It is known that Braid groups are linear. I am a bit confused while understanding its linear representation.
We say that a group $G$ has linear representation if the map $\alpha: G \rightarrow GL(n,F) $ is faithful, where $GL(n, F)$ is general linear group over the field $F$. But then in the Lawrence-Krammer representation $K : B_n \rightarrow GL(m, \mathbb{Z}[t^{\pm1}, q^{\pm1}])$, $\mathbb{Z}[t^{\pm1}, q^{\pm1}]$ is not a field.
By Krammer's article Braid groups are linear, it can be seen that the representation of the author's interest is $ρ: B_n → GL(V )$, where $V$ is an $m$-dimensional free module over some ring $R$, with $m = n(n − 1)/2$.
So, are Braid groups linear with respect to $GL(m,F)$ or $GL(V)$?
They both are the same. The link (https://groupprops.subwiki.org/wiki/Linear_representation) clarifies the representation of $G$ over $F$ as the left $F[G]$-module, where $F[G]$ represents a group ring. Further, representations of group ring at (https://en.wikipedia.org/wiki/Group_ring#Examples) shows that there is no particular reason for $R$ to be a field here.