Let $K = \mathbb F_2$ and $V= K^3$, $\phi \in End(V)$
How many possible characteristic Polynomials exist?
My attempt:
Since its a 3x3 - Matrix and the diagonals are either $0$ or $1$ i get $8 $ different polynomials:
$ a_3(x-1)^3 +a_0\qquad ; \qquad a_3(x-1)^3 +a_1(x-1) +a_0$
$a_3x(x-1)^2 +a_0 \qquad ; \qquad a_3(x-1)^2x + a_1(x-1) +a_1x +d$
$ax^2(x-1)+b \qquad ; \qquad a_3x^2(x-1) + a_1x +a_1(x-1) +d$
$ax^3 +b \qquad ; \qquad a_3x^3 + a_1x$
Is that correct?
As each characteristic polynomial in this case is a monic polynomial of degree $3$, there can be at most $8$ polynomials, which can be found by looking for all different combinations of coefficients of $x^2$, $x^1$ and $x^0$. Each coefficient must be in $\mathbb{F}_2$, thus we have: $$ \begin{eqnarray*} x^3+0x^2+0x+0 &=& x^3\\ x^3+0x^2+0x+1 &=& x^3+1\\ x^3+0x^2+1x+0 &=& x^3+x\\ x^3+0x^2+1x+1 &=& x^3+x+1\\ x^3+1x^2+0x+0 &=& x^3+x^2\\ x^3+1x^2+0x+1 &=& x^3+x^2+1\\ x^3+1x^2+1x+0 &=& x^3+x^2+x\\ x^3+1x^2+1x+1 &=& x^3+x^2+x+1 \end{eqnarray*} $$ In order to show that there are exactly 8 polynomials, we now need to show that each of those polynomials can actually occur as a characteristic polynomial of a $3\times 3$-Matrix over $\mathbb{F}_2$. This is trivial, we only need to take the companion matrices of the polynomials.