[Finite-Dimensional Linear Algebra] 2.1 Fields

Definition of a field

Let $F$ be a nonempty set with two defined operations for $\alpha , \beta \in F$ called addition and multiplication.

• addition: $\alpha + \beta \in F$
• multiplication: $\alpha\beta \in F$

 

$F$ is a field iff (if and only if) the operations satisfy the following properties: 

A. Axioms for Addition

1. Associative property of addition
   • for all $\alpha , \beta , \gamma \in F$, $$(\alpha + \beta) + \gamma = \alpha + (\beta + \gamma)$$
2. Commutative property of addition
   • for all $\alpha , \beta \in F$, $$\alpha + \beta = \beta + \alpha$$
3. Existence of an additive identity
   • there exists a unique element 0 of $F$ for all $\alpha \in F$ such that $$\alpha + 0 = \alpha$$
4. Existence of additive inverses
   • for each $\alpha \in F$, there exists a unique element $-\alpha \in F$ such that $$\alpha + (-\alpha ) = 0$$

B. Axioms for Multiplication

1. Associative property of multiplication
   • for all $\alpha , \beta , \gamma \in F$, $$(\alpha \beta )\gamma = \alpha (\beta \gamma)$$
2. Commutative property of multiplication
   • for all $\alpha , \beta \in F$, $$\alpha \beta = \beta \alpha$$
3. Existence of a multiplicative identity
   • there exists a unique nonzero element 1 of $F$ for all $\alpha \in F$ such that  $$\alpha \cdot 1 = \alpha$$
4. Existence of multiplicative inverses
   • for each $\alpha \in F$, $\alpha \neq 0$ there exists a unique element $\alpha^{-1} \in F$ with the property that $$\alpha \alpha^{-1} = 1$$

C. Distributive Axiom

1. Distributive property of multiplication over addition
   • for all $\alpha , \beta , \gamma \in F$, $$\alpha (\beta + \gamma ) = \alpha \beta + \alpha \gamma$$

Infinite Fields

$(\mathbb{Q} , +, \times ) \leq ( \mathbb{R} , +, \times ) \leq ( \mathbb{C} , +, \times )$

• $\mathbb{Q}  =  \left\{ \frac{n}{m} \mid m, n \in Z, m \neq 0 \right\}$, the set of all rational numbers
• $\mathbb{R}$ = the set of all real numbers
• $\mathbb{C}  =  \left\{ a + bi \mid a, b \in R \right\}$, the set of all complex numbers

 

Q. Why are $( \mathbb{Z} , +, \times )$ and $( \mathbb{N} , +, \times )$ NOT fields? 

• $\mathbb{Z}  = {..., -2, -1, 0, 1, 2, ...}$, the set of all integers
• $\mathbb{N}  = {1, 2, 3, ...}$, the set of all natural numbers 

A. The set $\mathbb{Z}$ of integers violates axiom (B4), the existence of multiplicative inverses; only integers 1 and -1 have multiplicative inverses that are also integers.

For each $n \in \mathbb{Z}, n \geq 2$ or $n \leq -2$,

$$\left| n \right| > 1$$

$$\Rightarrow 0 < \left| \frac{1}{n} \right| < 1$$

$$\Rightarrow \frac{1}{n} \notin \mathbb{Z}$$

The set of natural numbers $\mathbb{N} \subset \mathbb{Z}$ is also not a field. It violates axioms (A3), (A4), and (B4). 

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Basic Properties of Fields

Theorem (Cancellation Properties) Let $\alpha , \beta , \gamma$ be elements of field $F$. Then,

1. if $\alpha , \beta , \gamma \in F$ and $\alpha + \gamma = \beta + \gamma$, then $\alpha = \beta$
2. if $\alpha , \beta , \gamma \in F, \alpha \gamma = \beta \gamma$, and $\gamma \neq 0$, then $\alpha = \beta$

 

PROOF (1)

Given the following statement $$\alpha + \gamma = \beta + \gamma ,$$ Add the additive inverse of $\gamma$ on both sides: $$(\alpha + \gamma ) + (-\gamma ) = (\beta + \gamma ) + (-\gamma )$$ By associative property, $$\alpha + (\gamma  + (-\gamma )) = \beta + (\gamma  + (-\gamma ))$$ Since we know that $\gamma + (-\gamma ) = 0$, $$\alpha + 0 = \beta + 0$$ $$\Rightarrow \alpha = \beta$$

 

Theorem Let $F$ be a field. Then, for each $\alpha \in F$,

1. $0 \cdot \alpha = 0$
2. $-1 \cdot \alpha = -\alpha$

 

PROOF (1)

As stated by axiom (A3),  $$0 + \alpha = \alpha$$ Multiply both sides by $\alpha$: $$(0 + \alpha )\alpha = \alpha \cdot \alpha$$ The distributive property yields $$0 \cdot \alpha + \alpha \cdot \alpha = \alpha \cdot \alpha$$ By cancellation property, $$0 \cdot \alpha = 0$$

 

PROOF (2)

Assume the following statement: $$\alpha + (-1)\cdot \alpha = 0$$ Since $\alpha = 1 \cdot \alpha$, $$1 \cdot \alpha + (-1) \cdot \alpha = 0$$ By distributive property, $$(1 + (-1))\alpha = 0$$ $$\Rightarrow 0 \cdot \alpha = 0$$ Hence the statement is true and $-1 \cdot \alpha$ is a unique additive inverse of $\alpha$, $$-1 \cdot \alpha = -\alpha$$