determine the radical $\sqrt{m\mathbb{Z}}$

Question

Task: determine the radical $\sqrt{m\mathbb{Z}}$.

Obviously $m\mathbb{Z} \subseteq \sqrt{m\mathbb{Z}}$.

But how to systematically determine the elements of $\sqrt{m\mathbb{Z}}$?

I guess it has something to do with prime factors to a power greater than one in the prime factorization of $m$.

Answer 1

Let's just follow the definition. For $k\in \mathbb Z$ we have $k \in \sqrt{(m)}$, if for some $N \ge 0$, $k^N \in (m)$, that is $m\mid k^N$. Now let $m = \prod_p p^{\nu_p(n)}$ be the prime decomposition of $m$ and $k = \prod_p p^{n_p(k)}$ that of $k$. $m \mid k^N$ now reads $N\nu_p(k) \ge \nu_p(n)$ for all primes $p$. This is trivially true for the primes with $\nu_p(n) = 0$, and can be fulfilled for all $p$ iff $\nu_p(k) > 0$ for all $p$ with $\nu_p(n)>0$. That is we have $k \in \sqrt{(m)}$ iff $k$ is divided by each prime factor of $m$, so $\sqrt{(m)}$ is generated by $k := \prod_{\nu_p(n) > 0} p$.

Answer 2

Do you know the characterization of $\sqrt{I}$ as being the intersection of prime ideals containing $I$? If not, it would make a great exercise for you to prove :)

Since you know that $(m)\subseteq(p)$ iff $p$ divides $m$, you have a complete list of prime ideals containing $(m)$. Then their intersection is just $(\prod_{p|m}p)$.