(Bockstein Spectral Sequence) Why is $v_1$ also $p$-divisible?

Question

Why if $u=pv_1$, then $v_1$ is also $p$-divisible?

I am quite puzzled by the above. I can see that $v_1$ must also generate a copy of $\mathbb{Z}$, thus $v_1\notin\ker p^r$. I can't seem to proceed to conclude further.

Thanks for any help.