Fields

What numbers are we allowed to use when we do linear algebra? This sounds like a trivial question. Surely the answer is just "real numbers." But the moment you try to solve x2=−1x^2 = -1, or count solutions to equations modulo a prime, you realize the choice of number system is a genuine design decision. Every theorem about vector spaces, matrices, and linear maps secretly depends on which system of scalars we allow ourselves to add, subtract, multiply, and divide by. That system must behave well enough for the usual algebraic manipulations to go through without surprises. This includes clearing denominators, solving linear equations, and cancelling nonzero factors. A field is the precise algebraic structure that captures this idea. It describes "a number system where you can add, subtract, multiply, and divide (by nonzero elements) exactly as you expect." The rational numbers Q\mathbb{Q}, the real numbers R\mathbb{R}, and the complex numbers C\mathbb{C} are the fields most familiar from calculus and algebra. But there are also finite fields, such as Z/pZ\mathbb{Z}/p\mathbb{Z} for a prime pp. These look nothing like the number line, yet satisfy exactly the same axioms. This uniformity is the point. Once we know a structure is a field, every field axiom is available to us. This holds regardless of whether its elements are decimals, symbols, or residues. The reason this chapter exists inside a course on Systems of Linear Equations is simple. Gaussian elimination — the central algorithm of the course so far — only ever uses the field axioms. You add rows, multiply rows by nonzero scalars, and subtract multiples of one row from another. Nowhere do you use anything special about R\mathbb{R} beyond being a field. This means every fact about solving linear systems will automatically hold over any field. The same goes for everything to come about matrices, transposes, and determinants. This is the quiet, structural payoff of isolating the field axioms. We do this before building vector spaces on top of them. This article gives the complete list of field axioms. It proves the basic algebraic facts directly from the axioms: uniqueness of 00 and 11, cancellation, and the impossibility of dividing by zero. It also works through Q,R,C\mathbb{Q}, \mathbb{R}, \mathbb{C}, and the finite field F5\mathbb{F}_5 in full detail. It closes with a counterexample. This shows why a field — rather than a more general ring — is the right base structure for the linear algebra to come.

Key takeaways
  • A field is a set with ++ and ×\times satisfying commutativity, associativity, distributivity, identities 0,10,1, additive inverses, and multiplicative inverses for nonzero elements.
  • Q,R,C\mathbb{Q}, \mathbb{R}, \mathbb{C} are the classical infinite fields; Z/pZ\mathbb{Z}/p\mathbb{Z} (for prime pp) are the classical finite fields.
  • Every field axiom is used, at some point, in Gaussian elimination — division by a pivot requires multiplicative inverses.
  • Basic facts like 0⋅a=00 \cdot a = 0, uniqueness of inverses, and cancellation are theorems, not axioms — they must be (and are) proved from the axioms.
  • Z/nZ\mathbb{Z}/n\mathbb{Z} fails to be a field when nn is not prime, because nonzero elements can lack multiplicative inverses.
  • Key idea: linear algebra is built on top of whatever field of scalars you fix, and all its theorems transfer verbatim to any field.

Where this lives today

Fields are the invisible scaffolding beneath essentially all of linear algebra: vector spaces, matrices, determinants, and eigenvalues are all defined "over a field," and every theorem proved from the field axioms (rank-nullity, Gaussian elimination, Cramer's rule) transfers automatically whenever the scalar field changes. In cryptography, finite fields Fp\mathbb{F}_p and F2n\mathbb{F}_{2^n} underlie the arithmetic of AES encryption, elliptic-curve cryptography, and Reed–Solomon error-correcting codes used in QR codes and deep-space communication. In coding theory more broadly, linear codes are literally subspaces of Fqn\mathbb{F}_q^n for a finite field Fq\mathbb{F}_q, and their entire theory rests on the field axioms holding exactly as in R\mathbb{R}. In algebraic number theory and Galois theory, fields such as Q(2)\mathbb{Q}(\sqrt{2}) or Q(i)\mathbb{Q}(i) are studied via their symmetries, extending Galois's original nineteenth-century insight. In physics, the complex field C\mathbb{C} is the mandatory scalar field for quantum mechanics, since quantum amplitudes and the spectral theorem for unitary and Hermitian operators require complex, not merely real, scalars. In computer algebra and computational science, finite-field arithmetic is used for exact linear algebra (avoiding floating-point rounding) and for polynomial factorization algorithms.

Ask the tutor: Why does linear algebra require scalars to form a field rather than just a commutative ring?

