Simple Algebraic Field Extension consists of Polynomials in Algebraic Number
Let $F$ be a field.
Let $\theta \in \C$ be algebraic over $F$.
Let $\map F \theta$ be the simple field extension of $F$ by $\theta$.
Then $\map F \theta$ consists of polynomials that can be written in the form $\map f \theta$, where $\map f x$ is a polynomial over $F$.
Let $H$ be the set of all numbers which can be written in the form $\map f \theta$.
We have that:
- $H$ is closed under addition and multiplication.
- $H$ contains $0$ and $1$
Let $\map f \theta \ne 0$.
Then $\theta$ is not a root of $\map f x$.
Hence from Polynomial with Algebraic Number as Root is Multiple of Minimal Polynomial:
- the minimal polynomial $\map m x$ in $\theta$ does not divide $\map f x$.
From Minimal Polynomial is Irreducible, the GCD of $\map m x$ and $\map f x$ is $1$.
- $\exists \map s x, \map t x: \map s x \map m x + \map t x \map f x = 1$
Substituting for $\theta$:
- $\map s \theta \, \map m \theta + \map t \theta \, \map f \theta = 1$
Because $\map m \theta = 0$ it follows that:
- $\map t \theta \, \map f \theta = 1$
We have that $\map t \theta \in H$.
Thus $\map t \theta$ is the product inverse of $\map f x$ in $H$.
Thus $H$ is a field.
A field containing $F$ and $\theta$ must contain $1$ and all the powers of $\theta$ for positive integer index.
Hence such a field also contains all linear combinations of these, with coefficients in $F$.
So a field containing $F$ and $\theta$ contains all the elements of $H$:
- $H \subseteq \map F \theta$
But by definition, $\map F \theta$ is the smallest field containing $F$ and $\theta$.
- $\map F \theta \subseteq H$
- $\map F \theta = H$
and the result follows.
- 1969: C.R.J. Clapham: Introduction to Abstract Algebra ... (previous) ... (next): Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Theorem $72$