MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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 $\mathbb{Q}$, and not just any field (remark 9.7). They assert that if $Hom_k(Hom_{\mathbb{Z}}(G, k), k)=G$ 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 $k=\mathbb{Q}$ (and also that $G$ is finite dimensional as a $\mathbb{Q}$-vector space, which is clear enough). I can't quite see what property of $\mathbb{Q}$ this reduces to, and it really makes my head spin trying to think through it. The paper is now available online, just google "on pl de rham theory and rational homotopy type".

Any suggestions would be great.



share|cite|improve this question
Your assertion is false, take $k=\mathbb{F}_{p}$, $p$ a prime number, and $G$ any finite-dimensional $\mathbb{F}_{p}$-vector space. – Fernando Muro Apr 12 '12 at 20:35
The tag category-theory doesn't fit. – Martin Brandenburg Apr 12 '12 at 20:42
Although it is only tangentially related, I think it's worth noting that there is a similar theorem for the special role of $S^1$ in duality theory for locally compact abelian groups. See my comments to the question…. – KConrad Apr 13 '12 at 1:22
up vote 16 down vote accepted

First, $k$ needs to be assumed of characteristic $0$. The mysterious property is that $\mathbb{Q}$ is a prime field:

If $Hom_k(Hom_Z(G, k), k)=G$, $G$ has to be a $k$-vector space, say $G=kT$ for a basis $T$. $Hom_Z (kT;k)$ is a $k$-vector space.

Let me first argue that $T$ has to be finite. Since $Hom_k (kT;k) \to Hom_Z (kT;k)$ is injective, the dimension of $Hom_Z (kT;k)$ is at least the dimension of $(kT)^{\ast}$. Thus it follows that the dimension of $kT$ is at least the dimension of the bidual of $kT$. Thus can only be true if $T$ is finite. We can assume that $T$ has one element; therefore

$$dim_k (Hom_k (Hom_Z (k,k);k))=1.$$

If $k$ is not a prime field, then $dim (Hom_Z(k,k))>1$, which contradicts the above formula.

The same argument works in characteristic $p$ and shows that $k$ has to be $\mathbb{F}_p$.

share|cite|improve this answer
Ah, nice. Okay, I see now, Q is the prime subfield of any field of characteristic zero, and so Hom_Z(k,k)=1 requires k=Q for k characteristic zero. Perfect, thanks very much Johannes. – Brian Robertson Apr 13 '12 at 13:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.