LEC ALG3 15

Field extensions

Definition 119.1(Field extension).

A field extension is an injective ring homomorphism $F\to K$, where $F$ is called the base field, and $K$ is called the extension of $F$. We will denote the field extension $F\subseteq K$ by $K/F$.

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Definition 119.2(Algebraic elements).

Let $K/F$ be a field extension and $\alpha \in K$. We say that $\alpha$ is algebraic over $F$ if $\alpha$ satisfies a polynomial $f(x)\in F[x]$ ¹. $K$ is said to be algebraic if every $\alpha \in K$ is algebraic over $F$. $\alpha \in K$ is said to be transcendental over $F$ if it is not algebraic.

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Example 119.3(Example).

1. $\pi$, $e$ over $\mathbb{Q}$ are transcendental.
2. $2^{1/3}+2^{1/2}+1$ is algebraic over $\mathbb{Q}$; the minimal polynomial is $x^6-6x^5+9x^4+3x^2-42x+31$.

Proposition 119.4(Proposition).

Let $K/F$ be a field extension. Let $\alpha$ be algebraic over $F$. Then there is a unique monic irreducible polynomial in $F[x]$ such that $f(\alpha )=0$.

Proof.

Consider the ring homomorphism $\psi :F[x]\to K$ defined by $x\mapsto \alpha$. Since $F[x]$ is a PID, $\ker \psi =(f(x))$ for some irreducible monic $f$ such that $f(\alpha )=0$. If $g(x)$ is another irreducible monic polynomial in $F[x]$ such that $g(\alpha )=0$, then $f|g$. Since $g$ is irreducible, $f$ and $g$ must be associates, and since $g$ is monic, we have $f=g$.□

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Definition 119.5(Minimal polynomial).

The unique monic polynomial in Proposition 4 satisfied by $\alpha$ is denoted by $\text{Irr}(\alpha ;F)$.

Example 119.6(Example).

1. Consider $\mathbb{R}\subseteq \mathbb{C}$ and $\mathbb{Q}\subseteq \mathbb{C}$. $\text{Irr}(\sqrt{2},\mathbb{R})=x-\sqrt{2}$, $\text{Irr}(\sqrt{2},\mathbb{Q})=x^2-2$.
2. $\text{Irr}(e^{i2\pi j/n},\mathbb{R})=x^k-1$, where $k=\text{lcm}(j,n)/j$.

Definition 119.7(Definition).

Let $K/F$ be a field extension and let $\alpha \in K$. We denote the smallest subfield of $K$ containing $\alpha$ and $F$ by $F(\alpha )$.

The notation $F(\alpha )$ is suggestive. Indeed, it is easy to see that $F(\alpha )$ is the set of all rational functions in $\alpha$ with coefficients in $F$:

$$F(\alpha )=\left\{\frac{f(\alpha )}{g(\alpha )}:f,g\in F[x],g(\alpha )\ne 0\right\}.$$

Suppose $K/F$ and $\alpha \in K$. Clearly,

$$F[\alpha ]=\left\{\sum _{i=0}^n{a}_i{\alpha }^i:a_i\in F\right\}\subseteq F(\alpha ).$$

If $\alpha$ is algebraic over $F$, the reverse inclusion holds too.

Proposition 119.8(Proposition).

Let $K/F$ be a field extension and $\alpha \in K$ be algebraic over $F$.

$$F(\alpha )=F[\alpha ]\cong F[x]/\text{Irr}(\alpha ;F).$$

Proof.

Note that for $\psi :F[x]\to F[\alpha ]$ given by $x\mapsto \alpha$, we have $\ker (\psi )=(\text{Irr}(\alpha ,F))$. Since $\psi$ is surjective, we have $F[x]/\text{Irr}(\alpha ,F)\cong F[\alpha ]$. Thus, $F[\alpha ]$ is a field, and hence $F(\alpha )=F[\alpha ]$.□

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Remark 119.9(Constructing inverses in$F[\alpha ]$).

$F[\alpha ]$ is a field by Proposition 8; how can we obtain the inverse of $f(\alpha )\in F[\alpha ]$?

Let $m(x)=\text{Irr}(\alpha ,F)$. Since $m(x)$ is irreducible in $F[x]$, the gcd of $f(x)$ and $m(x)$ is $1$. Since $F[x]$ is a Euclidean domain, the euclidean algorithm supplies $g(x),h(x)\in F[x]$ such that $f(x)g(x)+m(x)h(x)=1$. Reducing modulo $m(x)$, we obtain $f(x)g(x)\equiv 1\mathrm{mod}m(x)$. Thus, $f(\alpha )g(\alpha )=1$.

Proposition 119.10(Proposition).

$F(\alpha )=F[\alpha ]$ iff $\alpha$ is algebraic over $F$.

Proof.

Let$\alpha \in K$ be transcendental over $F$. If $F(\alpha )=F[\alpha ]$, since ${\alpha }^{-1}\in F(\alpha )$, we have ${\alpha }^{-1}={\sum }_{i=0}^n{a}_i{\alpha }^i⟺0=\left({\sum }_{i=0}^n{a}_i{\alpha }^{i+1}\right)-1$. This implies $\alpha$ is algebraic over $F$, a contradiction.□

Proposition 119.11(Proposition).

Let $K/F$ be a field extension and $\alpha \in K$ be algebraic over $F$. Then, ${\text{dim}}_F(F(\alpha ))=n$, where $n$ is the degree of $\text{Irr}(\alpha ;F)$. Specifically, $B=\{1,\alpha ,\dots ,{\alpha }^{n-1}\}$ is an $F$-basis for $F(\alpha )$.

Proof.

If $B$ were not linearly independent, an equation of linear dependence would yield a polynomial of degree less than $n$ of which $\alpha$ is a root, contradicting Proposition 4.

Let $g(\alpha )\in F[\alpha ]$. Suppose $f=\text{Irr}(\alpha ;F)$. By the division algorithm, there exist unique $q(x),r(x)\in F[x]$ with $r(x)=0$ or $\text{deg}(r(x))<n$ such that

$$\begin{array}{rl}& g(x)=f(x)q(x)+r(x)\\ ⟹& g(\alpha )=r(\alpha ).\end{array}$$

If $r(x)=0$, then $r(\alpha )=0\in \text{span}(B)$. If $r(x)\ne 0$, then

$$r(x)=\sum _{i=0}^{n-1}{a}_i{x}^i$$

and hence $r(\alpha )\in \text{span}(B)$. Therefore, $F[\alpha ]\subseteq \text{span}(B)$. Since $\text{span}(B)\subseteq F[\alpha ]$, we get $F[\alpha ]=\text{span}(B)$.□

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If $K_1$ and $K_2$ are subfields of $K$ containing $\alpha$ and $F$, then $K_1\cap K_2$ is a subfield of $K$ containing $\alpha$ and $F$. In fact, if $\{K_i{\}}_{i\in \mathcal{A}}$ is a family of subfields of $K$ containing $\alpha$ and $F$, then $\tilde{K}:={\bigcap }_{i\in \mathcal{A}}{K}_i$ is a subfield of $K$ containing $\alpha$ and $F$ ².

Footnotes

1. Since $F$ is a field, we can assume that $f$ is monic. ↩

2. Let $\beta \in \tilde{K}\setminus \{0\}$. Then $\beta$ has a unique inverse each $K_i$, say ${\beta }_i$. We must have ${\beta }_i={\beta }_j$ for all $i,j\in \mathcal{A}$. ↩