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

Question

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 $$(p_1)\cap (p_2)\cap\cdots\cap(p_r)=(p_1p_2\cdots p_r)\quad\text{Why?}.$$

Thanks for your patience!

Answer 1

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”.