Timeline for Prove that the ideal of $\mathbb{C}G$ generated by a family of elements $\lbrace p_i\rbrace_{i=1}^n$ is equal to $\mathbb{C}G$
Current License: CC BY-SA 4.0
9 events
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| Sep 27, 2023 at 12:37 | comment | added | Simon Wadsley | The crucial thing for more general arguments along the lines I gave is the Artin-Wedderburn Theorem: en.wikipedia.org/wiki/Wedderburn%E2%80%93Artin_theorem. $\mathbb{F}G$ is a semisimple $\mathbb{F}$-algebra whenever $G$ is a finite group whose order is coprime to the characteristic of $\mathbb{F}$ (plus the case where the latter is $0$). As I hinted before the problem has two generalisations in the non-commutative setting depending on whether you wish 'ideal' to mean two-sided or one-sided ideals. | |
| Sep 27, 2023 at 11:58 | comment | added | John Wiltshire-Gordon | @Marcos Yes, and it will also work for any submodule of a free module. This is a good example of how staying organized about tensor and hom can make tricky calculations into simple ones. Try homming your right ideal to a right simple module and see what you get. (Alternatively, you could mod out by your ideal and tensor with a left simple). I'm happy to provide more details, but probably not in a comment. | |
| Sep 27, 2023 at 9:30 | comment | added | Marcos | @JohnWiltshire-Gordon can you elaborate more on that? I've tried to solve this problem with $\mathbb{F}=\mathbb{C}$ and $G$ finite abelian because it seemed to be the esiest possible case. But if there is a way to solve the same problem for a group algebra $\mathbb{F}G$ with $\mathbb{F}$ an arbitrary field and $G$ an arbitrary finite group I would be also interested. | |
| Sep 27, 2023 at 9:16 | vote | accept | Marcos | ||
| Sep 26, 2023 at 15:34 | comment | added | John Wiltshire-Gordon | Yes, good point, this is for one-sided ideals. I guess two-sided ideals will need to entirely zero out an irrep! | |
| Sep 26, 2023 at 15:27 | comment | added | Simon Wadsley | I guess that depends whether ideal means left ideal or two-sided ideal? | |
| Sep 26, 2023 at 15:19 | comment | added | John Wiltshire-Gordon | And if $G$ is non-abelian, the condition is just... there exists an irrep $\phi$ so that the block matrix with blocks $\phi(p_i)^T$ is not full rank. | |
| Sep 26, 2023 at 15:05 | history | edited | Simon Wadsley | CC BY-SA 4.0 |
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| Sep 26, 2023 at 11:16 | history | answered | Simon Wadsley | CC BY-SA 4.0 |