TITLE: Show the polynomial is irreducible in $\textbf Z_2$
QUESTION [5 upvotes]: Show that $p(\gamma) = \gamma^{4} + \gamma + 1$ is irreducible over $\textbf Z_2$.
Okay, I would just like to see if i'm on the right track for this problem.
$$\gamma^{4}+\gamma+1=(\gamma +b)(c\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)$$
So then $a = c = 1$.  
So $$\gamma^{4}+\gamma+1=(\gamma +b)(\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)$$
Let $b = 0$
$\gamma^{4} + \gamma+1=(\gamma)(\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)$ 
Let $\gamma = 0$
Then we see $1 = 0$ which is a contradiction.
Let $b = 1$
$\gamma^{4}+\gamma+1=(\gamma+1)(\gamma^{3}+d\gamma^{2}+e\gamma+\gamma)$
Take $\gamma = 1$
And again
$1=0$ 
So our polynomial is irreducible over $\textbf Z_2$
Is there another way I should be computing this?
Maybe with Long division of polynomials? 
I'm quite lost..

REPLY [2 votes]: Robert Lewis did a great job, but I think that his exposition could perhaps be shortened.  Consider
$$
p(x)=x^4+x+1 \in \mathbb{Z}_2[x]
$$
Since $p(0)=1$ and $p(1) = 1$, the original $p(x)$ has not root and thus cannot be factored into the product of a cubic and a first degree polynomial.
If it could be factored, then both quadratics would have to be of the form:
$$
x^2+ax+1
$$
because the first and last coefficients of both quadratics multiply up to unity in $\mathbb{Z}_2$.  If either of the two quadratics had $a=0$, then $p(1)=0$, contradicting the fact that $p(x)\in \mathbb{Z}_2[x]$ has no roots.  And if both quadratics had $a=1$, then $p(x)$ would have to have non-zero coefficients for the $x^3$ and $x^2$ terms, again a contradiction.