A question about localization of commutative rings

Given a commutative ring $$R$$ and a multiplicatively closed subset $$S$$ of $$R$$, there are two ways to consturct $$S^{-1}R$$:

1. define an equivalence relation $$\sim$$ on $$R\times S$$ and then take $$S^{-1}R := (R\times S)/\sim$$.

2. Let $$R[S]$$ be the free commutative algebra over $$R$$ generated by $$S$$ and $$i:S\to R[S]$$ the canonical embedding and $$I$$ the ideal of $$R[S]$$ generated by $$i(s)s-1$$ for each $$s$$ in $$S$$ and then take $$S^{-1}R := R[S]/I$$.

Suppose $$S$$ dosen't contain zero divisors, if one works in the first way, it's easy to prove that the canonical map $$j:R\to R^{-1}S$$ is injective. But if one works in the second way, then it's equivalent to state that $$I \cap R = \{ 0 \}$$. Can we directly prove this without considering $$(R\times S)/\sim$$ ?

• The statement “there's no nonzero $r$ such that $\operatorname{expression} =r$” where the expression does not depend on $r$, is bizarre: why not write $\operatorname{expression}=0$ then? Commented Jan 4, 2022 at 14:39
• It is bizarre indeed, also because the expression is generically nonzero. I think the question needs some rephrasing - as it is written, I cannot tell the correct form Commented Jan 4, 2022 at 14:44
• I think your question is equivalent to whether $(a_1x_1-1,+\dots+,a_nx_n-1)\cap R=\{0\}$ (with $(a_1x_1-1,\dots,a_nx_n-1)$ denoting the ideal generated by $a_1x_1-1,\dots,a_nx_n-1$ in $R[x_1,\dots,x_n]$), am I right? This is true if $R$ is a field, and, by passing to the fraction field, also if $R$ is a domain. Commented Jan 4, 2022 at 15:36
• if $r\in R$, $r\ne 0$, and $r=\sum (a_ix_i-i)f_i$, simply put $x_i=1/a_i$ (and multiply by a product of $a_i$ to a large power, so that you deal only with polynomial identity) to get 0 in RHS Commented Jan 4, 2022 at 16:54
• Why would you want to prove a statement about localisations without using localisations? Commented Jan 4, 2022 at 17:02

Maybe these are the arguments you're looking for: First reduce to the case of $$S$$ being finitely generated as a multiplicative subset of $$R$$. Next, reduce to the case $$S$$ where $$S$$ is generated by only one element $$s$$. The morphism $$R\rightarrow R[T]/(sT-1)$$ is injective for $$s$$ a nonzero divisor. As you said, it suffices to prove that $$(sT-1)\cap R=\{0\}.$$ Since $$s$$ is a nonzero divisor, $$\deg((sT-1)F)=\deg(sT-1)+\deg(F)= 1+\deg(F)$$ for all $$F\in R[T]$$. It follows that $$(sT-1)F\in R$$ only if $$\deg(F)=-\infty$$, i.e., $$F=0$$.
• Thanks! Now I realize that what's exatly my question: The canonical map $j:R\to S^{-1}R$ is characterized by the following property: If $f:R\to T$ is a ring homomorphism that maps every element in $S$ to a unit in $T$, then there's a unique ring homorphism $g:S^{-1}R\to T$ such that $f = g\circ j$. I'm wondering that how to prove $j$ is injective if $S$ contains no zero divisors? I know that one could prove this by define an equivalence relation on $R\times S$, but could we prove this without considering any certain construction? Commented Jan 7, 2022 at 5:11
Wlog assume $$n=m$$. Set $$S:=\{a_1^{k_1}\cdots a_n^{k_n} \mid k_i\in\mathbb{N}\}$$. Then $$R[x_1,\ldots,x_n]/I$$ is precisely the localisation $$S^{-1} R$$ that inverts $$a_1, \ldots, a_n$$ and your question is equivalent to asking whether or not the canonical map $$R\to S^{-1} R$$ is injective.
The kernel of the canonical map is well-known to be equal to $$\{r\in R \mid \exists s\in S: sr=0\}$$. In particular: It is zero iff none of the elements $$a_1,\ldots,a_n$$ are zero divisors of $$R$$.
• Yes. Given a multiplicatively closed set $S$ that contains $1$, then there are two ways to construct $S^{-1}R$. One is to define an equivalence relation on $R\times S$ and then take quotient, and the other is $R[S]/I$, where $R[S]$ is the free commutative algebra over $R$ generated by $S$ and, let $i:S\to R[S]$ the canonical embedding, $I$ the ideal of $R[S]$ generated by $i(s)s-1$ for each $s$ in $S$. It's very easy to prove that the canonical map $j:R\to S^{-1}R$ is injective while $S$ doesn't contain zero divisor if one consider the first construction. Commented Jan 4, 2022 at 17:29
• I mean that could one directly prove that, if $S$ doesn't contain zero divisors, then $I \cap R = \{ 0 \}$, without considering $(R\times S)/\sim$？ Commented Jan 4, 2022 at 17:41