# Abstract Algebra/Fields

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.

Examples:

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 integers.

### 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 $F$.

## 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.