Primitive element theorem

This fact is related to: Galois theory
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Name of theorem

This theorem is known as the primitive element theorem, or the theorem of the primitive element.

Statement

If a field extension $L/K$ is finite and separable, then it is a simple extension, i.e. $L=K(\theta )$ for some $\theta \in L$.

Proof

Write $L=K(\alpha _{1},\dots ,\alpha _{n})$ for some $\alpha _{i}\in L$. We must show that $L=K(\theta )$ for some $\theta \in L$.

It suffices to prove the case $n=2$, since the general case follows by induction on $n$.

Write $L=K(\alpha ,\beta )$. $f$ is the minimal polynomial of $\alpha$ over $K$, $g$ is the minimal polynomial of $\beta$ over $K$.

Let $M$ be a splitting field for $fg$ over $L$.

Write $f(X)=\prod _{i=1}^{r}(X-\alpha _{i})$, $\alpha _{i}\in M$. $\alpha =\alpha _{i}$. $g(X)=\prod _{i=1}^{s}(X-\beta _{i})$, $\beta _{i}\in M$, $\beta =\beta _{1}$.

So $L/K$ separable implies $\beta$ separable over $K$, implies $\beta _{1},\dots ,\beta _{s}$ are distinct.

We pick some $c\in K$, and let $\theta =\alpha +c\beta$.

Let $F(X)=f(\theta -cX)\in K(\theta )[X]$. Then $F(\beta )=f(\theta -c\beta )=f(\alpha )=0$. Also, $g(\beta )=0$.

• If $\beta _{2},\dots ,\beta _{s}$ are not roots of $F$, then $\gcd(F,g)=X-\beta$ in $M[X]$. Then $\gcd(F,g)=X-\beta \$in\$K(\theta )[X]$. Therefore $\beta \in K(\theta )$. Thus $\alpha =\theta -c\beta \in K(\theta )$, and thus $K(\alpha ,\beta )\subseteq K(\theta )$. So $K(\alpha ,\beta )=K(\theta )$ by how $\theta$ is defined.

• We are done unless $F(\beta _{j})=0$ for some $2\leq j\leq s$. In this problematic case, $f(\theta -c\beta _{j})=0$ for some $2\leq j\leq s$. Therefore, $\theta =\alpha +c\beta =\alpha _{i}+c\beta _{j}$ for some $1\leq i\leq r$, $2\leq j\leq s$. We can solve for $c$, and we only have finitely many values of $c$ which are a problem. If the field $K$ is infinite we can thus pick a value of $c$ that works. If $K$ is a finite field, then $L$ will be finite and thus $L^{\times }$ is cyclic. But then $L^{\times }$ is generated by $\theta$ and we get $L=K(\theta )$.