CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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$R^{m} \otimes_{R} R^{n} ≅ R^{m n}$ with the product basis, and $\dim_{F} (V \otimes_{F} W) = \dim_{F} V \dim_{F} W$

Statement

Let $R$ be a commutative ring and $m, n \in N$ . The standard finite free modules satisfy

$R^{m} \otimes_{R} R^{n} ≅ R^{m n},$

with the tensor products of standard basis vectors corresponding to the standard basis indexed by $m \times n$ .

If $F$ is a field and $V, W$ are finite-dimensional $F$ -vector spaces, then

$\dim_{F} (V \otimes_{F} W) = (\dim_{F} V) (\dim_{F} W) .$

Both assertions include a zero rank or zero-dimensional factor.

Facts & Assumptions

Given: Natural numbers $m, n$ , a commutative ring $R$ , and finite-dimensional vector spaces $V, W$ over a field $F$ .

[L1]

Tensor products of free modules with bases indexed by $I, J$ have basis indexed by $I \times J$ (The elementary tensors of two bases form the product basis of the tensor product).

[L2]

For any unital ring $R$ , the free module $R^{(m)}$ has its standard basis indexed by $m$ , including the empty basis at $m = 0$ (The free module on a set and its standard basis). For a field $F$ , this is the usual basis of $F^{m}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

[L3]

The dimension of a finite-dimensional vector space is the unique natural number equinumerous with a basis; the zero space has dimension zero (Finite-dimensional vector space, and its dimension $\dim_{F} V$ ; infinite-dimensional means having no finite basis).

Proof

technique · direct

1.1givenL1algebra

Apply [L1] to the standard bases indexed by $m$ and $n$ . Their product is indexed by $m \times n$ , which has $m n$ elements, giving the first isomorphism and its product basis.

1.2L1L3choose

Choose bases of $V$ and $W$ with respectively $p = \dim_{F} V$ and $q = \dim_{F} W$ elements. By [L1] their elementary tensors form a basis of $V \otimes_{F} W$ indexed by $p \times q$ , hence with $p q$ elements.

1.3L1L2L3

If $m = 0$ or $n = 0$ , or if $p = 0$ or $q = 0$ , [L2] makes one basis empty and [L1] makes the product basis empty, so both sides are the zero module or have dimension zero as asserted.

2.1step 1.2L3

By [L3], step 1.2 gives $\dim_{F} (V \otimes_{F} W) = p q = (\dim_{F} V) (\dim_{F} W)$ .

3.1step 1.1step 2.1step 1.3∎

Steps 1.1 through 2.1 prove both formulas, including their zero boundaries.

Depends on

Used by

Dependency tree · two levels

29 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

C. Dennis, Week 1 recap on tensor products (standard reference, not scraped)

Rm⊗RRn≅RmnRm⊗R​Rn≅Rmn with the product basis, and dim⁡F(V⊗FW)=dim⁡FV dim⁡FWdimF​(V⊗F​W)=dimF​VdimF​W