Let $R$ be a commutative ring, $\mathfrak{p} \subset R$ a prime ideal and $S \subset R/\mathfrak{p}$ be a multiplicative system. Let $A$ be an $R$-algebra.
I think that then $A \otimes_R S^{-1}R/\mathfrak{p} \simeq A/\mathfrak{p}A \otimes_{R/\mathfrak{p}} S^{-1}R/\mathfrak{p}$ (considering the right hand side as an $R$-module) by defining the obvious homomorphism and show that this is an isomorphism .
Is there an elementary way to see that this is true? I know that if we remove the localization by $S$ then this is true since both sides are canonically isomorpic to $A/\mathfrak{p}A$. But how do we take the localization into the equation?
Best Answer
Choose any $a\in A, p\in\mathfrak{p}, r\in R, s\in S$. Since the tensor product is over $R$, we have $ap\otimes\frac{\bar{r}}{s} = a\otimes p\frac{\bar{r}}{s} = a\otimes\frac{p\bar{r}}{s}=a\otimes\frac{\bar{r}}{s}$, where $\bar{r}$ is the element of $R/\mathfrak{p}$ corresponding to $r$. As products with the elements of $\mathfrak{p}$ are the same as the products with the unit, we may quotient out the ideal and identify $A\otimes_R S^{-1}R/\mathfrak{p}$ with $A/\mathfrak{p}A\otimes_{R/\mathfrak{p}}S^{-1}R/\mathfrak{p}$. In particular, there is no overlapping between the preimage $S'$ of $S$ (wrt. the canonical projection) and $\mathfrak{p}$, so this is consistent with the localisation. To see this, suppose there was an element $x\in S'\cap\mathfrak{p}$. Then $0=\bar{x}\in S$.