Definition and universal property of the direct limit of commutative unital rings

In this exercise, we develop the concept of direct limit. Let I be a nonempty paritally ordered set, and let \{A_i\}_I be a collection of abelian groups. We suppose also that I is directed; that is, for all i,j \in I, there exists k \in I with i,j \leq k.

Suppose that for every pair of indices i,j \in I with i \leq j, there is a map \rho_{i,j} : A_i \rightarrow A_j such that the following hold: (1) \rho_{j,k} \circ \rho_{i,j} = \rho_{i,k} whenever i \leq j \leq k and (2) \rho_{i,i} = 1 for all i \in I.

Let B = \bigcup_I A_i \times \{i\} be the disjoint union of the A_i. Define a relation \sigma on B as follows: (a,i) \ \sigma\ (b,j) if and only if there exists k \in I such that i,j \leq k and \rho_{i,k}(a) = \rho_{j,k}(b).

  1. Show that \sigma is an equivalence relation on B. We define \varinjlim A_i = B/\sigma.
  2. Let [x]_\sigma denote the class of x in \varinjlim A_i and define \rho_i : A_i \rightarrow \varinjlim A_i by \rho_i(a) = [(a,i)]_\sigma. Show that if each \rho_{i,j} is injective, then \rho_i is also injective for all i.
  3. Assume that the \rho_{i,j} are all group homomorphisms. For [(a,i)]_\sigma,[(b,j)]_\sigma \in \varinjlim A_i, show that the operation [(a,i)]_\sigma + [(b,j)]_\sigma = [(\rho_{i,k}(a) + \rho_{j,k}(b),k)]_\sigma, where k is any upper bound of i and j, is well defined and makes \varinjlim A_i an abelian group. Deduce that the \rho_i are group homomorphisms.
  4. Prove that if all the A_i are commutative rings with 1 \neq 0 and all the \rho_{i,j} are unital ring homomorphisms (that is, they send 1 to 1), then \varinjlim A_i may likewise be given the structure of a commutative ring with 1 \neq 0 such that the \rho_i are all ring homomorphisms.
  5. Under the hypotheses of part (c), prove that \varinjlim A_i has the following universal property: if C is any abelian group such that for each i \in I there is a homomorphism \varphi_i : A_i \rightarrow C with \varphi_i = \varphi_j \circ \rho_{i,j} whenever i \leq j, then there is a unique homomorphism \varphi : A \rightarrow C such that \varphi \circ \rho_i = \varphi_i for all i.

  1. To show that \sigma is an equivalence, we need to verify that it is reflexive, symmetric, and transitive.
    1. (\sigma is reflexive) Let (a,i) \in B. Note that i \leq i, and that \rho_{i,i}(a) = a = \rho_{i,i}(a). Thus (a,i) \ \sigma\ (a,i), and hence \sigma is reflexive.
    2. (\sigma is symmetric) Suppose (a,i) \ \sigma\ (b,j). Then there exists k \geq i,j such that \rho_{i,k}(a) = \rho_{j,k}(b). Certainly \rho_{j,k}(b) = \rho_{i,k}(a), so that (b,j)\ \sigma\ (a,i). Thus \sigma is symmetric.
    3. (\sigma is transitive) Suppose (a,i)\ \sigma\ (b,j) and (b,j)\ \sigma\ (c,k). Then there exist \ell \geq i,j such that \rho_{i,\ell}(a) = \rho_{j,\ell}(b) and m \geq j,k such that \rho_{j,m}(b) = \rho_{k,m}(c). Since I is a directed poset, there exists t \in I such that t \geq \ell,m. Now \rho_{i,t}(a) = \rho_{\ell,t}(\rho_{i,\ell}(a)) = \rho_{\ell,t}(\rho_{j,\ell}(b)) = \rho_{j,t}(b) = \rho_{m,t}(\rho_{j,m}(b)) = \rho_{m,t}(\rho_{k,m}(c)) = \rho_{k,t}(c). Thus (a,i)\ \sigma\ (c,k), and hence \sigma is transitive.

