GCD in unique factorization domain

Question

Let be D a unique factorization domain, k in D and $x=kd $ with $d$ a gcd of $a, b$ and $y$ a gcd of $ka, kb$ prove that x and y are associates; x divides y and y divides x. My attempt:

x=kd and a=dr, b=ds hence ka=kdr, kb=kds so ka=xr, kb=xs hence x divides y because y is gcd of ka and kb.

How to prove that y divides x?

since $d=\gcd(a,b)$ there are $s$ and $t$ such that $d=sa+tb$; if $y=\gcd(ka,kb)$ , then $y$ divides $ka$ and $kb$ and therefore $ska+tkb=kd=x$ — J. W. Tanner, Jun 23 '19 at 20:22
Duplicate of Prove that $(ma, mb) = |m|(a, b)\ $ [GCD Distributive Law] — Bill Dubuque, Jun 23 '19 at 20:26

score 1 · Accepted Answer · answered Jun 23 '19 at 21:49

1

Hint $\,\ k\mid (ka,kb)\mid ka,kb\,\Rightarrow\, (ka,kb)/k\mid a,b\,\Rightarrow\,(ka,kb)/k\mid (a,b)\,\Rightarrow\,(ka,kb)\mid k(a,b)$

Remark $ $ In fact the proof can be done more efficienitly bidirectionally as here.

answered Jun 23 '19 at 21:49

Bill Dubuque

272,048

GCD in unique factorization domain

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