$H_n(X, F_p) = 0$ implies $H_n (X,Q)=0$ for certain CW-complexes $X$ Consider a CW-complex $X$ with exactly one cell in each dimension. Suppose there is a prime $p$ such that $H_n(X, F_p) = 0$ for all $n\geq 1$. 
How can I show that this  implies $H_n (X,Q)=0$ for all $n\geq 1$? 
I thought about using the universal coefficient theorem for homology, but I do not know how to do it exactly. 
 A: By the UCT, we have a short exact sequence $0\to H_i(X; \mathbb{Z})\otimes \mathbb{F}_p \to H_i(X,\mathbb{F}_p)\to \mathrm{Tor}_1^\mathbb{Z}(H_{i-1}(X;\mathbb{Z}),\mathbb{F}_p)\to 0$
Thus your hypothesis yields that $H_i(X;\mathbb{Z})\otimes \mathbb{F}_p = 0$ for all $i\geq 1$, and similarly $\mathrm{Tor}_1^\mathbb{Z}(H_i(X;\mathbb{Z}),\mathbb{F}_p)=0$ for $i\geq 0$.
Note, moreover, that since $X$ has one cell in each dimension, cellular homology gives that for each $i$, $H_i(X;\mathbb{Z})$ is a quotient of $\mathbb{Z}$, that is $\mathbb{Z/nZ}$ for some (possibly $0,1$) $n$. 
Thus $H_i(X;\mathbb{Z})\otimes \mathbb{F}_p = 0$ tells us that $n$ is never $0$ except for $i=0$: indeed $\mathbb{Z}\otimes \mathbb{F}_p \simeq \mathbb{F}_p$. 
Therefore for all $i\geq 1$, $H_i(X;\mathbb{Z})$ is finite, so $H_i(X;\mathbb{Z})\otimes \mathbb{Q} = 0$. Moreoer, it's easy to compute $\mathrm{Tor}_1^\mathbb{Z}(\mathbb{Q, Z/nZ})  = 0$, no matter what $n$ is, so $\mathrm{Tor}_1^\mathbb{Z}(H_{i-1}(X;\mathbb{Z}), \mathbb{Q}) = 0$ for all $i\geq 1$. 
Applying the UCT once again with $\mathbb{Q}$ coefficients now yields $H_i(X;\mathbb{Q}) = 0$ for all $i\geq 0$. 
Note that the proof actually gives a much better result : it suffices that $X$ has a finite number of cells in any dimension ($X$ is locally finite). Indeed, in this case $H_i(X;\mathbb{Z})$ is a finitely generated abelian group for all $i$, and thus $H_i(X;\mathbb{Z})\otimes \mathbb{F}_p = 0$ still implies that $H_i(X;\mathbb{Z})$ is finite, and thus by easy computations, both $H_i(X;\mathbb{Z})\otimes \mathbb{Q} $ and $\mathrm{Tor}_1^\mathbb{Z}(H_{i-1}(X;\mathbb{Z}), \mathbb{Q})$ vanish. 
EDIT : here's a proof "by hand". It only works in the case where there's one cell per dimension though. 
The cellular complex of $X$ looks like $...\to \mathbb{Z}_2\to \mathbb{Z}_1\to \mathbb{Z}_0$ (the indices denote the position in the complex), with differentials $d_n : \mathbb{Z}_n\to \mathbb{Z}_{n-1}$ given by $d_n(1) = \alpha_n$. 
Tensor it with $\mathbb{F}_p$ to get a new complex, with $d_n(\overline{1}) = \overline{\alpha_n}$. Then since a map $\mathbb{F}_p\to \mathbb{F}_p$ is either an isomorphism or $0$, and $d^2 = 0$, it follows that $d_n\circ d_{n+1} = 0 \implies \overline{\alpha_n} \overline{\alpha_{n+1}} = 0$. Therefore one of $\alpha_n, \alpha_{n+1}$ is congruent to $0$ mod $p$. To see that the other is coprime to $p$, assume both are $0$ mod $p$. Then $d_n = d_{n+1} = 0$ on the mod $p$ complex, so $H_n(X;\mathbb{F}_p) = \ker d_n / \mathrm{im}(d_{n+1}) = \mathbb{F}_p/0 = \mathbb{F}_p$, which is absurd. Therefore exactly one of them is zero mod $p$, the other being invertible mod $p$. 
But now $d^2=0$ on the integral chain complex too, so $\alpha_n=0$ or $\alpha_{n+1}=0$ : the one that's $0$ must be the one congruent to $0$ mod $p$, and the other is nonzero. Tensor with $\mathbb{Q}$ to get that the rational complex is an alternance of isomorphisms and zeroes, which gives $0$ homology. 
A: Note if we can show that the homology is always finite for $n>0$ then $H_n(X,Q)=0$ for $n>1$ by the universal coefficient theorem. And $H_1(X,Q)=0$ by the universal coefficient theorem and the fact that $H_0(X)$ is free.
Now let us show that the integral homology is always finite. If $H_n(X,F_p)=0$, we have $H_n(X) \otimes F_p=0$ by the universal coefficient theorem. The equality implies $H_n(X)/pH_n(X)=0$, so $pH_n(X)=H_n(X)$. Since each dimension has 1 cell, $H_{n-1}(X)$ is a quotient of $\mathbb{Z}$. It is clear that if the quotient is $\mathbb{Z}$ then $p\mathbb{Z}\neq \mathbb{Z}$, so it must be a quotient by a nontrivial subgroup. This guarantees $H_n(X)$ finite, so we are done.
