# Nil radical of $(n)$ in $\mathbb{Z}$

I did not understand the following proof, from time to time the questions will be inserted at the critical point.

In the ring $$\mathbb{Z}$$, let us show that if $$n=p_1^{k_1}p_2^{k_2}\cdots p_r^{k_r}$$ is a factorization of the positive integer $$n\ne 1$$ into distinct primes $$p_j$$, then $$\sqrt(n)=(p_1\cdots p_r).$$

Indeed, if the integer $$a=p_1\cdots p_r$$ and $$k=\max\{k_1,k_2\dots,k_r\}$$, then we have $$a^k\in (n)$$ (Why?); this makes clear that $$(p_1p_2\cdots p_r)\subseteq \sqrt (n)$$ (Why?).

On the other hand, if some positive integral power of the integer $$m$$ is divisible by $$n$$ (that is, if $$m\in\sqrt (n)$$), then $$m$$ itself must be divisible by each of the primes $$p_1, p_2,\dots, p_r$$ (why?) and, hence, a member of the ideal $$(p_1)\cap (p_2)\cap\cdots\cap(p_r)=(p_1p_2\cdots p_r)\quad\text{Why?}.$$

Remember the definition of $$\sqrt{(n)}:=\{r\in \Bbb Z : r^s\in (n), s\in \Bbb Z (s\quad \text{varies with}\quad r\}.$$To show that $$(p_1p_2\cdots p_r)\subseteq \sqrt{(n)}$$ you can prove that $$p_1p_2\cdots p_r\in \sqrt{(n)}$$ and this follows by definition of $$\sqrt{(n)}$$. On the other hand, if $$m\in \sqrt {(n)}$$ then exist $$s\in \Bbb Z$$ such that $$m^s\in (n)$$. This implies that $$n | m$$ , $$p_1 |m \cdots p_r | m$$ and therefore $$m\in (p_1)\cap\cdots\cap (p_r)$$. I believe that you can prove the last “small proposition”.