    Thus \sigma is an equivalence relation.

  2. Suppose that the \rho_{i,j} are all injective. Choose i \in I, and let a,b \in A_i such that \rho_i(a) = \rho_i(b). Then we have [(a,i)]_\sigma = [(b,i)]_\sigma. That is, for some k \geq i, we have \rho_{i,k}(a) = \rho_{i,k}(b). Since \rho_{i,k} is injective, a = b. Thus \rho_i is injective.
  3. Suppose that the \rho_{i,j} are all group homomorphisms. First we show that + is well defined.

    Let [(a_1,i_1)]_\sigma = [(a_2,i_2)]_\sigma and [(b_1,j_1)]_\sigma = [(b_2,j_2)]_\sigma. Then there exists s \geq i_1,i_2 such that \rho_{i_1,s}(a_1) = \rho_{i_2,s}(a_2) and r \geq j_1,j_2 such that \rho_{j_1,r}(b_1) = \rho_{j_2,r}(b_2). Now (using the directedness of I) choose arbitrary k_1 \geq i_1,j_1 and k_2 \geq i_2,j_2. Finally, again choose t \geq k_1,k_2,r,s. Now note the following.

    \rho_{k_1,t}(\rho_{i_1,k_1}(a_1) + \rho_{j_1,k_1}(b_1))  =  \rho_{k_1,t}(\rho_{i_1,k_1}(a_1)) + \rho_{k_1,t}(\rho_{j_1,k_1}(b_1))
     =  \rho_{i_1,t}(a_1) + \rho_{j_1,t}(b_1)
     =  \rho_{s,t}(\rho_{i_1,s}(a_1)) + \rho_{r,t}(\rho_{j_1,r}(b_1))
     =  \rho_{s,t}(\rho_{i_2,s}(a_2)) + \rho_{r,t}(\rho_{j_2,r}(b_2))
     =  \rho_{i_2,t}(a_2) + \rho_{j_2,t}(b_2)
     =  \rho_{k_2,t}(\rho_{i_2,k_2}(a_2)) + \rho_{k_2,t}(\rho_{j_2,k_2}(b_2))
     =  \rho_{k_2,t}(\rho_{i_2,k_2}(a_2) + \rho_{j_2,k_2}(b_2))

    Thus (\rho_{i_1,k_1}(a_1) + \rho_{j_1,k_1}(b_1),k_1) \ \sigma\ (\rho_{i_2,k_2}(a_2) + \rho_{j_2,k_2}(b_2),k_2), and we have [(a_1,i_1)]_\sigma + [(b_1,j_1)]_\sigma = [(a_2,i_2)]_\sigma = [(b_2,j_2)]_\sigma. Thus + is well-defined.

    Next we show that (\varinjlim A_i, +) is an abelian group.

    1. (+ is associative) Let [(a,i)], [(b,j)], and [(c,k)] be in \varinjlim A_i, and let \ell \geq i,j, t \geq j,k, and m \geq \ell,t. Then we have the following.
      \left( [(a,i)] + [(b,j)] \right) + [(c,k)]  =  [\rho_{i,\ell}(a) + \rho_{j,\ell}(b)] + [(c,k)]
       =  [(\rho_{\ell,m}(\rho_{i,\ell}(a) + \rho_{j,\ell}(b)) + \rho_{k,m}(c),m)]
       =  [(\rho_{\ell,m}(\rho_{i,\ell}(a)) + \rho_{\ell,m}(\rho_{j,\ell}(b)) + \rho_{k,m}(c),m)]
       =  [(\rho_{i,m}(a) + \rho_{j,m}(b) + \rho_{k,m}(c),m)]
       =  [(\rho_{i,m}(a) + \rho_{t,m}(\rho_{j,t}(b)) + \rho_{t,m}(\rho_{j,t}(c)),m)]
       =  [(\rho_{i,m}(a) + \rho_{t,m}(\rho_{j,t}(b) + \rho_{k,t}(c)),m)]
       =  [(a,i)] + [(\rho_{j,t}(b) + \rho_{k,t}(c),t)]
       =  [(a,i)] + \left( [(b,j)] + [(c,k)] \right)

      So + is associative.

