If $(b,c) = 1$, then $(a,bc) = (a,b)(a,c)$

Question

I do not know how to solve the following:

Show that If $(b,c) = 1$, then $(a,bc) = (a,b)(a,c)$

Thanks.

Hi! Tell us what did you try! – DiegoMath Oct 17 '18 at 14:06 — DiegoMath, Oct 17 '18 at 14:06

Bill Dubuque · Answer 1 · 2018-10-17T14:13:51.977

2

Distributing: $\ (a,b)(a,c) = ((a,b)a,(a,b)c) = (aa,ab,ac,bc) = (a\!\!\!\!\!\overbrace{(a,\color{#c00}{b,c})}^{\large =1\ {\rm by}\ \color{#c00}{(b,c)=1}}\!\!\!\!\!,bc) = (a,bc)$

Remark $ $ We used only the GCD Distributive Law $\ x(y,z) = (xy,xz)\ $ and other universal gcd laws (associative, commutative) therefore the result holds true in any gcd domain (vs. proofs using Bezout or primes, which are not as general).

edited Oct 17 '18 at 14:13

answered Oct 17 '18 at 14:07

Bill Dubuque

272,048

Notice that the proof requires only the weaker hypothesis that $\ (a,b,c) = 1\ $ – Bill Dubuque Oct 17 '18 at 14:15

score 0 · Answer 2 · answered Oct 17 '18 at 14:08

0

Hint

Can you finish now?

answered Oct 17 '18 at 14:08

mfl

29,399

I don't think that division by primes is enough of a proof. Shouldn't you divide by $p^n$ for some $n$? – Jakobian Oct 17 '18 at 14:09
@Jakobian It is only a hint and not a complete answer. – mfl Oct 17 '18 at 14:12

If $(b,c) = 1$, then $(a,bc) = (a,b)(a,c)$

2 Answers2

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