Proving that $\operatorname {rad}(n) = (p_1 \dots p_l).$

Question

Here is the problem I am trying to solve:

Let $n \in \mathbb N = \{1,2, \dots\}, n = p_1^{a_1} \dots p_l^{a_l}, \, a_i \geq 1 \, , \, p_1 < p_2 < \dots ,$ so $\operatorname{rad}(n) = (p_1 \dots p_l) \,.$

Where $\operatorname{rad}(n)$ is the radical of the ideal $(n)$ which is defined as $$\{x \in A: x^k \in (n) \textbf{ for some } k > 0\}.$$

I do not know how to tackle this problem actually, I know how to tackle it if $n = 2 ^k$ but in the above general case I do not know what to do.

In case of $n = 2^k,$ we know that $\operatorname{rad}(n) = 2$ so we proceeded by showing that if you have a radical, then you have an even number and conversely if it is a radical then it is an even number. But this prove depended on knowing the prime itself.

Could someone help me please?

Answer 1

From the problem statement, you know that the answer should be rad$(n) = (p_1\cdots p_l)$ (by the way, no commas inside the RHS, we must multiply all of the primes together). This means we must show that given $m \in \Bbb{Z}$, we have $n | m^k$ for some $k \geq 0$ iff $p_1,\cdots,p_l | m$. I think the simplest approach is to just follow through the definitions.

Suppose $m \in \text{rad}(n)$. Then $n|m^k$ for some $k\geq 0$. Since $p_i | n$ for each $1 \leq i \leq l$, we have $p_i | m^k$, hence $p_i | m$ by Euclid's Lemma. As the $p_i$ are coprime, we have $p_1\cdots p_l | m$. Therefore $m \in (p_1\cdots p_l)$.

Now suppose $m \in (p_1\cdots p_l)$. Then $m = cp_1\cdots p_l$ for some $c \in \Bbb{Z}$. Let $a = \max\{a_1, \cdots, a_l\}$. Then we see $$ m^a = (cp_1\cdots p_l)^a = c^ap_1^a\cdots p_l^a = \left(c^ap_1^{a-a_1}\cdots p_l^{a-a_l}\right)\cdot\left(p_1^{a_1}\cdots p_l^{a_l}\right) = \left(c^ap_1^{a-a_1}\cdots p_l^{a-a_l}\right)n. $$ Thus $n | m^a$ and so $m \in \text{rad}(n)$ as desired.

There is a reason why the "radical" is called what it is. You might of noticed that taking the radical of $(n)$ essentially killed the powers on the prime factors of $n$. This is because we have to "power up" to get inside of rad$(n)$. This result is not unique to the integers either--it applies to any nonzero proper principal ideal in any unique factorization domain (UFD).