Let $A$ be a commutative ring with identity, $S\subset A$ a multiplicatively closed subset and $1\in S$. Does the equation $$S^{-1}A=\varinjlim_{s\in S}A_s$$ make sense? Here $A_s$ is the localization of $A$ at $s$. Note that to make sure the above equation make sense, we need to give a partial order to $S$ and make it to become a direct system.
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Sorry, I cannot enter your link, do you know why? – Lao-tzu Feb 05 '14 at 10:54
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I think the preorder $s\leq s' \Leftrightarrow s|s'$ in $A$ would be more interesting here. In the case of the trivial preorder, the direct limit is the coproduct. – zozoens Feb 05 '14 at 10:55
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But still this is not a partial order, since $a\leq b, b\leq a$ doesn't imply $a=b$. – Lao-tzu Feb 05 '14 at 10:59
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Hence the name preorder. You don't need to have an order to make a sense for inductive limit, only a preorder (here, it is actually a directed set because if $s,s'\in S$, $ss'\in S$ and we have $s|ss'$ and $s'|ss'$). – zozoens Feb 05 '14 at 11:04
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Do you mean a preorder is a binary relation that is reflexive and transitive but need not to be antisymmetric? – Lao-tzu Feb 05 '14 at 11:08
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Exactly: http://en.wikipedia.org/wiki/Preorder – zozoens Feb 05 '14 at 11:55
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Make $S$ into a category as follows: The object set is $S$. A morphism $s \to t$ is an element $u \in S$ such that $us=t$. For identities and composition use that $S$ is multiplicative. If $u : s \to t$ is a morphism, then there is a unique homomorphism of $A$-algebras $A_s \to A_t$ (because of the universal properties). The colimit $\mathrm{colim}_{s \in S} A_s$ has the same universal property as $S^{-1} A$, hence they are isomorphic. All this works for arbitrary commutative rings $A$.
Martin Brandenburg
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This seems odd to me. Given $s \mid t$, one can always define a map $A_s \to A_t$ by $\frac{a}{s^n} \to \frac{a}{s^n}$. More carefully written, it is $\frac{a}{s^n} \to \frac{a}{1} \cdot \left(\frac{s^n}{1}\right)^{-1}$. This makes sense because $\frac{s}{1}$ is a unit in $A_t$. In your construction, the map $A_s \to A_t$ seems to depend on the choice of $u$, which is by no means unique. – RghtHndSd Feb 25 '16 at 14:47
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@RghtHndSd The choice of $u$ merely shows that the map is the well-defined (by the universal property), but the map itself does not depend on $u$. – Sardines Feb 18 '24 at 04:22
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Let $A$ be a commutative ring and $S$ a multiplicative set. Then the family of rings $\left\{A_s \right\}_{s \in S}$ forms a directed family. To see this, first we define a partial order on $S$ by $s \le t$ if $t = u s$ for some $u \in S$. Next for $s \le t$ with $t = u s$, there exists a ring homomorphism $f_{s,t}: A_s \rightarrow A_t$, which is defined by $f(a/s^n) = a u^n / t^n$. Prove as an exercise that the transitive property of a directed family holds. Thus we can talk about $\lim_{s \in S} A_s$.
Recall that $\lim_{s \in S} A_s = \bigsqcup_{s \in S} A_s / \sim$, where $\bigsqcup_{s \in S} A_s = \left\{(\xi,s) : \, s \in S, \xi \in A_s \right\}$ is the disjoint union and $\sim$ is the following equivalence relation on $\bigsqcup_{s \in S} A_s$: Two elements $(a_1/s_1^{n_1},s_1)$ and $(a_2/s_2^{n_2},s_2)$ are equivalent if there exists $s_3 \in S$ such that $s_1 \le s_3, s_2 \le s_3$ and $f_{s_1,s_3}(a_1/s_1^{n_1}) = f_{s_2,s_3}(a_2/s_2^{n_2})$. We can denote an element of $\lim_{s \in S} A_s$ as $[a/s^n,s]$, i.e. the equivalence class of $(a/s^n,s)$.
Now define a map $\psi:\lim_{s \in S} A_s \rightarrow A_S$ by taking the element $[a/s^n,s]$ to $a/s^n \in A_S$. As an exercise show the following: 1) $\psi$ is well-defined, 2) $\psi$ is a ring homomorphism, 3) $\psi$ is bijective.
Manos
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If $A$ is a domain, it is clear that this is the case for the divisibility preorder in $A$.
Indeed, in that case, all localizations live in the fraction field of $A$, and the maps between them are just inclusion, so it is clear that $$S^{-1}A=\bigcup_{s\in S}A_s=\lim A_s.$$
I believe looking more closely at the categorical approach would give the same result when $A$ is not necessarily a domain.
zozoens
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