Sets and Fields

The set notation linear algebra is written in, and the field axioms its scalars must satisfy. Which familiar systems are fields, which fail and at which axiom, and what every result that says "over a field" is actually assuming.

Definition

Sets. A set is a collection determined by its members: ${1, 2, 3}$ and ${3, 1, 2, 2}$ are the same set, since order and repetition carry no information. Membership is written $x \in S$ , and set-builder notation ${x \in S : P (x)}$ names the members of $S$ satisfying $P$ . A set $A$ is a subset of $B$ , written $A \subseteq B$ , when every member of $A$ belongs to $B$ ; the two are equal exactly when $A \subseteq B$ and $B \subseteq A$ , which is how set identities are proved.

The operations are union $A \cup B$ , intersection $A \cap B$ , difference $A ∖ B$ and Cartesian product $A \times B = {(a, b) : a \in A, b \in B}$ . The last is what makes $R^{2}$ a set of ordered pairs, and $R^{n}$ a set of $n$ -tuples.

Fields. A field is a set $F$ with two operations, addition and multiplication, satisfying:

	Addition	Multiplication
Closure	$a + b \in F$	$a b \in F$
Associativity	$(a + b) + c = a + (b + c)$	$(a b) c = a (b c)$
Commutativity	$a + b = b + a$	$a b = b a$
Identity	$a + 0 = a$	$a \cdot 1 = a$ , with $1 \neq 0$
Inverses	$a + (- a) = 0$	$a a^{- 1} = 1$ for $a \neq 0$

together with distributivity, $a (b + c) = a b + a c$ , which is the only axiom linking the two operations.

The asymmetry is deliberate and is where most failures occur: every element has an additive inverse, but only the nonzero elements are required to have multiplicative inverses. Zero never has one, since $0 \cdot b = 0 \neq 1$ for every $b$ .

Consequences. From the axioms alone: the identities $0$ and $1$ are unique, inverses are unique, $0 \cdot a = 0$ , and $(- 1) a = - a$ . Most importantly, a field has no zero divisors, if $a b = 0$ and $a \neq 0$ then multiplying by $a^{- 1}$ gives $b = 0$ . So in a field a product vanishes only when a factor does, which is exactly what licenses cancellation, and cancellation is what Gaussian elimination and the determinant's expansion silently assume.

Why linear algebra says "over a field". A vector space is defined over a field: the scalars are field elements, and the space axioms use the field axioms directly. Division by a nonzero scalar, normalising a vector, dividing by a pivot, scaling an eigenvector, is multiplication by $a^{- 1}$ , and exists precisely because $F$ is a field.

Assumptions and scope

Multiplicative inverses are required only for nonzero elements. Demanding one for $0$ makes the axioms inconsistent, since $0 \cdot b = 0$ for every $b$ .
The requirement $1 \neq 0$ excludes the one-element set. Without it ${0}$ would satisfy every other axiom and would be a field in which every vector space is trivial.
Closure is necessary but nowhere near sufficient. $Z$ is closed under both operations and fails only at multiplicative inverses; $N$ is closed under both and fails at additive inverses too.
A field has no zero divisors, but the converse does not hold: $Z$ has none and is still not a field. An integral domain is the weaker structure.
$Z_{n}$ is a field exactly when $n$ is prime. For composite $n$ the factors of $n$ are nonzero elements whose product is $0$ .
Ordering is not a field axiom. $C$ is a field with no order compatible with its arithmetic, which is why complex numbers can be added and divided but not compared.

Worked material

Non-example

Structures that fail, and exactly where

Each of these satisfies most of the axioms. Naming the one that fails, and the element witnessing it, is the whole exercise.

$N = {0, 1, 2, \dots}$ — fails axiom 5. Closed under both operations, associative, commutative, distributive, with both identities present. But $2$ has no additive inverse: no natural number $b$ satisfies $2 + b = 0$ . Subtraction is unavailable, and axiom 6 fails too.

$Z$ — fails axiom 6 only. Every other axiom holds, which is what makes it the instructive case. The witness is $2$ : there is no integer $b$ with $2 b = 1$ , since $b$ would have to lie strictly between $0$ and $1$ . This single missing axiom is why elimination over the integers cannot divide by an arbitrary pivot. Adjoining the missing inverses produces $Q$ .

$Z_{6}$ — fails axiom 6, and visibly. Modular arithmetic supplies closure, associativity, commutativity, distributivity and both identities. But

2 \cdot 3 = 6 = 0 (\mod 6),

and two nonzero elements have multiplied to zero. Neither can have an inverse: if $2^{- 1}$ existed, multiplying $2 \cdot 3 = 0$ by it would give $3 = 0$ , which is false. Computing the products confirms that $2$ , $3$ and $4$ all lack inverses, while only $1$ and $5$ have them ( $5 \cdot 5 = 25 = 1 mod 6$ ).

The set of $2 \times 2$ real matrices — fails axioms 3 and 6. Multiplication is not commutative, and a nonzero matrix such as $(\begin{matrix} 1 & 0 \\ 0 & 0 \end{matrix})$ is singular, so has no inverse. It is also a second source of zero divisors, since that matrix times $(\begin{matrix} 0 & 0 \\ 0 & 1 \end{matrix})$ is the zero matrix.

${0}$ — fails axiom 4. Every other axiom holds vacuously, but $1 = 0$ here, which axiom 4 forbids. The exclusion exists precisely to rule this out; without it the trivial set would be a field and every vector space over it would collapse.

