Fields and Subfields

Fields ¶

Fields generalize the properties of familiar “continuous” number systems like the real numbers $ℝ$ and complex numbers $ℂ$ into arbitrary sets which follow a set of field axioms.

In an abstract sense, we need three “things” to define a field: a set of objects called scalars and two binary operations acting on those scalars.

Definition 1: Binary Operation

A function $𝑓$ is a binary operation on a set $𝑆$ iff:

𝑓 : 𝑆 \times 𝑆 \to 𝑆 .

This implicitly requires that any binary operation $𝑓$ on $𝑆$ is closed, meaning that applying $𝑓$ to any two elements of $𝑆$ results in another element of $𝑆$ .

Definition 2: Field

A field $(𝐹, +, \cdot)$ is a set $𝐹$ together with two binary operations $+$ (called addition) and $\cdot$ (called multiplication) such that the following properties hold:

Associativity of addition. $\forall 𝑎, 𝑏, 𝑐 \in 𝐹, (𝑎 + 𝑏) + 𝑐 = 𝑎 + (𝑏 + 𝑐) .$
Associativity of multiplication. $\forall 𝑎, 𝑏, 𝑐 \in 𝐹, (𝑎 \cdot 𝑏) \cdot 𝑐 = 𝑎 \cdot (𝑏 \cdot 𝑐) .$
Commutativity of addition. $\forall 𝑎, 𝑏 \in 𝐹, 𝑎 + 𝑏 = 𝑏 + 𝑎 .$
Commutativity of multiplication. $\forall 𝑎, 𝑏 \in 𝐹, 𝑎 \cdot 𝑏 = 𝑏 \cdot 𝑎 .$
Existence of additive identity. $\exists 0 \in 𝐹 s.t. \forall 𝑎 \in 𝐹, 𝑎 + 0 = 𝑎 .$
Existence of multiplicative identity. $\exists 1 \in 𝐹 s.t. \forall 𝑎 \in 𝐹, 𝑎 \cdot 1 = 𝑎 .$
Existence of additive inverses. $\forall 𝑎 \in 𝐹, \exists (- 𝑎) \in 𝐹 s.t. 𝑎 + (- 𝑎) = 0 .$
Existence of multiplicative inverses. $\forall 𝑎 \in 𝐹, 𝑎 \neq 0 \Rightarrow \exists 𝑎^{- 1} \in 𝐹 s.t. 𝑎 \cdot 𝑎^{- 1} = 1 .$
Distributivity over addition. $\forall 𝑎, 𝑏, 𝑐 \in 𝐹, 𝑎 \cdot (𝑏 + 𝑐) = 𝑎 \cdot 𝑏 + 𝑎 \cdot 𝑐 .$

Together, these are called the field axioms. The elements of $𝐹$ are called $𝑭$ -scalars, and with relevant context, $𝐹$ -scalars may be referred to as simply scalars.

For any $𝑎 \in 𝐹$ , the element $- 𝑎$ is called the additive inverse or the negative of $𝑎$ . For any $𝑎 \in 𝐹 \ {0}$ , the element $𝑎^{- 1}$ is called the multiplicative inverse of $𝑎$ .

Notation

Often, the binary operations associated with a field are not explicitly listed. A field $(𝐹, +, \cdot)$ is usually denoted simply as $𝐹$ , where the operations of addition and multiplication are implied. The notation $+_{𝐹}$ and $\cdot_{𝐹}$ may be used to refer to the addition and multiplication operations of $𝐹$ when there is ambiguity.

Notation

The additive and multiplicative identities of a field $𝐹$ may be denoted $0_{𝐹}$ and $1_{𝐹}$ , respectively, when there is ambiguity.

Notation

When unambiguous, field multiplication can be denoted by juxtaposition. For example, $𝑎 \cdot 𝑏$ may be written as $𝑎 𝑏$ .

Property: Additive and Multiplicative Identities are Unique

Let $𝐹$ be a field. Then there is exactly one additive identity in $𝐹$ , and exactly one multiplicative identity in $𝐹$ . That is, $\exists! 0 \in 𝐹$ and $\exists! 1 \in 𝐹$ .

Proof: Additive and Multiplicative Identities are Unique.

