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

2.$\mathbb {Z} _{p}$ , the integers modulo $p$ where $p$ is a prime, and $+$ and $\cdot$ are mod $p,$ is a family of finite fields.

## Fields and Homomorphisms

### Definition (embedding)

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 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. 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)

• The real numbers $\mathbb {R}$  can be extended into the complex numbers $\mathbb {C} .$
• 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)

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)

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 $f(\lambda )=0.$
• $K$  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)

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 $a_{1},a_{2},...,a_{n}$  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 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 $x^{p^{n}}-x$ )

let $F$  be a field such that $\left\vert F\right\vert =p^{n}$ , then every member is a root of the polynomial $x^{p^{n}}-x$ .

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 $a_{1},...a_{n}$  are distinct.