Timeline for If M is not a free A-module, can tensoring with a bigger field make it free?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 19, 2013 at 23:58 | vote | accept | Joshua Batson | ||
| Jan 17, 2013 at 3:52 | comment | added | user30180 | @Joshua Batson: Yes, the "circle" was even my original example. In the course of writing it up as an answer I planned to say at the end that I didn't use any too essential about $\mathbf{R}$, but I then decided that such a remark would be more convincing if I actually explained the construction in the general case. Perhaps in so doing I made it look too exotic, but indeed the circle example is the simplest special case (note you could use $K = \mathbf{Q}$ and $F = \mathbf{Q}(i)$ as well, maybe even "simpler" than with $K = \mathbf{R}$). | |
| Jan 16, 2013 at 18:16 | comment | added | Joshua Batson | I had a little trouble understanding this. Would an instance be $F:K = \mathbb{R}:\mathbb{C}$, with $A = \mathbb{R}[x,y]/x^2+y^2-1$ and $M = \langle x-1,y \rangle$? Then $M_\mathbb{C} = \langle z-1 \rangle$ (where $z = x+iy$) is principal, though $M$ is not. | |
| Jan 15, 2013 at 10:55 | comment | added | Piotr Pstrągowski | This is a very nice example. | |
| Jan 15, 2013 at 4:12 | history | answered | user30180 | CC BY-SA 3.0 |