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Let $M$ be a torsion abelian group. For a prime $p$, let $\Bbb Z_p$ be the $p$-adic integers (seen as an abelian group), and $M[p^{\infty}] := \{m \in M : p^n m= 0 \text{ for some } n \geq 0\}.$

  1. Is it true that $M[p^{\infty}] \cong M \otimes_{\Bbb Z} \Bbb Z_p$ ?

  2. If not, is it still possible to have a natural structure of $\Bbb Z_p$-module on $M[p^{\infty}]$ ?

I know for instance that $S^1[p^{\infty}] \cong \Bbb Q_p / \Bbb Z_p$ which is naturally a $\Bbb Z_p$-module, but I don't know about general $M$.

If $p^n m = 0$ and $a = ([a_k]_{p^k})_{k \geq 0} \in \Bbb Z_p$, we could define $a \cdot m := a_n m \in M$, but I'm not sure whether this is the right thing to do.

Alphonse
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  • This answer claims truth for 1) : https://math.stackexchange.com/q/538266 – Alphonse Jun 30 '18 at 15:47
  • I asked this: https://math.stackexchange.com/questions/2836890/isomorphism-between-bbb-r-otimes-bbb-z-bbb-z-p-and-p-prüfer-group – Alphonse Jun 30 '18 at 16:02
  • I don't know if $S^1 \otimes \Bbb Z_p$ is the Prüfer $p$-group $Q_p / Z_p$, but $$Q / Z \otimes Z_p = \bigoplus_q (Z[1/q]/Z \otimes Z_p) \cong \bigoplus_q Z/q \otimes Z_p \cong \bigoplus_q Z_p / q Z_p = Z_p / p Z_p = Z / p \cong Z[1/p]/Z = (Q / Z)[p^{\infty}]$$ is the Prüfer $p$-group – Alphonse Jun 30 '18 at 16:13
  • If $M$ is finite abelian, then $M(p) = M[p^{\infty}]$, the $p$-part of $M$, is the unique $p$-Sylow subgroup of $M$. – Alphonse Jul 14 '18 at 09:46
  • If $M$ is finitely generated abelian, then $M = \Bbb Z^n \oplus F$ so that $$M(p) = \bigoplus \Bbb Z / p_i^{e_i} \cong \bigoplus \Bbb Z_p / p_i^{e_i} \Bbb Z_p$$ is naturally a $\Bbb Z_p$-module. – Alphonse Oct 23 '18 at 12:18

1 Answers1

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Yes, it is true.

For a torsion group $M$ we have $M\cong \bigoplus_q M[q^\infty].$ It is easy to see that $M[q^\infty]\otimes \mathbb Z_p=0$ for $q\ne p.$ Hence, it is enough to prove that $$M[p^\infty]\otimes \mathbb Z_p\cong M[p^\infty].$$
Note that we have a similar isomorphism for the $p^n$-torsion subgroup $M[p^n]$ for a fixed $n$: $$M[p^n]\otimes \mathbb Z_p \cong M[p^n] \otimes \mathbb Z_p/p^n \cong M[p^n] \otimes \mathbb Z/p^n \cong M[p^n] \otimes \mathbb Z \cong M[p^n].$$ Further, we have $M[p^\infty]=\bigcup_n M[p^n].$ Hence $M[p^\infty]$ is the colimit of $M[p^n]$ $$M[p^\infty]={\sf colim} (M[p] \hookrightarrow M[p^2]\hookrightarrow M[p^3]\hookrightarrow \dots).$$ Finally, we combine this with the fact that the functor of tensor product $-\otimes \mathbb Z_p$ comutes with colimits (because it is left adjoint to the functor ${\sf Hom}(\mathbb Z_p,-)$): $$ M[p^\infty]\otimes \mathbb Z_p=({\sf colim}\ M[p^n])\otimes \mathbb Z_p \cong {\sf colim}\ (M[p^n]\otimes \mathbb Z_p ) \cong {\sf colim}\ M[p^n] \cong M[p^\infty].$$

Sergei Ivanov
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