DefinitionField

A field is a set FF together with two binary operations

+:F×F→F,⋅:F×F→F,+: F \times F \to F, \qquad \cdot : F \times F \to F,

called addition and multiplication, satisfying the following axioms for all a,b,c∈Fa, b, c \in F: 1. (Closure) a+b∈Fa + b \in F and a⋅b∈Fa \cdot b \in F. 2. (Commutativity) a+b=b+aa + b = b + a and a⋅b=b⋅aa \cdot b = b \cdot a. 3. (Associativity) (a+b)+c=a+(b+c)(a+b)+c = a+(b+c) and (a⋅b)⋅c=a⋅(b⋅c)(a\cdot b)\cdot c = a \cdot (b \cdot c). 4. (Distributivity) a⋅(b+c)=a⋅b+a⋅ca \cdot (b+c) = a\cdot b + a \cdot c. 5. (Additive identity) There exists an element 0∈F0 \in F with a+0=aa + 0 = a for all a∈Fa \in F. 6. (Additive inverses) For each a∈Fa \in F there exists −a∈F-a \in F with a+(−a)=0a + (-a) = 0. 7. (Multiplicative identity) There exists 1∈F1 \in F with 1≠01 \neq 0, and a⋅1=aa \cdot 1 = a for all a∈Fa \in F. 8. (Multiplicative inverses) For each a∈Fa \in F with a≠0a \neq 0, there exists a−1∈Fa^{-1} \in F with a⋅a−1=1a \cdot a^{-1} = 1. A set with +,⋅+, \cdot satisfying axioms 1–6 only (no multiplicative inverses required, and multiplicative identity optional) is called a commutative ring; a field is thus a commutative ring in which 1≠01 \neq 0 and every nonzero element is invertible.

Intuition

Think of a field as a minimal "arithmetic sandbox": whatever objects you put in it, you are guaranteed you can add, subtract, multiply, and — crucially — divide by anything nonzero, and the familiar rules (distributing multiplication over addition, commuting sums and products) all hold. The axioms are chosen to be exactly the properties used, over and over, when solving a system of linear equations by elimination: swapping two equations uses commutativity implicitly, combining equations uses distributivity, and — the step that truly requires a field — dividing a row by its pivot entry uses the existence of multiplicative inverses. The picture below shows two very different-looking fields sitting side by side: the familiar real number line R\mathbb{R}, and the finite field F5={0,1,2,3,4}\mathbb{F}_5 = \{0,1,2,3,4\} with arithmetic mod 55, drawn as a clock. Despite looking nothing alike — one is infinite and ordered, the other is finite and cyclic — both satisfy every single field axiom. This is exactly why a theorem proved from the field axioms alone (such as facts about linear systems) applies to both without change.

Figure
Two fields that look nothing alike: R\mathbb{R} as a line, F5\mathbb{F}_5 as a 5-cycle
Field ℝ (real numbers) -2 -1 0 1 2 0, 1 : identities Field 𝔽₅ = ℤ/5ℤ (clock arithmetic) 0 1 2 3 4 e.g. 3 + 4 = 7 ≡ 2 (mod 5) e.g. 3⁻¹ = 2, since 3·2 = 6 ≡ 1

Basic algebraic facts, proved from the axioms

Before stating the central theorem of this article, it is worth pausing to prove that several "obvious" facts about arithmetic are genuine consequences of the axioms, not extra assumptions. This is exactly the kind of rigor the axiomatic method demands: nothing is used unless it has been either postulated or proved.

LemmaUniqueness of identities and inverses

Let FF be a field. Then: (a) The additive identity 00 is unique. (b) The multiplicative identity 11 is unique. (c) For each a∈Fa \in F, the additive inverse −a-a is unique. (d) For each nonzero a∈Fa \in F, the multiplicative inverse a−1a^{-1} is unique.

Proof

(a) Suppose 00 and 0′0' both satisfy the additive identity axiom, so a+0=aa + 0 = a and a+0′=aa + 0' = a for all a∈Fa \in F. Then in particular, using a=0′a = 0' in the first statement and a=0a = 0 in the second (with commutativity, axiom 2):

0′=0′+0=0+0′=0.0' = 0' + 0 = 0 + 0' = 0.

Hence 0′=00' = 0, so the additive identity is unique.

(b) Identical argument with multiplication: if 11 and 1′1' both satisfy a⋅1=aa \cdot 1 = a and a⋅1′=aa \cdot 1' = a for all aa, then

1′=1′⋅1=1⋅1′=1,1' = 1' \cdot 1 = 1 \cdot 1' = 1,

using commutativity of multiplication (axiom 2) and substituting a=1′a = 1', then a=1a=1.

(c) Fix a∈Fa \in F and suppose bb and b′b' both satisfy a+b=0a + b = 0 and a+b′=0a + b' = 0. Then, using associativity (axiom 3), commutativity (axiom 2), and the additive identity property (axiom 5):

b=b+0=b+(a+b′)=(b+a)+b′=(a+b)+b′=0+b′=b′.b = b + 0 = b + (a + b') = (b+a) + b' = (a+b) + b' = 0 + b' = b'.

So b=b′b = b', proving additive inverses are unique.

