Set of Divisors of Integer

Theorem
Let $$n \in \Z: n > 1$$.

Let $$n$$ be expressed in its Prime Decomposition:

$$n = p_1^{k_1} p_2^{k_2} \ldots p_r^{k_r}$$ where $$p_1 < p_2 < \ldots < p_r$$ are distinct primes and $$k_1, k_2, \ldots, k_r$$ are positive integers.

The set of divisors of this integer can be found to be:

$$\left\{{p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}: 0 \le h_i \le k_i, i = 1, 2, \ldots, r}\right\}$$

Proof

 * Each integer in the given set is a divisor of $$n$$ because:


 * 1) $$\forall i: k_i - h_i \ge 0$$
 * 2) $$n = \left({p_1^{h_1} p_2^{h_2} \ldots p_r^{h_r}}\right) p_1^{k_1-h_1} p_2^{k_2-h_2} \ldots p_r^{k_r-h_r}$$

from Prime Decomposition Divisor. By the uniqueness of Prime Decomposition, these integers are distinct.


 * Now we need to show that the integers in this set are the only divisors of $$n$$.

Let $$d > 1$$ and let $$p \in \mathbb P: p \backslash d$$.

Now we need to show that $$\forall i: h_1 \le k_i$$.

First note that $$d \backslash n \Longrightarrow\forall i: p_i^{k_i} \backslash n$$.

Next, we know that all the primes $$p_i$$ are distinct, and therefore by Prime Not Divisor then Coprime:

$$p_1 \nmid p_2^{k_2} p_3^{k_3} \ldots p_r^{k_r} \Longrightarrow \gcd \left\{{p_1, p_2^{k_2} p_3^{k_3} \ldots p_r^{k_r}}\right\} = 1$$

So $$p_1^{h_1} \backslash n \Longrightarrow n = p_1^{k_1} \left({p_2^{k_2} p_3^{k_3} \ldots p_r^{k_r}}\right)$$ By Euclid's Lemma, $$p_1^{h_1} \backslash p_1^{k_1} \Longrightarrow h_1 \le k_1$$, and the same argument applies to each of the other prime factors of $$n$$.

The result follows.