Timeline for Reference for surjectivity of the canonical map $R^{G_1} \otimes R^{G_2} \to R^{G_1 \cap G_2}$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 13 at 14:37 | vote | accept | Zheming Xu | ||
| Aug 14 at 5:03 | comment | added | R. van Dobben de Bruyn | @AliTaghavi the question is whether the map $G \to \operatorname{Aut}(R)$ is injective, which is not true in the first example. | |
| Aug 13 at 13:48 | comment | added | Ali Taghavi | I wonder if it is necessary to consider "G is a subgroup of $Aut(R)$? I mean that according to the definition of action, it is implicity contained in the definition that $x\mapsto g.x$ is a Ring automorphism. On the other hand I think the first version of your answer satisfies the Ring automorphism property: the conjugation is a ring automorphism on the trancated polynomial ring $\frac{Z[x]}{x^2}$. Am I mistaken? | |
| Aug 13 at 7:21 | history | edited | HenrikRüping | CC BY-SA 4.0 |
modified the example to fit the additional assumption
|
| Aug 13 at 7:04 | comment | added | Zheming Xu | Thank you. Maybe I drop a condition that $G$ is a subgroup of $Aut(R)$. | |
| Aug 13 at 6:52 | history | answered | HenrikRüping | CC BY-SA 4.0 |