Talk:Artin's theorem on induced characters

Confusing proof

It's great that this page contains a version of Serre's proof, but it's confusing. It was also full of typos and misspellings, some of which I've fixed. John Baez (talk) 17:09, 6 March 2025 (UTC)

I have now completed work on the proof of Artin's theorem on induced characters. I don't like this step:

On the other hand, this map

$\mathbb {C} \otimes {\text{Ind}}\colon \mathbb {C} \otimes \bigoplus _{H\in X}{\mathcal {R}}(H)\to \mathbb {C} \otimes {\mathcal {R}}(G)$

is surjective, because otherwise ${\text{Ind}}$ would have an infinite cokernel, contradicting assumption 2.

I believe it's true that if the linear map $\mathbb {C} \otimes {\text{Ind}}$ is non-surjective, the original homomorphism of free abelian groups

${\text{Ind}}\colon \bigoplus _{H\in X}{\mathcal {R}}(H)\to {\mathcal {R}}(G)$

must have infinite cokernel. However, if it's really true, might be nice to give a one-sentence explanation of why.

Note the converse implication is used later. John Baez (talk) 22:43, 11 March 2025 (UTC)

Thank you for correcting the proof. As for the surjectivity of the map, here is a schetch of an argument :

take an element $x$ in $\mathcal{R}(G)$. The assumption on the cokernel of $Ind$ insure that one multiple of $x$, say $nx$ is contained within the image of the map. Thus $x=\frac{1}{n}nx$ is in $Im(\mathbb{C}\otimes Ind)$. Wéve just prove that the image of the tensored map contains $\mathcal{R}(G)$, thus contains $\mathbb{C}\otimes\mathcal{R}(G)$ which prove the surjectivity.

the only issue with this argument is its lenght… Galindo.maths (talk) 08:49, 14 March 2025 (UTC)