Degree of splitting field of $x^6-3$ over $\mathbb{Q}((-3)^{1/2})$ and also over $\mathbb{F}_5$ 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
 A: 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. 
A: The finite fields are often easier to deal with in this type of questions. A useful property of the finite fields is that all non-zero elements are roots of unity. The field $GF(5^m)$ contains all the roots of unity of order any factor of $5^m-1$. This is because the multiplicative group is cyclic. Another key fact to remember is that a given finite field has only one (up to isomorphism) extension of any given (finite) degree.
I hint at two solutions using these two facts. You are welcome to expand either or both of them to a full solution. Depending on how familiar you are with these properties (I cannot tell, because it depends how much of the basic properties of finite fields you have covered at this point).
A solution to this problem using the first fact would start as follows.
Here we see that $3^2\equiv -1 \pmod 5$ and $3^4\equiv 1\pmod 5$, so $3$ is a primitive root of unity of order $4$. Therefore any sixth root of $3$ in any extension of $F_5$ is a root of unity of order $n$, where $n$ is a factor of $24$. Therefore all the zeros of $x^3-6$ are in a field containing the $24^{th}$ roots of unity, which is... A remaining question then is: could the splitting field be a proper subfield of that field?
One solution using the second fact is to look for factorizations of $x^6-3$ over $F_5$.
We notice that $3=8=2^3$ in $F_5$, so 
$$
x^6-3=(x^2)^3-2^3=(x^2-2)(x^4+2x^2+4).
$$
The factor $x^2-2$ is easy to deal with. The second factor is trickier (which is why I slightly prefer the first solution). We can write it in $F_5[x]$ as
$$
x^4+2x^2+4=x^4+7x^2+9=(x^4+6x^2+9)+x^2=(x^2+3)^2-4x^2
$$
a difference of two squares, which leads to a factorization
$$
x^6-3=(x^2-2)(x^2-2x+3)(x^2+2x+3).
$$
But this is more than a bit ad hoc. Anyway, you can pick it up from here.
A third possibility (hinted at by Dylan Moreland) would be to combine these approaches, and determine the smallest extension field of $F_5$ that contains all the sixth roots of unity and one of the roots of $x^6-3$. Here I would use a zero of the easy factor $x^2-2$.
