Examples of nilpotent elements

An element x \in R, R a ring, is called nilpotent if x^m = 0 for some positive integer m.

  1. Show that if n = a^kb for some integers a,b, then \overline{ab} is nilpotent in \mathbb{Z}/(n).
  2. If a is an integer, show that the element \overline{a} \in \mathbb{Z}/(n) is nilpotent if and only if every prime divisor of n also divides a. Find the nilpotent elements of \mathbb{Z}/(72) explicitly.
  3. Let X be a nonempty set and F a field, and let R = {}^XF be the ring of functions X \rightarrow F. Prove that R contains no nonzero nilpotent elements.

  1. Suppose n = a^kb, where k \geq 1. Now (ab)^k = a^kb^k = (a^kb)b^{k-1} = nb^{k-1} \equiv 0 mod n.
  2. (\Rightarrow) Suppose \overline{a} \in \mathbb{Z}/(n) is nilpotent. Then a^m = nk for some m and k. Now if p is a prime dividing n, then p divides a^m, so that it divides a. Thus every prime dividing n divides a. (\Leftarrow) Let n = p_1^{e_1} \cdots p_k^{e_k} and a = p_1^{d_1} \cdots p_k^{d_k}m, where 1 \leq e_i, d_i for all i and m is some integer. Let t = \max \{ e_i \}. Then a^t = (p_1^{d_1} \cdots p_k^{d_k}m)^t = p_1^{d_it} \cdots p_k^{d_it}m^t, where d_it \geq e_i for each i. Thus a^t = nb for some integer b, and we have a^t \equiv 0 mod n.

    The nilpotent elements of \mathbb{Z}/(72), where 72 = 2^3 \cdot 3^2, are as follows: 0, 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 66.

  3. Suppose \alpha \in R is nilpotent. If \alpha \neq 0, then there exists x \in X such that \alpha(x) \neq 0. Let m be minimal such that \alpha(x)^m = 0; note that m \geq 1. Then \alpha(x) \alpha(x)^{m-1} = 0, where \alpha(x) \alpha(x)^{m-1} are not zero. Thus F contains zero divisors, a contradiction. Thus no nonzero element of R is nilpotent.
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  • curious  On February 7, 2011 at 10:19 pm

    Hello, don’t you mean $1 \leq e_{i} \leq d_{i}$ so that every prime divisor of $n$ divides $a$ ?

    • nbloomf  On February 7, 2011 at 10:41 pm

      Yes. Thanks!

  • Jason  On December 7, 2011 at 11:10 pm

    I think it’s not necessarily true that $1 \leq e_i \leq d_i$. For example, 6 is nilpotent in $Z/12Z$. However, I think the rest of the argument works without this assumption.

    Also a minor typo, I think you meant $a^t = (p_1^{d_1} \cdots p_k^{d_k}m)^t

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