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Apr 13, 2017 at 12:58 history edited CommunityBot
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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
Jan 23, 2013 at 20:07 history asked Martin Brandenburg CC BY-SA 3.0