Timeline for Rings over which every module is free
Current License: CC BY-SA 2.5
3 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2013 at 23:07 | comment | added | Manny Reyes | @o2genum: A basis for a right $R$-module is a linearly independent spanning set. In particular, linear independence implies that any element $x$ of a basis has right annihilator equal to zero. Thus the map $R \to xR$ sending $r \mapsto xr$ is an isomorphism. | |
| Dec 26, 2013 at 19:13 | comment | added | Andrei Moiseev | "and because x was a basis element we have xR ≅ R as right R-modules" — why is it true? What is the isomorphism? | |
| Oct 20, 2009 at 15:52 | history | answered | Manny Reyes | CC BY-SA 2.5 |