Galois Theory HT25, Primitive element theorem

Suppose:

• $K/F$ is a finite, separable extension

Then:

• $K/F$ is simple, i.e.
• There is an element $\alpha \in K$ such that $K = F(\alpha)$

Case $F$ infinite:

Since $K/F$ is finite, $K$ is generated over $F$ by a finite number of elements. By induction on the number of generators, it will be sufficient to prove that $K$ is generated by one element if it is generated by two elements.

Suppose $K = F(\beta, \gamma)$. For $d \in F$, we consider the intermediate field $F(\beta + d\gamma)$. By previous results, $K/F$ is contained in a Galois extension and hence there are only finitely many intermediate fields $F(\beta + d\gamma)$.

Since $K$ is infinite, we can thus find $d _ 1, d _ 2 \in F$ such that $d _ 1 \ne d _ 2$ and $F(\beta + d _ 1 \gamma) = F(\beta + d _ 2 \gamma)$, and thus by considering the sequence of isomorphisms

\[\frac{F[x]}{\langle m _ {\beta + d _ 1\gamma}(x) \rangle} \cong F(\beta + d _ 1 \gamma) = F(\beta + d _ 2 \gamma) \cong \frac{F[x]}{\langle m _ {\beta + d _ 2 \gamma}(x) \rangle}\]

there exists a polynomial $f \in F[x]$ such that $\beta + d _ 1 \gamma = f(\beta + d _ 2 \gamma)$. Hence

\[\gamma = \frac{f(\beta + d _ 2 \gamma) - (\beta + d _ 2 \gamma)}{d _ 1 - d _ 2}\]

and

\[\beta = (\beta + d _ 2 \gamma) - d _ 2 \frac{f(\beta + d _ 2 \gamma) - (\beta + d _ 2\gamma)}{d _ 1 - d _ 2}\]

and then in particular

\[F(\beta, \gamma) = F(\beta + d _ 2 \gamma)\]

as required.

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Case $F$ finite:

Note that $K^\times$ is a finite abelian group. Hence by the structure theorem,

\[K^\times \cong \left( \frac{\mathbb Z}{p _ i \mathbb Z} \right)^{k _ 1} \times \cdots \times \left( \frac{\mathbb Z}{p _ n \mathbb Z} \right)^{k _ n}\]

In fact, $k _ i = 1$ for all $i$. If $k _ i > 1$, then it would contain a subgroup of the form $\left(\frac{\mathbb Z}{p _ i \mathbb Z}\right)^2$. All elements of the subgroup must satisfy $x^{p _ i} = 1$. Therefore all elements are roots of $x^{p _ i} - 1$, but $x^{p _ i} - 1$ has at most $p _ i$ roots, not $p _ i^2$. Hence each $k _ i = 1$.

This implies that

\[K^\times \cong \frac{\mathbb Z}{p _ i \mathbb Z} \times \cdots \times \frac{\mathbb Z}{p _ n \mathbb Z}\]

and hence

\[K^\times \cong \frac{\mathbb Z}{p _ 1 \cdots p _ n \mathbb Z}\]

and so $K^\times = \langle z \rangle$ for some $z \in K$. But then the minimal polynomial of this element must have degree $m = [K : F]$ and hence $[F(z) : F] = [K : F]$, so that $F(z) = K$.