Finding roots in finite fields.

Question

On pg. 587 (in the finite fields chapter) of Abstract Algebra, 3rd ed. by Dummit and Foote, the following statement is made:

'If $f_1(x)=x^4+x^3+1$, $f_2(x)=x^4+x+1$ are two of the irreducible quartics over $\mathbb{F}_2$, then a simple computation verifies that $\alpha(x)=x^3+x^2$ is a root of $f_2(x)$ in $\mathbb{F}_{16}=\mathbb{F_2}/(x^4+x^3+1)$.'

I understand why $\alpha(x)$ is a root, but I just can't grasp how they came up with it. Its like it was pulled out of thin air, and they make no mention of it. I know that you could find all the elements of the quotient group and try each one individually, but that seems like an extremely weird thing not to mention. How did they come up with $\alpha(x)$?

Answer 1

In a situation where the set of possible solutions is finite and checking whether a candidate solution is indeed a solution involves simple calculations, the most straightforward algorithmic method to determine all solutions consists of trying all candidates in turn. This is especially true if the "finite" is not infeasibly large.

Answer 2

I think that @JyrkiLahtonen’s comment is most apposite.

The roots of $f_1$ are the reciprocals of the roots of $f_2$, as he points out. To clarify matters, let’s call $\xi$ the image of $x$ in $\Bbb F_2[x]/(f_1(x))$. The question asks to show that $\alpha=\xi^3+\xi^2$ is a root of $f_2$, in other words that $1/\alpha$ is a root of $f_1$.

But from $\xi\alpha=\xi^4+\xi^3=1$, we see that $1/\alpha$ is $\xi$ itself, certainly a root of $f_1$.