Number Fields and the Rigorous Definition of a Vector Space

Linear algebra is the study of a single algebraic structure — the vector space — and the maps between such structures that preserve it. Before that definition can mean anything, we need to fix what kind of numbers are allowed to scale vectors: this is the notion of a field. Everything in this block, from span and dimension to eigenvalues and Galois theory, rests on these two definitions being stated with complete precision, in the style of Axler's Linear Algebra Done Right.

Have to hand: basic set notation, and familiarity with $\mathbb{Q}$, $\mathbb{R}$, $\mathbb{C}$ as number systems you already compute in. We will not assume you know why they behave the way they do — that is exactly what the field axioms make explicit.

1. Fields

Definition (field). A field is a set $\mathbb{F}$ together with two binary operations $+ : \mathbb{F}\times\mathbb{F}\to\mathbb{F}$ and $\cdot : \mathbb{F}\times\mathbb{F}\to\mathbb{F}$, and two distinguished elements $0 \ne 1$, such that:

1. $(\mathbb{F}, +)$ is an abelian group: $+$ is associative and commutative, $0$ is an identity ($a+0=a$), and every $a$ has an additive inverse $-a$ with $a+(-a)=0$.
2. $(\mathbb{F}\setminus\{0\}, \cdot)$ is an abelian group: $\cdot$ is associative and commutative, $1$ is an identity, and every $a \ne 0$ has a multiplicative inverse $a^{-1}$ with $a\cdot a^{-1} = 1$.
3. Distributivity: $a\cdot(b+c) = a\cdot b + a\cdot c$ for all $a,b,c \in \mathbb{F}$.

The sets $\mathbb{Q}$, $\mathbb{R}$ and $\mathbb{C}$ are fields under ordinary addition and multiplication; so is $\mathbb{F}_p = \mathbb{Z}/p\mathbb{Z}$ for a prime $p$, with arithmetic modulo $p$. The set $\mathbb{Z}$ of integers is not a field: $2$ has no multiplicative inverse in $\mathbb{Z}$, so axiom 2 fails. This is the only reason $\mathbb{Z}$ is excluded, and it is precisely the reason division is not generally available there.

From the axioms alone, several familiar facts follow and must be proved, not assumed.

Proposition. In any field, $a\cdot 0 = 0$ for every $a$.

Proof. $a \cdot 0 = a\cdot(0+0) = a\cdot 0 + a\cdot 0$ by distributivity, since $0$ is the additive identity. Adding $-(a\cdot 0)$ to both sides gives $0 = a\cdot 0$. $\blacksquare$

This single line is why $0$ can never have a multiplicative inverse (which is why axiom 2 is stated on $\mathbb{F}\setminus\{0\}$): if $0^{-1}$ existed, $1 = 0\cdot 0^{-1} = 0$, contradicting $0 \ne 1$.

Throughout this block, $\mathbb{F}$ denotes an arbitrary field, and essentially all results are stated for a general $\mathbb{F}$ — most often you should read $\mathbb{F} = \mathbb{R}$ or $\mathbb{F} = \mathbb{C}$, but the proofs use only the field axioms, never any special property of the real or complex numbers (such as order or completeness).

2. Vector Spaces: the Axioms

Definition (vector space). Let $\mathbb{F}$ be a field. A vector space over $\mathbb{F}$ is a set $V$ together with an addition $+ : V \times V \to V$ and a scalar multiplication $\cdot : \mathbb{F}\times V \to V$ such that:

1. $(V,+)$ is an abelian group, with identity denoted $0 \in V$ (the zero vector) and inverse of $v$ denoted $-v$.
2. $1\cdot v = v$ for all $v \in V$.
3. $(ab)\cdot v = a\cdot(b\cdot v)$ for all $a,b\in\mathbb{F}$, $v\in V$ (associativity of scalar multiplication).
4. $a\cdot(u+v) = a\cdot u + a\cdot v$ for all $a \in \mathbb{F}$, $u,v \in V$.
5. $(a+b)\cdot v = a\cdot v + b\cdot v$ for all $a,b\in\mathbb{F}$, $v \in V$.

Elements of $V$ are called vectors, elements of $\mathbb{F}$ scalars. Note that two entirely different additions appear (in $\mathbb{F}$ and in $V$) and the symbol $0$ is overloaded between the field's zero and the vector space's zero; context always disambiguates, but the distinction matters in proofs.

A vector space is not "an arrow with magnitude and direction." That picture is one example ($\mathbb{R}^2$ or $\mathbb{R}^3$ with the usual arrows), but the definition says nothing about length or direction — only that eight axioms about $+$ and scalar multiplication hold. Polynomials, matrices and functions are vectors in exactly the same formal sense.

3. Canonical Examples

• $\mathbb{F}^n = \{(a_1,\dots,a_n) : a_i \in \mathbb{F}\}$, with componentwise addition and scaling. This is the model example and the one all finite-dimensional spaces are eventually identified with (Chapter 3).
• $M_{m\times n}(\mathbb{F})$, the set of $m\times n$ matrices with entries in $\mathbb{F}$, under entrywise addition and scaling.
• $\mathcal{P}(\mathbb{F})$, the set of all polynomials with coefficients in $\mathbb{F}$, and $\mathcal{P}_n(\mathbb{F})$, those of degree at most $n$.
• $\mathcal{F}(S,\mathbb{F})$, the set of all functions $f : S \to \mathbb{F}$ for a fixed set $S$, with $(f+g)(x) = f(x)+g(x)$ and $(af)(x) = a\,f(x)$.
• $\mathbb{C}$ itself is a vector space over $\mathbb{R}$ (forgetting complex multiplication, keeping only addition and real scaling) — this is a different vector space from $\mathbb{C}$ as a vector space over itself, and the distinction (real dimension $2$ versus complex dimension $1$) is a running theme of this course.

