Proof of product identities.

Question

The following Lemma is from P.M. Cohn: Algebra Volume 1 (2nd ed.), 1982

Given $a, b_1, ..., b_r\in\Bbb Z$ if $a$ is coprime to each of $b_1,...,b_r$, then $a$ is coprime to $b_1...b_r$.

Proof. By Bezout's identity, $ax_i + b_iy_i=1$ for some integers $x_i,\ y_i$, hence $b_1...b_ry_1...y_r\ =\ (1-ax_1)...(1-ax_r)$, but this is of the form $1-a\mathit z$, so a is coprime to $b_1...b_r$, as we had to show.

How do you prove that $\displaystyle \prod_{j=1}^r (1-ax_j)$ is of the form $1-a\mathit z$ ?

Answer 1

$1-az$ means that the expected answer is $1\pmod a$. So we have to prove that:

$$\prod_{j=1}^r (1-ax_j)\equiv 1\pmod a$$

which, because $a, x_i\in\mathbb{Z}$, is true.

Answer 2

Well, $\prod_j (1-ax_j) = 1 - az$ since if you multiply out, then you get $1$ by picking $1$ in each of the terms $1-ax_j$. In each other case, you have to pick at least one term $-ax_j$.

Answer 3

The claim is trivial if $r=1$. It's also true if $r=2$ because$$(1-ax_1)(1-ax_2)=1-a(x_1+x_2-ax_1x_2).$$Now we'll induct on $r$. If it works for $r=k\ge 2$, some integer $X$ satisfies$$\prod_{j=1}^k(1-ax_j)=1-aX\implies\prod_{j=1}^{k+1}(1-ax_j)=(1-aX)(1-ax_{k+1}),$$so the $r=2$ case completes the inductive step.