How can I find the order of the elements of the following group $\mathbb F_5[x]/\langle x^2+2 \rangle$?
Let denote the above field with $F$ then its multiplicative group has $24$ elements (excluding $0$) and I have to find for every divisor of $24$ one element. Is there any element of order $24$ ? Then the problem could be much easier. The elements are of the form, $c_1+c_2x$ with $c_i\in\{0,1,2,3,4\}$, but I want to avoid nasty computations. Is there a shorter way to do that ?
On
Proffering an alternative route (a slight shortcut?) to the final stage along route in paw88789's (+1) answer, where it was observed that the element $\alpha=x+\langle x^2+2\rangle$ is of order eight. To find an element of order $24$ we, indeed, need an element of order three.
A primitive third root of unity? What does that remind us of again? Within the complex numbers we remember that the mystery element $$ \omega=\frac{-1+\sqrt{-3}}2 $$ serves in that role. Note that we actually only need a square root of $-3$ for this formula to give us what we want. Getting warmer! Unfortunately we just adjoined $\sqrt{-2}$ instead of $\sqrt{-3}$ :-( No, problem! Because $2^2=4\equiv-1$, we already have a $\sqrt{-1}$. Also $-3\equiv2$, so (with a bit of abuse of notation) $$ \sqrt{-3}=\sqrt{-1}\sqrt{3}=\sqrt{-1}\sqrt{-2}=2\alpha. $$ Therefore (drums, please) $$ \omega=\frac{-1+2\alpha}2=\frac{4+2\alpha}2=2+\alpha $$ is of order three.
Thoughts:
You can fairly easily calculate that the element $\overline{x} = x+\langle x^2+2\rangle$ generates a subgroup of order $8$ that consists of all the monomials.
Also in a cyclic group of order $24$, there should be $\varphi(24)=8$ generators of that group. So I would look for a binomial ($\overline{ax+b}$ where $a,b\in\{1,2,3,4\}$) that cubes to an element of the form $cx$ where $c\in\{1,2,3,4\}$. This should not take too long.