Show that $x^2+x+1$ is irreducible over $GF(2^n)$ for each odd $n$.

Question

I am trying to show that $x^2+x+1$ is irreducible over $GF(2^n)$ for each odd $n$.

Clearly it's true for $n=1$ since then $GF(2) = \mathbb{Z}_2$.

I think the idea is to assume $u$ is a root of $x^2+x+1$ and then calculate $[GF(p^n)(u):GF(p^n)]$. But I am having trouble doing that.

You should "odd" at the end of your question's title, otherwise it misleads... — DonAntonio, May 27 '17 at 20:30

Zev Chonoles · Accepted Answer · 2017-05-27T20:15:57.923

4

Observe that if $x^2+x+1$ has roots in a field $F$, then they are cube roots of unity, i.e. they are elements of order $3$ in the multiplicative group $F^\times$. Also recall that $\mathbb{F}_{p^n}^\times$ is a cyclic group of size $p^n-1$.

When does the cyclic group $\mathbb{F}_{2^n}^\times$ have an element of order $3$?

edited May 27 '17 at 20:15

answered May 27 '17 at 20:15

Zev Chonoles

129,973

1

$3$ should divide $p^n -1$ right? – fosho May 27 '17 at 20:22
So then $n$ must be even! – fosho May 27 '17 at 20:28
Exactly right :) – Zev Chonoles May 27 '17 at 20:28
Great! Thanks for the answer :) – fosho May 27 '17 at 20:29

Show that $x^2+x+1$ is irreducible over $GF(2^n)$ for each odd $n$.

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