Let $𝐹$ be a field, and suppose that $0_{1}$ and $0_{2}$ are both additive identities of $𝐹$ . By the definition of additive identity, we have $0_{1} + 0_{2} = 0_{1}$ and $0_{1} + 0_{2} = 0_{2}$ . Thus, $0_{1} = 0_{2}$ , so there is exactly one additive identity in $𝐹$ .

Suppose that $1_{1}$ and $1_{2}$ are both multiplicative identities of $𝐹$ . By the definition of multiplicative identity, we have $1_{1} \cdot 1_{2} = 1_{1}$ and $1_{1} \cdot 1_{2} = 1_{2}$ . Thus, $1_{1} = 1_{2}$ , so there is exactly one multiplicative identity in $𝐹$ . ∎

Property: Additive and Multiplicative Inverses are Unique

Let $𝐹$ be a field, and let $𝑎 \in 𝐹$ . Then there is exactly one additive inverse of $𝑎$ in $𝐹$ . If $𝑎 \neq 0$ , then there is exactly one multiplicative inverse of $𝑎$ in $𝐹$ .

That is, $\exists! (- 𝑎) \in 𝐹$ and $𝑎 \neq 0 \Rightarrow \exists! 𝑎^{- 1} \in 𝐹$ .

Property: Products of Fields which are Zero

Let $𝐹$ be a field, and let $𝑎, 𝑏 \in 𝐹$ . If $𝑎 𝑏 = 0$ , then $𝑎 = 0$ or $𝑏 = 0$ .

Proof: Products of Fields which are Zero.

Let $𝐹$ be a field, and let $𝑎, 𝑏 \in 𝐹$ such that $𝑎 𝑏 = 0$ . If $𝑎 = 0$ , then we are done. Otherwise, if $𝑎 \neq 0$ , then by the existence of multiplicative inverses, there exists $𝑎^{- 1} \in 𝐹$ such that $𝑎 𝑎^{- 1} = 1$ . We can deduce:

\begin{matrix} 𝑎^{- 1} (𝑎 𝑏) & = 𝑎^{- 1} \cdot 0 & by assumption \\ (𝑎^{- 1} 𝑎) 𝑏 & = 0 & by associativity of multiplication \\ 1 𝑏 & = 0 & by definition of multiplicative inverse \\ 𝑏 & = 0 & by definition of multiplicative identity \end{matrix}

Thus, if $𝑎 𝑏 = 0$ and $𝑎 \neq 0$ , then $𝑏 = 0$ . ∎

Definition 3: Field Subtraction

Let $𝐹$ be a field. The binary operation of subtraction $-$ on $𝐹$ is defined as follows:

\forall 𝑎, 𝑏 \in 𝐹, 𝑎 - 𝑏 = 𝑎 + (- 𝑏) .

The quantity $𝑎 - 𝑏$ is called the difference of $𝑎$ and $𝑏$ .

Definition 4: Field Division

Let $𝐹$ be a field. The binary operation of division $/$ on $𝐹$ is defined as follows:

\forall 𝑎, 𝑏 \in 𝐹, 𝑎 / 𝑏 = 𝑎 \cdot 𝑏^{- 1} .

The quantity $𝑎 / 𝑏$ is called the quotient of $𝑎$ and $𝑏$ .

Notation

Field division can be denoted by the fraction notation $\div$ . For example, $𝑎 / 𝑏$ may be written as $\frac{𝑎}{𝑏}$ .

Examples of Fields ¶

Theorem 1

The set of all real numbers $ℝ$ is a field.

Proof: Theorem 1.

Let $𝑎, 𝑏, 𝑐$ be arbitrary elements of $ℝ$ . Choose $0 \in ℝ$ to be the additive identity and $1 \in ℝ$ to be the multiplicative identity.

Addition and multiplication of real numbers are associative and commutative, so the first four field axioms are satisfied.

The additive inverse of $𝑎$ is $- 𝑎$ , which is also a real number, so the axiom of existence of additive inverses is satisfied.

If $𝑎 \neq 0$ , choose the multiplicative inverse of $𝑎$ to be $1 / 𝑎$ , which is also a real number, so the axiom of existence of multiplicative inverses is satisfied.

We have $𝑎 (𝑏 + 𝑐) = 𝑎 𝑏 + 𝑎 𝑐$ , so the axiom of distributivity over addition is satisfied. Thus, all field axioms are satisfied, and $ℝ$ is a field. ∎

Proposition 1

The set of all complex numbers $ℂ$ is a field.

