Finite field extension

A finite extension of the finite field ${\mathbb{F}}_q$ of degree $n$ is a field with $q^n$ elements. As a vector space, it has $n$ coordinates with $q$ choices each. Up to an isomorphism fixing ${\mathbb{F}}_q$, there is exactly one such field, denoted ${\mathbb{F}}_{q^n}$.

For example, ${\mathbb{F}}_4={\mathbb{F}}_2[t]/(t^2+t+1)$ has four classes $0,1,t,1+t$. It is not the quotient ring $\mathbb{Z}/4\mathbb{Z}$, which has a nonzero zero divisor. Its Galois group over ${\mathbb{F}}_q$ is generated by $x\mapsto x^q$. The phrase here means an extension of a finite coefficient field, not an arbitrary finite-degree extension of any field.

For ${\mathbb{F}}_{q^2}/{\mathbb{F}}_q$, the field norm maps the nonzero elements onto ${\mathbb{F}}_q^{\ast }$: the multiplicative group of a finite field is cyclic, and the power $q+1$ has image of order $q-1$. The field trace is also onto. Indeed it is an ${\mathbb{F}}_q$-linear map into ${\mathbb{F}}_q$, and is not zero because its polynomial $X+X^q$ has degree $q$ and cannot vanish at all $q^2$ field elements. Consequently $a\mapsto \mathrm{Tr}(ac)$ is onto whenever $c\ne 0$.

The map $x\mapsto x^q$ is an automorphism of ${\mathbb{F}}_{q^n}$ fixing ${\mathbb{F}}_q$. Its powers are all the ${\mathbb{F}}_q$-automorphisms, and its order is $n$.

Over ${\mathbb{F}}_q$, the roots of the minimal polynomial of $\alpha \in {\mathbb{F}}_{q^n}$ are its distinct iterates $\alpha ,{\alpha }^q,{\alpha }^{q^2},\dots$ under the finite-field Frobenius automorphism.

Every finite field extension is a Galois extension, and