Help finding the minimal polynomial of $\beta$ over $GF(19)$ If we let $\beta$ be a root of $x^6+3$ over $GF(19)$.
Then I have two questions 
i) how can we find the minimal polynomial of $\beta$ over $GF(19) $?
ii)how can we determine the order of the splitting field ?
What I think so far;
i) as $\beta \notin GF(19) $ we are need of a field extension 
$$GF(19)(\beta):=\{a+b\sqrt[6]{3}w+c(\sqrt[6]{3}w)^2+e(\sqrt[6]{3}w)^3+f(\sqrt[6]{3}w)^4+g(\sqrt[6]{3}w)^5|a,b,c,e,f,g \in GF(19)\}$$
where w is the 6th root of unity 
(sorry this doesn't fit , its a long set :/) but i don't know where to go from here except that I know that to be the minimal polynomial it must be the least degree monic polynomial st $m(\sqrt[6]{3}w)=0$
ii)I feel like this is going to have something to do with $\beta^6=3, (\beta^6)^2=9$ etc until we reach some multiple of 19 , but i'm not sure if this is quite right ?
Any help would be greatly appreciated .
 A: The polynomial is $X^6+3$, over $\Bbb F_{19}$. Let me update you on the general facts that I’ll be using.
If $p$ is a prime, there is, for each $n>0$, a field with $p^n$ elements, most conventionally denoted $\Bbb F_{p^n}$, and up to isomorphism, there is only one such. It may be characterized as the set of roots of the $\Bbb F_p$-polynomial $X^{p^n}-X$. Its multiplicative group of all nonzero elements is of cardinality $p^n-1$, and this group is cyclic. The fact about cyclic groups that I’ll use here is that a cyclic group of order $m$ has, for each divisor $d$ of $m$, precisely one subgroup of order $d$.
Now for our specific case. As @WillJagy pointed out in a now-deleted post, since $X^6+3=0$ is the same as $X^6=16$ in our field $\Bbb F_{19}$, then of course a root $\beta$, satisfying $\beta^6=16$ certainly has either $\beta^3=4$ or $\beta^3=-4=15$. These two characterizations of a root $\beta$ are what I referred to in my comment. These two properties of your root are sufficient to call them $\beta,\beta'$: say $\beta^3=4$, $\beta'^3=15$.
From here on, I show you my own way of attacking these problems. Others’ methods will differ.
At this point, we have to look more closely at the multiplicative structure of our eighteen-element group $\Bbb F_{19}^\times$. I like to find a generator of the group (“primitive element”); in our case $2$ is a generator. It helps to write down the powers of the generator in order, $1,2,4,8,16,32\equiv13,26\equiv7,14,28\equiv9,18\equiv-1$, and I pause there to check that $-1$ appears halfway through the list, as it does, as $2^9$. If not, I’ve made an error. Anyway, you may continue with the list, $2\times18=36\equiv17\equiv-2,15,11,3,6,12,5,10,20\equiv1$.
This list is not essential, but I find that it is very reassuring to have beside me, especially when I need to raise something to a power or multiply two things. What it tells us is that $4=2^2$, so that $4$ is a primitive ninth root of unity in a cyclic group of order $18$ and certainly does not have a cube root in that group. Similarly, $15=2^{11}$, and therefore is another generator ($11$ is relatively prime to $18$), and equally certainly does not have a cube root in our group.
It follows that $X^3-15$ and $X^3-4$ are $\Bbb F_{19}$-irreducible, and both $\beta$ and $\beta'$ generate the field with $19^3=6859$ elements. In general, if $q$ is a power of $p$, and $f$ is of degree $m$ and irreducible as a $\Bbb F_q$-polynomial, any (and thus all) of its roots generates the field $\Bbb F_{q^m}$.
I guess the answer to your question, then, is that the minimal polynomial of a root $\beta$ of $X^6+3$ over $\Bbb F_{19}$ will be either $X^3-4$ or $X^3+4$. In either case, the splitting field is the cubic extension of the base field.
