the quotient $\overline{a} = 0$ proving that the quotient ring $R/ \mathfrak{R}$ has no nilpotent elements.

For question (c) I think you can do it by bashing out the algebra, but let me tell you that $\Bbb{Z}_{24}$ is guaranteed to have nilpotent elements because $24$ is not square free (it's prime factorisation is $24 = 2^{3} \cdot 3$).

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