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I am tasked to show that $x^{5}-2$ is irreducible over $\mathbb{F}_{11}$ the finite field of 11 elements. I've deduced that it has no linear factors by Fermat's little theorem. But showing it has no quadratic factors is proving harder.

My approach so far is assume it did. Factor the polynomial and remulitply and compare coefficients to obtain a contradiction. I'm having trouble doing that since there are so many cases. I was given then hint

"How many elements are in a quadratic extension of $\mathbb{F}_{11}$"? The answer is 121 but I don't know how that helps me. Any hints on dealing with the hint would be very nice. Thanks.

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  • $\begingroup$ I would regard the quadratic extension with respect to the only quadratic factor you mention in your post and have a look what happens to your polynomial there and how this relates to the number of elements. $\endgroup$ – Phira Dec 24 '13 at 23:06
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If $x^5 - 2$ has a quadratic factor, it has a root $\alpha$ in a quadratic extension $K$ of $\mathbb{F}_{11}$. Since the group of units of $K$ has $120$ elements, $\alpha^{120} = 1$. Now $\alpha^5 = 2$, so the order of $2$ must divide ...?

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  • $\begingroup$ Great. The order of 2 must divide 20. Which it doesn't by just checking, 2,4,5 and 20. Thanks. $\endgroup$ – TheNumber23 Dec 24 '13 at 23:10
  • $\begingroup$ @TheNumber23 No, the order must divide $120/5=24$. We know that $10$ is the order of $2$ in $\mathbb F_{11}$ $\endgroup$ – Thomas Andrews Dec 24 '13 at 23:18
  • $\begingroup$ yeah bad at dividing. Sorry. $\endgroup$ – TheNumber23 Dec 24 '13 at 23:20
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Absolutely nothing wrong with Daniel's solution. Just offering an alternative route.

We see that $4$ is a primitive fifth root of unity in $\Bbb{F}_{11}$. If $\alpha$ is a zero of $x^5-2$, then you already know that $\alpha$ is in some extension field $K=\Bbb{F}_{11}[\alpha]$ of degree $n>1$. Then the other zeros of the minimal polynomial of $\alpha$ are among $4^k\alpha, k=1,2,3,4$. Therefore there exists an automorphisms $\sigma\in Gal(K/\Bbb{F}_{11})$ such that $\sigma(\alpha)= 4^\ell\alpha$, $0<\ell<5$. But because $\sigma(4^\ell)=4^\ell$ this implies that the order of $\sigma$ is a multiple of five. The claim follows.

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$\, \color{#c00}{a^{\Large 5}\! = 2}\, \Rightarrow\, a^{\large{120}}\! = \dfrac{(\color{#c00}{a^{\Large{5}}})^{\large{25\!\!\!}}}{\color{#c00}{a^{\Large 5}}}\! = \dfrac{(\color{#c00}2^{\Large{5}})^{\large{5\!\!\!}}}{\color{#c00}2} = \dfrac{(-1)^{\Large{5}}\!\!\!}2\ne 1\ $ contra $\ X^{\Large{120}}\! = 1\,$ for $\ 0\ne X\in \Bbb F_{11^{\Large{2}}}\ \ $ QED

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