Section 29 Introduction to Extension Fields

Definition Extension Field

A First Course in Abstract Algebra, p.265

A field $E$ is an extension field of a field $F$ if $F\le E$.

Example: A tower of fields Here $F(x)$ is the field of all polynomials with coefficients in $F$ (with an indeterminate $x$) (see Intro)

29.3 Kronecker’s Theorem

A First Course in Abstract Algebra, p.266

Let $F$ be a field and let $f(x)$ be a nonconstant polynomial in $F[x]$. Then there exists an extension field $E$ of $F$ and an $\alpha \in E$ such that $f(\alpha )=0$.

In other words, every polynomial has some root once you allow yourself to enlarge your field

• Proof🔽

Simple Examples of Field Extensions

Example 1: Extend $\mathbb{R}$

• We must firstly find an irreducible polynomial. In this case we use $x^2+1$
• We find $\mathbb{C}=\mathbb{R}[x]/⟨x^2+1⟩$
• We force $p(x)=x^2+1=0$, and define an imaginary root $\alpha$ so that ${\alpha }^2=-1$ (the choice of $\alpha$ depends on the choice of the irreducible polynomial)
• We extend $\mathbb{R}$ with this new element (normally written as $i$)
• In the end we have $\mathbb{C}$

It’s interesting to know that according to Frobenius theorem every 2-dimensional extension of $\mathbb{R}$ is $\mathbb{C}$. Understanding this is simple:

• suppose we define $j$ as $j^2=-2$
• we can always substitute $j$ for $\sqrt{(}2)i$, and the result is still in the form $a+bi$

Example 2: Extend $\mathbb{Q}$ Similar to example 1 we find $\mathbb{Q}[x]/⟨x^2-2⟩$ The result is $\mathbb{Q}(\sqrt{(}2))$, which is just $\mathbb{Q}+\sqrt{(}2)$

Summary

• Extending $i$ from $\mathbb{R}$ is just an ordinary procedure. There is nothing special about $i$, we can do the same for $\mathbb{Q}$
• Note that unlike extending on $\mathbb{R}$, $\mathbb{Q}(\sqrt{(}2))$ is structurally different from $\mathbb{Q}(\sqrt{(}3))$

29.4 Example: Construct Field Extensions

An alternative approach of example 1: we want to find the field $\mathbb{E}=\mathbb{R}[x]/⟨x^2+1⟩$

In the factor ring $\mathbb{R}[x]/⟨x^2+1⟩$, every element is a coset of $⟨x^2+1⟩$ , so they are in the form of $p(x)+⟨x^2+1⟩$ If we project $x$ of $\mathbb{R}[x]$ (the indeterminate, see Intro) onto the factor ring, we get $\alpha =x+⟨x^2+1⟩$

NOTE

Consider the homomorphism: $\pi :\mathbb{R}[x]\to \mathbb{E}$ $\pi (p(x))=p(x)+⟨x^2+1⟩$

If we in insert x into $\pi$ , we get this:$\alpha =x+⟨x^2+1⟩$ ($\alpha$ is just for simplicity)

29.6 Definition: Algebraic and Transcendental Elements

A First Course in Abstract Algebra, p.267

An element α of an extension field $E$ of a field $F$ is algebraic over $F$ if $f(\alpha )=0$ for some nonzero $f(x)\in F[x]$. If $\alpha$ is not algebraic over $F$, then $\alpha$ is transcendental over $F$.

Examples:

• $\sqrt{2}$ is an algebraic element over $\mathbb{Q}$, because $\sqrt{2}$ is a zero of $x^2-2$
• Similarly, $i$ is an algebraic element over $\mathbb{Q}$
• $\pi$ and $e$ are transcendental over $\mathbb{Q}$
• BUT $\pi$ and $e$ are algebraic over $\mathbb{R}$, for they are zeros of $x-\pi$ and $x-e\in \mathbb{R}$ respectively.

29.11 Definition: Algebraic and Transcendental Numbers

A First Course in Abstract Algebra, p.268

An element of $\mathbb{C}$ that is algebraic over $\mathbb{Q}$ is an algebraic number. A transcendental number is an element of $\mathbb{C}$ that is transcendental over $\mathbb{Q}$.

