Structure of Simple Algebraic Field Extension
 | This article needs to be linked to other articles. In particular: throughout You can help $\mathsf{Pr} \infty \mathsf{fWiki}$ by adding these links. To discuss this page in more detail, feel free to use the talk page. When this work has been completed, you may remove this instance of {{MissingLinks}} from the code. |
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 $K \sqbrk X$ be the ring of polynomial functions.
Let $\gen {\mu_\alpha}$ be the ideal of $K \sqbrk X$ generated by $\mu_\alpha$.
Let $K \sqbrk \alpha$ be the subring of $F$ generated by $K \cup \set \alpha$.
Let $\map K \alpha$ be the subfield of $F$ generated by $K \cup \set \alpha$.
Then:
- $K \sqbrk \alpha = \map K \alpha \simeq K \sqbrk X / \gen {\mu_\alpha}$
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 Ring 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:
- $d = \deg \mu_\alpha = \index {K \sqbrk \alpha} K$
$\blacksquare$