# Splitting field over $\mathbb{F}_3$

The splitting field of $f(x)=x^8-1$ over $\mathbb{F}_3$ is $\mathbb{F}_{3^d}$ where $d=ord_{(\mathbb{Z}/8\mathbb{Z})^*}(3)=2$.

But $f(x)=(x^4+1)(x^4-1)$ and $x^4+1$ is irreducible over $\mathbb{F}_3$. By a "well known theorem" the splitting field of $f$ is $\mathbb{F}_{3^d}$ with $d\ge 4$. The "well known theorem" is the following: let $f\in\mathbb{F_p[x]}$ and let $f(x)=f_1(x)^{e_{1}}\cdots f_r(x)^{e_{r}}$ its factorization; if $\deg f_i=d_i$, the splitting field of $f$ over $\mathbb{F}_p$ is $\mathbb{F}_{p^d}$, where $d=lcm(d_1,\ldots d_r)$.

Where am I wrong?

You are wrong here: $x^4 + 1 = (x^2 + x + 2)(x^2 + 2x + 2)$.