Worked example — verifying the axioms for $\mathcal{F}(S,\mathbb{R})$. Take $S = [0,1]$ and $V = \mathcal{F}([0,1],\mathbb{R})$, all real functions on $[0,1]$. Axiom 1 ($(V,+)$ an abelian group) requires a zero vector: the constant function $0(x) = 0$ for all $x$, since $(f+0)(x) = f(x) + 0 = f(x)$ for every $f$. The additive inverse of $f$ is $(-f)(x) := -f(x)$, since $(f + (-f))(x) = f(x) - f(x) = 0$ for every $x$. Axioms 2–5 reduce pointwise to the field axioms of $\mathbb{R}$: for instance axiom 5, $((a+b)f)(x) = (a+b)f(x) = a f(x) + b f(x) = (af)(x) + (bf)(x)$, using distributivity in $\mathbb{R}$ at the single point $x$. The verification is routine precisely because it reduces, coordinate by coordinate (or point by point), to a field axiom — this is the recurring pattern for checking any function space or sequence space is a vector space.

4. Immediate Consequences of the Axioms

These are proved from the eight axioms alone, with no appeal to coordinates — the Axler-style discipline of this course.

Proposition (uniqueness of $0$). A vector space has exactly one zero vector.

Proof. Suppose $0$ and $0'$ both satisfy $v + 0 = v$ and $v + 0' = v$ for all $v$. Then $0' = 0' + 0 = 0 + 0' = 0$, using the defining property of $0$ with $v = 0'$, commutativity, and the defining property of $0'$ with $v=0$. $\blacksquare$

Proposition (uniqueness of additive inverse). For each $v \in V$ there is exactly one $w$ with $v+w=0$.

Proof. If $v+w=0$ and $v+w'=0$, then $w = w+0 = w+(v+w') = (w+v)+w' = 0+w' = w'$, using associativity and commutativity of $+$. $\blacksquare$

Proposition. For every $v \in V$: $0\cdot v = 0$ (scalar zero times $v$ equals the zero vector), and $(-1)\cdot v = -v$.

Proof. $0\cdot v = (0+0)\cdot v = 0\cdot v + 0\cdot v$ by axiom 5 (with $a=b=0$), since $0$ is the additive identity of $\mathbb{F}$. Adding $-(0\cdot v)$ to both sides gives $0 = 0\cdot v$. For the second claim, $v + (-1)\cdot v = 1\cdot v + (-1)\cdot v = (1+(-1))\cdot v = 0\cdot v = 0$, using axiom 2, axiom 5, and the result just proved; by uniqueness of the additive inverse, $(-1)\cdot v = -v$. $\blacksquare$

These two facts look obvious but are not axioms — they are theorems, and every step uses a named axiom. This is the hallmark of the rigorous treatment: "obviously $0v=0$" is not a proof.

Common Exam Pitfalls

• Treating $0 \in \mathbb{F}$ and $0 \in V$ as the same object. They live in different sets and satisfy different axioms; a proof that conflates them (e.g. writing "$0\cdot v = 0$" as if self-evident) earns no credit without the derivation in Section 4.
• Forgetting the field needs $0 \ne 1$. The trivial ring $\{0\}$ with $0=1$ satisfies every other field axiom vacuously; excluding it is deliberate and is why $1 \ne 0$ is part of the definition.
• Assuming every scalar is real. A vector space is always over a specified field; "is $\{(x,y) : x,y\in\mathbb{C}\}$ a vector space?" has no answer until you say over $\mathbb{C}$ or over $\mathbb{R}$ — the two give different structures with different dimension (Chapter 3).
• Citing a property of $\mathbb{R}$ (order, completeness, $\sqrt{\cdot}$) inside a proof meant to hold for general $\mathbb{F}$. Field axioms give you $+,\times$ and inverses only; $\mathbb{F}_p$ has none of those extra structures.
• Skipping the closure check. A subset of a known vector space is not automatically a vector space in its own right under the same operations unless addition and scalar multiplication do not leave the subset — this becomes the subspace test of Chapter 2.
• Confusing $\mathbb{Z}$ with a field. Integers fail invertibility under multiplication; many "vector space" intuitions silently assume division by scalars is legal, which requires a field, not merely a ring.

Summary

• A field $\mathbb{F}$ has two abelian group structures ($+$ and $\times$ on nonzero elements) linked by distributivity, with $0 \ne 1$; $\mathbb{Q}, \mathbb{R}, \mathbb{C}, \mathbb{F}_p$ are fields, $\mathbb{Z}$ is not.
• $a\cdot 0 = 0$ in any field, proved from distributivity and cancellation — never assumed.
• A vector space over $\mathbb{F}$ is a set $V$ with $+$ and scalar multiplication satisfying the eight axioms of Section 2; "vector" means nothing more or less than "element of a set satisfying these axioms."
• Canonical examples: $\mathbb{F}^n$, matrix spaces $M_{m\times n}(\mathbb{F})$, polynomial spaces $\mathcal{P}_n(\mathbb{F})$, function spaces $\mathcal{F}(S,\mathbb{F})$; the field of scalars must always be specified.
• The zero vector and additive inverses are unique, and $0\cdot v = 0$, $(-1)\cdot v = -v$ are theorems derived from the axioms, not additional axioms.