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Let $E$ be the splitting field of $x^{35}-1$ over the field $\mathbb{F}_8$. Determine $|E|$ and the number of subfields of $E$.

Attempt: I am confident that I computed $|E|$ correctly, but I am not sure if I have correctly determined the number of subfields of $E$.

Since $\mathbb{F}_8=\mathbb{F}_{2^3}$, $\mathbb{F}_8$ has characteristic $2$. Denote by $I_n(x)$ the $n$th cyclotomic polynomial, and denote by $\varphi$ Euler's Totient. Then $$x^{35}-1=I_1(x)I_5(x)I_7(x)I_{35}(x)$$ where $I_1(x)=x-1, I_5(x)=x^4+x^3+x^2+x+1, I_7(x)=x^6+x^5+\cdots +x+1$, and $I_{35}(x)$ can be computed by recursion but I don't believe the formula is needed. The root $1$ of $I_1$ gives us no new elements; the roots of $I_5$ give $\varphi(5)=4$ nonreal elements (which cannot be in $\mathbb{F}_8$ since $\mathbb{F}_8$ has characteristic $2$); the roots of $I_7$ give $\varphi(7)=6$ new elements; the roots of $I_{35}$ give $\varphi(35)=\varphi(5)\varphi(7)=24$ new roots. No roots were counted twice since the $I_n$'s are distinct irreducible polynomials. So $|E|=8+4+6+24=42$, and there are $5$ subfields (including $\mathbb{F}_8, E$) where the 3 nontrivial subfields are given by including the roots of $I_5, I_7, I_{35}$. Is the last statement true, and can it be made more rigorous using Galois theory?

The Substitute
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  • The standard notation for the $n$th cyclotomic polynomial is $\Phi_n$. $\Phi_7$ and $\Phi_{35}$ are not irreducible in characteristic $2$. The factorisation of $x^{35}-1$ over $\mathbf{F}_{8}$ can be obtained in Pari/GP with t = ffgen(8, 't); factor(x^35 - t^0). – fkraiem Sep 05 '15 at 12:24
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    May be this thread helps you understand why/how cyclotomic polynomials factor further when they are reduced modulo two. The theme has been discussed in other related threads as well. Ask, if you nee more. – Jyrki Lahtonen Sep 05 '15 at 13:04
  • $E$ is a field containing $\Bbb{F}_8$, so it is automatically a vector space over $\Bbb{F}_8$. Therefore its order must be a power of $8$. I'm afraid it looks like you have some lingering holes in your understanding of what extending a field means. When you adjoin an element, say $\zeta$, you must also add all the elements $p(\zeta)$, where $p(x)$ is any polynomial from $\Bbb{F}_8[x]$. This is because the end result, the extended field, must be a field itself. It needs to be closed under addtion and multiplication (in thes case of an algebraic extension you get closure under division gratis). – Jyrki Lahtonen Sep 06 '15 at 05:07

1 Answers1

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Hints:

  • All the non-zero elements of $\Bbb{F}_8$ are seventh roots of unity, all but one of them of order seven.
  • The smallest field of characteristic two that contains a primitive fifth root of unity if $\Bbb{F}_{16}$. Can you see why that is the case? Think cyclic groups!
  • If a field contains a primitive fifth root of unity and a primitive seventh root of unity, then it also contains a primitive $35$th root of unity. Why?
  • In view of the previous bullet $E$ is the smallest field that contains both $\Bbb{F}_8$ and $\Bbb{F}_{16}$ (as well as the field you are working over, but that happens to be in the list already). Review what you know about inclusions of finite fields, and determine $E$.
Jyrki Lahtonen
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  • Why are the nonzero elements of $\mathbb{F}_8$ seventh roots of unity? I would assume that none of these complex 7th roots of unity are in $\mathbb{F}_8$ since $\mathbb{F}_8$ has characteristic 2. – The Substitute Sep 06 '15 at 01:53
  • Can you expand on the second bullet. I don't see how 5th roots of unity would be in a field of characteristic 2. – The Substitute Sep 06 '15 at 01:55
  • Well the element of $\Bbb{F}8$ obviously aren't complex. But because the multiplicative group of $\Bbb{F}_8$ has seven elements by Lagrange's theorem they are all (except $1$, of course) of order seven. Similarly, the multiplicative group of $\Bbb{F}{16}$ is cyclic of order fifteen, so it has four elements of order $5$. Those are still called roots of unity! – Jyrki Lahtonen Sep 06 '15 at 04:59
  • Thank you for clarifying. Using my knowledge of field extensions, I deduce $E=\mathbb{F}_{12}$ since $12=\text{lcm}(3,4)$, where the 3 & 4 come from $8=2^3, 16=2^4$. – The Substitute Sep 06 '15 at 09:53
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    Good, if you mean $\Bbb{F}_{2^{12}}$. The least common multiple of 3 and 4 is the key anyway. – Jyrki Lahtonen Sep 06 '15 at 12:21