Structure of Simple Algebraic Field Extension

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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 $K[\alpha]$ (resp. $K(\alpha)$) be the subring (resp. subfield) of $F$ generated by $K \cup \{\alpha\}$.

Then


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

Where $\langle \mu_\alpha \rangle$ is the ideal of the ring of polynomial functions generated by $\mu_\alpha$.

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

By the corollary to Field Adjoined Set $\phi$ is surjective, so by the First Isomorphism Theorem,


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

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

Also $K[\alpha]$ is the smallest ring containing $K \cup \{\alpha\}$, so because a field is a ring it is also the smallest field containing $K \cup \{\alpha\}$.

This shows that $K[\alpha] = K(\alpha)$.

By Field Adjoined Set, $K[\alpha] = \{ f(\alpha) : f \in K[X] \}$.

where $K[X]$ is the Ring of Polynomial Functions over $K$.

By the Division Theorem for Polynomial Forms over a Field for each $f \in K[X]$ there are $q,r \in K[X]$ such that $f = q \mu_\alpha + r$ and $\operatorname{deg}r < d =: \operatorname{deg}\mu_\alpha$.

Therefore, since $\mu_\alpha(\alpha) = 0$, we have

Therefore $1,\ldots,\alpha^{d-1}$ span $K[\alpha]/K$ as a vector space.

Moreover $1,\ldots,\alpha^{d-1}$ are linearly independent: if $s \in K[X]$, $s(\alpha) = 0$ and $\operatorname{deg}s < \operatorname{deg}\mu_\alpha$ implies that $s = 0$.

Therefore no non-zero $K$-Definition:Linear Combination of $1,\ldots,\alpha^{d-1}$ is zero.

Therefore $\{1,\ldots,\alpha^{d-1}\}$ is a basis of $K[\alpha]/K$.

The degree of $K[\alpha]/K$ is by definition the number of elements of a basis for $K[\alpha]/K$.

Therefore $d = \operatorname{deg}\mu_\alpha = [K[\alpha] : K]$.