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 \(\alpha_i \neq 0\) and \(\beta_i \neq 0\) 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 \(\theta \in \mathbb{F}(\alpha, \beta)\).