    2. Note that for all i,j \in I, there exists k \geq i,j, and \rho_{i,k}(0) = \rho_{j,k}(0) since the \rho_{i,j} are group homomorphisms. Thus [(0,i)] = [(0,j)] for all i,j. Let 0 = [(0,i)]. Now let [(a,i)] \in \varinjlim A_i. Then 0 + [(a,i)] = [(0,i)] + [(a,i)] = [(\rho_{i,i}(0) + \rho_{i,i}(0), i)] = [(0+a,i)] = [(a,i)]. Similarly, [(a,i)] + 0 = [(a,i)]. Thus 0 = [(0,i)] is an additive identity element.
    3. Let [(a,i)] \in \varinjlim A_i. Note that [(a,i)] + [(-a,i)] = [(\rho_{i,i}(a) + \rho_{i,i}(-a),i)] = [(a-a,i)] = [(0,i)] = 0. Thus every element of \varinjlim A_i has an additive inverse.
    4. Let [(a,i)], [(b,j)] \in \varinjlim A_i, and let k \geq i,j. Then [(a,i)] + [(b,j)] = [(\rho_{i,k}(a) + \rho_{j,k}(b),k)] = [(\rho_{j,k}(b) + \rho_{i,k}(a),k)] = [(b,j)] + [(a,i)]. Thus + is commutative.

    Thus (\varinjlim A_i, +) is an abelian group. Finally, we show that each \rho_i : A_i \rightarrow \varinjlim A_i is a group homomorphism. Let a,b \in A_i. Then \rho_i(a+b) = [(a+b,i)] = [(\rho_{i,i}(a) + \rho_{i,i}(b),i)] = [(a,i)] + [(b,i)] = \rho_i(a) + \rho_i(b). Thus \rho_i is a group homomorphism for all i.

  4. Define an operator on \varinjlim A_i as follows: [(a,i)] \cdot [(b,j)] = [(\rho_{i,k}(a) \cdot \rho_{j,k}(b), k)], where k is any upper bound of i and j in I. Note the following.
    1. (\cdot is well defined) Let [(a_1,i_1)] = [(a_2,i_2)] and [(b_1,j_1)] = [(b_2,j_2)]. Then there exist r \geq i_1,i_2 and s \geq j_1,j_2 such that \rho_{i_1,r}(a_1) = \rho_{i_2,r}(a_2) and \rho_{j_1,s}(b_1) = \rho_{j_2,s}(b_2). Using the directedness of I, choose k_1 \geq i_1,j_1, k_2 \geq i_2,j_2, and t \geq k_1,k_2. Note the following.
      \rho_{k_1,t}(\rho_{i_1,k_1}(a_1) \cdot \rho_{j_1,k_1}(b_1))  =  \rho_{k_1,t}(\rho_{i_1,k_1}(a_1)) \cdot \rho_{k_1,t}(\rho_{j_1,k_1}(b_1))
       =  \rho_{i_1,t}(a_1) \cdot \rho_{j_1,t}(b_1)
       =  \rho_{r,t}(\rho_{i_1,r}(a_1)) \cdot \rho_{s,t}(\rho_{j_1,s}(b_1))
       =  \rho_{r,t}(\rho_{i_2,r}(a_2)) \cdot \rho_{s,t}(\rho_{j_2,s}(b_2))
       =  \rho_{i_2,t}(a_2) \cdot \rho_{j_2,t}(b_2)
       =  \rho_{k_2,t}(\rho_{i_2,k_2}(a_2)) \cdot \rho_{k_2,t}(\rho_{j_2,k_2}(b_2))
       =  \rho_{k_2,t}(\rho_{i_2,k_2}(a_2) \cdot \rho_{j_2,k_2}(b_2))

      Thus (\rho_{i_1,k_1}(a_1) \cdot \rho_{j_1,k_1}(b_1), t) \ \sigma\ (\rho_{i_2,k_2}(a_2) \cdot \rho_{j_2,k_2}(b_2),t), and in particular [(a_1,i_1)] \cdot [(b_1,j_1)] = [(a_2,i_2)] \cdot [(b_2,j_2)]. So \cdot is well-defined.