(d) Fix nonzero a∈Fa \in F and suppose c,c′c, c' both satisfy a⋅c=1a \cdot c = 1 and a⋅c′=1a \cdot c' = 1. Then

c=c⋅1=c⋅(a⋅c′)=(c⋅a)⋅c′=(a⋅c)⋅c′=1⋅c′=c′,c = c \cdot 1 = c \cdot (a \cdot c') = (c \cdot a) \cdot c' = (a \cdot c) \cdot c' = 1 \cdot c' = c',

using associativity and commutativity of multiplication as above. So c=c′c = c'. ■\blacksquare

∎
LemmaZero annihilates, and cancellation holds

Let FF be a field and let a,b,c∈Fa, b, c \in F. (a) a⋅0=0a \cdot 0 = 0. (b) If a⋅b=0a \cdot b = 0, then a=0a = 0 or b=0b = 0 (no zero divisors). (c) If a+c=b+ca + c = b + c, then a=ba = b (additive cancellation). (d) If a⋅c=b⋅ca \cdot c = b \cdot c and c≠0c \neq 0, then a=ba = b (multiplicative cancellation).

Proof

(a) Using the additive identity axiom, 0+0=00 + 0 = 0. Multiply both sides by aa and use distributivity (axiom 4):

a⋅0=a⋅(0+0)=a⋅0+a⋅0.a \cdot 0 = a \cdot (0 + 0) = a\cdot 0 + a \cdot 0.

So a⋅0=a⋅0+a⋅0a \cdot 0 = a\cdot 0 + a \cdot 0. Add −(a⋅0)-(a\cdot 0), the additive inverse of a⋅0a \cdot 0 (axiom 6), to both sides:

a⋅0+(−(a⋅0))=(a⋅0+a⋅0)+(−(a⋅0)).a \cdot 0 + \big(-(a\cdot 0)\big) = \big(a\cdot 0 + a\cdot 0\big) + \big(-(a \cdot 0)\big).

The left side is 00 by axiom 6. On the right, associativity (axiom 3) gives

a⋅0+(a⋅0+(−(a⋅0)))=a⋅0+0=a⋅0.a\cdot 0 + \Big(a \cdot 0 + \big(-(a\cdot 0)\big)\Big) = a \cdot 0 + 0 = a \cdot 0.

Hence 0=a⋅00 = a \cdot 0, i.e. a⋅0=0a \cdot 0 = 0.

(b) Suppose a⋅b=0a \cdot b = 0 and a≠0a \neq 0; we must show b=0b = 0. Since a≠0a \neq 0, axiom 8 gives a−1∈Fa^{-1} \in F with a−1⋅a=1a^{-1} \cdot a = 1. Multiply both sides of a⋅b=0a \cdot b = 0 on the left by a−1a^{-1}:

a−1⋅(a⋅b)=a−1⋅0.a^{-1} \cdot (a \cdot b) = a^{-1} \cdot 0.

By part (a), the right side is 00. By associativity (axiom 3) and axiom 8, the left side is

(a−1⋅a)⋅b=1⋅b=b.(a^{-1} \cdot a) \cdot b = 1 \cdot b = b.

Hence b=0b = 0.

(c) Suppose a+c=b+ca + c = b+c. Add −c-c (axiom 6) to both sides:

(a+c)+(−c)=(b+c)+(−c).(a+c) + (-c) = (b+c)+(-c).

By associativity, a+(c+(−c))=b+(c+(−c))a + (c + (-c)) = b + (c+(-c)), so a+0=b+0a + 0 = b+0, so a=ba = b by axiom 5.

(d) Suppose a⋅c=b⋅ca\cdot c = b \cdot c and c≠0c \neq 0. Since c≠0c \neq 0, c−1c^{-1} exists by axiom 8. Multiply both sides on the right by c−1c^{-1} and argue exactly as in (c), using associativity and axiom 8 in place of axioms 6 and 3: (a⋅c)⋅c−1=(b⋅c)⋅c−1(a\cdot c)\cdot c^{-1} = (b \cdot c)\cdot c^{-1} gives a⋅(c⋅c−1)=b⋅(c⋅c−1)a \cdot (c \cdot c^{-1}) = b\cdot (c\cdot c^{-1}), i.e. a⋅1=b⋅1a \cdot 1 = b \cdot 1, i.e. a=ba = b. ■\blacksquare

∎

These two lemmas already illustrate the method of the whole subject: every algebraic manipulation used in solving linear systems — clearing a term, dividing by a pivot, canceling a common factor — is licensed by one of these proofs, and hence works in any field whatsoever.

TheoremThe field axioms suffice for Gaussian elimination

Let FF be any field, and consider a system of mm linear equations in nn unknowns with coefficients in FF:

∑j=1naijxj=bi,i=1,…,m,aij,bi∈F.\sum_{j=1}^n a_{ij} x_j = b_i, \qquad i = 1, \dots, m, \qquad a_{ij}, b_i \in F.

