How to illustrate that

Question

there is $a, n \in \mathbb Z^{+}$ and prime number $p$, with relationship: $$p|a^{n}$$

It's straight forward that $p|a$, but I can't find a proper illustration of it.

Answer 1

Rewrite $a$ as the product of its prime factors:

• $a=\prod\limits_{i=1}^k{p_i}^{m}$

• $p|a^n \implies p\in\{p_1,\dots,p_k\} \implies p|a$

Answer 2

If $p$ does not divide $a$ then it does not divide $a^n$ which conclude the result.

Answer 3

Alternatively to the two other answers given, we can use the property that $p$ is prime to directly show this, without reference to the fundamental theorem of arithmetic.

Note that we begin with $p | (a a^{n-1})$. Since $p$ is prime, this implies that $p | a$ or $p | a^{n-1}$. In the former case, we're done. Otherwise, repeat with $p | a^{n-1}$. A simple inductive argument completes the proof.

Answer 4

Write: $a^n = a^n - a + a$. We know from Fermat's little theorem that: $p|a^n - a$, and $p|a^n$. It follows that $p|(a^n - (a^n - a))$, or $p|a$.

Answer 5

If you know about rings and ideals, you should use this to illustrate this fact.

(you may think, that this is abstract and therefore the wrong way, but believe me, here it will support your intuition)

So you know that $\mathbb{Z}$ is a prinicpal ideal domain. We can look at the ideals generated by $a$ and $p$. We know that in our ring, the prime ideals are precisely the maximal ideals. Therefore if the two ideals intersect, which they do since $(0)\neq(a^n)\subset (p)\cap (a)$, it holds that $(a)\subset (p)$. So as soon as an integer element "touches" a prime ideal, all multiples and roots (if they lie in $\mathbb{Z}$) are divisible by the prime spanning this ideal. (the latter fact with the roots uses that it is a pid, the other fact is obviously always true, but stresses what I want to say)

Hope this helps!