Timeline for Classification of symtrivial modules over a PID
Current License: CC BY-SA 3.0
8 events
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Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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Jan 26, 2013 at 10:31 | vote | accept | Martin Brandenburg | ||
Jan 23, 2013 at 22:39 | answer | added | Will Sawin | timeline score: 11 | |
Jan 23, 2013 at 20:45 | comment | added | Martin Brandenburg | If $A$ is a commutative ring with $\mathbb{Z}$-rank $\leq 1$, then $\mathbb{Z} \to A$ doesn't have to be an epimorphism and therefore $A$ is not symtrivial over $\mathbb{Z}$. For example, $A = \mathbb{Z}/2 \times \mathbb{Z}/2$. | |
Jan 23, 2013 at 20:34 | comment | added | Todd Trimble | Notice that $F: \mathbb{Q} \otimes_{\mathbb{Z}} -$ is a strong symmetric monoidal functor, so if $\sigma \neq 1: F(A) \otimes F(A) \to F(A) \otimes F(A)$, then $\sigma \neq 1: A \otimes A \to A \otimes A$. This means that if $A$ is of rank 2 of greater, then $A$ cannot be symtrivial. Thus being rank $0$ or $1$ is a necessary condition for symtriviality. I think it's also sufficient. | |
Jan 23, 2013 at 20:28 | comment | added | Todd Trimble | Quick guess is that they are the abelian groups of rank less than $2$, i.e., abelian groups such that the $\mathbb{Q}$-vector space $\mathbb{Q} \otimes_{\mathbb{Z}} A$ is of dimension $0$ or $1$. If so, would that be a satisfactory classification? | |
Jan 23, 2013 at 20:15 | history | edited | Martin Brandenburg | CC BY-SA 3.0 |
deleted 24 characters in body; added 158 characters in body
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Jan 23, 2013 at 20:07 | history | asked | Martin Brandenburg | CC BY-SA 3.0 |