I am trying to find the degree of the splitting field of this polynomial over
these two fields.
For the degree over $\mathbb{Q}((-3)^{1/2})$ I got 3. I am pretty sure this is correct.
For $\mathbb{F}_5$ I am not so sure though. I guess I am intimidated by characteristic p fields.
Here are my thoughts so far:
In $\mathbb{F}_5, x^6-3=x*x^5-3=x^2-3$.
I think that this polynomial is irreducible in $\mathbb{F}_5$ which would make the
degree 2. However I am not sure how to show this.
Any help would be very much appreciated.
Thanks
Best Answer
The roots of $x^6-3$ over the rationals are the numbers $\rho^m\root6\of3$ where $\rho$ is a primitive complex 6th root of unity and $m=0,1,\dots,5$, so the splitting field can be written $K={\bf Q}(\rho,\root6\of3)$. Now you've got a tower of fields ${\bf Q}\subset{\bf Q}(\rho)\subset K$. You ought to be able to show that ${\bf Q}(\rho)$ is ${\bf Q}(\sqrt{-3})$, and I think you'll find you got the wrong answer for the first question.