# Explicit isomorphism between a quartic extension of $\mathbb{F}_2$ and a quadratic extension of $\mathbb{F}_4$

I want to find an explicit isomorphism $\mathbb F _{2^4} \cong \mathbb F_{4^2}$, where $\mathbb{F}_{2^4}$ is a quartic extension of $\mathbb{F}_2$, and $\mathbb F_{4^2}$ is a quadratic extension of $\mathbb{F}_4$.

For sure it is $\mathbb F _{2^4} \cong \mathbb F_{4^2}$ by using the classification theorem of finitely generated Abelian groups.

My aim is, to construct $\mathbb F_2 [X] / (f)$ and $\mathbb F_4 [X] / (g)$ with irreducible polynomials $f \in \mathbb F_2 [X]$ and $g \in \mathbb F_4 [X]$ with degree 4, resp. 2, and to give a concrete ismorphism.

• What do you even mean by $\Bbb F_{4^2}$? – Naive Jul 10 '17 at 14:30
• I would say that $2^4=4^2\Longrightarrow\mathbb{F}_{2^4}=\mathbb{F}_{4^2}$. – José Carlos Santos Jul 10 '17 at 14:32
• Obviously, ${\Bbb F}_{16} \cong {\Bbb F}_{16}$, but your last sentence is a real question. The point is to realize that all fields of 16 elements are isomorphic. Can you, for starters, find an irreducble $f$ of degree $4$ over ${\Bbb F}_2$? – Magdiragdag Jul 10 '17 at 14:35
• Nice, now you need to find an explicit representation of ${\Bbb F}_4$ as ${\Bbb F}_{2}[x]/(h)$ for some $h$ of degree $2$. There is only one choice for $h$, so that is easy, and this gives ${\Bbb F}_4 = {\Bbb F}_2[\alpha]$ with $\alpha^2 + \alpha + 1 = 0$. Next, can you find an irreducble $g$ of degree $2$ over ${\Bbb F}_4 = {\Bbb F}_2[\alpha]$? After that, ${\Bbb F}_2[x]/(x^4+x^3+x^2+x+1)$ and ${\Bbb F}_4[x]/(g)$ must be isomorphic and to find an explicit isomorphism you need to find a root of $x^4 + x^3 + x^2 + x + 1$ in ${\Bbb F}_4[x]/(g)$. – Magdiragdag Jul 10 '17 at 14:45
• OP, I edited your post to make the question clearer. Feel free to rollback or edit it yourself if you don't like my changes. – André 3000 Jul 10 '17 at 21:38

## 1 Answer

Finding $f$ is easy. Just factor $x^{16}-x \bmod 2$ and take any factor of degree $4$: $$x^{16}-x = x (x + 1) (x^2 + x + 1) (x^4 + x + 1) (x^4 + x^3 + 1) (x^4 + x^3 + x^2 + x + 1)$$

• Why I can choose $X^{16}-X$ ? – Myrkuls JayKay Jul 10 '17 at 14:55
• @MyrkulsJayKay, because all elements in a finite field of order $q$ are roots of $X^q-X$. This is a consequence of Lagrange's theorem of group theory. – lhf Jul 10 '17 at 16:09