LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)verified 2026-07-28 (claude-fable-5)

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A field is a vector space over itself, and over any subfield $K \subseteq F$ every $F$ -vector space is a $K$ -vector space by restricting the scalars

Statement

Let $F$ be a field (Field).

$F$ is a vector space over itself (Vector space over a field): take the set to be $F$ , the vector addition to be the field addition, the zero vector to be $0_F$ , and the scalar multiplication $F \times F \to F$ to be the field multiplication.
Let $K \subseteq F$ be a subfield (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations) and let $V$ be a vector space over $F$ . Then $V$ , with the same addition and the same zero vector and with the scalar multiplication restricted to $K \times V$ , is a vector space over $K$ . This is called restricting the scalars from $F$ to $K$ .
In particular $F$ is a vector space over every subfield $K \subseteq F$ , with the field multiplication restricted to $K \times F$ as scalar multiplication.

Facts & Assumptions

Given: A field $F$ , a subfield $K \subseteq F$ , and a vector space $V$ over $F$ .

[L1]

The vector space axioms (V1)–(V5) (Vector space over a field): $(V,+,0_V)$ is an abelian group; $\lambda(u+v) = \lambda u + \lambda v$ ; $(\lambda+\mu)v = \lambda v + \mu v$ ; $(\lambda\mu)v = \lambda(\mu v)$ ; $1_F v = v$ .

[L2]

The field axioms of Field, read as they are read throughout this library: $(F,+,0_F)$ is an abelian group; multiplication on $F$ is associative and commutative with two-sided identity $1_F$ ; multiplication distributes over addition, so that $x(y+z) = xy+xz$ and $(y+z)x = yx+zx$ ; and every $x \ne 0_F$ has a multiplicative inverse.

[L3]

A subfield $K$ of $F$ is a subring of $F$ closed under inverses of its nonzero elements; equivalently, a subset containing $1_F$ with $a - b \in K$ and $ab \in K$ for all $a, b \in K$ and $x^{-1} \in K$ for every nonzero $x \in K$ . With the restricted operations $K$ is itself a field, its addition and multiplication being the restrictions of those of $F$ , and $0_K = 0_F$ , $1_K = 1_F$ , with the negatives and the inverses of $K$ those of $F$ (Subfield: a subring of a field closed under inverses of its nonzero elements, and therefore a field with the restricted operations).

Proof

technique · direct

1.1

Put $V_F := F$ as a set, let the vector addition be the field addition with zero vector $0_F$ , and let the scalar multiplication be the field multiplication, which is a map $F \times F \to F$ as required.

1.2

Axiom (V1) holds for $V_F$ : axiom (A) of a field says exactly that $(F,+,0_F)$ is an abelian group.

1.3

Axiom (V2) holds for $V_F$ : $\lambda(u+v) = \lambda u + \lambda v$ is distributivity of multiplication over addition.

1.4

Axiom (V3) holds for $V_F$ : $(\lambda+\mu)v = \lambda v + \mu v$ is distributivity on the other side.

1.5

Axiom (V4) holds for $V_F$ : $(\lambda\mu)v = \lambda(\mu v)$ is associativity of the field multiplication.

1.6

Axiom (V5) holds for $V_F$ : $1_F v = v$ is the multiplicative identity law.

1.7

For claim 2: since $K \subseteq F$ , restricting the scalar multiplication of $V$ to the subset $K \times V$ of $F \times V$ yields a map $K \times V \to V$ , which is the required datum.

1.8

The set $V$ , its addition and its zero vector are unchanged by the restriction, so $(V,+,0_V)$ is still an abelian group; this is axiom (V1) for the $K$ -structure.

1.9

For $\lambda, \mu \in K$ the sum $\lambda + \mu$ and the product $\lambda\mu$ formed in $K$ are the sum and the product formed in $F$ , and the multiplicative identity of $K$ is $1_F$ .

2.1

Steps 1.1 to 1.6 verify (V1)–(V5), so $F$ with the operations of step 1.1 is a vector space over itself: claim 1.

step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6L1

2.2

Let $\lambda, \mu \in K$ and $u, v \in V$ . Then $\lambda(u+v) = \lambda u + \lambda v$ and $(\lambda\mu)v = \lambda(\mu v)$ are the instances of (V2) and (V4) for these elements of $F$ , the product $\lambda\mu$ being the same whether formed in $K$ or in $F$ ; $(\lambda+\mu)v = \lambda v + \mu v$ is the instance of (V3), the sum $\lambda+\mu$ being likewise the same; and the identity of $K$ is $1_F$ , so $1_K v = 1_F v = v$ is the instance of (V5).

step 1.7step 1.9L1L3

3.1

With step 1.8, the restricted structure satisfies (V1)–(V5) over $K$ , so $V$ is a vector space over $K$ : claim 2.

step 1.8step 2.2L1

4.1

Claim 3 follows by applying claim 2 to the $F$ -vector space of claim 1: $F$ is a vector space over $K$ , its scalar multiplication being the field multiplication restricted to $K \times F$ .

step 2.1step 3.1∎

Remarks

On the field facts in [L2]. Axiom (M) of Field asserts that multiplication is associative and commutative on all of $F$ with $x \cdot 1_F = x$ for every $x \in F$ , the element $0_F$ included, and right distributivity then follows from axiom (D) by commuting, as Multiplication by zero: $0 \cdot a = 0$ already records. These unrestricted forms are spent in exactly three of the steps above, where axioms (V3), (V4) and (V5) are read off for arbitrary scalars including $0_F$ : step 1.4 needs distributivity on the right; step 1.5 needs associativity of the multiplication at $0_F$ as well; and step 1.6 needs $1_F x = x$ rather than the literal $x \cdot 1_F = x$ , which commutativity supplies, and needs it at $x = 0_F$ too. They are used nowhere else: step 1.2 is axiom (A) verbatim, step 1.3 is axiom (D) verbatim, and steps 1.7 to 4.1 use only the vector-space axioms of $V$ and the subfield facts of [L3].
Restricting the scalars changes the structure, not the set. The vectors, the addition and the zero are untouched; only the collection of scalars allowed to act shrinks. Everything that can be said about $V$ as a $K$ -vector space is therefore a statement about the same object with fewer operations available, and every $F$ -linear subspace of $V$ is in particular a $K$ -linear subspace (Linear subspace of a vector space). The converse fails, and that is the point of the construction.
The field is part of the data. Because of this lemma, a bare phrase like "the vector space $V$ " is incomplete: $\mathbb{R}$ is a vector space over $\mathbb{R}$ and also over the embedded copy of $\mathbb{Q}$ inside it, and these are different structures on one set. Every statement on this page names its field.
Nothing here is about dimension. How much smaller $K$ is than $F$ , and what that does to $V$ , is a question about bases and dimension, which are developed on a later page. This lemma asserts only that the restricted structure satisfies the five axioms.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 28 results over 16 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Restriction of scalars (Wikipedia) (standard reference, not scraped)
Vector space (Wikipedia) (standard reference, not scraped)

A field is a vector space over itself, and over any subfield K⊆FK \subseteq FK⊆F every FFF-vector space is a KKK-vector space by restricting the scalars