Suppose that X is a spherical complex with $H_q(X;F)=0$ for all $q>0$ where $F=\mathbb{Q}$ or $F=\mathbb{Z}/p\mathbb{Z}$, as $p$ runs over all primes. I want to prove $H_q(X;\mathbb{Z})=0$.
I know Singular Homology Theory and some of Singular Cohomology. I know that $H_q(X;\mathbb{Z})$ is a finitely generated $\mathbb Z$-module for every q, since $\mathbb{Z}$ is a Noetherian ring and $X$ is a spherical complex. But I don't know how to continue proving.
I give a definition of a spherical complex: Start with a finite discrete set of points, and successively attach cells, possibly of varying dimensions, but finite in number. The resulting Hausdorff space is called a spherical complex.
Thank you for your help!
Hint: by the universal coefficient theorem for homology, we have an isomorphism $$H_q(X, F) \cong \left(H_q(X, \mathbb Z) \otimes F\right) \oplus \text{Tor}(H_{i-1}(X,\mathbb Z), A).$$ Necessarily both terms vanish, and the fact that $H_q(X, \mathbb Z)$ is a finitely generated abelian group gives it an explicit form you can easily deal with.
EDIT: here's an approach with the UCT for cohomology. We know that $$H^i(X, F) \cong \text{Ext}_R^1(H_{i-1}(X,R),F) \oplus \text{Hom}_R(H_i(X,R),F)$$ Calculate $H^i(X,F)$ in two ways: first by using $R=F$, and then after seeing that $H^i(X,F)=0$ calculate it using $R=\mathbb Z$; again because $\text{Ext}$ and $\text{Hom}$ are well-behaved and you can explicitly write down what $H_i$ must be (by the classification of finitely generated abelian groups), the theorem you want falls out.