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I am trying to prove that the polynomial $P = X^5 + X^2 + 1 ∈ F_2 [X]$ is irreducible.

What I did:

I showed that $X^2+X+1∈F_2[X]$ is the only irreducible polynomial of degree 2. Is there a way to use this to prove that $P$ is irreducible without checking all the polynomial products giving polynomials of degree $5$?

Many thanks!

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marked as duplicate by Dietrich Burde abstract-algebra Apr 1 '18 at 16:07

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Well, sure. If you work $\pmod 2$ then either $P(x)$ has a root or it is divisible by the quadratic you mentioned. (or it's irreducible, of course). $\endgroup$ – lulu Apr 1 '18 at 15:38
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    $\begingroup$ The answer of JavaMan for degree $5$ is very instructive; see the duplicate. $\endgroup$ – Dietrich Burde Apr 1 '18 at 16:08
  • $\begingroup$ Thank you for your valuable answers ! $\endgroup$ – PerelMan Apr 1 '18 at 20:29
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Remark that if $P=Q_1Q_2$ the degree of $Q_1$ and $Q_2$ are different of $1$, then you can suppose that $Q_1=X^2+X+1$, but the roots of $X^2+X+1$ are third root of the unity, but a third root of the unity is not a root of $P$.

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  • $\begingroup$ The OP asks for the polynomial over $\Bbb F_2$ $\endgroup$ – Crostul Apr 1 '18 at 15:44
  • $\begingroup$ one can deine root of unity over finite fields. $\endgroup$ – Tsemo Aristide Apr 1 '18 at 15:45
  • $\begingroup$ @Crostul The trick with third roots of unity still works. If $a^3=1$, then $a^5+a^2=2a^2=0$, so $a$ is not a zero of $P$. I refrain from upvoting this (correct) solution, because the question has been handled many times already. $\endgroup$ – Jyrki Lahtonen Apr 1 '18 at 15:47

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