Closure is present in all five. It is the cheapest axiom and it decides nothing on its own. The failures live in the inverses, in commutativity, or in the requirement that the two identities be distinct.

Example

Fields of four different kinds

$Q$ , the rationals. The smallest field containing $Z$ . Every nonzero $p / q$ has inverse $q / p$ , which is where $Z$ fell short. Closure under division is exactly what was added.

$R$ , the reals. The field linear algebra defaults to. It carries an order compatible with its arithmetic, which is why real eigenvalues can be sorted and real quadratic forms can be called positive, but the order is extra structure, not a field axiom.

$C$ , the complex numbers. A field with no compatible order. Inverses come from the conjugate: for $z \neq 0$ ,

z^{- 1} = \frac{\bar{z}}{| z |^{2}},

which is defined precisely because $| z |^{2}$ is a nonzero real. It is also algebraically closed, which $R$ is not, and that is why characteristic polynomials always split over it.

$Z_{5}$ , a finite field. Five elements, and every nonzero one invertible:

$a$	1	2	3	4
$a^{- 1}$	1	3	2	4

Checking: $2 \cdot 3 = 6 = 1$ , $4 \cdot 4 = 16 = 1$ , and $1 \cdot 1 = 1$ , all modulo 5. No two nonzero elements multiply to zero, as the prime criterion guarantees. Linear algebra runs over this field unchanged, matrices, elimination, determinants and eigenvalues all work, with arithmetic done modulo 5.

$Q (\sqrt{2}) = {a + b \sqrt{2} : a, b \in Q}$ . Closed under multiplication, since

(a + b \sqrt{2}) (c + d \sqrt{2}) = (a c + 2 b d) + (a d + b c) \sqrt{2},

for instance $(1 + 2 \sqrt{2}) (3 - \sqrt{2}) = - 1 + 5 \sqrt{2}$ . Inverses come from the same conjugate trick as $C$ : multiplying by $a - b \sqrt{2}$ gives the rational norm $a^{2} - 2 b^{2}$ , so

(1 + 2 \sqrt{2})^{- 1} = \frac{1 - 2 \sqrt{2}}{1 - 8} = - \frac{1}{7} + \frac{2}{7} \sqrt{2},

and multiplying back returns exactly $1$ . The norm is never zero for a nonzero element, because $\sqrt{2}$ is irrational, so no rational $a / b$ can satisfy $a^{2} = 2 b^{2}$ .

These five differ in size, in whether they are ordered, and in whether polynomials split, and all satisfy the same seven axioms. That is what makes results stated "over a field" useful: they hold for every one of them at once.

Contrast

Integers against rationals, and $Z_{5}$ against $Z_{6}$

Two pairs, each differing in one axiom, which isolates what that axiom does.

$Z$ and $Q$ .

	$Z$	$Q$
Closure, associativity, commutativity, distributivity	✓
Identities $0$ , $1$ with $1 \neq 0$	✓
Additive inverses	✓
Multiplicative inverses for $a \neq 0$	✗ ( $2$ has none)	✓
Zero divisors	none	none

One axiom apart. The consequence for linear algebra is concrete: over $Q$ elimination divides by any nonzero pivot and reaches reduced row echelon form; over $Z$ that division may leave the set, so the theory of matrices over $Z$ is genuinely different. It is the theory of modules, where Smith normal form replaces row reduction.

Note that $Z$ has no zero divisors and still is not a field. Absence of zero divisors is necessary, not sufficient.

$Z_{5}$ and $Z_{6}$ .

	$Z_{5}$	$Z_{6}$
Size	5	6
Modulus	prime	composite, $6 = 2 \cdot 3$
Nonzero elements with inverses	all four: $1, 2, 3, 4$	only $1$ and $5$
Elements without	none	$2$ , $3$ , $4$
Zero divisors	none	$2 \cdot 3 = 0$
A field?	yes	no

The two are built by the same construction and differ only in the modulus. In $Z_{5}$ : $2 \cdot 3 = 6 = 1$ , so $2$ and $3$ are mutually inverse; $4 \cdot 4 = 16 = 1$ . In $Z_{6}$ the same product $2 \cdot 3$ equals $6 = 0$ instead, and the element that was an inverse in one is a zero divisor in the other.

Primality of the modulus is not an extra condition bolted onto the definition. It is exactly the condition under which no element shares a factor with $n$ , hence under which Bézout supplies an inverse for every nonzero element. Composite moduli hand their own factors back as zero divisors.

A caution about size. Finiteness is not the obstruction. $Z_{5}$ is finite and a field; $Z$ is infinite and is not. Finite fields exist for every prime power, and linear algebra over them is used throughout coding theory and cryptography.

Common errors

Common misconception

A set closed under addition and multiplication is a field, so $Z$ is a field because sums and products of integers are integers.

Related units

Connected

Vector Spaces and Subspaces (related)
Basis and Dimension (related)
Diagonalization (related)
Matrix Inverses and Elementary Matrices (related)
Complex Numbers (related)

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Sets and Fields

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Linear Algebra

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Introduction to Linear Algebra (2016)

Sets and Fields

Assumptions and scope

Worked material

Structures that fail, and exactly where

Fields of four different kinds

Integers against rationals, and $Z_{5}$ against $Z_{6}$

Common errors

Related units

Connected

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Sets and Fields

Assumptions and scope

Worked material

Structures that fail, and exactly where

Fields of four different kinds

Integers against rationals, and Z 5 against Z 6

Common errors

Related units

Connected

Learn this topic

Used in

Sources

Integers against rationals, and $Z_{5}$ against $Z_{6}$