How to prove that $\text{lcm}(zx, zy)=z \text{lcm}(x, y)$?

Question

In an integral domain $R$, when the lowest common multiple of $x$ and $y$ exists, is it true that the lowest common multiple of $zx$ and $zy$ also exists and $$ \text{lcm}(zx, zy)=z\:\text{lcm}(x, y) $$ holds?

Answer 1

You want to prove that

1. $z\operatorname{lcm}(x,y)$ is a multiple of $zx$ and $zy$;
2. if $a$ is a multiple of $zx$ and $zy$, then $a$ is a multiple of $z\operatorname{lcm}(x,y)$.

For simplicity, set $m=\operatorname{lcm}(x,y)$, which exists by assumption.

Let's prove 1. We know that $x\mid m$ and $y\mid m$; therefore $zx\mid zm$ and $zy\mid zm$.

Let's prove 2. Suppose $zx\mid a$ and $zy\mid a$, so $a=zxb$ and $a=zxc$. In particular, $a$ is divisible by $z$; if we set $a'=xb=yc$, then, by assumption, we have that $m\mid a'$. Therefore $zm\mid za'=a$.

Answer 2

Notice that $$lcm(a,b)=\dfrac{a\cdot b}{\gcd(a,b)}$$therefore$$lcm(zx,zy)=\dfrac{zx\cdot zy}{\gcd(zx,zy)}=\dfrac{zx\cdot zy}{z\gcd(x,y)}=\dfrac{zx\cdot y}{\gcd(x,y)}=z\cdot lcm(x,y)$$

Answer 3

The statement is tautological when $z=0$. Otherwise, since $R$ is an integral domain, for any $a,b\in R$ we have \begin{align*} a\mid b \quad\Longleftrightarrow\quad \exists r\colon ar=b \quad\Longleftrightarrow\quad \exists r\colon zar=zb \quad\Longleftrightarrow\quad za\mid zb. \end{align*} Now the statement follows from the definition of $\operatorname{lcm}$.