Commuting singular cohomology and tensor product Suppose $X$ is a complex quasiprojective variety (possibly singular). Is there an isomorphism
$$
H^i(X_{an}, \mathbb Z_\ell) \otimes_{\mathbb Z_\ell} \mathbb{Q}_\ell
\cong
H^i(X_{an}, \mathbb Q_\ell)
$$
of singular cohomology groups? $\mathbb Z_\ell$ denotes the ring of $\ell$-adic numbers.
More generally, what kind of theorems do we have for commuting tensor products with cohomology? I'm aware of the Universal Coefficients Theorem, but what I want seems to apply for homology, not cohomology.
Thank you.
 A: Here is a pedestrian way to see it:
Firstly, note that the homologies $H_{*}(X, \mathbb{Z}_{\ell})$ are finitely generated: this is a consequence of the fact that quasi-projective complex varieties can be modelled by finite CW complexes. This means, since $\mathbb{Z}_{\ell}$ is a pid, that we may decompose $H_{i}(X, \mathbb{Z}_{\ell})=F_i \oplus T_i$ where $F_i$ is a finite free $\mathbb{Z}_{\ell}$-module and $T_i$ is torsion, i.e. $\ell^NT_i=0$ for some $N$.
Now, there is a universal coeff. theorem for cohomology, which says that for a pid $R$ and an $R$-module $A$, one has for every $i$ the short exact sequence
$$0 \rightarrow \mathrm{Ext}^1_R(H_{i-1}(X, R),A)\rightarrow H^i(X, A)\rightarrow \mathrm{Hom}_R(H_{i}(X, R),A) \rightarrow 0.$$
We can use this fact twice: first for $A=\mathbb{Q}_{\ell}$. In that case, since $\mathbb{Q}_{\ell}$ is divisible hence injective, $\mathrm{Ext}^1_{\mathbb{Z}_{\ell}}(H_{i-1}(X, \mathbb{Z}_{\ell}),\mathbb{Q}_{\ell})=0$ and we get an isomorphism $H^i(X, \mathbb{Q}_{\ell})\simeq \mathrm{Hom}_{\mathbb{Z}_{\ell}}(H_{i}(X, \mathbb{Z}_{\ell}), \mathbb{Q}_{\ell})$.
Secondly, we can use it for $A=\mathbb{Z}_{\ell}$ itself. Now the first term $\mathrm{Ext}^1_{\mathbb{Z}_{\ell}}(H_{i-1}(X, \mathbb{Z}_{\ell}),\mathbb{Z}_{\ell})$ is not zero in general, but, it is equal to $\mathrm{Ext}^1_{\mathbb{Z}_{\ell}}(F_{i-1},\mathbb{Z}_{\ell})\oplus \mathrm{Ext}^1_{\mathbb{Z}_{\ell}}(T_{i-1},\mathbb{Z}_{\ell}),$where $\mathrm{Ext}^1_{\mathbb{Z}_{\ell}}(F_{i-1},\mathbb{Z}_{\ell})=0$ because $F_{i-1}$ is free and $\mathrm{Ext}^1_{\mathbb{Z}_{\ell}}(T_{i-1},\mathbb{Z}_{\ell})$ is torsion because $T_{i-1}$ is. So in this case, the map $H^i(X, \mathbb{Z}_{\ell})\rightarrow \mathrm{Hom}_{\mathbb{Z}_{\ell}}(H_{i}(X, \mathbb{Z}_{\ell}), \mathbb{Z}_{\ell})$ is only surjective with a trosion kernel, which, however, implies that $H^i(X, \mathbb{Z}_{\ell})\otimes_{\mathbb{Z}_{\ell}}\mathbb{Q}_{\ell}\simeq \mathrm{Hom}_{\mathbb{Z}_{\ell}}(H_{i}(X, \mathbb{Z}_{\ell}), \mathbb{Z}_{\ell})\otimes_{\mathbb{Z}_{\ell}}\mathbb{Q}_{\ell}$.
So now the last thing that remains is to show that $\mathrm{Hom}_{\mathbb{Z}_{\ell}}(H_{i}(X, \mathbb{Z}_{\ell}), \mathbb{Z}_{\ell})\otimes_{\mathbb{Z}_{\ell}}\mathbb{Q}_{\ell} \simeq \mathrm{Hom}_{\mathbb{Z}_{\ell}}(H_{i}(X, \mathbb{Z}_{\ell}), \mathbb{Q}_{\ell})$. Dissecting $H_{i}(X, \mathbb{Z}_{\ell})=F_i\oplus T_i$ again, similarly the torsion part contribution on both sides vanishes since the targets in the Hom groups are torsion free, and so we are left with showing that $\mathrm{Hom}_{\mathbb{Z}_{\ell}}(F_i, \mathbb{Z}_{\ell})\otimes_{\mathbb{Z}_{\ell}}\mathbb{Q}_{\ell} \simeq \mathrm{Hom}_{\mathbb{Z}_{\ell}}(F_i, \mathbb{Q}_{\ell})$. But this is now easy, $F_i \simeq \mathbb{Z}_\ell^{n}$ is finite free and thus both sides are easily seen to be isomorphic to $\mathbb{Q}_{\ell}^n$.
