Direct limit of modules & tensor product

Posted: February 16, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
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Theorem. Let R be a commutative ring, A an R-module and \{M_i, f_{ij} \} a direct system of R-modules. Then

\varinjlim (A \otimes_R M_i) \cong A \otimes_R \varinjlim M_i.

The above isomorphism is an isomorphism of R-modules. We will give the proof of this theorem at the end of this post. So let’s get prepared for the proof!

Notation 1R is a commutative ring, A is an R-module and \{M_i, f_{ij} \} is a direct system of R-modules over some partially ordered index set I. We will not assume that I is directed.

Notation 2. We proved in here that M=\varinjlim M_i exists. Recall that, as a part of the definition of direct limit, we also have canonical homomorphisms f_i : M_i \longrightarrow M satisfying the relations f_jf_{ij}=f_i, for all i \leq j. For modules we explicity defined f_i by f_i(x_i)=\rho_i(x_i) + N (see the theorem in here), but we will not be needing that.

Notation 3. Let A be an R-module. For all i,j \in I with i \leq j we will let M_i' = A \otimes_R M_i and f'_{ij}=\text{id}_A \otimes_R f_{ij}. By Example 5, \{M_i', f'_{ij} \} is a direct system of R-modules and so M'=\varinjlim (A \otimes_R M_i) exists. We also have canonical homomorphisms f_i' : M_i' \longrightarrow M' satisfying f_j'f_{ij}'=f_i', for all i \leq j.

Notation 4. For every i \in I we will let h_i = \text{id}_A \otimes f_i.

Remark 1.  Clearly h_i : M'_i \longrightarrow A \otimes_R M and h_jf_{ij}'=h_i, for all i \leq j, because

h_j f_{ij}' = (\text{id}_A \otimes f_j)(\text{id}_A \otimes f_{ij})=\text{id}_A \otimes f_jf_{ij}=\text{id}_A \otimes f_i = h_i.

 Lemma. Let X be an R-module. Suppose that for every i \in I there exists an R-module homomorphism g_i : M_i' \longrightarrow X such that g_jf_{ij}' = g_i, for all i \leq j. Let a \in A. Then

1) There exist R-module homomorphisms \nu_{i,a} : M_i \longrightarrow X such that \nu_{j,a} f_{ij}=\nu_{i,a}, for all i \leq j.

2) There exists a unique R-module homomorphism \nu_a : M \longrightarrow X such that \nu_a f_i = v_{i,a}, for all i \in I.

Proof. 1) \nu_{i,a} are defined very naturally: \nu_{i,a}(x_i)=g_i(a \otimes_R x_i), for all x_i \in M_i. Then

\nu_{j,a} f_{ij}(x_i)=g_j(a \otimes_R f_{ij}(x_i))=g_j f_{ij}'(a \otimes_R x_i)=g_i(a \otimes_R x_i)=\nu_{i,a}(x_i).

2) This part is obvious from the first part and the universal property of direct limit. (See Definition 1 in here) \Box

Proof of the Theorem. We will show that A \otimes_R M satisfies the conditions in the definition of M' = \varinjlim M_i' (see Definition 1 in here) and thus, by the uniqueness of direct limit, A \otimes_R M \cong M' and the theorem is proved. The first condition is satisfied by Remark 1. For the second condition (the universal property), suppose that X is an R-module and g_i : M_i' \longrightarrow X are R-module homomorphisms such that g_jf_{ij}'=g_i, whenever i \leq j. So the hypothesis in the above lemma is satisfied. For a \in A and i \in I let \nu_{i,a} and \nu_a be the maps in the lemma. Define the map \nu : A \times M \longrightarrow X by v(a,x)=v_a(x), for all a \in A and x \in M. See that \nu is R-bilinear and so it induces an R-module homomorphism f : A \otimes_R M \longrightarrow X defined by f(a \otimes_R x)=v_a(x). We also have

 fh_i(a \otimes_R x_i)=f(a \otimes_R f_i(x_i))=\nu_af_i(x_i)=\nu_{i,a}(x_i)=g_i(a \otimes_R x_i),

for all a \in A, \ i \in I and x_i \in M_i. Thus fh_i = g_i. So the only thing left is the uniqueness of f, which is obvious because, as we mentioned in the second part of the above lemma, \nu_a is uniquely defined for a given a \in A. \ \Box

Remark 2. If R is not commutative, M_i are left R-modules and A is a right R-module (resp. (R,R)-bimodule), then the isomorphism in the theorem is an isomorphism of abelian groups (resp. left R-modules).


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