Irreducible polynomial of degree $2$ over a finite field of characteristic $2$

Question

Let $F$ be a finite field of characteristic $2$. I need to find an irreducible polynomial of degree $2$ over $F$.

To that effect, I was given the following hint: "Find a polynomial $f$ and an element $a \in F$ such that $f(x) = f(x+a)$".

I thought about, perhaps letting $a = 2$. Then, if I let $f(x) = bx^{2} + cx + d$, I have that $f(x+2) = b(x+2)^{2}+c(x+2) + d = bx^{2} + 2b\cdot2 + cx + 2x+d = bx^{2} + 0 + cx + d = bx^{2}+cx + d = f(x),$

but I don't see how this gives me an irreducible polynomial of degree $2$.

Could somebody please help me figure this out? Thank you! :)

Answer 1

Let your field of characteristic two be $F$, and consider the mapping $g:F\to F$, by $z\mapsto z^2+z$. Notice that it’s additive, that is $g(a+b)=g(a)+g(b)$. Do you know a little group theory? What’s the kernel of this homomorphism? Finiteness of $F$ tells you that $g$ is not onto. Does that help?

Answer 2

Well, you can't use $a=2$, because in a field of characteristic 2, that means $a=0$ and you're back to square one.

Once you've found your $f$ and $a$, you can let $g(x) = f(x) + f(x+a) + 1$. Since $f(x) = f(x+a)$ (and $F$ chas characteristic 2), it follows that $g(x)=1$ for all $x$.