The elementary row operations — (i) swap two equations, (ii) multiply an equation by a nonzero scalar in FF, (iii) add a scalar multiple of one equation to another — are all well-defined operations on systems with coefficients in FF, each is invertible (undoing it is itself an elementary row operation over FF), and applying a finite sequence of them never changes the solution set of the system.

Proof

We use the method of direct verification from the field axioms, checking each operation in turn.

Well-definedness. Each new coefficient produced by an elementary row operation is obtained from the old coefficients using only ++, −- (i.e. addition of an additive inverse), and ⋅\cdot by field elements. Since FF is closed under ++ and ⋅\cdot (axiom 1) and every element has an additive inverse in FF (axiom 6), every new coefficient again lies in FF. So each operation produces another system with coefficients in FF. Operation (ii) requires multiplying by a nonzero scalar λ∈F\lambda \in F; this is legitimate on any field since FF contains nonzero elements (as 1≠01 \neq 0, axiom 7).

Invertibility. Operation (i) is its own inverse (swap back). For operation (ii), if equation ii is replaced by λ⋅(equation i)\lambda \cdot (\text{equation } i) for λ≠0\lambda \neq 0, then since λ≠0\lambda \neq 0 axiom 8 gives λ−1∈F\lambda^{-1} \in F, and multiplying by λ−1\lambda^{-1} recovers the original equation: by associativity and axiom 8, λ−1⋅(λ⋅aij)=(λ−1λ)aij=1⋅aij=aij\lambda^{-1}\cdot(\lambda \cdot a_{ij}) = (\lambda^{-1}\lambda) a_{ij} = 1\cdot a_{ij} = a_{ij}. So (ii) is invertible by another instance of (ii), using the field axiom that nonzero elements have inverses — this is precisely where "field" (rather than mere "ring") is essential. For operation (iii), if equation ii is replaced by (equation ii) +μ⋅+ \mu\cdot(equation kk) for k≠ik \neq i, then subtracting μ⋅\mu\cdot(equation kk) — i.e. adding (−μ)⋅(-\mu)\cdot(equation kk), using axiom 6 to form −μ-\mu — recovers the original equation ii by associativity and commutativity of addition (axioms 2, 3) together with the distributive law (axiom 4) applied to expand and re-collect terms.

Solution sets are preserved. Let S⊆FnS \subseteq F^n denote the solution set of the original system, i.e. the set of tuples (x1,…,xn)∈Fn(x_1,\ldots,x_n) \in F^n satisfying all mm equations. We show each operation produces a system with the same solution set SS. (i) Swapping the order of two equations does not change which tuples satisfy all of them (satisfying a set of equations does not depend on the order in which they are listed) — this uses commutativity of "and," not even an algebraic axiom. (ii) Suppose equation ii, namely ∑jaijxj=bi\sum_j a_{ij}x_j = b_i, is replaced by λ∑jaijxj=λbi\lambda \sum_j a_{ij} x_j = \lambda b_i for some λ≠0\lambda \neq 0 in FF. If (x1,…,xn)∈S(x_1,\ldots,x_n) \in S, then ∑jaijxj=bi\sum_j a_{ij}x_j = b_i holds, and multiplying both sides by λ\lambda (using that ⋅\cdot is a well-defined operation, axiom 1) gives λ∑jaijxj=λbi\lambda\sum_j a_{ij}x_j = \lambda b_i, i.e. the new equation holds; all other equations are unaffected. So S⊆S′S \subseteq S', where S′S' is the new solution set. Conversely if the new equation λ∑jaijxj=λbi\lambda \sum_j a_{ij}x_j = \lambda b_i holds, multiply both sides by λ−1\lambda^{-1} (which exists by axiom 8 since λ≠0\lambda \neq 0) to recover ∑jaijxj=bi\sum_j a_{ij} x_j = b_i by the same computation as in the invertibility check above. So S′⊆SS' \subseteq S, giving S=S′S = S'. (iii) Suppose equation ii is replaced by (equation ii) +μ⋅+\mu\cdot(equation kk), i.e. by

∑jaijxj+μ∑jakjxj=bi+μbk.\sum_j a_{ij}x_j + \mu\sum_j a_{kj}x_j = b_i + \mu b_k.

