**Question:** Let $p$ be an odd prime. Does there exist a closed manifold $M$ with $\widetilde H^\ast(M; \mathbb Q) = 0$ but $\widetilde H^\ast(M; \mathbb F\_p) \neq 0$?


When $p = 2$, an example is given by $\mathbb R \mathbb P^2$.


 After discussion, this question turns out to be equivalent to [this other question](https://mathoverflow.net/questions/410577/for-which-n-does-there-exist-a-closed-manifold-of-chromatic-type-n). That is, over at that question Saal Hardali explained that if $M$ is a closed manifold, then the chromatic type of $M$ at a prime $p$ is either 0 or 1. Both possibilities are realized at $p=2$; the question is whether chromatic type 1 is realized at odd primes. Chromatic type 0 just means having nonvanishing rational (co)homology. So the question is whether there exists a closed manifold $M$ which is rationally contractible but whose $p$-localization is nontrivial for an odd $p$.  This turned out to be mistaken, thanks to Ben Wieland for pointing this out at the other question.


**Side Question:** When $p=2$, what are some other examples of $M$ with $\widetilde H^\ast(M;\mathbb Q) = 0$ but $\widetilde H^\ast(M;\mathbb F\_2) \neq 0$ besides $M = \mathbb R \mathbb P^{2n}$ and products thereof?



Does there exist a closed manifold with vanishing reduced rational cohomology but nonvanishing odd torsion cohomology?
As mme noted in the comments, such examples cannot exist in odd dimensions, for Euler characteristic reasons. They can't exist in dimension 2 either, by classification. I claim that in all other dimensions $2n > 2$ we have (plenty of) examples.


Let $N$ be a rational homology $2n$-sphere, that is a $2n$-manifold with $H\_\*(N; \mathbb{Q}) = H\_\*(S^{2n}; \mathbb{Q})$. For every prime $p$ there exists a rational homology $2n$-sphere with $\dim\_{\mathbb{F}\_p} H\_\*(N;\mathbb{F}\_p) > 2$. For instance, you can take a spun lens space (any spun rational homology $2n-1$-sphere would do).


Now, the integral homology of $M = \mathbb{RP}^{2n} \# N$ splits as a direct sum of that of the two summands in all dimensions strictly between 0 and 2n, and it vanishes in dimension 2n (because $M$ is non-orientable) and it is $\mathbb{Z}$ in dimension 0 (because $M$ is connected). (This is Exercise 6 in Section 3.3 of Hatcher's *Algebraic topology*.) That is, $M$ is a rational homology ball, and its homology has as much $p$-torsion as that of $N$.


For the side question, if we choose $N$ to have no 2-torsion in its homology (e.g. spinning an odd lens space should do the trick), this gives plenty of examples.


