## 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 $\psi = \varphi$.