Question: I am trying to factor the polynomial $t^9-t$ in the field with three elements $0,1,2$. I think that the factors $(t)(t)(t)(t)(t)(t)(t)(t)(t-1)$ are all irreducible but for some reason this seemed too simple for a problem from the book I am working from. Any help or hints welcome!

Answer: 
Using the usual rules that apply in any field you can get down to $t^9-t=t(t^8-1)=t(t^4+1)(t^4-1)=t(t^4+1)(t^2+1)(t^2-1)=t(t^4+1)(t^2+1)(t+1)(t-1)$.

However, you aren't done at that point. While $t^2+1$ is irreducible ($\Bbb{F}\_3$ has no square root of $-1$), $t^4+1$ must factor into two quadratics over $\Bbb{F}\_3$ since the degree $2$ extension $\Bbb{F}\_9$ has fourth roots of $-1$. (All finite fields have cyclic multiplicative groups, so having $8$ non-zero elements means $\Bbb{F}\_9$ has primitive eighth roots of unity, and hence fourth roots of $-1$.) So the minimal polynomials for fourth roots of $-1$ have degree two and multiply together to get you $t^4+1$. So the hint is: factor $t^4+1$ into two quadratics over $\Bbb{F}\_3$.
Hint: factor it in stages
$$t^9-t$$
$$t(t^8-1)$$
$$t(t^4+1)(t^4-1)$$
$$t(t^4+1)(t^2-1)(t^2+1)$$
$$t(t-1)(t+1)(t^2+1)(t^4+1)$$