Math4302 Modern Algebra (Lecture 36)

Rings

Field Extension

If $F\subset E$ is a field extension, then $E$ is a vector space over $F$ .

We define the degree of the extension as

[E:F]\coloneqq \text{ the dimension of $E$ as a $F$ vector space}

Example

$[\mathbb C:\mathbb R]=2$ (take $i,1$ as basis)

$[\mathbb R:\mathbb Q]=\infty$ (no proof for now)

Definition for Algebraic Extension and Transcendental Extension

If $F\subset E$ is an extension, $\alpha\in E$ .

$\alpha$ is algebraic over $F$ if there is a non-zero polynomial $f(x)\in F[x]$ such that $f(\alpha)=0$ .
$\alpha$ is transcendental over $F$ if it is not algebraic over $F$ .

Example

$\mathbb Q\subset \mathbb R$

$\sqrt{2}$ is algebraic over $\mathbb Q$ , $f(x)=x^2-2$ , and $f(\sqrt{2})=0$ , and $f(x)\in \mathbb Q[x]$ .

$\sqrt{3}+\sqrt{2}$ is algebraic over $Q$ , $f(x)=x^4-10x+1$ , and $f(\sqrt{3}+\sqrt{2})=0$ .

$\pi$ and $e$ are transcendental over $\mathbb Q$ . (no proof here)

If $[E:F]<\infty$ , then every element $\alpha\in E$ is algebraic over $F$ .

Set $n=[E:F]$ , then $1,x,...,x^{n-1}$ are $n+1$ elemtns in $E$ , so they are linearly dependent, so there are $c_0,...,c_{n}$ in $F$ such that $\sum_{i=0}^n c_i x^i=0$ . where $\exists c_i\neq 0$ .

So $f(x)=\sum_{i=0}^n c_i x^i\in F[x]$ is a non-zero polynomial such that $f(\alpha)=0$ .

Degree of irreducible polynomial is the degree of the extension

If $f(x)\in F[x]$ is an irreducible polynomial, then if $E=F[x]/(f(x))$ , we have $[E:F]=\deg f(x)$ .

Proof

Recall from the last time, every element in $E$ can be written uniquely as $g(x)+I$ with $g=0$ or $\deg g<\deg f$ .

So every element in $E$ can be uniquely written as

(c_0+c_1x+\cdots+c_{n-1}x^{n-1})+I

which is equivalent to

c_0(1+I)+c_1x(x+I)+\cdots+c_{n-1}(x^{n-1}+I)

So $1+I,x+I,\cdots,x^{n-1}+I$ form a basis for $E$ over $F$ .

Therefore $[E:F]=\deg f(x)$ .

Definition of minimal polynomial

Let $F\subset E$ , $\alpha\in E$ is algebraic over $F$ .

let $I=\{f(x)|f(x)\in F[x],f(\alpha)=0\}\subset F[x]$

Then $I$ is clearly an ideal. We have $I=(p(x))$ with degree of $p(x)$ is the smallest among all polynomials in $I$ .

Every $f(x)\in I$ with $\deg f(x)=\deg p(x)$ is of the form $f=cp$ , where $c$ is a constant in $F$ .

So there is a unique polynomial $p(x)\in I$ such that $I=(p(x))$ . Equivalently, $p$ has the smallest possible degree among all non-zero polynomials in $I$ .

$p(x)$ is monic (leading coefficient is 1). $x^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$ .

We denote it by $\operatorname{irr}(\alpha, F)$ , calling it the minimal (irreducible) polynomial of $\alpha$ over $F$ .

Note that $p(x)$ is irreducible: If $p(x)=p_1(x)p_2(x)$ , then $p_1(\alpha)p_2(\alpha)=0$ . So $p_1(\alpha)=0$ or $p_2(\alpha)=0$ , so $p_1\in I$ or $p_2\in I$ , which is a contradiction.

Example

$\mathbb R\subset \mathbb C$ , $\operatorname{irr}(i,\mathbb R)=x^2+1$ .

If $F\subset E$ , $\alpha\in F$ is algebraic with $\operatorname{irr}(\alpha,F)=x-\alpha$ .

$\mathbb Q\subset \mathbb R$ , $\alpha=\sqrt{1+\sqrt{3}}\in \mathbb Q$ , $\operatorname{irr}(\alpha,\mathbb Q)$

$\alpah^2=1+\sqrt{3}$ , so $\alpha^4-2\alpha^2-2=0$ , then by Eisenstein, $x^4-2x^2-2$ is irreducible over $\mathbb Q$ , take $p=2$ , so $\operatorname{irr}(\alpha,\mathbb Q)=x^4-2x^2-2$ .

Proposition for tower of field extensions

If $F\subset E\subset K$ is a tower of field extensions, then $[K:F]=[E:F]\times [K:E]$ .

Corollary

If $F\subset E$ is a field extension, then the set of elements of $E$ that are algebraic over $F$ form a subfile of $E$ .

Proof

It is enough to show if $\alpha,\beta\in E$ are algebraic over $F$ , then $\alpha\neq \beta$ , $\alpha\beta,\frac{\alpha}{\beta}$ are also algebraic over $F$ .

Suppose $\alpha\in E$ is algebraic over $F$ ,

Let $n=\deg \operatorname{irr}(\alpha,F)$ .

Let $F(\alpha)\subset E$ be the smallest subfield of $E$ containing $F$ and $\alpha$ .

Some lemma: $[F(\alpha):F]=n$ where $n=\deg \operatorname{irr}(\alpha,F)$ .