LemmaStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-28 (claude-fable-5)

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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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; $λ (u + v) = λ u + λ v$ ; $(λ + μ) v = λ v + μ v$ ; $(λ μ) v = λ (μ v)$ ; $1_{F} v = v$ .

[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 $x (y + z) = x y + x z$ and $(y + z) x = y x + z x$ ; and every $x \neq 0_{F}$ 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 $a b \in K$ 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

1.1L2

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.

1.2L2

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

1.3L2

Axiom (V2) holds for $V_{F}$ : $λ (u + v) = λ u + λ v$ is distributivity of multiplication over addition.

1.4L2

Axiom (V3) holds for $V_{F}$ : $(λ + μ) v = λ v + μ v$ is distributivity on the other side.

1.5L2

Axiom (V4) holds for $V_{F}$ : $(λ μ) v = λ (μ v)$ is associativity of the field multiplication.

1.6L2

Axiom (V5) holds for $V_{F}$ : $1_{F} v = v$ is the multiplicative identity law.

1.7L3

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.

1.8L1

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.

1.9L3

For $λ, μ \in K$ the sum $λ + μ$ and the product $λ μ$ formed in $K$ are the sum and the product formed in $F$ , and the multiplicative identity of $K$ is $1_{F}$ .

2.1step 1.1step 1.2step 1.3step 1.4step 1.5step 1.6L1

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.

2.2step 1.7step 1.9L1L3

Let $λ, μ \in K$ and $u, v \in V$ . Then $λ (u + v) = λ u + λ v$ and $(λ μ) v = λ (μ v)$ are the instances of (V2) and (V4) for these elements of $F$ , the product $λ μ$ being the same whether formed in $K$ or in $F$ ; $(λ + μ) v = λ v + μ v$ is the instance of (V3), the sum $λ + μ$ 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).

3.1step 1.8step 2.2L1

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

4.1step 2.1step 3.1∎

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 $K \times F$ .

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: $R$ is a vector space over $R$ and also over the embedded copy of $Q$ 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.

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

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A field is a vector space over itself, and over any subfield K⊆FK⊆F every FF-vector space is a KK-vector space by restricting the scalars

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

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