Explanation on primal field I know that given a field $E$ and his primal field $\Bbb F_p$, then the splitting field of $T^q-T$ (with $q=|E|$) is exactly the extension $E/\Bbb F_p$.
Now, if I take the finite field $K=\Bbb Z/5\Bbb Z$, then $K$ is the primal field of $K[\sqrt 3]$, which has cardinality $5^2$. Now the polynomial $T^{25}-T$ over $K$ has as splitting field $K[\alpha]$ with $\alpha^{25}=1,\ \alpha\neq1$, so that $|K[\alpha]:K|=5$. However $K[\sqrt 3]$ and $K[\alpha]$ are not even isomorphic so I must have misunderstood something about this. I would be glad if someone could tell me where I'm wrong, thank you.
 A: Okay. A more precise way to put it would be that the polynomial $X^2 - 3$ in $K[X]$ has no root in $K$ and has degree $2$, therefore it is irreducible. One may then define the ring $L = K[X]/(X^2 - 3)$. The irreducibility of the polynomial we are taking the quotient with implies that $L$ is actually a field. We may describe $L$ in two different manners. 
First of all, by very definition, any element of $L$ can be described as the equivalent class of a polynomial, which we denote by $\overline{P}$, where $P \in K[X]$. The class of any polynomial $P$ also is the class of its remainder in the euclidian division by $X^2-3$ in $K[X]$, which is a polynomial of degree at most $1$. Hence, any element of $L$ can be described as $\overline{aX+b}$ where $a,b \in K$. This description is faithful: every element is represented by a unique polynomial of degree at most $2$. Hence, $L$ has cardinality $5\times 5 = 25$. Because $L$ is a $2$-dimensional vector space over $K$, the degree $[L:K]$ of the extension $L/K$ is $2$. By Lagrange formula, we know that all elements of $L$ must be roots of the polynomial $X^{25}-X \in K[X]$ and $L$ is of course generated by its own elements over $K$, whence $L$ also is the decomposition field of the polynomial $X^{25}-X \in K[X]$.
In other hand, we do know that every finite subgroup of the inversible group of a field is cyclic. This applies in particular to $L^{\star}$. Let $\alpha$ be a generator of the multiplicative group $L^{\star}$. In particular, $L$ must be generated as a $K$-algebra by the element $\alpha$, whence we may write $L = K[\alpha]$. Because $\alpha$ is a generator of the group $L^{\star}$ which has cardinality $24$, we have $\alpha^{24}=1$, so $\alpha$ is a root of $X^{25} - X \in K[X]$, and every root of this polynomial is a power of $\alpha$. We find again that $L$ is the decomposition field of this polynomial. 
Now, we may wonder how to write $\alpha$ (which is a priori not uniquely defined) in terms of the first description. Since we want $\alpha$ to generate $L$ as a $K$-algebra, we may wonder if $\beta := \overline{X}$ can play the role of $\alpha$. Indeed, we do have $L=K[\beta]$. However, note that by definition of $L$, $\beta$ is a root of $X^2-3$, hence $\beta^2 = 3$, which implies that $\beta^8 = 1$. Hence $\beta$ has order at most $8$ (in fact exactly $8$) in $L^{\star}$, so it isn't a generator. To find an explicit generator of the group $L^{\star}$ is in general not an easy problem. In order to build the field with cardinality $25$ (which is unique up to non-unique isomorphism), one may try to find another irreducible polynomial of degree $2$ over $K$ (other than $X^2-3$) so that the class of $X$ will be a generator of the multiplicative group of the quotient. This is often convenient for concrete computations.
