Direct limit of modules (1)

Posted: January 17, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes

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


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