4 Texified, since problem was on front-page anyway

# Mysterious property of Q$\mathbb{Q}$

Hi,

I am currently working through the paper by Bousfield and Gugenheim on rational homotopy theory, and have come to a point where they show why it is important to work over Q, $\mathbb{Q}$, and not just any field (remark 9.7). They assert that if Hom_k(Hom_Z(G, $Hom_k(Hom_{\mathbb{Z}}(G, k), k)=G k)=G$ for some non-trivial abelian group G, $G$, and some field k $k$ [edit: of characteristic 0] $0$] (thanks to Fernando Muro for pointing out that we need characteristic 0), $0$), then we must have k=Q $k=\mathbb{Q}$ (and also that G $G$ is finite dimensional as a Q-vector $\mathbb{Q}$-vector space, which is clear enough). I can't quite see what property of Q $\mathbb{Q}$ this reduces to, and it really makes my head spin trying to think through it. The paper has is now been available online, just google "on pl de rham theory and rational homotopy type".

Any suggestions would be great.

Thanks

Brian

3 added 101 characters in body

Hi,

I am currently working through the paper by Bousfield and Gugenheim on rational homotopy theory, and have come to a point where they show why it is important to work over Q, and not just any field (remark 9.7). They assert that if Hom_k(Hom_Z(G, k), k)=G for some non-trivial abelian group G, and some field k [edit: of characteristic 0] (thanks to Fernando Muro for pointing out that we need characteristic 0), then we must have k=Q (and also that G is finite dimensional as a Q-vector space, which is clear enough). I can't quite see what property of Q this reduces to, and it really makes my head spin trying to think through it. The paper has now been available online, just google "on pl de rham theory and rational homotopy type".

Any suggestions would be great.

Thanks

Brian

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