Mathematics.

field theory

Vector Spaces over General Fields

Abstract Algebra II30 minDifficulty6 out of 10

You should know: fields, vector space

Overview

The vector space axioms never actually require the scalars to be real numbers — they work verbatim for scalars drawn from any field F, whether that's ℚ, ℂ, a finite field 𝔽_p, or an abstract field arising in Galois theory. This generalization is not a mere curiosity: every field extension L/K is automatically a vector space over K (with the same field addition and multiplication restricted to scalars in K), and the degree [L:K] used throughout field theory is defined exactly as the dimension of this vector space. Working over general fields also produces genuinely new phenomena absent over ℝ or ℂ, such as finite-dimensional spaces with only finitely many vectors total (over 𝔽_p) and fields of positive characteristic where familiar facts about linear independence must be re-derived from the axioms rather than geometric intuition.

Intuition

Nothing in the definition of a vector space cares whether 'numbers' means real numbers — all that's needed is a field to scale by. Working over 𝔽_p instead of ℝ turns the familiar infinite plane ℝ² into a finite grid of p² points, yet linear independence, span, and basis all still make sense and behave the same way; this is exactly the setting used to count subspaces, error-correcting codes, and cryptographic constructions. The single most important instance is a field extension L/K itself: forgetting that L has its own multiplication and remembering only that K acts on it by scalars turns L into an ordinary K-vector space, so every fact about field-extension degree — the Tower Law, multiplicativity, additivity of bases — is secretly a fact about vector-space dimension, just phrased in field-theoretic language.

Formal Definition

Definition

Let F be any field. A vector space over F is a set V with addition and an F-scalar multiplication satisfying the same eight axioms as over ℝ:

(V,+) is an abelian group,a,b∈F, u,v∈V(V,+) \text{ is an abelian group}, \quad a,b \in F,\ u,v \in V
Additive structure
a(u+v)=au+av,(a+b)v=av+bv,a(bv)=(ab)v,1v=va(u+v) = au+av,\quad (a+b)v = av+bv,\quad a(bv)=(ab)v,\quad 1v=v
Scalar multiplication axioms
dim⁡F(V)=∣B∣ for any basis B of V (well-defined by the Steinitz exchange lemma)\dim_F(V) = |B| \text{ for any basis } B \text{ of } V \text{ (well-defined by the Steinitz exchange lemma)}
Dimension is basis-independent
L/K a field extension  ⟹  L is an K-vector space,[L:K]:=dim⁡K(L)L/K \text{ a field extension} \implies L \text{ is an } K\text{-vector space}, \quad [L:K] := \dim_K(L)
Field extensions as vector spaces

Properties

Field extensions are vector spaces

If K⊆L are fields, then L is a vector space over K, and [L:K]=dim⁡K(L).\text{If } K \subseteq L \text{ are fields, then } L \text{ is a vector space over } K \text{, and } [L:K] = \dim_K(L).

Finite fields give finite vector spaces

An n-dimensional vector space over Fq (q=pk) has exactly qn elements\text{An } n\text{-dimensional vector space over } \mathbb{F}_q \text{ (}q=p^k\text{) has exactly } q^n \text{ elements}

Steinitz exchange lemma

Any two bases of a finite-dimensional vector space over any field F have the same size, so dimension is well-defined regardless of F.\text{Any two bases of a finite-dimensional vector space over any field } F \text{ have the same size, so dimension is well-defined regardless of } F.

Characteristic affects linear algebra

Over a field of characteristic p, identities like the binomial or determinant formulas can vanish or collapse in ways that never happen over R or C.\text{Over a field of characteristic } p, \text{ identities like the binomial or determinant formulas can vanish or collapse in ways that never happen over } \mathbb{R} \text{ or } \mathbb{C}.

Worked Examples

  1. Each of the 3 coordinates independently takes one of 2 values (0 or 1) in 𝔽₂.

    ∣F23∣=23=8|\mathbb{F}_2^3| = 2^3 = 8

  2. Removing only the all-zero vector leaves the nonzero vectors.

    8−1=7 nonzero vectors8 - 1 = 7 \text{ nonzero vectors}

Answer: 𝔽₂³ has exactly 8 vectors total, of which 7 are nonzero — a genuinely finite vector space, impossible over ℝ or ℂ.

Practice Problems

Difficulty 5/10

How many vectors does an n-dimensional vector space over 𝔽_p (p prime) contain?

Difficulty 6/10

Explain why [K(α):K] (the field-extension degree, with α algebraic of minimal polynomial degree n) equals the K-vector-space dimension of K(α), and identify a basis.

Difficulty 7/10

Why can't you compare 'size' (cardinality) meaningfully between 𝔽₂⁴ and ℝ⁴ even though both are 4-dimensional vector spaces?

Common Mistakes

Common Mistake

Assuming all vector spaces are infinite because ℝⁿ and ℂⁿ are.

Vector spaces over finite fields, like 𝔽₂ⁿ, have exactly finitely many elements (2ⁿ in that case), even while retaining a well-defined finite dimension.

Common Mistake

Treating field-extension degree [L:K] and vector-space dimension dim_K(L) as merely analogous.

They are not analogous — they are literally the same number by definition; [L:K] is defined to be dim_K(L).

Quiz

The vector space axioms require the scalars to come from:
If K ⊆ L are fields, then L can be regarded as:
An n-dimensional vector space over the finite field 𝔽_p has exactly how many elements?

Summary

  • The vector space axioms hold verbatim for scalars from any field F, not just ℝ or ℂ.
  • Every field extension L/K is automatically a K-vector space, and [L:K] is defined as exactly dim_K(L).
  • Vector spaces over finite fields 𝔽_p have finitely many elements (p^n for dimension n), unlike ℝⁿ or ℂⁿ.
  • The Steinitz exchange lemma guarantees dimension is well-defined over any field, so basis-counting arguments from linear algebra transfer unchanged.
  • K(α) has K-basis {1, α, ..., α^{n-1}} where n is the degree of α's minimal polynomial, unifying field-extension and vector-space language.

References

  1. WebsiteWikipedia — Vector space
  2. WebsiteWikipedia — Field extension