I'm studying torsion-free abelian groups and I know (see Fuchs, "Infinite Abelian Groups", vol. $2$, pp $154$) that, if $\mathbb{Z}_p$ is the set of $p$- adic integers and $\mathbb{Z}_{(p)}$ denotes the subset of $\mathbb{Q}$ whose elements have denominator prime to $p$, then the following hold: for any torsion-free abelian group $A$ and $p$ prime number $$A=\bigcap_p \, (\mathbb{Z}_{(p)}\otimes A)$$ and $$\mathbb{Z}_{(p)}\otimes A=(\mathbb{Q}\otimes A)\cap (\mathbb{Z}_p\otimes A)$$.

Now, I'm considering the following question: can I say more (eventually assuming that $A$ has finite rank, say $n$)?

I mean that, using the facts I have stated above, $$A=(\mathbb{Q}\otimes A) \cap\bigcap_p (\mathbb{Z}_p\otimes A)$$ $$\subseteq \mathbb{Q}^{rank(A)}\cap\bigcap_p (\mathbb{Z}_p\otimes A)$$.

Does the equality hold at least in the finite rank case or this is always a strict inclusion?

Furthermore, can I write $A=\mathbb{Q}^n\cap(\mathbb{Z}_p\otimes A)$ in the finite rank case?

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    $\begingroup$ To take the intersection of groups, you need them all to be subgroups of a common group. The facts you quote from Fuchs make sense because in the first case the groups being intersected can all be naturally regarded as subgroups of $\mathbb{Q}\otimes A$, and in the second case of $\mathbb{Q}_p\otimes A$. But in which group do you want to take the intersection $(\mathbb{Q}\otimes A)\cap\bigcap_p(\mathbb{Z}_p\otimes A)$? $\endgroup$ – Jeremy Rickard Feb 21 at 7:59
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    $\begingroup$ Well, if $A$ has rank $n$, then $\mathbb{Q}\otimes A\cong \mathbb{Q}^n$ as abelian groups. However, to make sense of $\mathbb{Q}^n\cap(\mathbb{Z}_p\otimes A)$ as the intersection of two subgroups of $\mathbb{Q}_p\otimes A$, you need to know which subgroup $\mathbb{Q}^n$ is meant to be, and $\mathbb{Q}_p\otimes A$ has many subgroups isomorphic to $\mathbb{Q}^n$ which have very different intersections with $\mathbb{Z}_p\otimes A$. $\endgroup$ – Jeremy Rickard Feb 21 at 11:20
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    $\begingroup$ The fact that there are lots of subgroups isomorphic to $\mathbb{Q}^n$ is just linear algebra: $\mathbb{Q}_p\otimes A$ is an infinite dimensional vector space over $\mathbb{Q}$, so has lots of $n$-dimensional subspaces. If $A$ is $p$-local, you already know that $A = (\mathbb{Q}\otimes A)\cap(\mathbb{Z}_p\otimes A)$, regarding all the groups concerned as subgroups of $\mathbb{Q}_p\otimes A$. To decide whether $A=\mathbb{Q}^n \cap(\mathbb{Z}_p\otimes A)$ you need to know which copy of $\mathbb{Q}^n$ inside $\mathbb{Q}_p\otimes A$ you mean ... $\endgroup$ – Jeremy Rickard Feb 22 at 11:54
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    $\begingroup$ ... If the copy you mean is $\mathbb{Q}\otimes A$, then the equality is true, but it's just a restatement of what you already know. $\endgroup$ – Jeremy Rickard Feb 22 at 11:54
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    $\begingroup$ @JeremyRickard I see what you mean: your explanation is very plain $\endgroup$ – LBJFS Feb 22 at 12:13

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