Direct and inverse systems; basic examples

Posted: January 12, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
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Example 1. In the category of sets let S be a set and \{S_i \}_{i \in I} a collection of subsets of S. We also have i \leq j iff S_i \subseteq S_j. The map f_{ij}: S_i \longrightarrow S_j is defined to be the inclusion map. Clearly \{S_i, f_{ij} \} is a direct system.

Example 2. Let R be a commutative ring and let I be a non-empty subset of R such that no element of I is nilpotent. The partial order \leq is defined onI by i \leq j iff i \mid j in R, i.e. j=ri for some r \in R. For every i \in I let R_i be the localization of R at \{1,i,i^2, \cdots \}. We need now to define the morphisms f_{ij}. If i \leq j, then j=ri for some r \in R and so we can define f_{ij}: R_i \longrightarrow R_j by f_{ij}(a/i^n)=(r^na)/j^n for all a \in R, \ n \geq 0. It is easy to see that f_{ij} is a ring homomorphism. Clearlyf_{ii} is the identity map over R_i because in this case we may choose r=1. Also, if i \leq j \leq k with j=ri and k=sj, then k=rsi and hence

f_{jk}f_{ij}(a/i^n)=f_{jk}((r^na)/j^n)=(s^nr^na)/k^n=((rs)^na)/k^n = f_{ik}(a/i^n).

Thus f_{jk}f_{ij}=f_{ik} and so \{R_i, f_{ij} \} is a direct system in the category of rings.

Example 3. Let R be a ring and let L be a two-sided ideal of R. Let I be the set of positive integers and put R_i = R/L^i for every i \in I. For every i \leq j the ring homomorphism f_{ij}: R_j\longrightarrow R_i is defined naturally by f_{ij}(r + L^j)=r+L^i. This ring homomorphisms are well-defined because if i \leq j, then L^j \subseteq L^i. Clearly f_{ii} is the identity map of R_i and if i \leq j \leq k, then

f_{ij}f_{jk}(r + L^k)=f_{ij}(r + L^j)=r+L^i = f_{ik}(r+L^k).

Thus \{R_i, f_{ij} \} is an inverse system in the category of rings.

Similarly, if A is an R-module and we put A_i = A/L^iA, then we will have the inverse system \{A_i, g_{ij} \} in the category of R-modules. Here g_{ij}: A_j \longrightarrow A_i is defined by g_{ij}(a + L^jA)=a+L^i A whenever i \leq j.

Example 4. Let G be a group and let \{N_i \}_{i \in I} be a family of normal subgroups of G which have finite index in G. We define the partial order \leq on I by i \leq j if and only if N_j \subseteq N_i. We also define, whenever i \leq j, the group homomorphism f_{ij}: G/N_j \longrightarrow G/N_i by f_{ij}(gN_j)=gN_i. It is easily seen that \{G/N_i, f_{ij} \} is an inverse system in the category of finite groups.

Example 5. Let R be a commutative ring, let A be an R-module and let \{M_i, f_{ij} \} be a direct system of R-modules over some partially ordered index set I. For every i \in I let M_i'=A \otimes_R M_i and for all i,j \in I with i \leq j define f'_{ij} : M_i' \longrightarrow M_j' by f'_{ij}=\text{id}_A \otimes_R f_{ij}. That means f'_{ij}(a \otimes_R x)=a \otimes_R f_{ij}(x), for all a \in A and x \in M_i.

Claim . \{M_i', f_{ij}' \} is a direct system of R-modules.

Proof.  So we need to prove two things to show that \{M_i', f'_{ij} \} is a direct system:

1) f'_{ii} = \text{id}_{M_i'}, for all i \in I. This is obvious because f_{ii} = \text{id}_{M_i} and hence, for every a \in A and x \in M_i we have f'_{ii}(a \otimes_R x)=a \otimes_R f_{ii}(x)=a \otimes_R x.

2) f'_{jk}f'_{ij}=f'_{ik}, whenever i \leq j \leq k. This also follows very easily from f_{jk}f_{ij}=f_{ik}.

  1. rahul says:

    How do you show that the partial order in Example 2 is antisymmetric?

    • Yaghoub says:

      You are absolutely right, it is not antisymmetric but we do not need this property of posets to define a direct system (and direct limit). We only need our index set to be reflexive and transitive. Although this is clear, but still I should have mentioned it when I was defining direct systems. We do need elements of I to be non-nilpotent in order to be able to localize R.

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