    2. (\cdot is associative) Let [(a,i)], [(b,j)], [(c,k)] \in \varinjlim A_i. Let r \geq i,j, s \geq j,k, and t \geq r,s. Then we have the following.
      \left( [(a,i)] \cdot [(b,j)] \right) \cdot [(c,k)]  =  [(\rho_{i,r}(a) \cdot \rho_{j,r}(b), r)] \cdot [(c,k)]
       =  [(\rho_{r,t}(\rho_{i,r}(a) \cdot \rho_{j,r}(b)) \cdot \rho_{k,t}(c), t)]
       =  [(\rho_{r,t}(\rho_{i,r}(a)) \cdot \rho_{r,t}(\rho_{j,r}(b)) \cdot \rho_{k,t}(c), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{j,t}(b) \cdot \rho_{k,t}(c), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{s,t}(\rho_{j,s}(b)) \cdot \rho_{s,t}(\rho_{k,s}(c)), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{s,t}(\rho_{j,s}(b) \cdot \rho_{k,s}(c)), t)]
       =  [(a,i)] \cdot [(\rho_{j,s}(b) \cdot \rho_{k,s}(c), s)]
       =  [(a,i)] \cdot \left( [(b,j)] \cdot [(c,k)] \right)

      Thus \cdot is associative.

    3. (\cdot distributes over +) We will show that \cdot distributes over + on the left; distributivity on the right is similar. Let [(a,i)], [(b,j)], [(c,k)] \in \varinjlim A_i, let r \geq j,k, and let t \geq i,r. Then we have the following.
      [(a,i)] \cdot \left( [(b,j)] + [(c,k)] \right)  =  [(a,i)] \cdot [(\rho_{j,r}(b) + \rho_{k,r}(c) ,r)]
       =  [(\rho_{i,t}(a) \cdot \rho_{r,t}(\rho_{j,r}(b) + \rho_{k,r}(c)), t)]
       =  [(\rho_{i,t}(a) \cdot \left( \rho_{r,t}(\rho_{j,r}(b)) + \rho_{r,t}(\rho_{k,r}(c)) \right), t)]
       =  [(\rho_{i,t}(a) \cdot \left( \rho_{j,t}(b) + \rho_{k,t}(c) \right), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{j,t}(b) + \rho_{i,t}(a) \cdot \rho_{k,t}(c), t)]
       =  [(\rho_{t,t}(\rho_{i,t}(a) \cdot \rho_{j,t}(b)) + \rho_{t,t}(\rho_{i,t}(a) \cdot \rho_{k,t}(c)), t)]
       =  [(\rho_{i,t}(a) \cdot \rho_{j,t}(b), t)] + [(\rho_{i,t}(a) \cdot \rho_{k,t}(c), t)]
       =  [(a,i)] \cdot [(b,j)] + [(a,i)] \cdot [(c,k)]

      Thus \cdot distributes over +.

    Thus (\varinjlim A_i, +, \cdot) is a ring. Moreover, we have the following.

    1. (If all the A_i are commutative, then \varinjlim A_i is commutative) Suppose the A_i are all commutative. Let [(a,i)], [(b,j)] \in \varinjlim A_i, and let k \geq i,j. Then we have the following.
      [(a,i)] \cdot [(b,j)]  =  [(\rho_{i,k}(a) \cdot \rho_{j,k}(b), k)]
       =  [(\rho_{j,k}(b) \cdot \rho_{i,k}(a), k)]
       =  [(b,j)] \cdot [(a,i)]

      Thus \varinjlim A_i is a commutative ring.

