Posts Tagged ‘direct limit’

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

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.