Proposition 2

The set of all integers $ℤ$ is not a field.

Proof: Proposition 2.

Let

𝑎 = 2 \in ℤ

. Then there is no multiplicative inverse of

𝑎

ℤ

, since there is no integer

𝑎^{- 1}

such that

𝑎 \cdot 𝑎^{- 1} = 2 𝑎^{- 1} = 1

. Thus,

ℤ

does not satisfy the field axiom of existence of multiplicative inverses, so

ℤ

is not a field. ∎

Proposition 3: Additional Properties of Fields

For the following, let $𝐹$ be a field with additive identity $0_{𝐹}$ .

Distributivity over Multiplication. $\forall 𝑎, 𝑏, 𝑐 \in 𝐹, (𝑎 + 𝑏) 𝑐 = 𝑎 𝑐 + 𝑏 𝑐$ .
Multiplication by Zero. $\forall 𝑎 \in 𝐹, 𝑎 \cdot 0_{𝐹} = 0_{𝐹}$ .
Double Negative Property. $\forall 𝑎 \in 𝐹, - (- 𝑎) = 𝑎$ .
Product of Negatives. $\forall 𝑎, 𝑏 \in 𝐹, (- 𝑎) (- 𝑏) = 𝑎 𝑏$ .
No Zero Divisors. $\forall 𝑎, 𝑏 \in 𝐹, if 𝑎 𝑏 = 0_{𝐹} then 𝑎 = 0_{𝐹} or 𝑏 = 0_{𝐹}$ .

Subfields ¶

A subfield is a subset of a field that is itself a field under the same operations.

Definition 5: Subfield

Let $𝐹$ be a field, and let $𝐻$ be a subset of $𝐹$ . Then $𝐻$ is a subfield of $𝐹$ iff $𝐻$ is itself a field under the same operations of addition and multiplication as $𝐹$ .

If we already know that some set $𝐻$ is a subset of a field $𝐹$ , then only three criteria need to be checked (rather than rechecking all axioms):

Theorem 2: Subfield Criteria

Let $𝐹$ be a field, and let $𝐻$ be a subset of $𝐹$ . Let $0_{𝐹}$ and $1_{𝐹}$ denote the additive and multiplicative identities of $𝐹$ , respectively. Then $𝐻$ is a subfield of $𝐹$ if and only if the following criteria are met:

Existence of identities. $0_{𝐹} \in 𝐻$ and $1_{𝐹} \in 𝐻$ .
Closure under subtraction. $\forall 𝑎, 𝑏 \in 𝐻, 𝑎 - 𝑏 \in 𝐻$ .
Closure under division. $\forall 𝑎, 𝑏 \in 𝐻, 𝑏 \neq 0 \Rightarrow 𝑎 / 𝑏 \in 𝐻 .$

Proof: Subfield Criteria.

Suppose $𝐻$ is a subset of $𝐹$ . Let $𝑝$ be the property that $𝐻$ is a subfield of $𝐹$ , and let $𝑞$ be the property that $𝐻$ satisfies the three conditions listed in Theorem 2.

$𝒑 \Rightarrow 𝒒$ . Assume $𝐻$ is a subfield of $𝐹$ . Then $𝐻$ is a field under the same operations as $𝐹$ .

Existence of identities. By the definition of a field, $𝐻$ contains the additive and multiplicative identities of $𝐹$ , i.e. $0_{𝐹} \in 𝐻$ and $1_{𝐹} \in 𝐻$ .
Closure under subtraction. By the definition of a field, $𝐻$ must contain additive inverses, such that for any $𝑏 \in 𝐻$ , there must be a $- 𝑏 \in 𝐻$ . Since $𝐻$ is closed under addition, for any $𝑎 \in 𝐻$ , $𝑎 + (- 𝑏) = 𝑎 - 𝑏 \in 𝐻$ .
Closure under division. By the definition of a field, $𝐻$ must contain multiplicative inverses, such that for any $𝑏 \in 𝐻$ with $𝑏 \neq 0$ , there must be a $𝑏^{- 1} \in 𝐻$ . Since $𝐻$ is closed under multiplication, for any $𝑎 \in 𝐻$ , $𝑎 \cdot 𝑏^{- 1} = 𝑎 / 𝑏 \in 𝐻$ .

