## 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 1$R$ 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).