Abstract Algebra/Fields

We will first define a field.

Definition. A field is a non empty set $F$ with two binary operations $+$ and $\cdot$ such that $(F,+,\cdot )$ has commutative unitary ring structure and satisfy the following property:

$\forall x\in F-\{0\}\exists y\in F:x\cdot y=1$

This means that every element in $F$ except for $0$ has a multiplicative inverse.

Essentially, a field is a commutative division ring.

Examples:

1. ${\mathbb {Q}},{\mathbb {R}},{\mathbb {C}}$ (rational, real and complex numbers) with standard $+$ and $\cdot$ operations have field structure. These are examples with infinite cardinality.

2. ${\mathbb {Z}}_{p}$ , the integers modulo $p$ where $p$ is a prime, and $+$ and $\cdot$ are mod $p,$ is a family of finite fields.

Fields and Homomorphisms

Definition (embedding)

An embedding is a ring homomorphism $f:F\rightarrow G$ from a field $F$ to a field $G$ . Since the kernel of a homomorphism is an ideal, a field's only ideals are ${0}$ and the field itself, and $f(1_{F})=1_{G}$ , we must have the kernel equal to ${0}$ , so that $f$ is injective and $F$ is isometric to its image under $f$ . Thus, the embedding deserves its name.

Field Extensions

Definition (Field Extension and Degree of Extension)

Let F and G be fields. If $F\subseteq G$ and there is an embedding from F into G, then G is a field extension of F.
Let G be an extension of F. Consider G as a vector space over the field F. The dimension of this vector space is the degree of the extension, $[G:F]$ . If the degree is finite, then $G$ is a finite extension of $F$ , and $G$ is of degree $n=[G:F]$ over F.

Examples (of field extensions)

The real numbers ${\mathbb {R}}$ can be extended into the complex numbers ${\mathbb {C}}.$
Similarly, one can add the imaginary number $i$ to the field of rational numbers to form the field of Gaussian rationals.

Theorem (Existence of Unique embedding from the integers into a field)

Let F be a field, then there exists a unique homomorphism $\alpha :\mathbb {Z} \rightarrow F.$

Proof: Define $\alpha$ such that $\alpha (1)=1_{F}$ , $\alpha (2)=1_{f}+1_{F}$ etc. This provides the relevant homomorphism.

Note: The Kernel of $\alpha$ is an ideal of $\mathbb {Z}$ . Hence, it is generated by some integer $m$ . Suppose $m=ab$ for some $a,b\in \mathbb {Z}$ then $0=\alpha (m)=\alpha (a)\alpha (b)$ and, since $F$ is a field and so also an integral domain, $\alpha (a)=0$ or $\alpha (b)=0$ . This cannot be the case since the kernel is generated by $m$ and hence $m$ must be prime or equal 0.

Definition (Characteristic of Field)

The characteristic of a field can be defined to be the generator of the kernel of the homomorphism, as described in the note above.

Algebraic Extensions

Definition (Algebraic Elements and Algebraic Extension)

Let $K$ be an extension of $F$ then $\lambda \in K$ is algebraic over $F$ if there exists a non-zero polynomial $f(x)\in F[x]$ such that $f(\lambda )=0.$
$K$ is an algebraic extension of $F$ if $K$ is an extension of $F$ , such that every element of $K$ is algebraic over $F$ .

Definition (Minimal Polynomial)

If $x$ is algebraic over $F$ then the set of polynomials in $F[x]$ which have $x$ as a root is an ideal of $F[x]$ . This is a principle ideal domain and so the ideal is generated by a unique monic non-zero polynomial, $m(x)$ . We define the $m(x)$ to be the minimal polynomial.

Splitting Fields

Definition (Splitting Field)

Let $F$ be a field, $f(x)\in F[x]$ and $a_{1},a_{2},...,a_{n}$ are roots of $F$ . Then a smallest Field Extension of $F$ which contains $a_{1},...,a_{n}$ is called a splitting field of $f(x)$ over $F$ .

Finite Fields

Theorem (Order of any finite field)

Let F be a finite field, then $\left\vert F\right\vert =p^{n}$ for some prime p and $n\in\mathbb{N}$ .

proof: The field of integers mod $p$ is a subfield of $F$ where $p$ is the characteristic of $F$ . Hence we can view $F$ as a vector space over ${\mathbb {Z}}_{p}$ . Further this must be a finite dimensional vector space because $F$ is finite. Hence any $x\in F$ can be expressed as a linear combination of $n$ members of $F$ with scalers in ${\mathbb {Z}}_{p}$ and any such linear combination is a member of $F$ . Hence $\left\vert F\right\vert =p^{n}$ .

Theorem (every member of F is a root of $x^{p}-x$ )

let $F$ be a field such that $\left\vert F\right\vert =p^{n}$ , then every member is a root of the polynomial $x^{p}-x$ .

proof: Consider $F^{*}=F/{0}$ as a the multiplicative group. Then by la grange's theorem $\forall x\in F^{*},x^{p^{n}-1}=1$ . So multiplying by $x$ gives $x^{p^{n}}=x$ , which is true for all $x\in F$ , including $0$ .

Theorem (roots of $x^{p}-x$ are distinct)

Let $x^{p}-x$ be a polynomial in a splitting field $E$ over $\mathbb {Z} _{p}$ then the roots $a_{1},...a_{n}$ are distinct.

This article is issued from Wikibooks. The text is licensed under Creative Commons - Attribution - Sharealike. Additional terms may apply for the media files.