What is the splitting field of $x^4-x^2-2$ over GF(3), where GF(3) is the finite field with 3 elements.
I know how to get the splitting field of polynomials over $Q$ but I’m not sure how to do this since I’m not even sure what the elements of GF(3) are...
I would greatly appreciate if anyone can help. Thanks!
Note $$(x^2+1)^2=x^4+2x^2+1$$ In $GF(3)=\mathbb{Z}/3\mathbb{Z}$ we have $2=-1$ and $1=-2$. So $$(x^2+1)^2=x^4-x^2-2$$
Now you just have to find the spliting field of $x^2+1$ over $GF(3)$. This is easier since it can only have linear factors. A quick check that $0^2=0$, $1^2=1$ and $2^2=1$ verifies that $x^2+1$ is irreducible. So the extension $GF(3)[x]/(x^2+1)=GF(3)[w]$ where $w$ is a solution to $x^2+1$ should be the splitting field (note the other root is $-w$ which will be in there for free). This field is in-fact isomorphic to $GF(9)$