# Irreducibility in $\mathbb{F_2}[X]$ and field extensions

Here's an exercise I found:

1) Show that $$f(X):=X^3+X+1$$ $$g(X):=X^3+X^2+1$$ are irreducible in $\mathbb{F_2}[X]$.

2) Let $a, b \in \overline{\mathbb{F_2}}$ with $f(a)=g(b)=0$. $$\mathbb{F_2}\subseteq\mathbb{F_2}(a),\mathbb{F_2}(b)$$ are field extensions. Show that $\mathbb{F_2}(a)$ and $\mathbb{F_2}(b)$ are isomorphic $\mathbb{F_2}$-algebras.

1) is pretty easy to show: Since $\mathbb{F_2}$ is a field, a polynomial of degree $3$ can only be divided into a polynomial of degree $2$ and another one of degree $1$. So it's enough to show that $f(X)$ and $g(X)$ have no roots in $\mathbb{F_2}$. And so $f(1),g(1)=1$ and $f(0),g(0)=1$. Is there anything more to show?

2) I think we only know one theorem about this topic. It says:

If $\mathbb{F_2}(a)$ and $\mathbb{F_2}(b)$ are algebraic closures of $\mathbb{F_2}$, then an isomorphism between $\mathbb{F_2}$(a) and $\mathbb{F_2}(b)$ exists.

This would easily proof 2), but I don't know how to show that these are algebraic closures of $\mathbb{F_2}$.

It'd be great, if anyone could help me.

• Those fields aren't algebraic closures, but they are both of order $8$, which means they're isomorphic as fields. – user363624 Dec 7 '16 at 22:07
• Two ideas: (1) Show that both fields are the splitting field of $X^8 - X$. Then they are necessarily isomorphic. (2) Try to find an explicit isomorphism. Find an element of $\mathbb{F}_2(a)$ whose minimal polynomial is $X^3 + X^2 + 1$. Map $b$ to this element. – André 3000 Dec 7 '16 at 22:10
• An algebraic closure has to be an infinite field. Further, there's a problem with 2): 1) asserts that $f$ and $g$ are irreducible over $\mathbf F_2$, hence they can't have have roots in $\mathbf F_2$. – Bernard Dec 7 '16 at 23:58
• @Xeny: I'm afraid I don't see how this is supposed to have the order 8... – Tobi92sr Dec 8 '16 at 20:38
• Hint: Show that $f(1/b)=0$. Or $g(1/a)=0$. – Jyrki Lahtonen Dec 9 '16 at 15:11