# Structure of Simple Algebraic Field Extension

## Theorem

Let $F / K$ be a field extension.

Let $\alpha \in F$ be algebraic over $K$.

Let $\mu_\alpha$ be the minimal polynomial of $\alpha$ over $K$.

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

Then:

- $K \left[{\alpha}\right] = K \left({\alpha}\right) \simeq K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle$

where $\left\langle{\mu_\alpha}\right\rangle$ is the ideal of the ring of polynomial functions generated by $\mu_\alpha$.

Moreover:

- $n := \left[{K \left({\alpha}\right) : K}\right] = \deg \mu_\alpha$

and:

- $1, \alpha, \dotsc, \alpha^{n - 1}$ is a basis of $K \left({\alpha}\right)$ over $K$.

## Proof

Define $\phi: K \left[{X}\right] \to K \left[{\alpha}\right]$ by:

- $\phi \left({f}\right) = f \left({\alpha}\right)$.

We have

- $\phi \left({f}\right) = 0 \iff f \left({\alpha}\right) = 0 \iff \mu_\alpha | f$

where the last equivalence is proved in Minimal Polynomial.

Thus:

- $\ker \phi = \left\{ {f \in K \left[{X}\right]: \mu_\alpha | f}\right\} =: \left\langle{\mu_\alpha}\right\rangle$

By the corollary to Field Adjoined Set $\phi$ is surjective.

So by the First Isomorphism Theorem,

- $K \left[{X}\right] / \left\langle{\mu_\alpha}\right\rangle \simeq K \left[{\alpha}\right]$

By Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal:

- $\left\langle{\mu_\alpha}\right\rangle$ is maximal.

So $K \left[{\alpha}\right]$ is a field by Maximal Ideal iff Quotient Ring is Field.

Also $K \left[{\alpha}\right]$ is the smallest ring containing $K \cup \left\{ {\alpha}\right\}$.

So because a field is a ring it is also the smallest field containing $K \cup \left\{ {\alpha}\right\}$.

This shows that $K \left[{\alpha}\right] = K \left({\alpha}\right)$.

- $K \left[{\alpha}\right] = \left\{ {f \left({\alpha}\right) : f \in K \left[{X}\right] }\right\}$

where $K \left[{X}\right]$ is the ring of polynomial functions over $K$.

From Division Theorem for Polynomial Forms over Field, for each $f \in K \left[{X}\right]$ there are $q, r \in K \left[{X}\right]$ such that:

- $f = q \mu_\alpha + r$

and:

- $\deg r < d =: \deg \mu_\alpha$

Therefore, since $\mu_\alpha \left({\alpha}\right) = 0$, we have

\(\displaystyle K \left[{\alpha}\right]\) | \(=\) | \(\displaystyle \left\{ {r \left({\alpha}\right): r \in K \left[{X}\right], \ \deg r < \deg \mu_\alpha}\right\}\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \left\{ {a_0 + a_1 \alpha + \dotsb + a_{d - 1} \alpha^{d - 1}: \alpha_0, \dotsc, \alpha_d \in K}\right\}\) |

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

Moreover if $s \in K \left[{X}\right]$, then $s \left({\alpha}\right) = 0$ and $\deg s < \deg \mu_\alpha$ implies that $s = 0$.

Thus $1, \dotsc, \alpha^{d - 1}$ are linearly independent.

Therefore no non-zero $K$-combination of $1, \dotsc, \alpha^{d-1}$ is zero.

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

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

Therefore:

- $d = \deg \mu_\alpha = \left[{K \left[{\alpha}\right] : K}\right]$

$\blacksquare$