Divisors of $z$ and $m$ compared to divisors of $z$ and $mn$, where $(m,n) = 1$.

Question

Suppose $(m,n)=1$. Then is it true that the common divisors of $z$ and $m$ are exactly the common divisors of $z$ and $mn$? If not, is it true that the second set is a subset of the first? This is probably a very simple question, but I cannot get my mind around it atm. Thanks!

Edit: We have $z<m$, $z<n$ and $(z,n)=1$

Did you mean to also assume that $\gcd(z,n)=1$? Otherwise, you should easily be able to find counterexamples. — lulu, Oct 11 '20 at 13:47
You don't need to make the assumptions regarding inequalities. Since $\gcd(z,n)=1$ we see that $\gcd(z,mn)=\gcd(z,m)$ (see, e.g., this question) and the desired claim follows at once. — lulu, Oct 11 '20 at 13:56

score 0 · Answer 1 · answered Oct 11 '20 at 22:25

0

Note $\ d\mid z,m\iff d\mid (z,m) := g\ $ by the gcd Universal Property

and $\ d\mid z,nm\iff d\mid (z,nm) = g (z/g,\, n\:\!m/g) = g(z/g,n),\, $ by $(z/g,m/g)=1\,$ & Euclid.

So they're equivalent $\iff (z/g,n) = 1,\,$ which is true when $(z,n)=1$.

answered Oct 11 '20 at 22:25

Bill Dubuque

272,048

Divisors of $z$ and $m$ compared to divisors of $z$ and $mn$, where $(m,n) = 1$.

1 Answers1