如何从二元泛性质结合结合律证明n重张量积的泛性质？

Alright, let's walk through this carefully, sticking to the setup from Dummit & Foote's §10.4. We'll start by grounding ourselves in the binary tensor product's universal property (from Corollary 12), then use induction and the associativity isomorphism we already proved to extend this to the n-fold case.

First, let's restate the key result we already have for two modules—this is our foundation:

Given R-modules (M) and (N), the tensor product (M \otimes_R N) is the "universal" object for R-bilinear maps from (M \times N). More precisely:

• There’s a canonical R-bilinear map (\iota: M \times N \to M \otimes_R N) sending ((m, n) \mapsto m \otimes n).
• For any R-module (P) and any R-bilinear map (\varphi: M \times N \to P), there exists a unique R-module homomorphism (\Phi: M \otimes_R N \to P) such that (\Phi \circ \iota = \varphi) (i.e., (\Phi(m \otimes n) = \varphi(m, n)) for all (m \in M, n \in N)).

In plain terms: every bilinear map from the product factors uniquely through the tensor product.

Before jumping to n-fold tensors, we need the associativity result from §10.4: for any three R-modules (M, N, L), there’s a natural R-module isomorphism:
[
\alpha_{M,N,L}: (M \otimes_R N) \otimes_R L \to M \otimes_R (N \otimes_R L)
]
This map sends ((m \otimes n) \otimes l) to (m \otimes (n \otimes l)), and crucially, it preserves the multilinear structure—it commutes with the canonical maps from the product (M \times N \times L) to either parenthesized tensor product. This means we can treat all parenthesizations of the tensor product as equivalent for our purposes.

We’ll use mathematical induction here, since the n-fold tensor product is just an iteration of binary products, and associativity lets us glue them together consistently.

3.1 Inductive Definition of the n-Fold Tensor Product #

First, let's define the n-fold tensor product recursively:

• For (n=1): (T_1(M_1) = M_1), with the canonical map (\iota_1: M_1 \to T_1(M_1)) being the identity map.
• For (n > 1): Define (T_n(M_1, M_2, ..., M_n)) as (T_{n-1}(M_1, ..., M_{n-1}) \otimes_R M_n). Thanks to the associativity isomorphism, this is naturally isomorphic to any other parenthesization (like (M_1 \otimes_R (M_2 \otimes_R ... \otimes_R M_n))), so we don’t have to worry about which way we bracket the product.

3.2 Statement of the n-Fold Universal Property #

We want to prove this:

Let (M_1, M_2, ..., M_n) be R-modules. There exists a canonical n-linear map (\iota_n: M_1 \times M_2 \times ... \times M_n \to T_n(M_1,...,M_n)) sending ((m_1, ..., m_n) \mapsto (...((m_1 \otimes m_2) \otimes m_3) ...) \otimes m_n). For any R-module (P) and any n-linear map (\varphi: M_1 \times ... \times M_n \to P), there exists a unique R-module homomorphism (\Phi: T_n(M_1,...,M_n) \to P) such that (\Phi \circ \iota_n = \varphi).

In short: every n-linear map from the product factors uniquely through the n-fold tensor product.

3.3 Inductive Proof #

Base Cases ((n=1, 2)) #

• (n=1): Trivial. The 1-linear maps are just R-module homomorphisms, and the identity map (\iota_1) lets us factor any such homomorphism uniquely through (T_1(M_1) = M_1).
• (n=2): Exactly Corollary 12, which we already know is true.

Inductive Step: Assume true for (n=k), prove for (n=k+1) #

Suppose the universal property holds for k-fold tensor products. Now take an arbitrary (k+1)-linear map (\varphi: M_1 \times ... \times M_k \times M_{k+1} \to P) (where (P) is any R-module).

