Krdytkyu

We have an irreducible polynomial $x^2 - 2 in mathbb{F}_5[x]$, and I have to find the primitive $12^{text{th}}$ roots of unity in $mathbb{F}_{5^2}$ and then compute their minimal polynomials over $mathbb{F}_5$, and then the factorisation of $Phi_{12}(x)$ in $mathbb{F}_5[x]$.

Now, I sort of went in the "reverse direction". We have $$Phi_{12}(x) = x^4 - x^2 + 1 implies overline{Phi}_{12}(x) = (x^2 - 2x - 1)(x^2 + 2x - 1) pmod{5}$$

Which is the factorisation of $Phi_{12}(x)$ in $mathbb{F}_5[x]$, and also clearly the minimal polynomials of the primitive $12^{text{th}}$ roots of unity over $mathbb{F}_5$

Now, clearly $mathbb{F}_{5^2}=mathbb{F}[x]/langle x^2 - 2 rangle=mathbb{F}_5[sqrt{2}]$. Therefore the roots of the factorised polynomails above are our primitive roots of unity, which are precisely: $1+ sqrt{2}, 1- sqrt{2}, -1- sqrt{2}, -1+ sqrt{2}$

Now, I was wondering if there's a way to go about the question in a "non-reverse fasion" i.e. first compute the primitive $12^{text{th}}$ roots of unity in $mathbb{F}_{5^2}$. Now, I understand we have $|mathbb{F}^{times}_{5^2}|=24$, and since $overline{zeta}_{12} neq 0 implies overline{zeta} _{12} in mathbb{F}^{times}_{5^2}$

Now how do I compute the primitive $12^{text{th}}$ roots of unity assuming that the only information I have at my disposal is the information provided in the paragraph above?

I'm not positive about what exactly you want to do, but the following may fit.

In $Bbb{F}_5$ the element $2$ is a primitive fourth root of unity. Let's record this fact in the form $zeta_4=2$, $zeta_4^4=1$.

Remember how in the field of complex numbers $zeta_3=(-1+sqrt{-3})/2$ is a primitive third root of unity? You should check that the relation $zeta_3^3=1$ only depends on that form of the number. Basically because it is a zero of $x^2+x+1=0$ (quadratic formula!), and $x^3-1=(x-1)(x^2+x+1)$. This means that if we can locate an element serving in the role of $sqrt{-3}$, then we can also find a third root of unity (barring the exceptional that the recipe yields $zeta_3=1$, but that happens only in characteristic three so does not concern us).

A point is that in $Bbb{F}_5$ we have $-3=2$. Therefore a square root of $2$ will also be a square root of $-3$. To get the extension $Bbb{F}_{25}$ you alread adjoined $sqrt2$. Just what the doctor ordered! We can use $zeta_3=(-1+sqrt2)/2$ as a primitive third root of unity. You are invited to verify the relation $zeta_3^3=1$ just to remove any doubts!

To finish off we recall the fact that if in an abelian group the element $a$ has order $m$, the element $b$ has order $n$, and $gcd(m,n)=1$, then the product $ab$ has order $mn$. In the present case we observe that $gcd(4,3)=1$, and thus may fully expect that
$$
zeta_4zeta_3=2cdotleft(frac{-1+sqrt2}2right)=-1+sqrt2
$$
must have order twelve.

You get the other primitive roots of order twelve by using different combinations of $zeta_4=pm2$ and $zeta_3=(-1pmsqrt{-3})/2$.

I leave it as an extra exercise to construct an eighth root of unity in the same way,
using the corresponding complex root of unity
$$zeta_8=cosfracpi4+isinfracpi4=frac{1+i}{sqrt2}$$
as a model.