If (x1,…,xn)∈S(x_1,\ldots,x_n) \in S, both ∑jaijxj=bi\sum_j a_{ij}x_j = b_i and ∑jakjxj=bk\sum_j a_{kj}x_j = b_k hold; multiplying the second by μ\mu and adding to the first (using axioms 1, 2, 3, 4 to justify that the sum of two true equalities, combined termwise, is a true equality) gives exactly the new equation. So S⊆S′S \subseteq S'. Conversely, given the new equation and the (unchanged) equation kk, subtracting μ⋅\mu\cdot(equation kk) — i.e. adding (−μ)⋅(equation k)(-\mu)\cdot(\text{equation }k) — from the new equation recovers the original equation ii, by the same distributive/associative bookkeeping. So S′⊆SS' \subseteq S, giving S=S′S = S'. In all three cases the solution set is unchanged, and by induction on the number of operations applied, any finite sequence of elementary row operations preserves the solution set. ■\blacksquare

∎
CorollaryElimination works over any field

Gaussian elimination and back-substitution, as algorithms, can be carried out over Q\mathbb{Q}, R\mathbb{R}, C\mathbb{C}, or any finite field Fp\mathbb{F}_p with pp prime, with identical justification. In particular, a linear system over a field FF has a unique solution, no solution, or infinitely many solutions, exactly as over R\mathbb{R} — the trichotomy theorem from the theory of linear systems depends only on the field axioms and hence transfers verbatim.

Worked examples

Example 1: verifying Q\mathbb{Q} satisfies all eight axioms, and pinpointing where Z\mathbb{Z} fails. Let Q={p/q:p,q∈Z,q≠0}\mathbb{Q} = \{p/q : p, q \in \mathbb{Z}, q \neq 0\} with the usual addition and multiplication of fractions, where p/qp/q is understood as an equivalence class under p/q∼p′/q′p/q \sim p'/q' iff pq′=p′qpq' = p'q. Closure, commutativity, associativity, and distributivity are routine (and inherited from the corresponding properties of Z\mathbb{Z}, checked by clearing denominators). The additive identity is 0=0/10 = 0/1, and the multiplicative identity is 1=1/11 = 1/1, with 1≠01 \neq 0 since 1⋅1≠0⋅11 \cdot 1 \neq 0\cdot 1 in Z\mathbb{Z}. Every p/q∈Qp/q \in \mathbb{Q} has additive inverse (−p)/q∈Q(-p)/q \in \mathbb{Q}. The nontrivial axiom to check is multiplicative inverses: if p/q≠0p/q \neq 0, then p≠0p \neq 0, so q/pq/p is a well-defined element of Q\mathbb{Q}, and

pq⋅qp=pqqp=1,\frac{p}{q}\cdot\frac{q}{p} = \frac{pq}{qp} = 1,

using commutativity of multiplication in Z\mathbb{Z}. So Q\mathbb{Q} satisfies axiom 8, and hence all eight field axioms: Q\mathbb{Q} is a field. Contrast this with Z\mathbb{Z} itself, with its usual ++ and ⋅\cdot. Axioms 1–7 all hold for Z\mathbb{Z} (closure, commutativity, associativity, distributivity, identities 0,10, 1, additive inverses −n-n all exist in Z\mathbb{Z}). But axiom 8 fails: take a=2∈Za = 2 \in \mathbb{Z}, a≠0a \neq 0. If 2−12^{-1} existed in Z\mathbb{Z}, it would be an integer cc with 2c=12c = 1; but 2c2c is even for every integer cc, while 11 is odd, so no such cc exists. Hence Z\mathbb{Z} is a commutative ring but not a field — this is exactly the gap that forces us to pass from integer coefficients to rational (or real) coefficients when we want to divide by a pivot in Gaussian elimination.

Example 2: the finite field F5=Z/5Z\mathbb{F}_5 = \mathbb{Z}/5\mathbb{Z}, with explicit inverse table and a full linear system solved inside it. Let F5={0,1,2,3,4}\mathbb{F}_5 = \{0,1,2,3,4\} with addition and multiplication performed modulo 55: e.g. 3+4=7≡2(mod5)3 + 4 = 7 \equiv 2 \pmod 5, and 3⋅4=12≡2(mod5)3 \cdot 4 = 12 \equiv 2 \pmod 5. Axioms 1–7 hold because addition and multiplication mod 55 inherit commutativity, associativity, and distributivity from Z\mathbb{Z} (reduction mod 55 respects sums and products: if a≡a′a \equiv a' and b≡b′(mod5)b \equiv b'\pmod 5 then a+b≡a′+b′a+b \equiv a'+b' and ab≡a′b′(mod5)ab \equiv a'b' \pmod 5), 00 and 11 act as identities, and −a≡5−a(mod5)-a \equiv 5-a \pmod 5 is the additive inverse of a≠0a \neq 0 (with −0=0-0 = 0). For axiom 8, we must check every nonzero element of F5\mathbb{F}_5 has a multiplicative inverse inside F5\mathbb{F}_5. We compute the full multiplication table restricted to nonzero elements:

⋅123411234224133314244321\begin{array}{c|cccc} \cdot & 1 & 2 & 3 & 4\\ \hline 1 & 1 & 2 & 3 & 4\\ 2 & 2 & 4 & 1 & 3\\ 3 & 3 & 1 & 4 & 2\\ 4 & 4 & 3 & 2 & 1 \end{array}