29.12 Theorem

A First Course in Abstract Algebra, p.268

Let $E$ be an extension field of a field $F$ and let $\alpha \in E$. Let ${\varphi }_{\alpha }:F[x]\to E$ be the evaluation homomorphism of $F[x]$ into $E$. e.g.

• ${\varphi }_{\alpha }(a)=a$ for $a\in F$
• ${\varphi }_{\alpha }(x)=\alpha$

Then $\alpha$ is transcendental over $F$ if and only if ${\varphi }_{\alpha }$ gives an isomorphism of $F[x]$ with a subdomain of $E$, that is, if and only if ${\varphi }_{\alpha }$ is a one-to-one map

Proof: The element $\alpha$ is transcendental over $F$ if and only if $f(\alpha )\ne 0$ for all nonzero $f(x)\in F[x]$, which is true (by definition) if and only if ${\varphi }_{\alpha }(f(x))\ne 0$ for all nonzero $f(x)\in F[x]$, which is true if and only if the kernel of ${\varphi }_{\alpha }$ is $\{0\}$ (remember 13.13 Definition Kernel), that is, if and only if ${\varphi }_{\alpha }$ is a one-to-one map.

Notes: Remember The Fundamental Homomorphism Theorem: $F[x]/ker{\varphi }_{\alpha }\cong Im({\varphi }_{\alpha })\subseteq E$

• If $\alpha$ is transcendental → $ker{\varphi }_{\alpha }=\{0\}$ → $F[x]\cong Im({\varphi }_{\alpha })$
• If $\alpha$ is algebraic → $ker{\varphi }_{\alpha }\ne \{0\}$ → ${\varphi }_{\alpha }$ is not a one-to-one map

29.13 Theorem: Minimal Polynomial

If $\alpha$ is algebraic over $F$ then there’s a “smallest” irreducible polynomial $p(x)$ in $F[x]$ (unique up to scaling) such that $p(\alpha )=0$. Moreover, any other polynomial $f(x)$ in $F[x]$ that vanishes at $\alpha$ must be divisible by this $p(x)$.

Proof

• We define firstly ${\varphi }_{\alpha }$ as the evaluation homomorphism of $F[x]$ into $E$ (see homomorphism example).
• $ker{\varphi }_{\alpha }$ is an ideal and by 27.24 Theorem Ideal in polynomial Field, it must be a principal ideal generated by some $p(x)\in F[x]$
• we write this ideal as $⟨p(x)⟩$, every other $f(x)$ with $f(\alpha )=0$ is a multiple of $p(x)$, and there exists a irreducible $p(x)$ of minimal degree

Definition: monic polynomial

If we multiply a polynomial by a suitable constant so that the coefficient of the highest power of x appearing in $p(x)$ of 29.13 Theorem Minimal Polynomial is $1$, then this polynomial is called a monic polynomial

Definition: irr() & deg()

A First Course in Abstract Algebra, p.269

Let $E$ be an extension field of a field $F$, and let $\alpha \in E$ be algebraic over $F$. The unique monic polynomial $p(x)$ having the property described in 29.13 Theorem Minimal Polynomial is the irreducible polynomial for $\alpha$ over $F$ and will be denoted by $irr(\alpha ,F)$. The degree of $irr(\alpha ,F)$ is the degree of $\alpha$ over $F$, denoted by $deg(\alpha ,F)$.

Example: $irr(\sqrt{2},\mathbb{Q})=x^2-2$ It’s easy to check that $\alpha =\sqrt{1+\sqrt{3}}$ in $\mathbb{R}$, meaning that $\alpha$ is also a zero of $x^4-2x^2-2$ in $\mathbb{Q}$. Using the Eisenstein divisibility check (where we pick $p=2$), the $x^4-2x^2-2$ is irreducible in $\mathbb{Q}$ $irr(\sqrt{1+\sqrt{3}},\mathbb{Q})=x^4-2x^2-2$ Thus, we have $deg(\sqrt{1+\sqrt{3}},\mathbb{Q})=4$

Warning

We must always specific the field when we are talking about degree or algebraic. Example: $\sqrt{2}\in \mathbb{R}$ is algebraic of degree 2 over $\mathbb{Q}$ but algebraic of degree 1 over $\mathbb{R}$, for $irr(\sqrt{2},\mathbb{R})=x-\sqrt{2}$.

