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Can someone provide a reference of the proof (or the proof itself) of this statement?

The polynomial $1+x+\dots+x^{n-1}$ is irreducible over $\mathbf{F}_2[x]$ if and only if $n$ is an odd prime and $2$ is a primitive element of $\mathbf{F}_{n}$.

Vinícius
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  • by $\mathbb{F}_n$ what do you exactly mean? Because $\mathbb{F}_n$ is only a field if n is a prime number – Abellan Mar 13 '15 at 15:36
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    $\mathbb{F}_n$ is not field unless $n$ is prime. And $2$ can't be primitive element for any odd prime $n$. – Fardad Pouran Mar 13 '15 at 15:37
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    @FardadPouran that depends on what $\Bbb F_n$ means here; there are fields of order $p^k$. – Ben Grossmann Mar 13 '15 at 15:39
  • Oh, thanks for pointing that, I had forgotten to express one of the conditions. Fixed. – Vinícius Mar 13 '15 at 15:41
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    Hint: Show that if $h$ is a irreducible factor of $f$ in $\mathbf F_2[x]$ (with $f(x) = 1+x+\dots+x^{n-1}$), then $\operatorname{deg}(h) = \operatorname{ord}(2,n)$ – Stefan Mar 13 '15 at 15:42
  • The statement appears to be wrong. First, you mean either in $\mathbf F_2[x]$ or over $\mathbf F_2$ (what amounts to the same). Second, what is wrong with the even primes? – Marc van Leeuwen Mar 13 '15 at 17:57
  • Closely related. When $n>2$ this result is combination of the facts that A) your polynomial is irreducible in $\Bbb{Z}[x]$ if and only if $n$ is a prime, B) a root of unity $\alpha$ (here of order $n$) shares its minimal polynomial over $\mathbf{F}_2$ with $\alpha^2$ and consequently with $\alpha^4$, $\alpha^8$, $\alpha^{16}$ et cetera. That set consists of all the primitive $n$th roots of unity if and only if $2$ is a primitive root modulo $n$. I'm a bit surprised that I didn't find an exact duplicate. – Jyrki Lahtonen Mar 14 '15 at 17:12

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The polynomial in question is irreducible if and only if the smallest field of characteristic $2$, which admits a $n$-th root of unity, is $\mathbb F_{2^{n-1}}$.

This is equivalent of saying $n | 2^k-1$ holds for $k = n-1$ but not for $0< k < n-1$. This implies that $n$ is prime and that $2$ is a primitive root of $(\mathbb Z/n\mathbb Z)^*$.

Let me clarify the first part of the proof (I think the second part is clear):

$f = 1+ x + \dotsb + x^{n-1}$ is irreducible iff it is the minimal polynomial of its roots. Its roots are exactly the non-trivial $n$-th roots of unity. Let $\zeta$ be such a $n$-th root of unity. Then $f$ is the minimal polynomial of $\zeta$ iff $[\mathbb F_2(\zeta):\mathbb F_2]=n-1$, which is the same as saying $\mathbb F_2(\zeta) = \mathbb F_{2^{n-1}}$.

By the way: It is a priori clear that $n$ must be prime, because if $n$ is not prime, then $f = 1+ x + \dotsb + x^{n-1}$ is even reducible over $\mathbb Z$. So we can assume $n$ to be prime and only have to care about the primitive root part. This makes life a little bit easier.

MooS
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  • Can you add some references for a proof or provide a proof (your post does not answer the question but clarifies it) – Elaqqad Mar 13 '15 at 16:58
  • My post is a proof, isn't it? – MooS Mar 13 '15 at 17:38
  • when I say that it's not a proof I mean: every person who can ask the question may understand the answer, right? for example I don't see why is it necessary to involve the degree of the extension $\mathbb{F}_2(\zeta)$? – Elaqqad Mar 13 '15 at 18:26
  • If your answer is a proof, then how come you missed $n=2$? – Marc van Leeuwen Mar 13 '15 at 18:36
  • The case $n=2$ is trivial. – MooS Mar 13 '15 at 19:26
  • But that does not make your opening sentence true. – Marc van Leeuwen Mar 13 '15 at 19:49
  • The theory of the cyclotomic polynomials and $n$-th roots of unity does only work, if the characteristic does not divide $n$. This is because $x^n-1$ is not separable if $char(k)|n$. So we do only consider the case where $n$ is odd and note that the case $n=2$ is trivial. I really do not know what your point is. My proof works as well as a proof should work. – MooS Mar 13 '15 at 21:13