# Structure of Simple Algebraic Field Extension

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## 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 $K \sqbrk \alpha$ (resp. $\map K \alpha$) be the subring (resp. subfield) of $F$ generated by $K \cup \set \alpha$.

Then:

- $K \sqbrk \alpha = \map K \alpha \simeq K \sqbrk X / \gen {\mu_\alpha}$

where $\gen {\mu_\alpha}$ is the ideal of the ring of polynomial functions generated by $\mu_\alpha$.

Moreover:

- $n := \index {\map K \alpha} K = \deg \mu_\alpha$

and:

- $1, \alpha, \dotsc, \alpha^{n - 1}$ is a basis of $\map K \alpha$ over $K$.

## Proof

Define $\phi: K \sqbrk X \to K \sqbrk \alpha$ by:

- $\map \phi f = \map f \alpha$

We have

- $\map \phi f = 0 \iff \map f \alpha = 0 \iff \mu_\alpha \divides f$

where the last equivalence follows from $\mu_\alpha$ being the minimal polynomial of $\alpha$.

Thus:

- $\ker \phi = \set {f \in K \sqbrk X: \mu_\alpha \divides f} =: \gen {\mu_\alpha}$

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

So by the First Isomorphism Theorem,

- $K \sqbrk X / \gen {\mu_\alpha} \simeq K \sqbrk \alpha$

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

- $\gen {\mu_\alpha}$ is maximal.

So $K \sqbrk \alpha$ is a field by Maximal Ideal iff Quotient Ring is Field.

Also $K \sqbrk \alpha$ is the smallest ring containing $K \cup \set \alpha$.

So because a field is a ring it is also the smallest field containing $K \cup \set \alpha$.

This shows that $K \sqbrk \alpha = \map K \alpha$.

- $K \sqbrk \alpha = \set {\map f \alpha: f \in K \sqbrk X}$

where $K \sqbrk X$ is the ring of polynomial functions over $K$.

From Division Theorem for Polynomial Forms over Field, for each $f \in K \sqbrk X$ there are $q, r \in K \sqbrk X$ such that:

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

and:

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

Therefore, since $\map {\mu_\alpha} \alpha = 0$, we have

\(\ds K \sqbrk \alpha\) | \(=\) | \(\ds \set {\map r \alpha: r \in K \sqbrk X, \ \deg r < \deg \mu_\alpha}\) | ||||||||||||

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

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

Moreover if $s \in K \sqbrk X$, then $\map s \alpha = 0$ 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 $\set {1, \dotsc, \alpha^{d - 1} }$ is a basis of $K \sqbrk \alpha / K$.

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

Therefore:

- $d = \deg \mu_\alpha = \index {K \sqbrk \alpha} K$

$\blacksquare$