I asked this question and in response realised that splitting field of $p(x)=x^4+1$ over $F_3$ is $F_9$, which has degree $2$ over $F_3$.
But I want to understand what is wrong with this approach as this doesn't match with the answer. Let $x^4=-1=2$ in $F_3$, so to find splitting field I adjoin all the roots of $x^4+1$ with $F_3$. Now all the roos of $x^4=2$ in some extension are $\{\pm \sqrt[4]{2}, \pm \sqrt[4]{2}i\}$ but $-1=2 $ and $i=\sqrt2$ in $F_3$ so the set of roots become $\{2^{\frac{1}{4}}, 2^{\frac{3}{4}}, 2^{\frac{5}{4}}, 2^{\frac{7}{4}}\}$ and adjoining all these roots with $F_3$ gives splitting field to be $F_3(2^{\frac{1}{4}})= \{a+b2^{\frac{1}{4}}+c2^{\frac{2}{4}}+d2^{\frac{3}{4}}\ |\ a,b,c,d \in F_3\}$ so it has $81 $ elements , and has degree $4$ over $F_3$.
I know this approach is wrong somehow. But what is wrong with this approach?
Best Answer
If $u$ is a fourth root of $-1$ over $\Bbb F_3$, then $1$, $u$, $u^2$ and $u^3$ are not linearly independent over $\Bbb F_3$, so the $a+bu+cu^2+du^3$ don't give $3^4$ distinct values.
Over $\Bbb F_3$, $$x^4+1=(x^2+x-1)(x^2-x-1)$$ so that either $u^2=u+1$, or $u^2=-u+1$.