LEC ALG3 16

Definition 322.1(Algebraic extension).

A field extension $K/F$ is called an algebraic extension if every $\alpha \in K$ is algebraic over $F$.

Proposition 322.2(Proposition).

Let $K/F$ be a field extension. Assume that $\alpha \in K$ is algebraic over $F$. Then $F(\alpha )/F$ is an algebraic extension.

Proof.

Let $n=\text{deg}(\text{Irr}(\alpha ;F))$. Let $\beta \in F(\alpha )$. Then $\{1,\beta ,\dots ,{\beta }^n\}\subseteq F(\alpha )$ is linearly dependent over $F$ by Proposition 119.11. Thus, there exist coefficients $a_0,\dots ,a_n\in F$ not all zero such that ${\sum }_{i=o}^n{a}_i{\beta }^i=0$. Thus, $\beta$ is algebraic over $F$.□

Proposition 322.3(Proposition).

Let $K/F$ be a field extension and $\alpha \in K$ be algebraic over $F$. Suppose $\text{deg}(\text{irr}(\alpha ;F))=n$. If $\beta \in F(\alpha )$, then $\text{deg}(\text{irr}(\beta ;F))\le n$.

Proof.

Since$\beta \in F(\alpha )$ and since $F(\beta )$ is the smallest subfield of $F(\alpha )$ containing $\beta$ and $F$, we have $F(\beta )\subseteq F(\alpha )$. Since ${\text{dim}}_F(F(\alpha ))=n$, we get ${\text{dim}}_F(F(\beta ))\le n$. We know that $\text{deg}(\text{irr}(\beta ;F))={\text{dim}}_F(F(\beta ))\le n$.□

Proposition 322.4(Proposition).

Let $F\subseteq K$ be a field extension. Suppose $\alpha ,\beta \in K$ be algebraic over $F$. If $\text{irr}(\alpha ;F)=\text{irr}(\beta ;F)$ then there is an isomorphism $\psi :F(\alpha )\to F(\beta )$ fixing elements of $F$ such that $\psi (\alpha )=\beta$.

Proof.

Consider$\psi$ as in the statement. We want to show that $\psi$ is an isomorphism of rings. Note that by definition, $\psi$ is a ring homomorphism (this actually requires the irreducible polynomials to be equal!). Since $\ker \psi \ne (1)$, and $F[\alpha ]$ is a field, we must have $\ker (\psi )=(0)$, that is, $\psi$ is injective. Note that the $F$-basis $\{1,\alpha ,\dots ,{\alpha }^{n-1}\}$ of $F[\alpha ]$ maps to the $F$-basis $\{1,\beta ,\dots ,{\beta }^{n-1}\}$ of $F[\beta ]$. Since $\psi$ is a $F$-linear map, we get $\psi$ is surjective.□

The converse is not true if we do not assume $\psi (\alpha )=\beta$. Consider $F=\mathbb{R}$, $K=\mathbb{C}$, $\alpha =i$, $\beta =2i$.

“Replace field by a ring, then need monic for finite extension.”

[Proposition]
Let $K,K^{\prime }$ be field extensions of $F$ and $\psi :K\to K^{\prime }$ be a ring homomorphism ==fixing $F$==($\psi$ can be extended to a map $\overline{\psi }:K[X]\to K^{\prime }[X]$). Let $\alpha \in K$ be a root of $f(x)\in F[X]$. Then $\psi (\alpha )$ is a root of $\psi (f)$.

[!Proof]-
Let $f(x)={\sum }_{i=0}^n{a}_i{x}^i$. We have $f(\alpha )=0$. Then,

$$\begin{array}{rl}\psi (f)(\psi (\alpha ))& =f(\psi (\alpha ))\\ & =\sum _{i=0}^n{a}_i(\psi (\alpha {)}^i)\\ & =\psi (f(\alpha ))\\ & =\psi (0)=0.\end{array}$$

For example, this shows that if $\alpha$ is a root of $f(x)\in \mathbb{R}[x]$, then $\overline{\alpha }$ is also a root of $f(x)$.

Definition 322.5(Definition).

We say ${\text{dim}}_F(K)$ is the degree of the field extension $K/F$ and denote it by $[K:F]$.
If $[K:F]$ is finite, we say $K/F$ is a finite (field) extension.

By Proposition 119.11, If $K/F$ and $\alpha \in K$ is algebraic, then $[F(\alpha ):F]<\infty$.

Lemma 322.6(Lemma).

If $[K:F]$ is finite, then $K$ is algebraic over $F$.

Proof.

Suppose $[K:F]=n$. Choose $\alpha \in K$. Then the elements $1,\alpha ,\dots ,{\alpha }^n$ are linearly dependent over $F$. A relation of linear dependence now gives the desired polynomial in $F[x]$ that $\alpha$ must satisfy.□