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?

  • $\begingroup$ 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? $\endgroup$ – Lubin Dec 23 '18 at 21:54
  • $\begingroup$ $x^2-5$ is not irreducible in $\bf F_5[x]$ since it is the same as $x^5$. $\endgroup$ – Bernard Dec 23 '18 at 21:56
  • $\begingroup$ Made a terrible mistake, it was $2$, not $5$. Corrected $\endgroup$ – Naweed G. Seldon Dec 23 '18 at 21:56
  • $\begingroup$ @Lubin I'm not sure, I was hoping there was an alternate method $\endgroup$ – Naweed G. Seldon Dec 23 '18 at 21:57
  • $\begingroup$ 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, $\endgroup$ – 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.


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