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Prove that each irreducible factor of $f(x)=x^{2^n}+x+1$ in $\mathbb Z_2[x]$ has degree $k$, where $k\mid 2n$.

Edit. I know I should somehow relate the question to an extension of $\mathbb Z_2$ of degree $2n$, say $GF(2^{2n})$. By this way I will be able to correspond each irreducible factor of $f(x)$ to a subfield of $GF(2^{2n})$ that obviously has degree $k$, where $k\mid 2n$.

Xam
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user97635
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    Do you have any thoughts on the problem? What have you tried? – RghtHndSd Sep 29 '13 at 15:29
  • To prove that you are trying hard you could give the factors for small values of $n$. If you are stuck, it's always a good idea to try small cases!! Undoubtedly you know that with $n=1$ and $n=2$ your polynomial is irreducible (fitting the claim). So $n=3$ is the first real test. Have you tried whether $x^8+x+1$ has a quadratic factor? (Hint: There is only one irreducible quadratic over $\Bbb{Z}_2$ and we already saw that.) What about cubic ones? – Jyrki Lahtonen Sep 29 '13 at 15:49
  • @rghthndsd & Yan Yau : I added my thoughts in Edit notes. – user97635 Sep 29 '13 at 16:42
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    @JyrkiLahtonen, I have not tried the small cases, but are you sure they actually point to a general pattern? This problem seems easier to solve with "$n$" which forces consideration of what tools are available, how to rewrite the equation in more congenial terms and so on. And this is true of many other problems. Or one should not so much solve the special cases but explore them for generalizable features. – zyx Sep 29 '13 at 16:56
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    Agree, @zyx. I just wanted to get the OP to show any work done, and made the suggestion in that spirit. Well, they responded with some comments, so I'm happier. The good answer by YACP is now available (hopefully Thomas also undeletes). – Jyrki Lahtonen Sep 29 '13 at 17:03
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    @JyrkiLahtonen, since we "go there": yes, providing a hint is actually useful unlike the blank W*YT comments. But reasonable persons who do opt for this useful format seem to be put in the place of intermediaries between the OP and the unreflective wrath of the "show effort for the sake of showing effort, or else" question closing faction. I think some sort of WHYT reform along lines similar to what you proposed on meta would be a good thing. Close/delete are somehow beyond the range of worry or evaluation, being buttons one presses in the software. The WHYT stuff is quite deliberate. – zyx Sep 29 '13 at 17:45
  • @JyrkiLahtonen I hope you remember I asked the same question in bounty. Now , I have a doubt in accepted answer but I can't ask OP as he is inactivce for a long time or answerer as his account is deleted. So, I am asking you. If you are wiling to answer kindly tell me . In the 2nd line of accepted answer how does $a^{2^n} = a+1$ implies $a^{2}^{2^n} =a$ ? and how does it implies $\mathbb{F_2}(a) \subseteq \mathbb{F_2^{2n}}$ ? –  May 17 '21 at 05:51
  • @No-One Sure! You start with $$a^{2^n}=a+1.$$ Raise both sides to power $2^n$ to arrive at $$a^{2^{2n}}=a^{2^n}+1^{2^n}=a+1+1=a.$$ In an algebraic closure of $\Bbb{F}2$ the elements of $\Bbb{F}{2^\ell}$ are exactly the solutions of $x^{2^\ell}=x$. Apply that to $\ell=2n$. – Jyrki Lahtonen May 17 '21 at 07:39

2 Answers2

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Let $p(x)\mid x^{2^n}+x+1$, $p$ irreducible, and $a\in\overline{\Bbb F}_2$ a root of $f$. Then $a^{2^n}+a+1=0$, equivalently $a^{2^n}=a+1$. It follows that $a^{2^{2n}}=a$, so $\Bbb F_2(a)\subset\Bbb F_{2^{2n}}$. This shows that $\deg p\mid 2n$.

Xam
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An answer along the idea you had is that $t(x) = x^{2^n} + x$ is the function that computes the trace of the quadratic extension $\mathbb{F}_{2^{2n}} / \mathbb{F}_{2^n}$. Thus every element of $\mathbb{F}_{2^{2n}}$ of trace $1$ over $\mathbb{F}_{2^n}$ is a root of $f(x)$. As there are $2^n$ of these, they are all of the roots of $f(x)$.