# Abstract Algebra/Fields

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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 Homomorphisms

### Definition (embedding)

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 Extensions

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

• 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)

• 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)

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)

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 ${\displaystyle K}$ 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)

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 Fields

### Definition (Splitting Field)

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 Fields

### Theorem (Order of any finite field)

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 ${\displaystyle x^{p}-x}$)

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)

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.