Timeline for Question on localization technique
Current License: CC BY-SA 3.0
6 events
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Dec 4, 2012 at 18:37 | history | edited | Fred Rohrer | CC BY-SA 3.0 |
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Dec 4, 2012 at 16:57 | vote | accept | Axy | ||
Dec 4, 2012 at 15:26 | comment | added | Fred Rohrer | ... If the result holds in case we have a local base ring then it holds in this situation, and thus the claim will be proven once we prove it for the case of a local base ring. | |
Dec 4, 2012 at 15:25 | comment | added | Fred Rohrer | Dear @Axy, concerning scalar restriction you should take a look at Bourbaki's Algèbre II.1.13. Concerning the proof, I explained that if $M_{\mathfrak{p}}=N_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ in $R_0$, then $M=N$. So, if we want to show that $M=N$, then we have to show that $M_{\mathfrak{p}}=N_{\mathfrak{p}}$ for every prime ideal $\mathfrak{p}$ in $R_0$. To do this, we take such a prime ideal and localise everything at it, obtaining a similar situation as before but this time over a local base ring... | |
Dec 4, 2012 at 14:56 | comment | added | Axy | @Fred Rohrer: Dear Fred, what do you mean by the $B$ modules obtain from $M$ and $N$ by scalar restriction along $h$ are equal ? can you make it more precise? As you said we can conclude that $M_{\mathfrak_{p}}=N_{\mathfrak{p}}$ on $R_0$ and therefore $M=N$ on $R_0$. So, why prooving the problem in the local case makes the proof complete ? | |
Dec 4, 2012 at 13:35 | history | answered | Fred Rohrer | CC BY-SA 3.0 |