Primitive Element Theorem

The proof of this theorem is constructive. As such, we will prove the simple case, with two known generator of $\mathbb{K}$, showing exactly how the primitive element can be calculated, and then prove the general result by induction.

Proof

Let $\mathbb{K}=\mathbb{F}(\alpha ,\beta )$ be a finite extension and let $f$ and $g$ be the minimal polynomials over $\mathbb{F}$ over $\alpha$ and $\beta$ respectively.

Let ${\alpha }_1,\dots ,{\alpha }_n$ be the conjugates of $\alpha$ (the roots of $f$) and ${\beta }_1,\dots ,{\beta }_m$ be the conjugates of $\beta$ (the roots of $g$) with $\alpha ={\alpha }_1$ and $\beta ={\beta }_1$.

Now, let $c$ be such that

for all $i\in \{1,\dots ,n\}$ and $j\in \{2,\dots ,m\}$.

Such a $c$ exists because there are only finitely many equations, and ${\alpha }_i\ne 0$ and ${\beta }_i\ne 0$ unless $\alpha$ or $\beta$ is zero, a trivial case.

Notice that $f(\alpha )=0$ and therefore $f(\theta -c\beta )=0$. Hence $\beta$ is a root of $f(\theta -cX)$ and $g(X)$. This is the only common root since $g$ has only roots of the form ${\beta }_k$, and if ${\beta }_k$ is a root of $f(\theta -cX)$ then $\theta -c{\beta }_k={\alpha }_i$ (since ${\alpha }_i$ are only roots of $f$) for some ${\alpha }_i$, a contradiction unless ${\beta }_k=\beta$.

Therefore $f(\theta -cX)$ and $g(X)$ has a unique common root, $\beta$.

Let $h$ be the minimal polynomial of $\beta$ over $\mathbb{F}(\theta )$, and consider $f(\theta -cX)$ and $g(X)$ as polynomials in $\mathbb{F}(\theta )[X]$. Since the minimal polynomial divides any polynomial with has the same root, we can deduce that

This means that $h$ is linear, since the two polynomials only have one common root and neither have repeated roots because we are working over a separable extension. Since $h$ is monic, we can then conclude that $h(X)=X-\beta$ which is in $\mathbb{F}(\theta )[X]$ and thus $\beta \in \mathbb{F}(\theta )$.

With $\theta -c\beta =\alpha$ we have that $\alpha \in \mathbb{F}(\theta )$, and hence $\mathbb{F}(\alpha ,\beta )\subseteq \mathbb{F}(\theta )$. The other inclusion follows simply because $\theta \in \mathbb{F}(\alpha ,\beta )$.