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$A$ be a commutative ring with $1$ with finitely many minimal primes $\{p_1,\ldots,p_n\}.$ Then how can I show that $S^{-1}A \cong A_{p_1} \times\cdots \times A_{p_n},$ where $S=A \setminus \bigcup_{i=1}^{n}p_{i}.$

So far I could prove that $\text{Spec}(S^{-1}A)=\text{Max}(S^{-1}A)=\{A_{p_1},\ldots,A_{p_n}\}$ and there exists a natural map $S^{-1}A \to A_{p_1} \times \cdots \times A_{p_n}$ , which exists by universal property of localization. How can I show that the natural map in injective as well as surjective?

user26857
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user371231
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1 Answers1

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Very generally, if $B$ is a commutative ring with only finitely many prime ideals $q_1,\dots,q_n$, all of which are maximal, then $B\cong B_{q_1}\times\dots\times B_{q_n}$. This follows immediately from basic scheme theory: $X=\operatorname{Spec} B$ is covered by the disjoint open sets $\{q_1\},\dots,\{q_n\}$, so by the gluing property for the structure sheaf, $$B\cong\mathcal{O}_X(X)\cong \mathcal{O}_X(\{q_1\})\times\dots\mathcal{O}_X(\{q_n\})$$ and $\mathcal{O}_X(\{q_i\})$ is just $B_{q_i}$.

In your case, we can apply this with $B=S^{-1}A$ and $q_i=p_iB$. The localization $B_{q_i}$ is naturally isomorphic to $A_{p_i}$: to form $B_{q_i}$, we first invert every element of $S$ and then invert every element of $B\setminus q_i$. But an element of $B\setminus q_i$ is just an element of $A\setminus p_i$ divided by an element of $S$, and $S\subseteq A\setminus p_i$, so this is equivalent to just inverting every element of $A\setminus p_i$.

Eric Wofsey
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