Coprimality with a product of integers
If and are such that is separately coprime with each , then is coprime with their product .
Proof: Through Bézout. being coprime with each , we can write, for each :
If we take the product of each of these expressions, we get:
Among the terms of this product, all but one contain as a factor. The one that doesn’t have as a factor is . Hence we obtain an expression of the form:
Hence and are coprime.
In the case of polynomials
The same result holds with polynomials: if and are polynomials over a field , such that is coprime with each , then is coprime with their product .
The proof is identical.