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I have some question on this answer. In the proof of $\Bbb F$ is a characteristic $p$ field, he assumed $H_k(X;\Bbb Z) = \Bbb Z^{b_k}\oplus(\Bbb Z/p)^{c^p_k}\oplus T^p_k$ where $T^p_k$ is not $p$-torsion. I wonder if I can write like that. It's possible that the torsion part of $H_k(X;\Bbb Z)$ contains $\Bbb Z/p^l$ for some $l>1$. Then it contains a $p$-torsion subgroup. Could you explain why I can write like that? I don't believe $(\Bbb Z/p^l)\otimes\Bbb F\simeq\Bbb F$.

one potato two potato
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    Well, any bilinear map $\mathbb{Z}/(p^n) \times F \rightarrow A$ factors uniquely through $\mathbb{Z}/(p) \times F$ – and any bilinear map $\mathbb{Z}/(p) \times F$ comes from some bilinear map $\mathbb{Z}/(p^n)\times F$. So $\mathbb{Z}/(p^n) \otimes F \cong \mathbb{Z}/(p) \otimes_{\mathbb{Z}} F = \mathbb{Z}/(p) \otimes_{\mathbb{Z}/(p)} F=F$. – Aphelli Sep 08 '21 at 08:11
  • @Mindlack Why any bilinear map $\Bbb Z/p \times F$ comes from bilinear map $\Bbb Z/p^n\times F$ ? – one potato two potato Sep 08 '21 at 08:38
  • @love_sodam For any $\mathbb{Z}$-bilinear $B\colon\mathbb{Z}/p\times F\to A$, define map $\mathbb{Z}/p^n\times F\to A$ by $(m+p^n\mathbb{Z},y)\mapsto B(m+p\mathbb{Z},y)$. Easy to check this is $\mathbb{Z}$-bilinear. – user10354138 Sep 08 '21 at 09:07

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