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Mainly, I would like to check if I am correct. Over $\mathbb{Q}$, the result is shown in Sec. 13.4 or 13.6 of Dummit and Foote. I think the result is essentially the same here.

Since $x-1$ is a root in $\mathbb{F}_2$, factor that out. This leaves: $\frac{x^{17}-1}{x-1} = x^{16}+x^{15}+ \dots + x + 1$, which is irreducible in $\mathbb{F}_2$. Therefore this is the minimal polynomial of $\zeta_{17}$ (the primitive 17th root of unity) over $\mathbb{F}_2$, so $[K:\mathbb{F}_2] = 16$.

Is this correct? It seems rather straightforward, which makes me suspicious.

Thanks

Edit: Related Question.

In the question I linked above, the argument is as follows: The splitting field will be $\mathbb{F}_{2^r}$ for the smallest $r$ such that there are elements of order $17$. The order of the multiplicative group of $\mathbb{F}_{2^r}$ is $2^r-1$. The smallest $r$ such that $17 |2^r-1$ is $r = 8$. This would suggest that the polynomial above could be factored (I believe the factors are $x^8+x^7+x^6+x^4+x^2+x+1$ and $x^8+x^5+x^4+x^3+1$).

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Steve
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  • How do you show that $x^{16}+x^{15}+\ldots+1$ is irreducible in $\mathbb{F}_2$? – Steven Stadnicki Dec 01 '13 at 23:49
  • In this case you could just test both elements. Both evaluate to $1 \neq 0$ – Steve Dec 02 '13 at 00:04
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    I've changed my mind. I think the answer may actually be 8. I'm going to try to write that up. – Steve Dec 02 '13 at 00:44
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    Steve: that just means that it has no linear factors. That doesn't imply irreducibility over $\mathbb{F}_2$, any more than e.g. not having any integer roots implies irreducibility over $\mathbb{Z}$. – Steven Stadnicki Dec 02 '13 at 00:52
  • Yeah, thanks. I realized that after I sent my reply. I've edited the original post. – Steve Dec 02 '13 at 01:22

1 Answers1

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The solution you suggest in your edit is correct.

Bruno Joyal
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