Abstract Algebra/Fields

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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.


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

2.\Bbb{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 HomomorphismsEdit

Definition (embedding)Edit

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 ExtensionsEdit

Definition (Field Extension and Degree of Extension)Edit

  • 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)Edit

  • The real numbers \Bbb{R} can be extended into the complex numbers \Bbb{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)Edit

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)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 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)Edit

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 FieldsEdit

Definition (Splitting Field)Edit

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.

Theorem (Existence of Splitting Fields)Edit

Finite FieldsEdit

Theorem (Order of any finite field)Edit

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)Edit

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)Edit

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.