Posts Tagged ‘direct limit of modules’

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).


In this post we will keep the notation in the previous part. We will assume that I is a directed set and \{M_i, f_{ij} \} is a direct systen of R-modules. In Remark 2 in there we proved that

\varinjlim M_i=\{ \rho_i(x_i) + N : \ i \in I, x_i \in M_i \}.

Our goal now is to find a necessary and sufficient condition for an element of \varinjlim M_i to be 0, i.e. we want to see when we will have \rho_i(x_i) \in N. Let’s start with a notation which will simlify our work.

Notation. For any j,k \in I with j \leq k and y_j \in M_j we let \alpha(j,k,y_j)=\rho_k f_{jk}(y_j)-\rho_j(y_j). Note that \alpha(j,k,y_j) are just the generators of N as an R-submodule of M = \bigoplus M_i.

Now two easy properties of \alpha:

Lemma 1. If r \in R, then

1) r\alpha(j,k,y_j)=\alpha(j,k,ry_j),

2)  \alpha(j,k,y_k) + \alpha(j,k,y'_k)=\alpha(j,k,y_k+y'_k),

Proof. Trivial because \rho_j, \rho_k and f_{jk} are all R-module homomorphisms. \Box.

Lemma 2. If j \leq k \leq t and y_j \in M_j, then \alpha(j,k,y_j)=\alpha(j,t,y_j)+ \alpha(k,t, - f_{jk}(y_j).

Proof. This is also a trivial result of this fact that f_{kt}f_{jk}=f_{jt}. \ \Box

And the main result of this section:

Theorem. \rho_i(x_i) \in N if and only if f_{it}(x_i)=0 for some t \in I with i \leq t.

Proof. If f_{it}(x_i)=0 for some t with i \leq t, then \rho_i(x_i)=\rho_i(x_i) - \rho_t f_{it}(x_i) \in N. Conversely, supppose that \rho_i(x_i) \in N. Then by Lemma 1, part 1, we can write

\rho_i(x_i) = \sum \alpha(j,k,y_j) \ \ \ \ \ \ \ (1).

Let t \in I be such that i,j,k \leq t for all j,k occuring in (1). Since  \rho_t f_{it}(x_i)=\rho_t f_{it}(x_i) - \rho_i(x_i) + \rho_i(x_i), we will have

\rho_t f_{it}(x_i) = \alpha(i,t,x_i) + \sum \alpha(j,k,y_j) \ \ \ \ \ \ \ \ \ (2),

by (1). Applying Lemma 2 to (2) we will get

\rho_t f_{it}(x_i) = \sum \alpha(\ell, t, z_{\ell}) \ \ \ \ \ \ \ \ \ (3),

where the sum runs over \ell. Note that by Lemma 1, part 2, we may assume that all the indices \ell in (3) are distinct. We can rewrite (3) as

\rho_t f_{it}(x_i) = \rho_t \left (\sum f_{\ell t}(z_{\ell}) \right) - \sum \rho_{\ell}(z_{\ell}) \ \ \ \ \ \ \ \ \ (4).

But since every element of \bigoplus M_i is uniquely written as \sum \rho_j(x_j), in (4) we must have \rho_{\ell}(z_{\ell})=0, and hence z_{\ell}=0, for all \ell \neq t. Thus (4) becomes

\rho_t f_{it}(x_i) = \rho_t f_{tt}(z_t) - \rho_t(z_t) = 0,

which gives us f_{it}(x_i)=0. \ \Box

Corollary. Let \{M_i, f_{ij} \} be a direct system of rings. We proved in Remark 3 in this post that M = \varinjlim M_i is also a ring. If each M_i is a domain, then M is also a domain.

Proof. Suppose that a = \rho_i(x_i) + N, \ b = \rho_i(y_i) + N are two elements of \varinjlim M_i and ab = 0. That means \rho_i(x_iy_i) \in N and hence, by the theorem, there exists some t with i \leq t such that


Therefore, since M_t is a domain, we have either f_{it}(x_i)=0 or f_{it}(y_i)=0. Applying the theorem again, we get either a=0 or b = 0. \ \Box

Theorem. Let R be a ring and let \{M_i, f_{ij} \} be a direct system of left R-modules over some partially ordered index set I. Let M= \bigoplus_{i \in I} M_i and for every i \in I, let \rho_i : M_i \longrightarrow M be the natural injection map, i.e. \rho_i(x_i)=(x_j)_{j \in I}, where x_j=0 for all j \neq i. Define the R-submodule N of M by

N = \langle \rho_jf_{ij}(x_i) - \rho_i(x_i): \ i \leq j, \ x_i \in M_i \rangle.

