Structure of Simple Algebraic Field Extension

Theorem
Let $F/K$ be a field extension, and let $\alpha \in F$ be algebraic over $K$.

Let $\mu_\alpha$ be the Minimal Polynomial of $\alpha$ over $K$.

Let $K(\alpha)$ be the subfield of $F$ generated by $K \cup \{\alpha\}$.

Then


 * $K[\alpha] = K(\alpha) \simeq K[X]/\langle \mu_\alpha \rangle$

Moreover $n := [K(\alpha) : K] = \operatorname{deg}\mu_\alpha$ and $1,\alpha,\ldots,\alpha^{n-1}$ is a basis of $K(\alpha)$ over $K$.

Proof
Define $\phi : K[X] \to K[\alpha]$ by $\phi(f) = f(\alpha)$.

We have
 * $\phi(f) = 0 \Leftrightarrow f(\alpha) = 0 \Leftrightarrow \mu_\alpha | f$

where the last equivalence is proved in Minimal Polynomial.

Thus $\operatorname{ker}\phi = \{ f \in K[X] : \mu_\alpha | f\} =: \langle \mu_\alpha \rangle$.

Evidently $\phi$ is surjective, so by the First Isomorphism Theorem,


 * $K[X]/\langle \mu_\alpha \rangle \simeq K[\alpha]$.

Now by Principle Ideal Generated by Irreducible Element is Maximal, $\langle \mu_\alpha \rangle$ is maximal, so $K[\alpha]$ is a field by Maximal Ideal iff Quotient Ring is Field.

Since $K(\alpha) \supseteq K[\alpha]$ and $K(\alpha)$ is the smallest field containing $K \cup \alpha$, it follows that $K[\alpha] = K(\alpha)$.