Prove that $\mathbb F_8=\mathbb F_2[X]/(X^3+X+1)$

Question

I'm new to field extensions and I can neither see why or how to prove this statement.

If I'm not mistaken, to construct a splitting field, you construct rupture fields one after the other, using that the $8$ element field is the splitting field of $X^8-X$ over $\mathbb F_2$, since $2$ is the only prime divisor of 8.

Over $\mathbb F_2$ $X^8-X$ has two roots : $0$ and $1$. Therefore $X^8-X=X(X-1)(1+X+...+X^6)$. The last factor, call it $P$, has no roots. So let $\alpha$ be a root in some extension $\mathbb F_2(\alpha)$ of $\mathbb F_2$.

We now have $P=(X-\alpha)(X^5+\alpha X^4+...+\alpha^4X+\alpha^5)$.

I suspect $\alpha$ might not be the way to go and if so, my second plan would be to reduce $P$. I don't know how to do this but assuming it to be done, hence having $P=P_1...P_n$, I could start using $P_1$'s rupture field, and then my best guess is that if $\mathbb F_2[X]/(P_1)$ is not sufficient, I re-reduce $P$ over this new field and consider the rupture field of one of the factors, and so on.

Now assuming I wasn't given the statement itself, is it elementary to construct the $8$ element field using $\mathbb F_2$ ?

Answer 1

In characteristic $2$ $$X^8-X=X(X-1)(X^3+X+1)(X^3+X^2+1).$$ If $\alpha$ is a zero of $X^3+X+1$ then $$X^3+X+1=(X-\alpha)(X-\alpha^2)(X-\alpha^4)$$ and $$X^3+X^2+1=(X-\alpha^3)(X-\alpha^5)(X-\alpha^6).$$ This implies that $$\Bbb F_8=\Bbb F_2(\alpha)\cong\frac{\Bbb F_2[X]}{\left<X^3+X+1\right>}.$$

Answer 2

Show that the polynomial is irreducible over the field $F_2$. From that it will follow that the corresponding quotient ring is indeed a field.

Compute its dimension over $F_2$ and using that you will immediately know how many elements it has: eight.

Now, to show that it is isomorphic to $F_8$ what you do depends on what you know about this last field. For example, you might know that it is the unique field with eight elements up to isomorphism, and then of course you are done.