Timeline for Absolute integral closure of Noetherian local domain
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| May 19, 2023 at 18:25 | comment | added | Karl Schwede | If $R$ is complete local then $R^+$ is still local. Indeed, $R^+$ is local if and only if $R$ is Henselian (stacks.math.columbia.edu/tag/04GE) | |
| May 19, 2023 at 18:21 | comment | added | Wojowu | @CARLO It's not true that $R^+$ is local, nor that $mR^+$ is the maximal ideal of $R^+$. Take for instance $R$ to be the localization $\Bbb Z_{(2)}$. Its maximal ideals correspond to ideals lying over $2$ in $\overline{\Bbb Z}$, of which there are uncountably many. | |
| May 19, 2023 at 18:21 | comment | added | Karl Schwede | An integral integral-domain extension of a field is a field. To see this, note your tensor product (call it $T$) is an integral domain since it is a localization of $R^+$. $T$ also is an integral extension of $K(R)$. Hence all primes of $T$ lie over $(0)$ and by incomparability, they are all minimal (see going up). Thus $(0)$ is the only prime (since it's obviously the smallest) and you have a field. | |
| May 19, 2023 at 18:13 | comment | added | Wojowu | $K(R)\otimes_R R^+$ is the localization of $R^+$ at nonzero elements of $R$. Any element of the algebraic closure has an $R$-multiple in $R^+$ by "clearing denominators" from its minimal polynomial, so will lie in that localization. | |
| May 19, 2023 at 16:42 | comment | added | pinaki | If $f \in R^+$, then from an integral eqn $f^s + \sum_ja_jf^{s-j} = 0$ of $f$ over $R$, you have $f(f^{s-1}+a_{s-1}f^{s-2}+\cdots+a_1) = -a_0$, which is a unit in $K(R) \otimes_R R^+$ | |
| May 19, 2023 at 16:12 | history | edited | CARLO | CC BY-SA 4.0 |
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| May 19, 2023 at 16:04 | comment | added | CARLO | I don't believe that $K(R)\otimes_{R}R^{+}$ be a field. $R^{+}$ is a local ring with maximal ideal $\mathfrak{m}R^{+}$. | |
| May 19, 2023 at 15:58 | history | edited | CARLO | CC BY-SA 4.0 |
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| May 19, 2023 at 7:33 | comment | added | Wojowu | Isn't $K(R)\otimes_RR^+$ simply the algebraic closure of $K(R)$? It being a field then trivially implies the equality you ask about. | |
| May 19, 2023 at 4:53 | history | edited | YCor | CC BY-SA 4.0 |
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| May 19, 2023 at 2:46 | comment | added | CARLO | Is the integral closure in the algebraic closure of the field K(R). | |
| May 19, 2023 at 2:43 | comment | added | abx | What is the absolute integral closure? | |
| May 19, 2023 at 2:25 | history | asked | CARLO | CC BY-SA 4.0 |