0
$\begingroup$

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 $ ?

Improve this question
$\endgroup$
7
  • $\begingroup$ 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? $\endgroup$
    – Gro-Tsen
    Commented Jan 4, 2022 at 14:39
  • $\begingroup$ 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 $\endgroup$
    – Andrea Ferretti
    Commented Jan 4, 2022 at 14:44
  • $\begingroup$ 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. $\endgroup$
    – Uriya First
    Commented Jan 4, 2022 at 15:36
  • 1
    $\begingroup$ 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 $\endgroup$
    – Fedor Petrov
    Commented Jan 4, 2022 at 16:54
  • 1
    $\begingroup$ Why would you want to prove a statement about localisations without using localisations? $\endgroup$
    – Johannes Hahn
    Commented Jan 4, 2022 at 17:02

2 Answers 2

Reset to default
3
$\begingroup$

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., $F=0$.

Improve this answer
$\endgroup$
1
  • $\begingroup$ 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? $\endgroup$
    – Yu Li
    Commented Jan 7, 2022 at 5:11
2
$\begingroup$

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 $R$.

Improve this answer
$\endgroup$
3
  • $\begingroup$ 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. $\endgroup$
    – Yu Li
    Commented Jan 4, 2022 at 17:29
  • $\begingroup$ 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 $？ $\endgroup$
    – Yu Li
    Commented Jan 4, 2022 at 17:41
  • $\begingroup$ Now I realize that what's exatly my question: The canonical map j:R→S−1R is characterized by the following property: If f:R→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−1R→T such that f=g∘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×S, but could we prove this without considering any certain construction? $\endgroup$
    – Yu Li
    Commented Jan 7, 2022 at 5:11

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .