Find the splitting field of $f(x)=x^8-1$ over $\mathbb{F}_2$, $\mathbb{F}_3$ and $\mathbb{F}_{16}$.
I tried this: We claim that the field with $q=p^m$ elements is unique. A field with $q$ elements is the smallest field containing $\mathbb{F}_p$ and all the roots of $x^q−x$. Such a field is termed a splitting field of the polynomial $x^q−x$ over $\mathbb{F}_p$, that is, the smallest extension field of $\mathbb{F}_p$ containing all the roots of the polynomial. So, $x^9-x$ is of the form $x^{3^2}-x$, hence the splitting field is $\mathbb{F}_9[x]=\dfrac{\mathbb{F}_3[x]}{\left (x^2+1\right )}$. On the other hand, we know that the splitting field of $f(x)=x^8-1$ over $\mathbb{F}_2$ is of the form $\mathbb{F}_{q^t}$, where $q^t\equiv 1\pmod n$, so in this example $2^t\equiv 1\pmod 8$, which is impossible. And I don't know how to find the splitting field over $\mathbb{F}_{16}$.