Coprimality with a product of integers

olivierd

Aug 30, 2018 2 min read

Coprimality with a product of integers

The theorem

If $n > 0$ and $a, b_1, \ldots b_n \in \Z$ are such that $a$ is separately coprime with each $b_i$ , then $a$ is coprime with their product $b_1 b_2 \ldots b_n$ .

The proof

Proof: Through Bézout. $a$ being coprime with each $b_i$ , we can write, for each $i = 1, \ldots n$ :

$p_i a + q_i b_i = 1$

If we take the product of each of these expressions, we get:

$\prod_{i = 1}^n (p_i a + q_i b_i) = 1$

Among the $2^n$ terms of this product, all but one contain $a$ as a factor. The one that doesn’t have $a$ as a factor is $q_1 b_1 q_2 b_2 \ldots q_n b_n$ . Hence we obtain an expression of the form:

$(\ldots) a + (q_1 \ldots q_n) (b_1 \ldots b_n) = 1$

Hence $a$ and $b_1 b_2 \ldots b_n$ are coprime. $\blacksquare$

In the case of polynomials

The same result holds with polynomials: if $P$ and $Q_1, \ldots Q_n$ are polynomials over a field $\K$ , such that $P$ is coprime with each $Q_i$ , then $P$ is coprime with their product $Q_1 Q_2 \ldots Q_n$ .

The proof is identical.