Let $\mathfrak g$ be a Kac-Moody algebra (symmetric, or hyperbolic, or whatever other assumptions you need) with simple roots $\alpha\_i$. For $\alpha$ a root, write $\alpha$ in the basis of simple roots $\alpha = \sum\_i a\_i \alpha\_i$. I am interested in when the gcd of all the $a\_i$ is 1 (in which case I will say $\alpha$ is "coprime").


It seems that any real root of $\mathfrak g$ is coprime, at least for the cases that I've tested numerically -- namely, $E\_{10}$, $E\_{11}$, and $\begin{bmatrix}2&-3\\-3&2\end{bmatrix}$. More generally, the Weyl group seems to preserve gcd's in the sense that if $w$ is in the Weyl group and $w\alpha = \sum\_i b\_i\alpha\_i$, then the gcd of the $a\_i$ is the same as the gcd of the $b\_i$. Is this true?



Does the Weyl group preserve coprimality in Kac-Moody algebras?
Note that, in your notation, if $d\_i \in \mathbb Z$ are such that $\sum\_i d\_i a\_i = \gcd\_i a\_i$, then also $\sum\_{i \ne j} (d\_i + \langle\alpha\_j^\vee, \alpha\_i\rangle d\_j)a\_i + d\_j(a\_j - \sum\_i \langle\alpha\_j^\vee, \alpha\_i\rangle a\_i) = \sum\_i d\_i a\_i = \gcd\_i a\_i$, so that the simple reflexion $w$ in $\alpha\_j$ preserves gcd's. The general result follows.