Reading off where each row hits 11: 1−1=11^{-1}=1, 2−1=32^{-1}=3 (since 2⋅3=6≡12\cdot 3 = 6\equiv 1), 3−1=23^{-1}=2, 4−1=44^{-1}=4 (since 4⋅4=16≡14\cdot 4=16\equiv 1). Every nonzero element has an inverse in F5\mathbb{F}_5, verifying axiom 8, so F5\mathbb{F}_5 is a field. (This works because 55 is prime; see the counterexample below for what fails when the modulus is composite.) Now solve, entirely inside F5\mathbb{F}_5, the system

2x+3y=4,x+4y=1,2x + 3y = 4, \qquad x + 4y = 1,

with x,y∈F5x, y \in \mathbb{F}_5. Use elimination exactly as justified by the theorem above: multiply the second equation by 2−1⋅22^{-1}\cdot 2 — more directly, eliminate xx by multiplying the second equation by 22 (subtracting it from suitably scaled first equation), or simply multiply the first equation by 1−1⋅1^{-1}\cdot (a convenient scalar). We multiply equation 2 by 33 (an allowed axiom-8-free operation since 33 need not be inverted here, only used as a scalar multiplier):

3x+12y=3  ⟹  3x+2y=3(mod5).3x + 12 y = 3 \;\Longrightarrow\; 3x + 2y = 3 \pmod 5.

Subtract this from equation 1 multiplied by... instead, cleanly: multiply equation 2 by 22: 2x+8y=22x + 8y = 2, i.e. 2x+3y=2(mod5)2x + 3y = 2 \pmod 5. Subtract from equation 1 (2x+3y=42x+3y=4):

(2x+3y)−(2x+3y)=4−2  ⟹  0=2,(2x+3y) - (2x+3y) = 4 - 2 \;\Longrightarrow\; 0 = 2,

which is impossible unless we made an arithmetic slip — indeed we subtracted the same left side from itself, which signals we scaled equation 2 to match equation 1's xx-coefficient and its yy-coefficient simultaneously, meaning the two equations are proportional and we should instead eliminate directly. Let's redo this cleanly using 2−1=32^{-1}=3. Multiply equation 1 by 2−1=32^{-1}=3:

3⋅(2x+3y)=3⋅4  ⟹  6x+9y=12  ⟹  x+4y=2(mod5).3\cdot(2x+3y) = 3\cdot 4 \;\Longrightarrow\; 6x + 9y = 12 \;\Longrightarrow\; x + 4y = 2 \pmod 5.

Now subtract equation 2 (x+4y=1x+4y=1) from this:

(x+4y)−(x+4y)=2−1  ⟹  0=1(mod5),(x+4y) - (x+4y) = 2 - 1 \;\Longrightarrow\; 0 = 1 \pmod 5,

which is false. This shows the two original equations are inconsistent over F5\mathbb{F}_5: after scaling equation 1 by 3=2−13=2^{-1} we obtained exactly equation 2's left-hand side, x+4yx+4y, but with right-hand sides 2≠12 \neq 1. Hence the system

2x+3y=4,x+4y=12x+3y=4,\qquad x+4y=1

has no solution in F5\mathbb{F}_5 — the coefficient rows are proportional (row 1 is 2×2\times row 2 in the coefficients 2,32,3 vs 1,41,4: indeed 2⋅1=2≠32\cdot 1=2\neq 3... let us double check proportionality directly: is (2,3)=λ(1,4)(2,3) = \lambda(1,4) for some λ∈F5\lambda\in\mathbb{F}_5? From the first coordinate λ=2\lambda = 2; check second coordinate 2⋅4=8≡3(mod5)2\cdot 4=8\equiv3\pmod5. Yes, matches.) while the right-hand sides 44 and 2⋅1=22\cdot 1=2 disagree (4≠24\neq 2), confirming inconsistency by the trichotomy for linear systems: proportional coefficient rows with non-proportional constants force the "no solution" case, exactly as it would over R\mathbb{R}. This example is genuinely useful: it shows the same elimination algorithm, the same trichotomy (unique / none / infinite), and the same notion of proportional rows detecting inconsistency, transplanted without modification into a finite field — precisely the content of the corollary above.

