# Abstract Algebra/Fields

We will first define a field.

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

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

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

Essentially, a field is a commutative division ring.

Examples:

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

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

## Fields and HomomorphismsEdit

### Definition (embedding)Edit

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

## Field ExtensionsEdit

### Definition (Field Extension and Degree of Extension)Edit

• Let F and G be fields. If ${\displaystyle F\subseteq G}$  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, ${\displaystyle [G:F]}$ . If the degree is finite, then ${\displaystyle G}$  is a finite extension of ${\displaystyle F}$ , and ${\displaystyle G}$  is of degree ${\displaystyle n=[G:F]}$  over F.

### Examples (of field extensions)Edit

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

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

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

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

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

### Definition (Characteristic of Field)Edit

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 ExtensionsEdit

### Definition (Algebraic Elements and Algebraic Extension)Edit

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

### Definition (Minimal Polynomial)Edit

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

## Splitting FieldsEdit

### Definition (Splitting Field)Edit

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

## Finite FieldsEdit

### Theorem (Order of any finite field)Edit

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

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

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

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

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

### Theorem (roots of ${\displaystyle x^{p}-x}$  are distinct)Edit

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