## 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 $f_if=g_i. \ \Box$

## Direct and inverse limits; definition & uniqueness

Posted: January 12, 2011 in Direct and Inverse Limit, Noncommutative Ring Theory Notes
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Definition 1. Let $\{M_i, f_{ij} \}$ be a direct system in a category $\mathcal{C}.$ Suppose there exists an object $M$ and morphisms $f_i : M_i \longrightarrow M,$ for every $i,$ such that the following conditions are satisfied:

1) $f_j f_{ij}=f_i,$ whenever $i \leq j,$

2) (The universal property) Suppose that there exist an object $X$ and morphisms $g_i : M_i \longrightarrow X, \ i \in I,$ such that $g_j f_{ij}=g_i$ whenever $i \leq j.$ Then there exists a unique morphism $f: M \longrightarrow X$ such that $ff_i = g_i, \ i \in I.$

Then $M$ is called the direct limit of our direct system and we will just write $\varinjlim M_i = M.$

Definition 2. Let $\{M_i, f_{ij} \}$ be an inverse system in a category $\mathcal{C}.$ Suppose that there exists an object $M$ and morphisms $f_i : M \longrightarrow M_i,$ for every $i,$ such that the following conditions are satisfied:

1) $f_{ij}f_j=f_i,$ whenever $i \leq j,$

2) If there exists some object $X$ and morphisms $g_i : X \longrightarrow M_i,$ for every $i,$ such that $f_{ij}g_j=g_i$ whenever $i \leq j,$ then there must exist a unique morphism $f : X \longrightarrow M$ such that $f_if = g_i$ for all $i.$

Then $M$ is called the inverse limit of our inverse system and we will just write $\varprojlim M_i = M.$

In our definitions I wrote “the” direct limit and “the” inverse limit. That is because “the” direct (resp. inverse) limit of a direct (resp. inverse) system, if it exists, is unique, up to isomorphism of course, as we are going to see:

Fact. The direct (resp. inverse) limit of a direct (resp. inverse) system is unique up to isomorphism if it exists.

Proof. I will prove the fact for direct limit only. The proof for inverse limit is similar. Suppose that both $\{M, f_i \}$ and $\{M', f'_i \}$ satisfy the conditions in Dedinition 1. Then, by condition 2), there exist (unique) morphisms $f: M \longrightarrow M'$ and $f': M' \longrightarrow M$ such that $ff_i=f'_i$ and $f'f'_i=f_i.$ Thus the morphism $f'f : M \longrightarrow M$ satisfies $(f'f)f_i=f_i.$ But we also have that the identity morphism $\text{id}_M : M \longrightarrow M$ satisfies $\text{id}_M f_i = f_i.$ Thus, by the uniqueness condition in 2), we must have $f'f = \text{id}_M.$ Similarly $ff'= \text{id}_{M'}$ and hence $f$ is an isomorphism and $f'$ is its inverse.

Remark. You might have found the relationships between the morphisms in the definition of direct (resp. inverse) systems and limits complicated but they are not! If you draw diagrams, then you will see that they just mean that those diagrams are commutative.