## 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 on$I$ 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. Clearly$f_{ii}$ 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 $f_{jk}f_{ij}=f_{ik}.$

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 $R.$