12.12.4 Tensor Extension Uniqueness Condition

Tensor Extension Uniqueness Condition is the requirement, arising from the universal property of scalar extension, that any two S-linear maps out of the extended module V ⊗_R S which agree on the image of V under the canonical map v ↦ v ⊗ 1 must be equal as maps on the whole of V ⊗_R S. Equivalently, it is the statement that an S-linear map with domain V ⊗_R S is completely and uniquely determined by its restriction to the generating set {v ⊗ 1 : v ∈ V}. This condition is the half of the universal property responsible for rigidity: it guarantees that extending a map by scalars, once possible at all, can be done in exactly one way, ruling out any ambiguity in how the extension is carried out.

Statement of the Condition

The Formal Requirement

Let g and h be two S-linear maps from the extended module V ⊗_R S into an arbitrary S-module M. The uniqueness condition asserts that if

for every v ∈ V, then g = h identically on V ⊗_R S, with no exceptions and no additional hypotheses required beyond S-linearity of g and h.

Why the Generating Set Suffices

Every element of V ⊗_R S is a finite sum Σ vₖ ⊗ sₖ. Because sₖ · (vₖ ⊗ 1) = vₖ ⊗ sₖ by the defining bilinearity of the tensor product over R, and because g, h are assumed S-linear, agreement of g and h on elements of the form v ⊗ 1 forces agreement on sₖ · (vₖ ⊗ 1) = vₖ ⊗ sₖ for every scalar sₖ ∈ S, and hence on the finite sum Σ vₖ ⊗ sₖ by additivity. The uniqueness condition is therefore a direct consequence of {v ⊗ 1 : v ∈ V} generating V ⊗_R S as an S-module.

Proof of the Uniqueness Condition

Reduction to a Single Difference Map

Define k = g − h, which is S-linear since g and h are. The hypothesis becomes k(v ⊗ 1) = 0 for all v ∈ V, and the goal is to show k = 0 on all of V ⊗_R S.

Extending the Vanishing by Linearity

For an arbitrary element Σ vₖ ⊗ sₖ of V ⊗_R S:

using S-linearity of k to pull each scalar sₖ out and then applying the hypothesis to each term. Since every element of V ⊗_R S has this form, k vanishes identically, so g = h.

Where Each Hypothesis Is Used

S-linearity of g and h is used twice: once to form the difference k as an S-linear map, and once to distribute k over the scalar multiples sₖ in the finite sum. No property of M beyond being an S-module is required, and no assumption about R → S beyond it being a ring homomorphism is needed, since the argument only uses the algebraic form of elements of V ⊗_R S.

Diagram of the Uniqueness Condition

Role Alongside the Existence Condition

Complementary Half of the Universal Property

The universal property of scalar extension pairs an existence condition — every R-linear map V → M (with M restricted to an R-module via R → S) extends to some S-linear map V ⊗_R S → M — with the uniqueness condition described here. Existence alone would permit several different extensions of the same map; uniqueness eliminates that possibility, together giving a genuine bijective correspondence rather than a mere existence statement.

Consequence: The Extension Functor Is Well-Defined

Because the extended map is unique whenever it exists, the assignment V ↦ V ⊗_R S together with the extension of morphisms is a well-defined functor rather than a multi-valued correspondence. Any construction that produced more than one valid extension for the same input map would violate this condition and would not qualify as an instance of scalar extension.

Consequence: Compatibility with Composition

The uniqueness condition also underlies the fact that extending a composite map equals the composite of the extended maps, (f ∘ g) ⊗ id_S = (f ⊗ id_S) ∘ (g ⊗ id_S). Both sides are S-linear maps that agree on the generating set {v ⊗ 1}, so the uniqueness condition identifies them without any further computation.

Failure Modes When Uniqueness Fails

Symptom: Ambiguous Extensions

If a construction violates this condition, an R-linear map V → M could admit two distinct S-linear extensions to V ⊗_R S, and no canonical choice between them would exist. This would make any formula or algorithm relying on "the" extension of a map ill-defined, since the result would depend on an arbitrary selection.

Symptom: Breakdown of Naturality Squares

Naturality arguments in tensor algebra routinely rely on two maps being shown equal by checking agreement on a generating set. If the uniqueness condition failed, such arguments would be invalid, since agreement on generators would no longer imply agreement everywhere, undermining the broader structural theory built on top of scalar extension.