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Question: Given commutative rings $(R_i)_{i \in I}$ with multiplicative sets $(S_i)_{i \in I}$ (it is implicitly implied that $\forall i \in I \ S_i \subseteq R_i$), is it true that the rings $$\prod_{i \in I} S_i^{-1}R_i$$ and $$\big(\prod_{i \in I} S_i\big)^{-1}\big(\prod_{i \in I} R_i\big)$$ are always canonically isomorphic?

Note that $I$ may be finite or infinite here, unlike localizations of direct products of modules at a fixed multiplicative subset, where the indexing set must be finite for the localization to be isomorphic to the direct product of the localizations.

Consider the category $C$ whose objects are pairs $(R, S)$ with $R$ being a commutative ring and $S$ being a multiplicative subset of $R$, and whose morphisms $(R_1, S_1) \to (R_2, S_2)$ are ring homomorphisms $f:R_1 \to R_2$ such that $f(S_1) \subseteq S_2$. Then, we have a functor $\mathbf{CRing} \to C$ that sends a commutative ring $R$ to $(R, R^{\times})$, where $R^{\times}$ is the group of units of $R$. Moreover, this functor has a left adjoint that sends a pair $(R, S)$ to the localization $S^{-1}R$, so the question is asking whether that left adjoint preserves (possibly infinite) products.

The idea, of course, is to consider the homomorphism from the localization of the product to the product of the localizations that sends $(a_i)_{i \in I}/(s_i)_{i \in I}$ to $(a_i/s_i)_{i \in I}$, and show that this is an isomorphism.

The axiom of choice is implicitly assumed (namely, proving surjectivity requires choosing numerators and denominators for each fraction, while proving injectivity (i.e. that the kernel vanishes) requires choosing an annihilator of the $i$-th numerator in $S_i$ for each $i \in I$).

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As you mentioned, consider the natural function $f:\big(\prod_{i \in I} S_i\big)^{-1}\big(\prod_{i \in I} R_i\big)\rightarrow \prod_{i \in I} S_i^{-1}R_i$ by $f((a_i)_{i \in I}/(s_i)_{i \in I})=(a_i/s_i)_{i \in I}$. Clearly $f$ is well defined and a ring homomorphis, it is sufficient to show that $f$ is an isomorphism.

Let $f((a_i)_{i \in I}/(s_i)_{i \in I})=(a_i/s_i)_{i \in I}=0_{\prod_{i \in I} S_i^{-1}R_i}$, so for each $i$, $a_i/s_i=0_{ S_i^{-1}R_i}$. Then for each $i$, there exists $t_i\in S_i$ such that $t_ia_i=0_{ S_i^{-1}R_i}$. Hence $(t_i)_{i \in I}\in \prod_{i \in I} S_i$, and we have $(a_i)_{i \in I}/(s_i)_{i \in I}=((t_i)_{i \in I}/(t_i)_{i \in I})((a_i)_{i \in I}/(s_i)_{i \in I})=((t_ia_i)_{i \in I}/(t_is_i)_{i \in I})=0_{\big(\prod_{i \in I} S_i\big)^{-1}\big(\prod_{i \in I} R_i\big)}$. Thus $f$ is injective.

Now let $(a_i/s_i)_{i \in I}\in\prod_{i \in I} S_i^{-1}R_i$. Since $f((a_i)_{i \in I}/(s_i)_{i \in I})=(a_i/s_i)_{i \in I}$, $f$ is surjective, and so $\big(\prod_{i \in I} S_i\big)^{-1}\big(\prod_{i \in I} R_i\big)\cong \prod_{i \in I} S_i^{-1}R_i$.

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  • $\begingroup$ I don't know why my answer was deleted though. $\endgroup$
    – Olórin
    Aug 27 '19 at 19:17

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