Then \varinjlim M_i = M/N.

Proof. For every i \in I define the R-module homomorphism f_i : M_i \longrightarrow M/N by f_i(x_i)=\rho_i(x_i) + N. Clearly if i \leq j, then \rho_j f_{ij}(x_i) - \rho_i(x_i) \in N and thus

f_j f_{ij}(x_i) = \rho_j f_{ij}(x_i) + N = \rho_i(x_i) + N = f_i(x_i).

So f_j f_{ij} = f_i. Suppose now that there exists a left R-module X and R-module homomorphisms g_i : M_i \longrightarrow X such that g_j f_{ij}=g_i whenever i \leq j. We need to prove that there exists a unique R-module homomorphism f: M/N \longrightarrow X such that ff_i = g_i for all i. Let’s see how this f must be defined. Well, we have g_i(x_i)=ff_i(x_i)=f(\rho_i(x_i) + N). Now, since every element of M/N is in the form \sum \rho_i(x_i) + N we have to define f in this form: f(\sum \rho_i(x_i) + N)=\sum g_i(x_i). The only thing left is to prove that f is well-defined. To do so, we define the homomorphism \hat{f} : M \longrightarrow X by \hat{f}(\sum \rho_i(x_i)) = \sum g_i(x_i). So I only need to prove that N \subseteq \ker \hat{f}. So suppose that i \leq j and let x_i \in M_i. Then

\hat{f}(\rho_j f_{ij}(x_i) - \rho_i (x_i))=g_jf_{ij}(x_i) - g_i(x_i)=0,

because g_j f_{ij}=g_i for all i \leq j. \ \Box

Remark 1. The direct limit of a direct system of abelian groups always exists by the theorem because abelian groups are just \mathbb{Z}-modules.

We now show that if I is a directed set, then \varinjlim M_i will look much simpler.

Remark 2. If, in the theorem, I is directed, then \varinjlim M_i = \{\rho_i(x_i) + N : \ i \in I, \ x_i \in M_i \}.

Proof. Let a = \sum_{i \in J} \rho_i(x_i) + N be any element of \varinjlim M_i. Since I is directed and J is finite, there exists some k \in I such that i \leq k for all i \in J. Then for every i \in J we’ll have

\rho_i(x_i) +N = \rho_k f_{ik}(x_i) + \rho_i(x_i) - \rho_k f_{ik}(x_i) + N = \rho_k f_{ik}(x_i) + N.

So if we let y_k = \sum_{i \in J} f_{ik}(x_i) \in M_k, then a = \rho_k (y_k) + N. \ \Box

Remark 3. If, in the theorem, I is directed and \{M_i, f_{ij} \} is a direct system of rings, then \varinjlim M_i is also a ring.

Proof. Since M_i are all abelian groups, \varinjlim M_i exists and it is an abelian group by the theorem. So we just need to define a multiplication on \varinjlim M_i. Let a,b \in \varinjlim M_i. By Remark 2 there exist i,j \in I such that a = \rho_i(x_i) + N, \ b = \rho_j(x_j) + N. Choose k \in I with i,j \leq k. Then by the definition of N

a=\rho_i(x_i) + N = \rho_k f_{ik}(x_k) + N

and thus a= \rho_k (y_k) + N. Similarly b = \rho_k(z_k) + N. Now we define

ab = \rho_k(y_k z_k) + N.

We need to show that the multiplication we’ve defined is well-defined. So suppose that a = \rho_{\ell}(y_{\ell}) + N and b = \rho_{\ell}(z_{\ell}) + N is another way of representing a,b. Choose n \in I such that k, \ell \leq n. Then

\rho_k(y_k z_k) + N = \rho_n f_{kn}(y_k z_k) + N = (\rho_nf_{kn}(y_k) + N)(\rho_n f_{kn}(z_k) + N). \ \ \ \ (1)

But we also have

\rho_n f_{kn}(y_k) + N = \rho_k(y_k) + N=a= \rho_{\ell}(y_{\ell}) + N = \rho_n f_{\ell n}(y_{\ell}) + N \ \ \ \ \ \ (2)

and similarly

\rho_n f_{kn}(z_k) = \rho_n f_{\ell n}(z_{\ell}) + N. \ \ \ \ \ (3)

Plugging (2) and (3) into (1) will give us

\rho_k(y_kz_k) + N =\rho_n f_{\ell n}(y_{\ell} z_{\ell}) + N = \rho_{\ell}(y_{\ell} z_{\ell}) + N,

which proves that the multiplication we defined is well-defined. \Box