Timeline for In this special situation, does $M \otimes B=0$ imply $M=0$?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 14, 2014 at 23:34 | vote | accept | Louis | ||
| May 14, 2014 at 20:59 | comment | added | Jason Starr | @Louis: In the same way that $A/\langle 1-x \rangle$ is also the localization $A[y]/\langle yx - 1 \rangle$, also $A/\langle x \rangle$ is also the localization $A[z]/\langle z(1-x) - 1 \rangle$. As a localization, $A[z]/\langle z(1-x)-1 \rangle$ is flat over $A$. | |
| May 14, 2014 at 20:46 | comment | added | Louis | I've read and mostly understood what you wrote. Can you maybe explain how to see that $B$ is flat as $A$-algebra? Thank you! | |
| S May 14, 2014 at 20:33 | history | answered | Jason Starr | CC BY-SA 3.0 | |
| S May 14, 2014 at 20:33 | history | made wiki | Post Made Community Wiki by Jason Starr |