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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? $$\operatorname{gcd}(x, \operatorname{lcm}(y,z)) = \operatorname{lcm}(\operatorname{gcd}(x,y),\operatorname{gcd}(x,z))$$

Vik78
  • 3,877
  • Yes, it does. See the "distributive lattice" claim in https://en.wikipedia.org/wiki/GCD_domain . It can be derived from Proposition 2.3 of https://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 06 '19 at 22:14
  • Also, I think that Bill Dubuque's method from https://math.stackexchange.com/questions/2161615/about-a-lcm-and-gcd-formula (with $\gcd$ and $\operatorname{lcm}$ interchanged) works in every gcd domain. (I don't know enough about gcd domains to check.) – darij grinberg Mar 06 '19 at 22:16

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