Abstract Algebra/Fields

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:

$\forall x\in F-\{0\},\ \exists y\in F : x\cdot y=1$ (every element in F except for 0 has a multiplicative inverse)

Essentially, a field is a commutative division ring.

Examples:

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

2.- $\Bbb{Z}_p$, the integer set modulo p (p a prime positive integer number) with $+_{(mod\ p)}$ and $\cdot_{(mod\ p)}$ operations 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 a field's only ideals are 0 and the field itself, the kernel of a homomorphism is an ideal and f(1_F) = 1_G, F is isometric to its image. Thus, it 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. Now 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 $<\infty$, then G is a finite extension of F, and that G is of degree n=(G:F) over F.

Examples (of field extensions)

• The field of real numbers can be embedded into the complex numbers by taking the image of the real like to be the real axis.
• Similarly, one can add the imaginary number i to the field of rational numbers to form the field of Gaussian integers.

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 α such that α(1) = 1_F, α(2) = 1_f + 1_F etc. This provides the relevant homomorphism.

Note: The Kernel of α 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 = α(m) = α(a)α(b) and, since F is a field and so also an integral domain, α(a) = 0 or α(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(λ) = 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 a₁,a₂,...,a_n are roots of F. Then a smallest Field Extension of F which contains a₁,...,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 a₁,...a_n are distinct.