Two ways of simple extensions

${\varphi }_{\alpha }$ is defined as the evaluation homomorphism just as before

Case I

Suppose $\alpha$ is algebraic over $F$. Then the kernel of ${\varphi }_{\alpha }$ is $⟨irr(\alpha ,F)⟩$, which by 27.25 Theorem is a maximal ideal of $F[x]$. Therefore, $F[x]/irr(\alpha ,F)$ is a field and is isomorphic to the image ${\varphi }_{\alpha }[F[x]]$ in $E$. This subfield ${\varphi }_{\alpha }[F[x]]$ is then the smallest subfield of $E$ containing $F$ and $\alpha$. We shall denote this by $F(\alpha )$

Case II

Suppose $\alpha$ is transcendental over $F$. Then ${\varphi }_{\alpha }$ gives an isomorphism of $F[x]$ with a subdomain of $E$. Thus, ${\varphi }_{\alpha }[F[x]]$ is not a field but an integral domain (denoted as $F[\alpha ]$) $E$ contains a field of quotients of $F[\alpha ]$ (just like in Case I, it is denoted by $F(\alpha )$), which is thus the smallest subfield of $E$ containing $F$ and $\alpha$.

29.16 Example

A First Course in Abstract Algebra, p.270

Since π is transcendental over $\mathbb{Q}$, the field $\mathbb{Q}(\pi )$ is isomorphic to the field $\mathbb{Q}(\pi )$ of rational functions over $\mathbb{Q}$ in the indeterminate $x$.

Thus from a structural viewpoint, an element that is transcendental over a field $F$ behaves as though it were an indeterminate over $F$.

It makes sense, because you can construct ${\pi }^2,{\pi }^3,\dots$ from $\pi$, which can all be viewed as basis. Therefore, you get an infinite basis.

Feature	Q(√2)	Q(x) (or Q(π))
Type of adjoined element	Algebraic ($\sqrt{2}$ satisfies the polynomial $x^2-2=0$)	Transcendental ($x$ satisfies no nonzero polynomial over $\mathbb{Q}$)
Vector-space dimension	Finite (2‑dimensional, basis $\{1,\sqrt{2}\}$)	Infinite (basis $\{1,x,x^2,\dots \}$)
Typical element form	$a+b\sqrt{2}$, with $a,b\in \mathbb{Q}$	$\frac{p(x)}{q(x)}$, with $p,q\in \mathbb{Q}[x],q\ne 0$
“Size” of the extension	Small—just one polynomial relation	Huge—no polynomial relations, so you get endlessly many powers

29.17 Definition: simple extension

A First Course in Abstract Algebra, p.270

An extension field $E$ of a field $F$ is a simple extension of $F$ if $E=F(\alpha )$ for some $\alpha \in E$.

29.18 Theorem

A First Course in Abstract Algebra, p.270

Let $E$ be a simple extension $F(\alpha )$ of a field $F$, and let $\alpha$ be algebraic over $F$. Let the degree of $irr(\alpha ,F)$ be $n\ge 1$. Then every element $\beta$ of $E=F(\alpha )$ can be uniquely expressed in the form $\beta =b_0+b_1\alpha +\cdots +b_{n-1}{\alpha }^{n-1}$ where the $b_i$ are in $F$.

29.19 Example: Field GF(4)

• ${\mathbb{Z}}_2[x]$ is a field
• consider ${\mathbb{Z}}_2[x]/⟨x^2+x+1⟩$, we want to find ${\mathbb{Z}}_2(\alpha )$
• We know from 29.18 Theorem ${\mathbb{Z}}_2(\alpha )$ has elements 0 + 0α, 1 + 0α, 0 + 1α, and 1 + 1α OR 0, 1, α, and 1 + α
• THIS is a new finite field of four elements!!! called ${\mathbb{F}}_4$ or $GF(4)$
• The structure of this field is shown below

For example: ${\alpha }^2=-\alpha -1=\alpha +1$, (note that in ${\mathbb{Z}}_2$ , $-1=1$ and $-\alpha =\alpha$)

We can do the same to $\mathbb{R}$ and we get $\mathbb{C}$ as the smallest possible extension field.