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: 
We have:

$$\begin{align}
t^9-t&=t(t^8-1)\\&=t(t^4+1)(t^4-1)\\&=t(t^4+1)(t^2+1)(t^2-1)\\&=t(t^4+1)(t^2+1)(t+1)(t-1)\end{align}$$

It is easy to see (right?) that $t^4+1$ and $t^2+1$ have no linear factors in $\Bbb F\_3$. So it remains to try and factorise $t^4+1$ into two quadratic terms:
$$t^4+1=(t^2+at+b)(t^2+ct+d)$$

By equating coefficients modulo $3$, you get four equations in $a,b,c,d$ which are easily seen to be soluble. Try it and see!
Hint: factor it in stages
$$t^9-t$$
$$t(t^8-1)$$
$$t(t^4+1)(t^4-1)$$
$$t(t^4+1)(t^2-1)(t^2+1)$$
$$t(t-1)(t+1)(t^2+1)(t^4+1)$$