Posts Tagged ‘localization of commutative ring’

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