CounterexampleWhy the field axioms — not just "commutative ring" — are needed

Consider Z/6Z={0,1,2,3,4,5}\mathbb{Z}/6\mathbb{Z} = \{0,1,2,3,4,5\} with addition and multiplication mod 66. This structure satisfies axioms 1–7 (closure, commutativity, associativity, distributivity all inherited from Z\mathbb{Z} as before; 0,10,1 are identities; every element has an additive inverse mod 66). So Z/6Z\mathbb{Z}/6\mathbb{Z} is a perfectly good commutative ring. But it is not a field, because axiom 8 fails: take a=2a = 2. If 2−12^{-1} existed in Z/6Z\mathbb{Z}/6\mathbb{Z}, there would be c∈{0,…,5}c \in \{0,\ldots,5\} with 2c≡1(mod6)2c \equiv 1 \pmod 6. But 2c(mod6)∈{0,2,4}2c \pmod 6 \in \{0,2,4\} for every integer cc (it is always even), and 11 is odd, so no such cc exists. Worse, Z/6Z\mathbb{Z}/6\mathbb{Z} has zero divisors: 2⋅3=6≡0(mod6)2 \cdot 3 = 6 \equiv 0 \pmod 6, yet 2≠02 \neq 0 and 3≠03 \neq 0. This directly breaks the cancellation lemma proved above: from 2⋅3=2⋅0  (=0)2\cdot 3 = 2 \cdot 0 \;(= 0) we cannot cancel the common factor 22 to conclude 3=03 = 0, because 22 has no multiplicative inverse to cancel with. The consequence for linear algebra is severe. Consider the "linear system" over Z/6Z\mathbb{Z}/6\mathbb{Z}:

2x=4.2x = 4.

Over a genuine field, this would have the unique solution x=2−1⋅4x = 2^{-1}\cdot 4. But in Z/6Z\mathbb{Z}/6\mathbb{Z}, checking all six values shows x=2x = 2 and x=5x = 5 both satisfy 2x≡4(mod6)2x\equiv 4\pmod 6 (since 2⋅2=42\cdot 2=4 and 2⋅5=10≡42\cdot 5=10\equiv 4), so the equation has two solutions rather than the unique solution guaranteed by the corollary above whenever the coefficient is nonzero in a field. The entire trichotomy (unique/none/infinite) proved for fields simply does not hold in this form over Z/6Z\mathbb{Z}/6\mathbb{Z}: a single linear equation in one unknown, with nonzero coefficient, can have more than one but not all solutions. This is precisely why linear algebra is built over fields, not merely commutative rings: dropping axiom 8 destroys cancellation, and with it, the uniqueness theory of linear systems.

Scope, edge cases & common traps

SituationWhat to checkCommon error
Is Z/nZ\mathbb{Z}/n\mathbb{Z} a field?nn must be prime; then every nonzero residue is invertible (Bézout's identity)Assuming any modulus works; Z/6Z\mathbb{Z}/6\mathbb{Z} fails axiom 8
Solving ax=bax=b over a fieldIf a≠0a \neq 0, unique solution x=a−1bx=a^{-1}b; if a=0a=0, either no solution (b≠0b\neq0) or all of FF (b=0b=0)Forgetting to split into cases on whether a=0a=0
Ring vs. fieldA commutative ring need not have multiplicative inversesSilently assuming "1/a1/a" makes sense in an arbitrary ring
Ordered fieldsR,Q\mathbb{R}, \mathbb{Q} are ordered; C,Fp\mathbb{C}, \mathbb{F}_p are not ordered fieldsTrying to say "i>0i > 0" or compare elements of F5\mathbb{F}_5 by size
CharacteristicIn Fp\mathbb{F}_p, 1+1+⋯+1⏟p=0\underbrace{1+1+\cdots+1}_{p} = 0Assuming n⋅1≠0n\cdot 1 \neq 0 for all positive integers nn, as in Q\mathbb{Q} or R\mathbb{R}

The mathematician behind it

Origins · 1 minÉvariste Galois and how this was discovered
ÉG
Évariste Galois
1811–1832 · Founder of what is now called Galois theory, and (in effect) the theory of fields and their symmetries

Working in Paris in the final years of his short life, Galois studied when polynomial equations can be solved "by radicals," and in doing so uncovered the deep link between the symmetries of the roots of a polynomial and the structure of the number systems (fields) those roots generate. Although the modern axiomatic definition of a field was formalized later — largely through the work of Heinrich Weber and Ernst Steinitz at the turn of the twentieth century — Galois's insight that different "number systems" could be compared, extended, and classified by their symmetry groups is the conceptual seed from which field theory grew. He died at twenty, from wounds suffered in a duel, having written down in a single feverish night before the duel a summary of ideas that would occupy mathematicians for the rest of the century.

History
- 1832 — Évariste Galois, on the eve of a fatal duel, writes a letter outlining the ideas that become Galois theory, implicitly comparing different fields of "constructible" numbers.
- 1871 — Richard Dedekind introduces the term Körper (German for "field," literally "body") for number systems closed under the four operations, in his treatment of algebraic number theory.
- 1893 — Heinrich Weber gives one of the first fully explicit axiomatic definitions of an abstract field, independent of any specific number system.
- 1910 — Ernst Steinitz publishes Algebraische Theorie der Körper, systematically developing the abstract theory of fields, including finite fields and field extensions, in the modern axiomatic style.
- 1930s — Emmy Noether and Bartel van der Waerden's development of abstract algebra cements the field axioms as the standard foundation for linear algebra, in exactly the form used here.
Is every finite set with ++ and ×\times automatically a field?

