## Direct and inverse limits; basic examples

Posted: January 16, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
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Definition. A partially ordered set $(I, \leq )$ is called a directed set if for every $i,j \in I$ there exists some $k \in I$ such that $i \leq k$ and $j \leq k.$

Example 1. Let $\{S_i \}_{i \in I}$ be a collection of subsets of a set, where $I$ is a directed set and $i \leq j$ iff $S_i \subseteq S_j.$ Let $\{S_i, f_{ij} \}$ be the direct system defined in Example 1 in this post. Then $\varinjlim S_i=\bigcup S_i.$

Proof. Let $T:=\bigcup_{i \in I} S_i.$ For every $i \in I$ we define $f_i : S_i \longrightarrow T$ to be the inclusion map.

1) If $s \in S_i$ and $i \leq j,$ then $f_j f_{ij}(s)=f_j(s)=s=f_i(s).$ Thus $f_j f_{ij}=f_i.$

2) Suppose that there exists a set $X$ and the maps $g_i: S_i \longrightarrow X$ such that $g_jf_{ij}=g_i$ for all $i \leq j.$ Let’s see how we have to define $f.$ Suppose that $f: T \longrightarrow X$ is a map with $ff_i = g_i,$ for all $i.$ Let $s \in T.$ Then $s \in S_i$ for some $i$ and hence $f(s)=ff_i(s)=g_i(s).$ Now, if $s$ is also in $S_j,$ then choosing $k \in I$ with $i \leq k$ and $j \leq k$ we will have  $g_i(s)=g_k f_{ik}(s)=g_k(s)$ and $g_j(s)=g_kf_{jk}(s)=g_k(s).$ Thus $g_i(s)=g_j(s).$ So we must define $f(s)=g_i(s)$ whenever $s \in S_i.$ We just proved that $f$ is well-defined. Clearly $ff_i=g_i$ for all $i. \ \Box$

Remark. Clearly the result in Example 1 holds if $S_i$ were groups, rings, modules, etc.

Example 2. Let $S$ be a multiplicatively closed subset of a commutative ring $R$ with $1 \in S$ and $0 \notin S.$ Let $\{R_i, f_{ij} \}, \ i,j \in S,$ be the direct system defined in Example 2 in this post. Then $\varinjlim R_i =S^{-1}R.$

Proof. Let $T:=S^{-1}R$ and, for every $i \in S,$ define $f_i : R_i \longrightarrow T$ to be the inclusion map.

1) If $i \leq j,$ then $j=ri$ for some $r \in R$ and hence $\displaystyle \frac{r^na}{j^n}=\frac{a}{i^n}$ for all $a \in R$ and $n \geq 1.$ That means $f_jf_{ij}=f_i.$

2) Suppose that there exists a ring $X$ and the ring homomorphisms $g_i: R_i \longrightarrow X$ such that $g_jf_{ij}=g_i$ for all $i \leq j.$ Let $f: T \longrightarrow X$ be a ring homomorphism such that $ff_i = g_i,$ for all $i.$ Let’s see how we have to define $f.$  Let $t = a/i \in T.$ Then $t \in R_i$ and thus $f(t)=ff_i(t)=g_i(t).$ Also if $t = a/i = b/j,$ then  choosing $k=ij$ we’ll have $i \leq k, \ j \leq k.$ Thus

$g_i(t)=g_k f_{ik}(t)=g_k(ja/k)=g_k(a/i)=g_k(t)$

and $g_j(t)=g_k f_{jk}(t)=g_k(ib/k)=g_k(b/j)=g_k(t).$ So we must define $f(a/i)=g_i(a/i).$ We just proved $f$ is well-defined and clearly $f$ is a ring homomorphism and $ff_i=g_i$ for all $i. \ \Box$

Example 3. Let $R$ be a commutative ring with 1 and let $R[x]$ be the polynomial ring in the indeterminate $x.$ Let $L = \langle x \rangle,$ the ideal of $R[x]$ generated by $x.$ For every positive integer $i$ let $R_i = R[x]/L^i.$ Let $\{R_i, f_{ij} \}$ be the inverse system in Example 3 in this post. Then $\varprojlim R_i = R[[x]].$

Proof. Let $T:=R[[x]]$ and define the map $f_i : T \longrightarrow R_i$ by $f_i(h)=h + L^i$ for all $i$ and $h \in T.$

1) If $i \leq j$ and $h \in T,$ then $f_{ij}f_j(h)=f_{ij}(h + L^j)=h+ L^i=f_i(h).$ Thus $f_{ij}f_j = f_i.$

2) Suppose first that there exists a ring $X$ and the ring homomorphisms $g_i : X \longrightarrow R_i$ such that $f_{ij}g_j=g_i$ for all $i \leq j.$ Suppose also that there exists a ring homomorphism $f: X \longrightarrow T$ such that $f_if = g_i,$ for all $i.$ Then for any positive integer $i$ and all $u \in X$ we will have $g_i(u)=f_i f(u)=f(u) + L^i.$ So here is how we have to define $f$: for every positive integer $i$ we have

$g_i(u)=\sum_{m=0}^{i-1} a_{mi}(u)x^{i-1} + L^i,$

where $a_{mi}(u) \in R.$ Now we define

$f(u)=\sum_{m=0}^{\infty} a_{m, m+1}(u) x^m.$

Clearly $f$ is well-defined and $f(u+v)=f(u)+f(v)$ for all $u,v \in X.$ So,  to prove that $f$ is a ring homomorphism we must show that $f(uv)=f(u)f(v)$ for all $u,v \in X.$ To prove this, we only need to show that for every integere $k \geq 0,$ the coefficient of $x^k$ in both $f(uv)$ and $f(u)f(v)$ are equal. To do so, we begin with the fact that $f_{ij}g_j=g_i$ for all $i \leq j,$which gives us

$a_{i, i+1}(u)=a_{i, j+1}(u) \ \ \ \ \ \ \ \ \ \ (1)$

for all $i \leq j.$ Now, the coefficient of $x^k$ in $f(u)f(v)$ is

$\sum_{i=0}^k a_{i,i+1}(u)a_{k-i, k-i+1}(v). \ \ \ \ \ \ \ \ \ \ (2)$

On the other hand, the coefficient of $x^k$ in $f(uv)$ is $a_{k,k+1}(uv).$ But, since $g_{k+1}(uv)=g_{k+1}(u)g_{k+1}(v),$ we will have

$a_{k,k+1}(uv)=\sum_{i=0}^k a_{i,k+1}(u)a_{k-i,k+1}(v). \ \ \ \ \ \ \ \ \ (3).$

It is clear from (1), (2) and (3) that the coefficients of $x^k$ in both $f(uv)$ and $f(u)f(v)$ are equal.

Finally, using (1), we have

$f_if(u)=f(u) + L^i=\sum_{m=0}^{i-1}a_{m,m+1}(u)x^m + L^i=\sum_{m=0}^{i-1}a_{mi}(u)x^m + L^i = g_i(u)$

and thus $f_if=g_i. \ \Box$