Suppose $M$ is a compact smooth manifold with Morse's polynomial $\mathcal{M}(t)$ and Poincaré's polynomial $\mathcal{P}(t)$ satisfying $\mathcal{M}(t)=\mathcal{P}(t)$ for any coefficient field $\mathbb{F}$.
I want to prove that the integral homology $H_i(M,\mathbb{Z})$ has no torsion.
I know that for any fields $K$ and $L$, I have $\dim_K(H_i(M,K))=\dim_L(H_i(M,L))$. I think the idea is to show that these dimensions are zero and conclude that $H_i(M,\mathbb{Z})$ has no torsion, but dont know how to formalize both steps.
The last result you mention is enough to prove it. Since $M$ is compact, all the $H_p(M) = H_p(M, \mathbb{Z})$ are finitely generated. Hence $H_p(M) = \mathbb{Z}^{n_p} \oplus T_p$ for some finite abelian group $T_p$. Suppose some $T_p$ is nontrivial, and choose a prime $\pi | \# T_p$. By the universal coefficient theorem, $$H_{p+1}(X, A) = (H_{p+1}(X) \otimes A)\oplus \text{Tor}(H_p(X), A)$$ for any $\mathbb{Z}$-module $A$. Over $A = \mathbb{Q}$, we have $$H_{p+1}(X, \mathbb{Q}) = H_{p+1}(X)\otimes \mathbb{Q} = \mathbb{Q}^{n_p}.$$ Over $A = \mathbb{F}_\pi$, we have $$\dim_{\mathbb{F}_p} H_{p+1}(X, \mathbb{F}_\pi) = n_p + \dim_{\mathbb{F}_\pi} \text{Tor}(T_p, \mathbb{F}_\pi) > n_p = \dim_{\mathbb{Q}} H_{p+1}(X, \mathbb{Q})$$ since $\text{Tor}(T_p, \mathbb{F}_\pi)\not = 0$ (as $T_p$ contains a summand of the form $\mathbb{Z}_q$ with $\pi | q$). The result follows.