$𝒒 \Rightarrow 𝒑$ . Assume that $𝐻$ satisfies the three conditions listed in Theorem 2. Since $𝐻$ is a subset of $𝐹$ , the operations of addition and multiplication on $𝐻$ are inherited from $𝐹$ .

Associativity of addition and multiplication. Since $𝐻$ is a subset of $𝐹$ and the operations on $𝐻$ are inherited from $𝐹$ , the associativity of addition and multiplication in $𝐹$ implies the associativity of addition and multiplication in $𝐻$ .

Since $𝑝 \Rightarrow 𝑞$ and $𝑞 \Rightarrow 𝑝$ , we have $𝑝 \Leftrightarrow 𝑞$ . ∎

Proposition 4: A Finite Field

For any prime number $𝑝$ , the set of integers between $0$ and $𝑝 - 1$ , inclusive, is a field under addition and multiplication modulo $𝑝$ .

Proof: Proposition 4.

Let $𝑝$ be a prime, and let $𝐹 = {0, 1, 2, \dots, 𝑝 - 1}$ . Define addition and multiplication on $𝐹$ as follows:

\begin{matrix} \forall 𝑎, 𝑏 \in 𝐹, 𝑎 + 𝑏 & = (𝑎 + 𝑏) \mod 𝑝 \\ 𝑎 \cdot 𝑏 & = (𝑎 \cdot 𝑏) \mod 𝑝 . \end{matrix}

It is trivial to see that $𝐹 \subseteq ℝ$ , so we can check the three conditions of Theorem 2:

Existence of identities. Since $𝑝$ is prime, $𝑝 \geq 2$ , so $𝑝 - 1 \geq 1$ . By the definition of $𝐹$ , $0 \in 𝐹$ and $1 \in 𝐹$ .
Closure under subtraction. Let $𝑎, 𝑏 \in 𝐹$ .
- If $𝑎 \geq 𝑏$ , then $𝑎 - 𝑏 \geq 0$ and since $𝑏$ is nonnegative, $𝑎 - 𝑏 < 𝑎 < 𝑝$ , so $𝑎 - 𝑏 \in 𝐹$ .
- If $𝑎 < 𝑏$ , then $𝑎 - 𝑏 < 0$ , but $𝑎 - 𝑏 + 𝑝 \geq 0$ and $𝑎 - 𝑏 + 𝑝 < 𝑝$ , so $𝑎 - 𝑏 + 𝑝 \in 𝐹$ . Since $𝑎 - 𝑏 + 𝑝 \equiv 𝑎 - 𝑏 (\mod 𝑝)$ , $𝑎 - 𝑏 \in 𝐹$ .
Closure under division. Let $𝑎, 𝑏 \in 𝐹$ with $𝑏 \neq 0$ . Since $𝑝$ is prime, $𝑏$ and $𝑝$ are coprime, so there exist integers $𝑥, 𝑦$ such that $𝑏 𝑥 + 𝑝 𝑦 = 1$ . Taking this equation modulo $𝑝$ , we have $𝑏 𝑥 \equiv 1 (\mod 𝑝)$ , so $𝑏 𝑥 \mod 𝑝 = 1$ .
Thus, the multiplicative inverse of $𝑏$ is $𝑥 \mod 𝑝$ , which is an element of $𝐹$ . Since $𝐹$ is closed under multiplication, for any $𝑎 \in 𝐹$ , $𝑎 \cdot (𝑥 \mod 𝑝) = 𝑎 / 𝑏 \in 𝐹$ . ∎

Exercises ¶

Exercise 1.

Show that the set of rational numbers

ℚ

is a subfield of

ℝ

Exercise 2.

Show that the Boolean algebra

{0, 1}

is a field under Boolean addition (XOR) and Boolean multiplication (AND).

Exercise 3.

Prove Proposition 1.

Exercise 4.

Prove that Additive and Multiplicative Inverses are Unique.

Exercise 5.

Prove the Additional Properties of Fields.

Exercise 6.

Let

𝐹

be a field, and let

𝐻_{1}, 𝐻_{2}, \dots, 𝐻_{𝑛}

be a finite collection of subfields of

𝐹

. Show that the intersection of

𝐻_{1} \cap 𝐻_{2} \cap \dots \cap 𝐻_{𝑛}

is also a subfield of

𝐹