proof of primitive element theorem

Theorem.

Let $F$ and $K$ be arbitrary fields, and let $K$ be an extension of $F$ of finite degree. Then there exists an element $\alpha \mathrm{\in }K$ such that $K\mathrm{=}F\mathit{}\mathrm{(}\alpha \mathrm{)}$ if and only if there are finitely many fields $L$ with $F\mathrm{\subseteq }L\mathrm{\subseteq }K$.

Proof.

Let $F$ and $K$ be fields, and let $[K:F]=n$ be finite.

Suppose first that $K=F(\alpha )$. Since $K/F$ is finite, $\alpha$ is algebraic over $F$. Let $m(x)$ be the minimal polynomial of $\alpha$ over $F$. Now, let $L$ be an intermediary field with $F\subseteq L\subseteq K$ and let $m^{\prime }(x)$ be the minimal polynomial of $\alpha$ over $L$. Also, let $L^{\prime }$ be the field generated by the coefficients of the polynomial $m^{\prime }(x)$. Thus, the minimal polynomial of $\alpha$ over $L^{\prime }$ is still $m^{\prime }(x)$ and $L^{\prime }\subseteq L$. By the properties of the minimal polynomial, and since $m(\alpha )=0$, we have a divisibility $m^{\prime }(x)|m(x)$, and so:

$$[K:L]=\mathrm{deg}(m^{\prime }(x))=[K:L^{\prime }].$$

Since we know that $L^{\prime }\subseteq L$, this implies that $L^{\prime }=L$. Thus, this shows that each intermediary subfield $F\subseteq L\subseteq K$ corresponds with the field of definition of a (monic) factor of $m(x)$. Since the polynomial $m(x)$ has only finitely many monic factors, we conclude that there can be only finitely many subfields of $K$ containing $F$.

Now suppose conversely that there are only finitely many such intermediary fields $L$. If $F$ is a finite field, then so is $K$, and we have an explicit description of all such possibilities; all such extensions are generated by a single element. So assume $F$ (and therefore $K$) are infinite. Let ${\alpha }_1,{\alpha }_2,\mathrm{\dots },{\alpha }_n$ be a basis for $K$ over $F$. Then $K=F({\alpha }_1,\mathrm{\dots },{\alpha }_n)$. So if we can show that any field extension generated by two elements is also generated by one element, we will be done: simply apply the result to the last two elements ${\alpha }_{j-1}$ and ${\alpha }_j$ repeatedly until only one is left.

So assume $K=F(\beta ,\gamma )$. Consider the set of elements $\{\beta +a\gamma \}$ for $a\in F^{\times }$. By assumption, this set is infinite, but there are only finitely many fields intermediate between $K$ and $F$; so two values must generate the same extension $L$ of $F$, say $\beta +a\gamma$ and $\beta +b\gamma$. This field $L$ contains

$$\frac{(\beta +a\gamma )-(\beta +b\gamma )}{a-b}=\gamma$$

and

$$\frac{(\beta +a\gamma )/a-(\beta +b\gamma )/b}{1/a-1/b}=\beta$$

and so letting $\alpha =\beta +a\gamma$, we see that

$$F(\alpha )=L=F(\beta ,\gamma )=K.$$

∎