Exercise 5.4.2

Proof. Let $F$ a finite field, and $F\subset L$ a finite extension.

(a)

As $n=[L:F]<\infty$, there exists a basis $(l_1,\cdots ,l_n)$ of $L$ over $F$, thus every element $\alpha \in L$ is of the form $\alpha ={\gamma }_1{l}_1+\cdots +{\gamma }_n{l}_n$, with a unique $({\gamma }_1,\dots ,{\gamma }_n)\in F^n$. Therefore $L$ is isomorphic to $F^n$ as a vector space, thus $|L|=|F{\text{|}}^n<\infty$. $L$ is a finite field.

(b)

$L^{\text{∗}}$ being the finite multiplicative group of a field is cyclic (Proposition A.5.3), with a generator $\alpha \in L$: $$L^{\text{∗}}=\{1,\alpha ,{\alpha }^2,\cdots ,{\alpha }^{m-1}\},\text{where }m=|L|-1.$$

So every $\gamma$ in $L^{\text{∗}}$ is of the form $\gamma ={\alpha }^k,k\in \mathbb{N}$, thus $L^{\text{∗}}\subset F(\alpha )$, and $0\in F[\alpha ]$, thus $L\subset F(\alpha )$. Moreover $F\subset L$, and $\alpha \in L$, thus $F(\alpha )\subset L$.

$$L=F(\alpha ).$$

(c)

As $(L^{\text{∗}},\times )$ is a group of cardinality $m=|L|-1$, Lagrange’s Theorem shows that every $\gamma \in L^{\text{∗}}=\{1,\alpha ,{\alpha }^2,\cdots ,{\alpha }^{m-1}\}$ satisfies ${\gamma }^m=1$, and so is a root of $x^m-1$.

Since the order of $\alpha$ is $m$, ${\alpha }^i\ne {\alpha }^j$ if $0\le i<j\le m-1$, thus the polynomial $p=(x-1)(x-\alpha )\cdots (x-{\alpha }^{m-1})$ divides $x^m-1$. The degree of the quotient is 0, so this quotient is a constant $c\in F^{\text{∗}}$. Since $p$ and $x^m-1$ are monic, $c=1$.

$$x^m-1=(x-1)(x-\alpha )\cdots (x-{\alpha }^{m-1}).$$

(d)

The minimal polynomial $f$ of $\alpha$ over $F$ divides $x^m-1$, which is separable by part (c). Thus $f$ is also separable. Therefore $\alpha$ is separable, and $L=F(\alpha )$: the Theorem of the Primitive Element is proved in the case of a finite extension of a finite field.