Timeline for Why is quadratic integer ring defined in that way?
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 7, 2021 at 9:03 | history | edited | user26857 | CC BY-SA 4.0 | added 2 characters in body |
| Sep 28, 2015 at 0:03 | history | edited | darij grinberg | CC BY-SA 3.0 | added 2 characters in body |
| Mar 20, 2015 at 21:58 | comment | added | Adam Hughes | I see, so you're using the fact that there exists a basis, and then noting an advantageous one. Thanks for clearing that up. | |
| Mar 20, 2015 at 21:57 | comment | added | Bernard | The denominator $2$ appears naturally as a a basis vector of the integral closure, that's why. Moreover this integral closure is a euclidean ring, while $\mathbf Z[\sqrt 5]$ is not even a Dedekind domain. | |
| Mar 20, 2015 at 21:46 | comment | added | Adam Hughes | Just out of curiosity, why do you mention the integral closure being a free $\Bbb Z$-module, $\Bbb Z[\sqrt{5}]$ is also a free $\Bbb Z$ module, and it's not clear to me what relevance you attach to that fact. | |
| Mar 20, 2015 at 11:03 | history | answered | Bernard | CC BY-SA 3.0 |