    2. (If all the A_i have 1 \neq 0 and the \rho_{i,j} are unital, then \varinjlim A_i has a multiplicative identity) Note first that – because the \rho_{i,j} are unital ring homomorphisms – \rho_{i,k}(1) = 1 = \rho_{j,k}(1) whenever k \geq i,j. Thus [(1,i)] = [(1,j)] for all i,j. Define 1 = [(1,i)]. Now let [(a,i)] \in \varinjlim A_i. Then 1 \cdot [(a,i)] = [(1,i)] \cdot [(a,i)] = [\rho_{i,i}(1) \cdot \rho_{i,i}(a), i)] = [(1 \cdot a,i)] = [(a,i)]. Similarly, [(a,i)] \cdot 1 = [(a,i)]. Thus 1 is a multiplicative identity in \varinjlim A_i.
    3. (If 1 \neq 0 for all A_1, then 1 \neq 0 in \varinjlim A_i) Suppose [(0,i)] = [(1,i)]. Then there exists j \geq i such that \rho_{i,j}(0) = \rho_{i,j}(1), so that 0 = 1 in A_j – a contradiction. Thus 1 \neq 0 in \varinjlim A_i.

    Thus if the A_i are commutative rings with 1 \neq 0, then \varinjlim A_i is a commutative ring with 1 \neq 0.

  5. Suppose now that C is an abelian group and that we have an indexed family of group homomorphisms \varphi_i : A_i \rightarrow C such that \varphi_i = \varphi_j \circ \rho_{i,j} for all i,j \in I.
    1. (Existence) Define \varphi : \varinjlim A_i \rightarrow C by \varphi([(a,i)]) = \varphi_i(a). We need to show that \varphi is a well defined group homomorphism.
      1. (Well-defined) Suppose [(a,i)] = [(b,j)]. Then there exists k \geq i,j such that \rho_{i,k}(a) = \rho_{j,k}(b). Since each latex \rho_{i,j}$ is well defined, \varphi_k(\rho_{i,k}(a)) = \varphi_k(\rho_{j,k}(b)). Then (\varphi_k \circ \rho_{i,k})(a) = (\varphi_k \circ \rho_{j,k})(b), and we have \varphi_i(a) = \varphi_j(b). Hence \varphi([(a,i)]) = \varphi([b,j)]), and \varphi is well defined.
      2. (Group Homomorphism) Let [(a,i)], [(b,j)] \in \varinjlim A_i, and let k \geq i,j. Then we have the following.
        \varphi([(a,i)] + [(b,i)])  =  \varphi([(\rho_{i,k}(a) + \rho_{j,k}(b), k)])
         =  \varphi_k(\rho_{i,k}(a) + \rho_{j,k}(b))
         =  \varphi_k(\rho_{i,k}(a)) + \varphi_k(\rho_{j,k}(b))
         =  (\varphi_k \circ \rho_{i,k})(a) + (\varphi_k \circ \rho_{j,k})(b)
         =  \varphi_i(a) + \varphi_j(b)
         =  \varphi([(a,i)]) + \varphi([(b,j)])

      So \varphi is a group homomorphism. Finally, note that for all i and all a \in A_i, (\varphi \circ \rho_i)(a) = \varphi(\rho_i(a)) = \varphi([(a,i)]) = \varphi_i(a). Thus \varphi \circ \rho_i = \varphi_i for all i.

    2. (Uniqueness) Suppose that we have a group homomorphism \psi : \varinjlim A_i \rightarrow C which also satisfies \psi \circ \rho_i = \varphi_i for all i \in I. Then for all i and all a \in A_i, we have \psi([(a,i)]) = \psi(\rho_i(a)) = (\psi \circ \rho_i)(a) = \varphi_i(a) = (\varphi \circ \rho_i)(a) = \varphi(\rho_i(a)) = \varphi([(a,i)]). Thus we have \psi = \varphi.
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