A divisor of a product of integers is a product of divisors

Question

Generally speaking, if $m \in \mathbb{N}$ and $m=n_1\cdot n_2\cdot \cdots \cdot n_r$ and $k\mid m$ why do there exist $k_1\mid n_1,k_2 \mid n_2,...k_r\mid n_r$ with $k=k_1\cdot k_2 \cdots k_r$?

Answer 1

For what it's worth: a proof by prime factorization.

Proceed by strong induction on $k$. If $k = 1$, the statement is trivial.

Now, suppose that the statement holds for all $1 \leq k < K$. We have $K \mid n_1\cdots n_r$. $K$ must have a prime factor $p$. $p$ divides $K$ and therefore divides the product $n_1\cdots n_r$. However, if $p$ divides the product, then it must divide one of the factors; WLOG suppose $p \mid n_1$. We see that $\frac Kp$ divides $\frac {n_1}{p}n_2 \cdots n_r$ and that $\frac Kp < K$. By our inductive hypothesis, there exist $h_1,\dots,h_r$ with $$ \frac Kp = h_1 \cdots h_r, \qquad h_1 \mid \frac{n_1}p, \quad h_j \mid n_j \text{ for } j = 2,\dots,r. $$ It now suffices to take $k_1 = ph_1$ and $k_j = h_j$ for $j = 2,\dots,r$.

The conclusion follows.