Timeline for A question about localization of commutative rings
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
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| Jan 5, 2022 at 12:10 | answer | added | Johannes Huisman | timeline score: 3 | |
| Jan 4, 2022 at 18:12 | comment | added | Yu Li | @Johannes Hahn: Since I want to find some other proofs of this. | |
| Jan 4, 2022 at 17:39 | history | edited | Yu Li | CC BY-SA 4.0 |
added 490 characters in body; edited title
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| Jan 4, 2022 at 17:02 | comment | added | Johannes Hahn | Why would you want to prove a statement about localisations without using localisations? | |
| Jan 4, 2022 at 16:54 | comment | added | Fedor Petrov | 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 | |
| Jan 4, 2022 at 16:52 | answer | added | Johannes Hahn | timeline score: 2 | |
| Jan 4, 2022 at 16:49 | comment | added | Yu Li | @Uriya First: Yes, I exactly mean that, thank you. In fact, I'm thinking about the localization of a commutative ring and I found that one could construct $ S^{-1}R $ by taking $ R[S]/I $, where $ R[S] $ is the free commutative algebra over $ R $ generated by a multiplicatively closed subset $ S $ of $ R $ and $ I $ the ideal that contains $ s i(s)-1 $ for each $ s $ in $ S $, where $ i:S\to R[S] $ the canonical embedding. So I wonder that can we prove this proposition without using localization. | |
| Jan 4, 2022 at 16:38 | history | edited | Yu Li | CC BY-SA 4.0 |
deleted 33 characters in body
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| Jan 4, 2022 at 15:36 | comment | added | Uriya First | 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. | |
| Jan 4, 2022 at 14:44 | comment | added | Andrea Ferretti | 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 | |
| Jan 4, 2022 at 14:39 | comment | added | Gro-Tsen | 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? | |
| Jan 4, 2022 at 14:17 | history | edited | YCor | CC BY-SA 4.0 |
fixed typo
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| Jan 4, 2022 at 14:16 | history | edited | Yu Li | CC BY-SA 4.0 |
edited body
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| S Jan 4, 2022 at 14:04 | review | First questions | |||
| Jan 4, 2022 at 14:52 | |||||
| S Jan 4, 2022 at 14:04 | history | asked | Yu Li | CC BY-SA 4.0 |