## Note on p-divisibility in Bockstein Spectral Sequence

If $u\in H_n(X)$ generates a copy of $\mathbb{Z}$, then $u\notin\ker p^r$.

Write $u=a+b$, where $a\in pH_n(X)$, $b\in\ker p^r$. Note that since $b=u-a\in\ker p^r$, hence $b=u-a$ does not generate a copy of $\mathbb{Z}$. The only way that is possible is when $b=u-a=0$, i.e $u=a$.