No. Finiteness alone guarantees nothing about invertibility. Z/nZ\mathbb{Z}/n\mathbb{Z} with the usual mod-nn arithmetic is a field if and only if nn is prime. When nn is composite, say n=abn = ab with 1<a,b<n1<a,b<n, the product a⋅b≡0(modn)a\cdot b \equiv 0 \pmod n exhibits zero divisors, and one can show aa has no multiplicative inverse in Z/nZ\mathbb{Z}/n\mathbb{Z}. When n=pn=p is prime, Bézout's identity from elementary number theory guarantees that for each a≢0a \not\equiv 0, there exist integers x,yx,y with ax+py=1ax+py=1, so xx is the desired inverse mod pp. So primality of the modulus is exactly the condition needed.

Why do we require 1≠01 \neq 0 in the field axioms?

Without this requirement, the one-element set {0}\{0\}, with 0+0=00+0=0 and 0⋅0=00\cdot 0=0, would trivially satisfy every other axiom (vacuously, since there is only one element to check, and it is its own additive and multiplicative identity). But this "zero field" is useless: it contains no nonzero elements at all, so every vector space built over it would collapse to the zero vector space, and no interesting linear algebra could be done. Requiring 1≠01\neq0 rules out this degenerate case and guarantees a field has at least two elements.

Are R\mathbb{R} and C\mathbb{C} the only infinite fields relevant to linear algebra courses?

They are the most common ones in introductory courses, but far from the only fields in use. Q\mathbb{Q} is essential whenever exact (rounding-free) computation matters, such as in symbolic computation or number theory. Fields like Q(2)\mathbb{Q}(\sqrt{2}) (rational numbers extended by 2\sqrt 2) arise in Galois theory and in studying which geometric constructions are possible with straightedge and compass. Function fields, such as the field of rational functions F(t)F(t) over a base field FF, are used throughout algebraic geometry. Every one of these satisfies the same eight axioms, so all of linear algebra transfers to them unchanged.

Does C\mathbb{C} satisfy every field axiom, given that it is not ordered like R\mathbb{R}?

Yes — orderability is not one of the field axioms. C={a+bi:a,b∈R}\mathbb{C} = \{a+bi : a,b \in \mathbb{R}\} with the usual complex addition and multiplication satisfies all eight axioms: closure, commutativity, associativity, and distributivity follow from direct computation with i2=−1i^2=-1; the additive identity is 0+0i0+0i; the multiplicative identity is 1+0i1+0i; additive inverses are −a−bi-a-bi; and every nonzero a+bia+bi has multiplicative inverse a−bia2+b2\frac{a-bi}{a^2+b^2}, which is well-defined precisely because a2+b2≠0a^2+b^2 \neq 0 whenever a+bi≠0a+bi\neq 0. The lack of a compatible order (there is no way to declare C\mathbb{C} "positive" or "negative" consistently with the field operations) is a separate structural fact, unrelated to the field axioms themselves.

Why does this article's theorem talk about Gaussian elimination rather than vector spaces directly?

Because the field axioms are logically prior to, and independent of, the notion of a vector space: a field is exactly the set of allowable scalars, and a vector space is defined as an abelian group of vectors equipped with a scalar multiplication by a chosen field. Since Systems of Linear Equations already involves solving equations with coefficients drawn from some number system, it is the most concrete and immediate setting in which to see that all the needed algebraic properties are exactly the field axioms, before the more abstract vector space machinery is introduced.

Can a field have characteristic other than 0 or a prime?

No. The characteristic of a field FF is the smallest positive integer nn such that 1+1+⋯+1⏟n=0\underbrace{1+1+\cdots+1}_{n} = 0, or 00 if no such nn exists (as in Q,R,C\mathbb{Q}, \mathbb{R}, \mathbb{C}, all of characteristic 00). One can show, using the no-zero-divisors lemma proved above, that if such an nn exists it must be prime: if n=kℓn = k\ell with 1<k,ℓ<n1<k,\ell<n, then (1+⋯+1⏟k)(1+⋯+1⏟ℓ)=1+⋯+1⏟n=0(\underbrace{1+\cdots+1}_k)(\underbrace{1+\cdots+1}_\ell) = \underbrace{1+\cdots+1}_n = 0 forces one of the two factors to be 00 by the no-zero-divisors property, contradicting minimality of nn. Hence every field has characteristic 00 or a prime pp; Fp\mathbb{F}_p has characteristic exactly pp.

Is "field" the same thing as "number system"?

Not quite — "field" is more general and more precise. Every field axiom concerns only the algebraic operations +,⋅+, \cdot and says nothing about size, order, geometry, or decimal expansions, so a field can be a completely abstract finite set like F5\mathbb{F}_5 that has no natural notion of "bigger" or "smaller." Conversely, not every familiar number system is a field: Z\mathbb{Z}, the "whole numbers," fails axiom 8 and is only a commutative ring. So "field" precisely isolates the algebraic properties that make division work, independent of whatever intuitive picture of "numbers" one starts with.

Before this

Where this leads