(a) Show that if \(n = a^k b\) for some integers \(a\) and \(b\) then \(\overline{ab}\) is a nilpotent element of \(\mathbb{Z}/n\mathbb{Z}\).
(b) If \(a \in \mathbb{Z}\), show that \(\overline{a}\) is nilpotent if and only if every prime divisor of \(n\) is also a divisor of \(a\). In particular, determine the nilpotent elements of \(\mathbb{Z}/72\mathbb{Z}\) explicitly.
(c) Let \(R\) be the ring of functions from a nonempty set \(X\) to a field \(F\). Prove that \(R\) contains no nonzero nilpotent elements.
(a) \(\overline{ab}^k = \overline{a^k b} \cdot \overline{b^{k-1}} = \overline{0} \cdot \overline{b^{k-1}} = \overline{0}\)
(b) Suppose \(\overline{a}\) is nilpotent; then \(a^m = kn\) for integers \(k\) and \(m\). We then have \(p \mid n \implies p \mid kn \implies p \mid a^m \implies p \mid a\). The last implication holds because \(p\) is prime (the reader can convince himself of this fact).
Suppose the converse is true; let \(p_1, \ldots, p_k\) be the prime divisors of \(n\) (which are also prime divisors of \(a\)), and \(p_{k+1}, \ldots, p_n\) be the prime divisors of \(a\) that do not divide \(n\). Then we have \(n = \prod_{i=1}^k {p_i}^{a_i}\) and \(a = \prod_{j=1}^n {p_j}^{b_j}\) for integers \(a_i\), \(b_j\). Notice that \(a^m = \prod_{j=1}^n {p_j}^{mb_j}\), and in fact we can choose \(m\) such that \(mb_i \ge a_i\) for all \(1 \le i \le k\). This implies \(n \mid a\) (we implicitly assumed that \(a\) is positive, but the proof clearly holds for \(a < 0\) as well with slight modifications).
The nilpotent elements of \(\mathbb{Z}/72\mathbb{Z}\) are precisely the elements \(\overline{6n}\) where \(n \in \mathbb{Z}\).
(c) A field \(F\) contains no zero divisors, so it follows that all nonzero elements are not nilpotent.
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