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I am studying a paper which uses the following lemma. The context is irrelevant, as the lemma is only used as a technical trick and has no pointer to a reference or hint in the proof but its link to theory of heights in $p$-torsion abelian groups (which I ignore).

There exists a sequence $B_\bullet = (B_\alpha)$ of $p$-torsion abelian groups, one for each ordinal number $\alpha\in \text{Ord}$, satisfying the following properties:

  • each $B_\alpha$ contains an element $x_\alpha$ such that $p x_\alpha = 0$;
  • when $\alpha < \beta$ every function of sets $f_{\alpha\beta} \colon B_\alpha \to B_\beta$ such that $f(px)=p f(x)$ sends $x_\alpha$ to zero.

Also, I would like to know if the following refinement of the result is true.

There exists a sequence $B_\bullet = (B_\alpha)$ of $p$-torsion abelian groups, one for each ordinal number $\alpha\in \text{Ord}$, satisfying the following properties:

  • each $B_\alpha$ contains an element $x_\alpha$ such that $p x_\alpha = 0$;
  • when $\alpha < \beta$ every function of sets $f_{\alpha\beta} \colon B_\alpha \to B_\beta$ such that $f(px)=p f(x)$ sends $x_\alpha$ to zero.
  • There exists B such that each $B_\alpha$ has a map $B_\alpha \to B$ of groups that does not vanish on $x_\alpha$.

I am pretty sure that function of sets is not needed, so if someone has the answer with the hypotesis morphisms of group, please don't be shy.

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    $\begingroup$ I assume from context that by $\mathbb{Z}_p$ you mean a cyclic group of order $p$ rather than the $p$-adic integers? The refinement can't be true, since with the extra condition $B_\alpha$ contains a cyclic direct summand of order $p$ generated by $x_\alpha$, so there will always be a homomorphism $B_\alpha\to B_\beta$ sending $x_\alpha$ to $x_\beta$. $\endgroup$ – Jeremy Rickard Mar 18 '17 at 11:19
  • $\begingroup$ What if I replace $\mathbb{Z}_p$ with any other abelian group? (I edited question). $\endgroup$ – Ivan Di Liberti Mar 18 '17 at 12:09
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    $\begingroup$ In the edited version of this question the refined statement is equivalent to the unrefined one: you can take $B=\mathbb{Q}/\mathbb{Z}$, which is an injective cogenerator of the category of abelian groups. $\endgroup$ – Neil Strickland Mar 18 '17 at 12:13
  • $\begingroup$ That is perfect (and trivial, lol), thanks! $\endgroup$ – Ivan Di Liberti Mar 18 '17 at 12:33
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    $\begingroup$ Is $x_{\alpha} \neq 0$ missing? $\endgroup$ – HeinrichD Mar 18 '17 at 13:17

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