# Prove $x^4 +x+1$ is irreducible in $\mathbb{Z}/2\mathbb{Z}[x]$ [duplicate]

$$x^4 +x+1$$ 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.

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

• Well how many ways can you split up a 4th degree without getting a zero? – Sorfosh Oct 17 '18 at 18:30
• There is only one irreducible polynomial of degree $2$: $x^2+x+1$. – lhf Oct 17 '18 at 18:33
• A solution is also contained in this older thread. – Jyrki Lahtonen Oct 17 '18 at 18:36
• 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! – Wesley Strik Oct 17 '18 at 18:41
• Yes, that is it. :) – Cornman Oct 17 '18 at 18:42

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 $$\mathbb{Z}/2\mathbb{Z}$$
• 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 $a,b,c,d$! – Jyrki Lahtonen Oct 17 '18 at 18:37
• @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. – Cornman Oct 17 '18 at 18:40