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I am thinking about this question: Find all the conjugates of primitive $35th$ root of unity over $\mathbb{F_{13}}$

I nearly finished it, but I was stuck in the last step.

Let $K$ be the splitting field of the polynomial $x^{28}+x^{21}+x^{14}+x^{7}+1$ over $\mathbb{F_{13}}$. Notice that $x^{28}+x^{21}+x^{14}+x^{7}+1=\frac{x^{35}-1}{x^{7}-1}$, so this will give us the primitive $35th$ root of unity. Denote primitive $35th$ root of unity to be $\zeta$.

Now, consider $[K:\mathbb{F_{13}}]=d$. Then by Lagrange, since we have a subgroup of $K^{*}$ of order 35, we have $35||K^{*}|$, but $|K^{*}|=13^{d}-1$, so we have $35|13^{d}-1$. Equivalently, we have $13^{d}=1$$(mod35)$.

By several trying, we could get the smallest $d$ is $d=4$. We will choose this smallest one since the splitting field of the polynomial is the smallest field in which it splits.

Then, we have $K=\mathbb{F_{13^{4}}}$, and $[K:\mathbb{F_{13}}]=4$. The converse of Lagrange holds in cyclic groups, so that means that there is an element of order $35$ in $\mathbb{F_{13^{4}}}$, and hence $\zeta$. After we get $\zeta$, we will get all the root of the polynomial.

Thus, $Gal(\mathbb{F_{13^{4}}}/\mathbb{F_{13}})\cong\mathbb{Z_{4}}$, since every Galois extension over a finite field is cyclic, and the degree of the extension is 4.

Now, we could consider the conjugates. So since this is an extension over finite field, it should be not that complicated. We have identity map $id:\zeta\rightarrow\zeta$, and we have the Frobinus Map $\phi$ with order 4. In other word, $\phi^{4}(\zeta)=\zeta^{i^{4}}=\zeta$. Thus $i^{4}=1 (mod35)$.

This will give us, $i=6,8,13,22,27,29,34$.

Note that we cannot use $i=6,29,34$ at the beginning of the map, since by some calculation you will see that if $\phi(\zeta)=\zeta^{i}$ where $i=6,29,34$, then $\phi$ will be only of order $2$.

Thus, we have following four conditions.

$A:\zeta\rightarrow\zeta^{8}\rightarrow\zeta^{29}\rightarrow\zeta^{22}\rightarrow\zeta$

$B:\zeta\rightarrow\zeta^{13}\rightarrow\zeta^{29}\rightarrow\zeta^{27}\rightarrow\zeta$

$C:\zeta\rightarrow\zeta^{22}\rightarrow\zeta^{29}\rightarrow\zeta^{8}\rightarrow\zeta$

$D:\zeta\rightarrow\zeta^{27}\rightarrow\zeta^{29}\rightarrow\zeta^{13}\rightarrow\zeta$

Notice that $A$ and $C$ will give us same conjugates of $\zeta$, so do $B$ and $D$.

Thus, we have two possibilities of the conjugates of $\zeta$ in the splitting field.

$\zeta,\zeta^{8},\zeta^{29},\zeta^{22}$ or $\zeta,\zeta^{13},\zeta^{29},\zeta^{27}$

I am stuck here. Is that possible to eliminate one of those two case? or it is true that we can have two possibilities and that's it?

I have one more question, is there any other way to find all the conjugates? I believe it is necessary to get the Galois group, but is there any other way to figure out the conjugates? Since I am also finding the conjugates of $\zeta$ over $\mathbb{F_{17}}$, the Galois group is $\mathbb{Z_{12}}$ if I am not wrong. Finally, it will involve $i^{12}=1(mod35)$, which is really irritating.

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    $\begingroup$ Shouldn't we have to divide it further by $\frac{x^5-1}{x-1}$? $\endgroup$ – Berci Apr 24 '18 at 21:27
  • $\begingroup$ @Berci That's okay since if you keep doing this, you will get finally it is $\phi_{5}(x)\phi_{35}(x)$, where $\phi_{n}(x)$ is the cyclotomic polynomial, and the root of $\phi_{35}(x)$ generates everything. $\endgroup$ – JacobsonRadical Apr 24 '18 at 21:29
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All that’s missing from your argument is the realization that for an algebraic extension of $\Bbb F_q$, the Galois group is generated by Frobenius, $z\mapsto z^q$, valid for all $z$ in that algebraic extension.

In particular, since here $q=13$, the conjugates of your primitive $35$-th root of unity $\zeta$ are $\{\zeta,\zeta^{13},\zeta^{169}=\zeta^{29},\zeta^{2197}=\zeta^{27}\}$ .

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  • $\begingroup$ Oh!!!!! Thank you so much! $\endgroup$ – JacobsonRadical Apr 25 '18 at 4:39
  • $\begingroup$ In case you’re interested, I found, after much floundering, a quartic polynomial whose roots are a primitive $35$-th root of unity and its conjugates. I got $1 + 9x + 7x^2 + 11x^3 + x^4$ . $\endgroup$ – Lubin Apr 25 '18 at 4:43
  • $\begingroup$ Yea. Interesting! Thank you so much!! $\endgroup$ – JacobsonRadical Apr 25 '18 at 5:15

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