First, fix an element (m_{k+1} \in M_{k+1}). The map (\varphi(-, ..., -, m_{k+1}): M_1 \times ... \times M_k \to P) is k-linear (since (\varphi) is (k+1)-linear). By our inductive hypothesis, there’s a unique homomorphism (\Phi_{m_{k+1}}: T_k(M_1,...,M_k) \to P) such that (\Phi_{m_{k+1}}(\iota_k(m_1,...,m_k)) = \varphi(m_1,...,m_k, m_{k+1})).

Next, define a map (\psi: T_k(M_1,...,M_k) \times M_{k+1} \to P) by (\psi(t, m_{k+1}) = \Phi_{m_{k+1}}(t)) for (t \in T_k) and (m_{k+1} \in M_{k+1}). We need to confirm (\psi) is R-bilinear:

• Additive/R-linear in the first argument: Follows directly from (\Phi_{m_{k+1}}) being an R-module homomorphism.
• Additive in the second argument: Since (\varphi) is (k+1)-linear, (\varphi(-,..., -, m_1 + m_2) = \varphi(-,..., -, m_1) + \varphi(-,..., -, m_2)). By uniqueness of the k-fold homomorphism, (\Phi_{m_1 + m_2} = \Phi_{m_1} + \Phi_{m_2}), so (\psi(t, m_1 + m_2) = \psi(t, m_1) + \psi(t, m_2)).
• R-linear in the second argument: Similar to the additive case—(\varphi(-,..., -, r m) = r \varphi(-,..., -, m)), so (\Phi_{r m} = r \Phi_m), hence (\psi(t, r m) = r \psi(t, m)).

Since (\psi) is bilinear, we can apply the binary universal property (Corollary 12) to get a unique homomorphism (\Phi: T_k(M_1,...,M_k) \otimes_R M_{k+1} = T_{k+1}(M_1,...,M_{k+1}) \to P) such that (\Phi(t \otimes m_{k+1}) = \psi(t, m_{k+1})).

Now check that (\Phi \circ \iota_{k+1} = \varphi):
For any tuple ((m_1,...,m_{k+1})), (\iota_{k+1}(m_1,...,m_{k+1}) = \iota_k(m_1,...,m_k) \otimes m_{k+1}). Then:
[
\Phi(\iota_{k+1}(m_1,...,m_{k+1})) = \Phi(\iota_k(m_1,...,m_k) \otimes m_{k+1}) = \psi(\iota_k(m_1,...,m_k), m_{k+1}) = \Phi_{m_{k+1}}(\iota_k(m_1,...,m_k)) = \varphi(m_1,...,m_{k+1})
]
That’s exactly the commutativity we need.

Uniqueness of (\Phi) #

Suppose there’s another homomorphism (\Phi': T_{k+1} \to P) such that (\Phi' \circ \iota_{k+1} = \varphi). For any generator (t \otimes m_{k+1}) of (T_{k+1}) (where (t = \iota_k(m_1,...,m_k))), we have:
[
\Phi'(t \otimes m_{k+1}) = \Phi'(\iota_{k+1}(m_1,...,m_{k+1})) = \varphi(m_1,...,m_{k+1}) = \Phi(t \otimes m_{k+1})
]
Since these generators span (T_{k+1}), (\Phi' = \Phi) everywhere. So (\Phi) is unique.

3.4 Why Parenthesization Doesn’t Matter #

What if we’d defined the n-fold tensor product as (M_1 \otimes_R (M_2 \otimes_R ... \otimes_R M_n)) instead? The associativity isomorphism (\alpha) lets us translate between this and our recursive definition, and since (\alpha) commutes with the canonical n-linear maps, the universal property holds regardless of how we bracket the product. All n-fold tensor products are naturally isomorphic, so we can safely refer to "the" n-fold tensor product without ambiguity.

The core idea is that we build the n-fold case by stacking the binary universal property, using induction to extend it from 2 to n. The associativity isomorphism ensures our choice of parenthesization doesn’t break the universal property, making the result consistent across all ways to form the n-fold tensor product. This is exactly the path Dummit & Foote hints at after proving Corollary 12 and the associativity isomorphism.

内容的提问来源于stack exchange，提问作者Hugo Jenkins