$\mathbb F_4 \otimes_{\mathbb F_2}\mathbb F_8$ is isomorphic to $\mathbb F_{64}.$ How can I prove that $\mathbb F_4 \otimes_{\mathbb F_2}\mathbb F_8$ is isomorphic to $\mathbb F_{64}.$ I don't even know how to approach. I need some help. Thanks.
 A: This is a corollary of how we construct compositum fields in general (without assuming a common field contains the two of them). For two fields algebraic over a common field, $k\subseteq K, L$ and $[L:k]$ coprime to $[K:k]$, we have that
$$K\otimes_k L\cong KL\subseteq \overline{k}$$
But then in your case it's a simple matter of noting that since $\Bbb F_4\cap\Bbb F_8=\Bbb F_2$--because the degrees are coprime--that the compositum field has degree $2\cdot 3=[\Bbb F_4:\Bbb F_2][\Bbb F_8:\Bbb F_2]$ over the base field, and this of course corresponds to the field $\Bbb F_{64}$.
There are not a lot of details to the general theorem I quote, but far too long to include in a MSE post when they're basically textbook. The interested parties can read them here.
A: As a vector space over $\mathbb{F}_2$ we have a basis for $\mathbb{F}_4$ consisting of the two elements $1$ and $\alpha$ where $\alpha^2 = -\alpha - 1$ and we have an element $\overline{\alpha}$ such that $\alpha + \overline{\alpha} = -1$ and $\alpha\overline{\alpha} = 1$.
The given tensor product can then be identified with the set of $\mathbb{F}_8$-linear combinations of $1$ and $\alpha$ with multiplication induced from this.
Now, note that $(a + b\alpha)(a + b\overline{\alpha}) = a^2 + b^2 - ab$ so the claim that this is isomorphic to $\mathbb{F}_{64}$ is equivalent to the fact that the only solution to $a^2 + b^2 - ab = 0$ in $\mathbb{F}_8$ is $a = b = 0$.
Unfortunately, as I write this I realize that I don't have a good argument for why this should be, but I will make this community wiki and maybe someone else can finish it.
