Proof $\gcd(b,c)=1$ and $a \mid b \implies \gcd(a,c)=1$

Question

Proof $\gcd(b,c)=1$ and $a \mid b \implies \gcd(a,c)=1$. Here are my thoughts. Let the $\gcd(b,c)=1$. So $1=bx+cy$ and $c,y$ are in $\mathbb{Z}$. So $1=a(fx)+c(y)$ is true. So the $\gcd(a,c)=1$. I feel like my proof isn't very good and might not actually prove anything. Any help would be great!

Answer 1

Your proof is basically fine. However, there are three relatively small changes I suggest:

1. State you're using Bézout's identity to get that $1 = bx + cy$ for some integers $x$ and $y$.

2. Correct a typo where you should write $x, y$ are in $\mathbb{Z}$, instead of $c, y$ being in $\mathbb{Z}$.

3. Explicitly state where $f$ comes from. In particular, you should say something like that since $a \mid b$, then $b = af$ for some integer $f$.

Note there are alternate ways to prove this, with a few being somewhat simpler, such as what lulu's comment indicates.

Answer 2

That idea works, and it actually shows $\,(a,c)\mid (a\bar a,c\bar c),\,$ which we can also show directly, namely

$$(a,c)\mid a\bar a,c\bar c\,\Rightarrow\, (a,c)\mid (a\bar a,c\bar c)$$

by using the GCD Universal Property $\ d\mid m,n\iff d\mid (m,n).\, $ This proof is more general since it also works in UFDs that don't have gcd Bezout equations, e.g. polynomial rings like $\,\Bbb Z[x],\ \Bbb Q[x,y]$.

Remark $ $ Note that the proof that you gave using Bezout is simply a special case of one direction of the linked proof of the GCD Universal Property. It is this property (not Bezout) that is more at the heart of the matter regarding gcds and divisor theory so it is well-worth mastering,