To be specific, my question is about specializations $\beta \colon B_n \to GL_\text{n}\left( \mathbb C \right)$ of the unreduced Burau representation given by \begin{array}{cr} \beta \left( \sigma_{i} \right)= I_{i-1}\oplus\begin{pmatrix} 1-z & z \\ 1 & 0 \end{pmatrix}\oplus I_{n-i-1} & 1\leq i\leq{n-1},~z\in \mathbb C^{*} \end{array} where $B_n$ is the braid group on n strands and $\{\sigma_1,...,\sigma_{n-1}\}$ are the standard generators.
Is this representation completely reducible or not?
I think that the answer is no, but I can't find an argument to prove it. I'm first and foremost interested in just knowing the answer, and where to find a reference of it. Secondly I would like to be able to prove it, if possible without use of homological arguments and the definition of the braid group as a mapping class group, as I'm not altogether very familiar with that theory.
In the book "Braid groups" by Kassel and Turaev it is shown that in a suitable basis the representation can be written as \begin{array}{c} \beta ( \sigma_{i} )=\begin{pmatrix} \hat{\beta} (\sigma_{i}) & 0 \\ \star_{i} & 1 \end{pmatrix} \end{array} where $\hat{\beta}$ is the reduced Burau representation and $\star_{i}$ is a row of length $n-1$ equal to zero if $i=1,...,n-2$ and equal to $(0,...,0,1)$ if $i=n-1$. Also, $n$ should be greater than or equal to $3$.
It also says in the book that "the Burau representation can be reduced but is not a direct sum of its reduced form with a one-dimensional representation", but I'm unsure whether it is meant that this is true in general, or if it's somehow specific for the case above. In either case, I'm looking for a proof showing that $\beta$ is, or is not, completely reducible.
Thanks in advance for any help with this.
When $n=2$, the braid group $B_n$ is just the free group on one generator $\sigma$, and the representation is determined by the matrix via which $\sigma$ acts on $\mathbf{C}^2$; for $z=-1$ this is $$\sigma \mapsto \left( \begin{matrix} 2 & -1 \\ 1 & 0 \end{matrix} \right). $$ This matrix is conjugate to $$\left( \begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix} \right),$$ and therefore the representation is not completely reducible: its unique submodule is the (genuine) eigenspace for the eigenvalue $1$, of dimension $1$.
The general fact here is that the quotient of the Burau representation by the span of the all $1$'s vector is also known as the Specht module (indexed by the partition $(n-1,1,0,\dots,0)$, deforming the reflection representation) for the finite Hecke algebra of $S_n$ . This is irreducible unless the parameter $z$ is a non-trivial $n$th root of $1$, in which case it is reducible and it is not semisimple.
This material is covered for instance in Andrew Mathas' book here
http://www.ams.org/bookstore-getitem/item=ULECT-15