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This question already has an answer here:

$ x^4 +x+1$ in $\mathbb{Z}/2\mathbb{Z}[x]$ is an irreducible polynomial.

So far we have only treated quadratic and cubic polynomials, which are irreducible if they do not have any zeros. However, now I want to show that $x^4 + x+1$ in $\mathbb{Z}/2\mathbb{Z}[x]$ is irreducible, I cannot go about checking if it has any zeros, this does not guarantee irreducibility. Is there any clever approach or do I need to determine all the polynomials of lower degree that are irreducible and show that upon division there is always a remainder? $$ \{x, x+1 ,x^2+x+1, x^3 +x+1, x^3 +x^2+1\} $$ are the polynomials I immediately thought of.

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marked as duplicate by Dietrich Burde, Sil, Jyrki Lahtonen, Community Oct 17 '18 at 18:37

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    $\begingroup$ Well how many ways can you split up a 4th degree without getting a zero? $\endgroup$ – Sorfosh Oct 17 '18 at 18:30
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    $\begingroup$ There is only one irreducible polynomial of degree $2$: $x^2+x+1$. $\endgroup$ – lhf Oct 17 '18 at 18:33
  • $\begingroup$ A solution is also contained in this older thread. $\endgroup$ – Jyrki Lahtonen Oct 17 '18 at 18:36
  • $\begingroup$ That was very helpful! so the zero argument basically checks if we can split off a linear term and write it as a cubic and linear polynomial. There are no zeros, so now we need to also check the other possibility, it is a product of degree $2$ polynomials, we show this leads to a contradiction and we are done. That was very nice, thanks guys! $\endgroup$ – Wesley Strik Oct 17 '18 at 18:41
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    $\begingroup$ Yes, that is it. :) $\endgroup$ – Cornman Oct 17 '18 at 18:42
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You can check easily, that your polynomial has no roots in $\mathbb{Z}/2\mathbb{Z}[X]$ by setting $X=0,1$.

Since it has no roots it has to be $X^4+X+1=(X^2+aX+b)(X^2+cX+d)$ now go ahead and compare the coefficients and deduce a contradiction. Which means that one of a,b,c,d is not an element of $\mathbb{Z}/2\mathbb{Z}$

Edit: To be more specific a contradiction arises, because one of the equalities you obtain does not hold.

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    $\begingroup$ In this case a simpler way forward is to use the fact that $x^2+x+1$ is the only irreducible quadratic. Implying that there is no freedom whatsoever in the choice of $a,b,c,d$! $\endgroup$ – Jyrki Lahtonen Oct 17 '18 at 18:37
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    $\begingroup$ @JyrkiLahtonen That is a nice approach. I did not think about that and had not in mind, that $x^2+x+1$ is the only irreducible quadratic in $\mathbb{Z}/(2)$. I will keep that in mind. Anyways, I think it is helpful to calculate it all the way through, since this method is a basic approach to such problems. $\endgroup$ – Cornman Oct 17 '18 at 18:40

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