I've been looking over past exam papers and I came across a question I could not solve , it says ;
Consider $F:=GF(17^4),$
i)if $\alpha\in F$ what are the possible orders of $\Bbb Z_{17}(\alpha)$
ii) If $\alpha \in F$ is a root of the irreducible polynomial $x^4-5$ over $\Bbb Z_{17}$ what are the other 3 roots.
I'm really not sure how to approach the question though . I have soe inklings but I don't know if they will prove fruitful or not, my thughts are ;
i) if $\alpha \in F$ , then the order of the element should divide the order of the group $F^*$. So the possible orders of $\alpha$ are 1,2,3,4,5,6,8,9,10 etc.... but it seems like there are two many possible orders. So the second thing I think is that perhaps because $\Bbb Z_{17}$ has order 17 and F has order $17^4$, that theres some way to calculate this using GF(17) instead ?
ii)check all elements of $GF(17^4)$ but again there are far oo many to do this
Could anyone please help ?
$\mathbf F_{17^4}$, as a vector space over $\mathbf F_{17}$ has degree $4$, and $\mathbf F_{17}(\alpha)$ has degree a divisor of $4$. So its order can be $17$, $17^2$ or $17^4$.
Hint for ii): What are the fourth roots of $1$ in $\mathbf F_{17}$?