Timeline for Localizations as free, finite rank modules
Current License: CC BY-SA 2.5
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 10, 2010 at 0:28 | comment | added | KConrad | There is also not even any reason to mention the second prime $Q$. Assume for some prime $P$ over $p$ in $O_K$ that $O_P$ is a f.g. module over ${\mathbf Z}_{(p)}$. Then $O_{P}$ is in the integral closure of ${\mathbf Z}_{(p)}$ in $K$, and any number in $K$ that's integral over ${\mathbf Z}_{(p)}$ is certainly integral over the larger ring $O_P$, so it must lie in $O_P$ since $O_P$ is int. closed. Thus the int. closure of ${\mathbf Z}_{(p)}$ in $K$ is $O_P$. The number of primes over $p$ in $O_K$ is the number of primes over $p$ in the int. closure of ${\mathbf Z}_{(p)}$ and $O_P$ has just 1. | |
| Jul 9, 2010 at 20:53 | comment | added | Keenan Kidwell | That's a good point. +1 | |
| Jul 9, 2010 at 19:55 | comment | added | KConrad | No need to introduce $\beta$. Let $P$ and $Q$ be primes lying over $p$ in $O_K$. If $O_P$ is integral over ${\mathbf Z}_{(p)}$ then it's inside $O_p$ and hence inside $O_Q$. Both $O_P$ and $O_Q$ are maximal subrings of $K$, so containment implies equality and then intersecting $O_P$ and $O_Q$ with $O_K$ shows $P = Q$. | |
| Jul 9, 2010 at 14:35 | history | edited | Kevin Buzzard | CC BY-SA 2.5 |
fixed latex with backticks
|
| Jul 9, 2010 at 12:53 | vote | accept | Pedro Martins Rodrigues | ||
| Jul 9, 2010 at 12:52 | comment | added | Pedro Martins Rodrigues | Thank you, Georges and Keenan, for your answers. My question was exactly as Keenan put it and I was getting to the same conclusion. But if $pO=P^{d}$ then everything is allright, right? | |
| Jul 9, 2010 at 12:46 | history | answered | Keenan Kidwell | CC BY-SA 2.5 |