Let given abelian group $A$ of finite rank (without torsion), such that $\cap_{i\geq 0}p^iA = 0$, for prime number $p$. Is it true that $\mathbf{Z}_p$-module $\mathbf{Z}_p\otimes_{\mathbb{Z}}A$ has finite rank, too?

Furthermore is it true that $\mathbf{Z}_p\otimes_{\mathbb{Z}}A$ is a finitely generated $\mathbf{Z}_p$-module?

Where $\mathbf{Z}_p$ denotes the $p$-adic integers.

  • $\begingroup$ What ddo you denote $\mathbb Z_p$? Integers modulo ˆpˆ, or the ring of fractions with denominator a power of $p$? $\endgroup$
    – Bernard
    Oct 2 '16 at 16:31
  • $\begingroup$ No, $p$- adic integers $\endgroup$
    – Ask
    Oct 2 '16 at 17:11
  • $\begingroup$ Probably the closing vote comes from a misinterpretation of the question: finite rank means that there is a bound on the size of $R$-free, where $R$ is the base ring ($\mathbf{Z}$ or $\mathbf{Z}_p$ here). Equivalently in the torsion-free case, this means that $A$ is isomorphic to an $R$-submodule of $K^r$ for some $r$, where $K$ is the field of fractions (anyway this shows that the first question clearly has a positive answer). $\endgroup$
    – YCor
    Oct 3 '16 at 3:59

I answer the second question (I said in a comment that the first question is clearly true). Here is a counterexample.

Let $A$ be an infinitely generated subgroup of $\mathbf{Z}[1/p]^2$ such that $\bigcap p^nA=0$. I'll justify later the existence of $A$ by a counting argument.

So we have an exact sequence $$0\to\mathbf{Z}^2\to A\to\mathbf{Q}_p/\mathbf{Z}_p\to 0.$$ Tensoring (over $\mathbf{Z}$) with $\mathbf{Z}_p$ (which is flat), we get $$0\to\mathbf{Z}_p^2\to \mathbf{Z}_p\otimes A\to\mathbf{Q}_p/\mathbf{Z}_p\to 0,$$ so $\mathbf{Z}_p\otimes A$ is not finitely generated (it can actually be shown to be isomorphic to $\mathbf{Z}_p\times\mathbf{Q}_p$).

Here's now the counting argument: (a) the set of f.g. subgroups of $\mathbf{Z}[1/p]^2$ is countable (clear). (b) the set of subgroups $A$ of $\mathbf{Z}[1/p]^2$ with $\bigcap p^nA\neq 0$ is countable: indeed each subgroup is image by some element of the countable group $\mathrm{GL}_2(\mathbf{Z}[1/p])$ of a subgroup containing $\mathbf{Z}[1/p]\times\{0\}$, and such subgroup are in bijection with the set of subgroups of $\mathbf{Z}[1/p]$, which is countable (as it has countably many f.g. subgroups, countably many finite index subgroups, and they exhaust). (c) The set of subgroups of $\mathbf{Z}[1/p]^2$ is uncountable: for the set of subgroups of $\mathbf{Z}[1/p]^2/\mathbf{Z}^2$ is uncountable. This can be seen since this group is isomorphic to $\mathbf{Q}_p^2/\mathbf{Z}_p^2$, and the map mapping a 1-dimensional subspace of $\mathbf{Q}_p^2$ to $(D+\mathbf{Z}_p^2)/\mathbf{Z}_p^2$ yields an injection of $\mathbb{P}^1(\mathbf{Q}_p)$ into the set of subgroups of $\mathbf{Q}_p^2/\mathbf{Z}_p^2$, and hence of $\mathbf{Z}[1/p]^2$. (d) Combining, we get that $\mathbf{Z}[1/p]^2$ has uncountably many infinitely generated subgroups with $\bigcap p^nA=0$. (If we wish to avoid this non-explicit proof, a more careful argument shows that whenever $D$ is not in the countable set $\mathbb{P}^1(\mathbf{Q})$, the subgroup of $\mathbf{Z}[1/p]^2$ it defines by the previous construction satisfies this condition.)


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