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 B├ęzout. 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.

Leave a Reply

Your email address will not be published. Required fields are marked *