Timeline for Length of a module over different rings
Current License: CC BY-SA 2.5
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 20, 2010 at 18:53 | vote | accept | TonyS | ||
| Sep 20, 2010 at 15:39 | comment | added | TonyS | Yes, i have this article on my desk. I think if $R$ is a complete d.v.r., $D$ a skew field over K=Quot(R) and $S$ is the unique maximal $R$-order in $D$, then $M=S/rad(S)$ is the unique simple $S$ module. We can compute $\ell_R(M)$ after an etale base change $R \rightarrow R'$, so we can assume $S$ is a hereditary order in $M_n(K)$, where $n^2=[D:K]$.This should give $\ell_R(M)=n$. | |
| Sep 20, 2010 at 15:11 | comment | added | Hailong Dao | Dear Tony, you probably knew this already, but just in case: if $S$ is a maximal order in some $M_n(K)$ where $K$ is the quotient field of $R$, and $S$ is free as $R$-module, then $S=M_a(R)$ for some $a$. This can be found in Auslander-Goldman article "Maximal orders". | |
| Sep 20, 2010 at 13:28 | comment | added | TonyS | Yes that is what i meant. Thanks for pointing that out. | |
| Sep 20, 2010 at 13:27 | history | edited | TonyS | CC BY-SA 2.5 |
added 27 characters in body
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| Sep 19, 2010 at 16:40 | comment | added | Peter Samuelson | By "free R-algebra" do you mean "free algebra over R" or "algebra over R that is free as an R-module"? From your example $S = M_n(R)$ it seems that you mean the second, but this isn't clear (to me) from the phrasing of the first sentence. | |
| Sep 18, 2010 at 22:55 | answer | added | Manny Reyes | timeline score: 5 | |
| Sep 18, 2010 at 17:48 | history | asked | TonyS | CC BY-SA 2.5 |