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Let $R$ be a commutative ring with unity. Let $U \subseteq R$ be a multiplicatively closed subset. Recently, in my Algebra I class, we learned about the localization $U^{-1} R$. We constructed this as a set of equivalence classes in $R \times U$, viewing the pair $(a, s) \in R \times U$ as $\frac{a}{s} \in U^{-1} S$.

My question is: Can $U^{-1} R$ also be constructed as $\newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} \quotient{R[x_s \ : \ s \ \in \ U]}{\langle s x_s - 1 \ : \ s \ \in \ U\rangle}$? Certainly, there is a natural homomorphism $$ \varphi: R \to \quotient{R[x_s \ : \ s \ \in \ U]}{\langle s x_s - 1 \ : \ s \ \in \ U\rangle} $$ such that for every $s \in U$, $\varphi(s)$ is invertible. Hence, by the universal property of $U^{-1} R$, there is a unique homomorphism $\overline{\varphi}$ such that the following diagram commutes: $$ \begin{array}{ccc} R & \varphi & \\ \downarrow & \searrow & \\ U^{-1} R & \rightarrow & \quotient{R[x_s \ : \ s \ \in \ U]}{\langle s x_s - 1 \ : \ s \ \in \ U\rangle} \\ & \overline{\varphi}& \\ \end{array} $$ The question is now whether $\overline{\varphi}$ is an isomorphism.

Ben Steffan
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Smiley1000
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1 Answers1

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We can explicitely construct an inverse for $\overline{\varphi}$.

Consider the map $\psi: R[x_s : s \in U] \to U^{-1} R, x_s \mapsto \frac{1}{s}$. This is defined via the universal property of the polynomial algebra. Now we have $\psi(s x_s - 1) = \psi(s) \psi(x_s) - \psi(1) = s \frac{1}{s} - 1 = 0$. Hence $\langle s x_s - 1 : s \in U \rangle \subseteq \operatorname{ker}(\psi)$. By the universal property of the quotient, we obtain a factorization $$ \newcommand\quotient[2]{{^{\Large #1}}/{_{ \Large #2}}} \begin{array}{ccc} R[x_s : s \in U] & \psi & \\ \downarrow & \searrow & \\ \quotient{R[x_s \ : \ s \ \in \ U]}{\langle s x_s - 1 \ : \ s \ \in \ U\rangle} & \rightarrow & U^{-1} R \\ & \overline{\psi}& \\ \end{array} $$ It remains to show that $\overline{\varphi}$ and $\overline{\psi}$ are mutually inverse.

Take any $\frac{a}{s} \in U^{-1} R$, with $a \in R$ and $s \in U$. Then we have $\overline{\psi}\left(\overline{\varphi}\left(\frac{a}{s}\right)\right) = \overline{\psi}([a x_s]) = a \frac{1}{s} = \frac{a}{s}$. Hence $\overline{\psi} \circ \overline{\varphi} = \operatorname{id}$.

We also have $\overline{\varphi}(\overline{\psi}([x_s])) = \overline{\varphi}\left(\frac{1}{s}\right) = [x_s]$. But now $\overline{\psi} \circ \overline{\varphi}$ and $\operatorname{id}$ are $R$-algebra homomorphisms which agree on the $[x_s]$, which generate $\quotient{R[x_s \ : \ s \ \in \ U]}{\langle s x_s - 1 \ : \ s \ \in \ U\rangle}$ as an $R$-algebra. Hence $\overline{\psi} \circ \overline{\varphi} = \operatorname{id}$.

I've posted a similar answer here:

Smiley1000
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