Computing primitive roots of unity in a Finite Field extension

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 sure I understand what you’re looking for. You say, “…first compute the primitive $12$-th roots of unity in $\Bbb F_{25}$.” How were you thinking of doing this, other than by the method you already used above that? – Lubin Dec 23 '18 at 21:54
• $x^2-5$ is not irreducible in $\bf F_5[x]$ since it is the same as $x^5$. – Bernard Dec 23 '18 at 21:56
• Made a terrible mistake, it was $2$, not $5$. Corrected – Naweed G. Seldon Dec 23 '18 at 21:56
• @Lubin I'm not sure, I was hoping there was an alternate method – Naweed G. Seldon Dec 23 '18 at 21:57
• We can find a generator $\alpha$ for $\mathbb F_{5^2}$. Then all powers of $\alpha$ that are co-prime with 24 are 24th roots of unity. Their squares are the 12th roots of unity, – Klaas van Aarsen Dec 23 '18 at 22:55

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 $$\zeta_4\zeta_3=2\cdot\left(\frac{-1+\sqrt2}2\right)=-1+\sqrt2$$ must have order twelve.

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

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 $$\zeta_8=\cos\frac\pi4+i\sin\frac\pi4=\frac{1+i}{\sqrt2}$$ as a model.