# Abstract Algebra/Fields

We will first define a field.

**Definition.** A *field* is a non empty set with two binary operations and such that has commutative unitary ring structure and satisfy the following property:

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

Essentially, a field is a commutative division ring.

**Examples:**

1. (rational, real and complex numbers) with standard and operations have field structure. These are examples with infinite cardinality.

2., the integers modulo where is a prime, and and are mod is a family of finite fields.

## Fields and Homomorphisms[edit]

### Definition (embedding)[edit]

An **embedding** is a ring homomorphism from a field to a field . Since the kernel of a homomorphism is an ideal, a field's only ideals are and the field itself, and , we must have the kernel equal to , so that is injective and is isometric to its image under . Thus, the embedding deserves its name.

## Field Extensions[edit]

### Definition (Field Extension and Degree of Extension)[edit]

- Let F and G be fields. If 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, . If the degree is finite, then is a**finite extension**of , and is of degree over F.

### Examples (of field extensions)[edit]

- The real numbers can be extended into the complex numbers
- Similarly, one can add the imaginary number to the field of rational numbers to form the field of Gaussian integers.

### Theorem (Existence of Unique embedding from the integers into a field)[edit]

Let F be a field, then there exists a unique homomorphism

Proof: Define such that , etc. This provides the relevant homomorphism.

Note: The Kernel of is an ideal of . Hence, it is generated by some integer . Suppose for some then and, since is a field and so also an integral domain, or . This cannot be the case since the kernel is generated by and hence 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 Extensions[edit]

### Definition (Algebraic Elements and Algebraic Extension)[edit]

- Let be an extension of then is
**algebraic**over if there exists a non-zero polynomial such that - is an
**algebraic extension**of if is an extension of , such that every element of is algebraic over .

### Definition (Minimal Polynomial)[edit]

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

## Splitting Fields[edit]

### Definition (Splitting Field)[edit]

Let be a field, and are roots of . Then a smallest Field Extension of which contains is called a splitting field of over .

### Theorem (Existence of Splitting Fields)[edit]

## Finite Fields[edit]

### Theorem (Order of any finite field)[edit]

Let F be a finite field, then for some prime p and .

proof: The field of integers mod is a subfield of where is the characteristic of . Hence we can view as a vector space over . Further this must be a finite dimensional vector space because is finite. Hence any can be expressed as a linear combination of members of with scalers in and any such linear combination is a member of . Hence .

### Theorem (every member of F is a root of )[edit]

let be a field such that , then every member is a root of the polynomial .

proof: Consider as a the multiplicative group. Then by la grange's theorem . So multiplying by gives , which is true for all , including .

### Theorem (roots of are distinct)[edit]

Let be a polynomial in a splitting field over then the roots are distinct.