# The natural numbers form a distributive lattice under gcd and lcm. In arbitrary gcd domains, does gcd distribute over lcm?

Basically what it says in the title. If $A$ is a $\operatorname{gcd}$ domain, for any $x, y, z \in A$, does this identity hold? $$\operatorname{gcd}(x, \operatorname{lcm}(y,z)) = \operatorname{lcm}(\operatorname{gcd}(x,y),\operatorname{gcd}(x,z))$$

• Yes, it does. See the "distributive lattice" claim in en.wikipedia.org/wiki/GCD_domain . It can be derived from Proposition 2.3 of ncatlab.org/nlab/show/distributive+lattice , because if $x, y, z$ are three nonzero elements of a gcd domain satisfying $\gcd\left(x,y\right) = \gcd\left(x,z\right)$ and $\operatorname{lcm}\left(x,y\right) = \operatorname{lcm}\left(x,z\right)$, then $xy = xz$. (All equalities here are understood as equalities up to units.) – darij grinberg Mar 6 at 22:14
• Also, I think that Bill Dubuque's method from math.stackexchange.com/questions/2161615/… (with $\gcd$ and $\operatorname{lcm}$ interchanged) works in every gcd domain. (I don't know enough about gcd domains to check.) – darij grinberg Mar 6 at 22:16