Linear Algebra: Diagonalisability 
Problem
Let A be the matrix
$\Bigg(\begin{matrix} 0&0&1\\ 1&0&0 \\0&1&0 \end{matrix} \Bigg)$
Giving brief justifications, determine whether A is diagonalizable
  over (a) the complex field; (b) the field $Z_2$ with two elements; (c)
  the field $Z_3$ with three elements; (d) the field $Z_7$ with 7
  elements.

Progress
The characteristic polynomial $\chi_A(x)=x^3-1$, which in this case is equal to the minimum polynomial, $m_A(x)$. We make use of the fact that A is diagonalisable $\Leftrightarrow$ $m_A(x)$ can be expressed as the product of disjoint linear factors.
(a) In $\mathbb{C}$, $m_A(x)=x^3-1=(x-1)(x+\alpha)(x-\beta)$, where $\alpha \neq \beta$ and so A is diagonalisable.
Not sure how to apply the argument to the fields $\mathbb{Z}_n$ for $n=2,3,7$ however. Any help would be appreciated. Regards.
 A: Your minimal polynomial $x^3-1$ factors $x^3-1=(x-1)(x^2+x+1)$ over any field.  The second factor $x^2+x+1$ certainly splits over $\mathbb C$ (roots $-\frac12\pm\frac{\sqrt3}2\mathbb i$), over $\mathbb Z/2\mathbb Z$ it is irreducible, over $\mathbb Z/3\mathbb Z$ it splits $(x-1)^2$, and over $\mathbb Z/7\mathbb Z$ it splits $(x-2)(x-4)$. So the matrix is diagonalizable over $\mathbb C$ and over $\mathbb Z/7\mathbb Z$ (three distinct roots in both cases), not over the other two fields. It does diagonalize over the extension field $\mathbb F_4$ over $\mathbb Z/2\mathbb Z$ where $x^2+x+1$ has two distinct roots, but not over any extension field of $\mathbb Z/3\mathbb Z$ since the minimal polynomial $x^3-1=(x-1)^3$ has a triple root there.
A: Over the complex numbers and over $\mathbb{Z}_7$, the characteristic polynomial has three distinct roots, so our matrix is diagonalizable.
Over $\mathbb{Z}_2$, the only eigenvalue is $1$. A short computation shows that the only eigenvector is $(1,1,1)^T$, so the matrix is not diagonalizable.  Over $\mathbb{Z}_3$, again the only eigenvalue is $1$, and again the eigenspace is $1$-dimensional.  Indeed over any field $K$, the eigenspace for any fixed eigenvalue is $1$-dimensional.
