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I'm uncertain if I am doing this problem correctly. This is an old algebra prelim problem regarding Galois theory. We have to find the Galois group of the irreducible polynomial (the problem already assumes so) $f(x)=x^5 + x^2 + 1$ over the field of two elements, i.e. $\mathbb{F}_{2}$.

My attempt at a solution:
Let $F$ be a splitting field of $f(x)= x^5 + x^2 + 1$ over $\mathbb{F}_{2}$. Now let $\alpha$ be a root of $f(x)$. Since any finite extension of $\mathbb{F}_{p}$ is Galois for $p$ prime, in this case we have $F= \mathbb{F}_{2}(\alpha)$. In the particular case of this problem, then the Galois extension $F/\mathbb{F}_{2}$ has degree $[F:\mathbb{F}_{2}]=5$ since $f(x)$ is irreducible of degree $5$ over $\mathbb{F}_{2}$. I want to say that the Galois group of $f(x)$ over $\mathbb{F}_{2}$ is cyclic of order $5$, but I don't think that's correct.

My reasoning:
If I ended up with the result that $\textit{Gal}(F/\mathbb{F}_{2}) \cong \mathbb{Z}_{5}$, that would have meant that $F$ was isomorphic to the splitting field of $x^{2^5}-x$, i.e. $\mathbb{F}_{2^5}$. Because of this, it would follow that $$ \textit{Gal}(F/\mathbb{F}_{2})= \langle \sigma_{2} \rangle \cong \mathbb{Z}_{5},$$ where $\sigma_{2}$ is the Frobenius automorphism for $p=2$. So my question is, is it true that $F= \mathbb{F}_{2^5}$, or is it isomorphic to it? Am I on the right track?

Libertron
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    Every finite extension of a finite field is a Galois extension, as you already observed, so $F/\mathbf F_5$ is Galois and its degree is $\deg f = 5$, so of course (?) the order of the Galois group is $5$. What groups have order $5$? – KCd Dec 31 '14 at 00:24
  • Cyclic groups of order $5$. – Libertron Dec 31 '14 at 00:25
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    Ok, and you evidently already know a generator of the Galois group for any finite extension of $\mathbf F_p$. Why exactly are you unsure of yourself? Have you ever computed a Galois group of a concrete polynomial over $\mathbf F_p$ before? – KCd Dec 31 '14 at 00:28
  • My only concern was whether it is possible for a splitting field to be the same for two different polynomials, in this case $x^5 + x^2 +1$ and $x^{2^5}-x$. – Libertron Dec 31 '14 at 00:29
  • Yes I have, and please read my concern I just mentioned above this comment. – Libertron Dec 31 '14 at 00:30
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    Hmm, what about $x^2+1$ and $((x+3)^2+1)(x-5)(x-8)(x+11)$ over $\mathbf Q$? – KCd Dec 31 '14 at 00:31
  • Ah I see now. I guess that puts my concern to rest. Thanks KCd. – Libertron Dec 31 '14 at 00:33

1 Answers1

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Yes. Your argument is correct.

  • As the comments from KCd and Libertron explained, every finite extension of a finite field is Galois with a cyclic Galois group.
  • There is no mystery attached to the fact that the polynomials $x^5+x^2+1$ and $x^{32}-x$ share the same splitting field over $\Bbb{F}_2$. For more details about this specific instance you can read my answer here, where it is explained (when read with a suitable goal in mind) that actually all the zeros of $(x^{32}-x)/(x^2-x)$ can be gotten from those of $x^5+x^2+1$ by some sequence of functions $a\mapsto a+1$, $a\mapsto 1/a$, $a\mapsto a^2$. Thar is very specific to this field though.
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Jyrki Lahtonen
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