if $gcd(x,y) = 1$ and $xa = yb$ then $xa = lcm(a,b)$

Question

The problem says: Let $D$ an UFD and $a,b,x,y \in D$ all distinct than 0. If $gcd(x,y)$ = 1 and $xa = yb$ then $xa = lcm(a,b)$.

I have tried to use a previous result that I proved saying:

If $D$ is a UFD and $a,b,d \in D, ab \ne 0$ such $gcd(d,a) = 1$ then $d|ab \implies d|b$

But I haven't reach anything non-trivial.

Answer 1

I don't remember if this formula is valid in all UFD's: $ab=\gcd(a, b) \mathrm{lcm}(a,b)$, but if that's the case, then: $ax\gcd(a, b) = a \gcd(xa, xb) = a \gcd(yb, xb) = a b \gcd(x, y) = ab = \gcd(a, b) \mathrm{lcm}(a,b)$ from where you get your result canceling out the gcd.

Answer 2

Apply your previous result, since $x \mid yb$ and $y \mid xa$, we have $x \mid b$ and $y \mid a$. Combining with the assumption that $xa=yb$, one can conclude that $a=ym$ and $b=xm$ for some nonzero $m \in D$. Thus, $lcm(a,b)=xym=xa$.

Answer 3

By FTA

$$(x,y)=1 \iff x=\prod p_i^{\alpha_i} \quad y=\prod q_i^{\beta_i}$$

thus

$$xa=yb \iff a=k\prod q_i^{\beta_i} \quad b=k\prod p_i^{\alpha_i} $$

therefore

$$lcm(a,b)=k\prod p_i^{\alpha_i}\prod q_i^{\beta_i}=xa$$