Is the field $\mathbb{F}_{64}$ the splitting field of $x^8-x$ over $\mathbb{F}_4$? What I find is:
Let $E$ be the splitting field of $x^8-x$ over $\mathbb{F}_4.$ Thus, we have $E$ is isomorphic to $\mathbb{F}_8.$ Since $64=2^6, 8=2^3,$ $\mathbb{F}_{64}$ contains a subfield isomorphic to $\mathbb{F}_8.$ But how can I show that they are equal?
It suffices to check that $ \mathbb{F}_{64} $ is the smallest field containing $ \mathbb{F}_4 $ in which $ X^8 - X $ has $ 8 $ roots. First, note that any element of $ \mathbb{F}_{64} $ is a root of $ X^{64} - X = X(X^{63} - 1)$, and the latter factor is divisible by $ X^7 - 1 $ so that $ X^8 - X $ indeed splits in $ \mathbb{F}_{64} $. On the other hand, the only field extension of $ \mathbb{F}_4 $ smaller than $ \mathbb{F}_{64} $ is $ \mathbb{F}_{16} $, and $ \mathbb{F}_{16}^{\times} $ cannot have an element of order $ 7 $ by Lagrange, meaning that $ X^7 - 1 $ cannot split. Therefore, $ \mathbb{F}_{64} $ is the smallest such field, and is by definition the splitting field.