# infinitely $p$-divisible elements in $A\otimes \mathbb{Z}_p$

Let $A$ be a (possibly non-finitely generated) torsion-free abelian group. Suppose that $A$ contains no infinitely $p$-divisible elements, then does the same hold for $A\otimes \mathbb Z_p$, where $\mathbb Z_p$ denotes the $p$-adic integers?

Edit: This question has been put on hold for not providing enough context, so let me give some. I encountered the problem in algebraic geometry, namely: If $A,B$ be abelian varieties, then it is well-known that the map $\mathrm{Hom} (A,B)\otimes \mathbb Z_p\rightarrow\mathrm{Hom}(T_pA,T_pB)$ is injective for all $p\neq \mathrm{char}k$, where $k$ is the ground field. I noticed that the proof of this fact would be much easier if the answer to the question were positive, which intuitively seemed plausible to me. (Of course $\mathrm{Hom} (A,B)$ is in fact finitely generated, but this is a corollary and can't be assumed in the proof). I will admit though that I don't have much confidence in my intuition in infinitely-generated situations.

Here are a few thoughts I had about some intrinsic (commutative-algebra) motivation to the question: For any abelian group, we can consider the $p$-adic topology on it, just as for any $R$-module $M$ we can consider the $I$-adic topology, where $R$ is a commutative ring, and $I$ an ideal. For finitely generated modules over noetherian rings, there is a nice theory of completions. The completion $\hat M$ is itself complete and coincides with the base change to the completed ring: $\hat M\cong \hat R\otimes M$.

For non-finitely-generated modules, we can still consider the $I$-adic topology, but it doesn't seem to be very well-behaved: the $p$-adic topology in this sense on $\mathbb Q$ as a $\mathbb Z$-module is trivial. But there is another way of putting a topology on these modules: consider them as a direct limit over their finitely generated submodules, endow these with the $I$-adic topology, and look at the limit topology on $M$. In this way, $\mathbb Q$ attains the "right" $p$-adic topology.

Since the completion of a noetherian ring is flat over the ring itself, we can also make $M\otimes \hat R$ into a topological $\hat R$-module in the same way, by writing it as the direct limit of finitely generated submodules of $M$, each tensored with $\hat R$. In the example of before we get, unsurprisingly, $\mathbb Q_p$.

So now a natural question is how these objects all fit together: We have $M$ with the naive $I$-adic topology, $M$ with the limit topology, and $M\otimes \hat R$. Is $M\otimes \hat R$ the completion of $M$ with the limit topology, and is said completion itself complete? Here the completion may be defined as a projective limit over $M/U$, where $U$ ranges through open subgroups containing 0.

In the question, $R=\mathbb{Z}$, $I=(p)$ and $M$ contains no infinitely $p$-divisible elements, which I think should make the naive and the limit topology on $M$ coincide and the whole situation can probably be regarded as the easiest case. If the completion is to be equal to $M\otimes \mathbb Z_p$, then there had better not be any infinitely $p$-divisible elements in there. I think this would be equivalent to the statement that $M\otimes \mathbb Z_p$ with the naive $p$-adic topology is Hausdorff.

So this is more than I thought I'd write and given the speculative nature of the thing, I'm bound to have made some embarassing mistake along the way.. In any case, I'd also appreciate references about completions and the like for infinitely-generated modules. And given the above discussion, I'm also putting the commutative-algebra tag back in.

• If there are no answers, you can post your question also on mathoverflow.net. That's a side for research level maths where many hard questions had been answered that remained unanswered on math.stackexchange.com. – Todd Leason Apr 3 '16 at 18:27
• The answer is no. If I get some time I will write up an answer. – H.Durham Aug 23 '16 at 13:45

This is not true in general. For instance, let $A=\mathbb{Z}_p$. Pick any element $a\in \mathbb{Z}_p\setminus\mathbb{Z}_{(p)}$ and consider $x=1\otimes a-a\otimes 1\in \mathbb{Z}_p\otimes\mathbb{Z}_p$. Note that $x\neq 0$, since $a$ and $1$ are linearly independent over $\mathbb{Z}$. But I claim $x$ is infinitely divisible by $p$.
Indeed, for any $n$, we have $(\mathbb{Z}_p\otimes\mathbb{Z}_p)\otimes \mathbb{Z}/(p^n)\cong \mathbb{Z}_p\otimes\mathbb{Z}/(p^n)\cong\mathbb{Z}/(p^n)$. Under this isomorphism, both $1\otimes a$ and $a\otimes 1$ map to the residue of $a$ mod $p^n$. It follows that $x$ maps to $0$, which means exactly that $x$ is divisible by $p^n$.
(If you prefer a more explicit computation, write $a=b+p^nc$ where $b\in\mathbb{Z}$ and $c\in\mathbb{Z}_p$. Then $x=1\otimes b+p^n(1\otimes c)-b\otimes 1-p^n(c\otimes 1)$. Since $b\in \mathbb{Z}$, $1\otimes b=b\otimes 1$, so $x=p^n(1\otimes c-c\otimes 1)$.)
• This may be a naive question, but why is $1$ and $a$ being linearly independent imply that $1\otimes a - a\otimes 1 \neq 0$? I'm trying to use something like Martin's answer in math.stackexchange.com/questions/288431/… but I can't see the end of it – Andrei.B Oct 28 '18 at 3:18
• @Andrei.B: Consider the subgroup $A\subset\mathbb{Z}_p$ generated by $1$ and $a$. Since $A$ and $\mathbb{Z}_p$ are both torsion-free and hence flat, the natural map $A\otimes A\to\mathbb{Z}_p\otimes\mathbb{Z}_p$ is injective (it is the composition of the two injective maps $A\otimes A\to A\otimes\mathbb{Z}_p$ and $A\otimes\mathbb{Z}_p\to \mathbb{Z}_p\otimes\mathbb{Z}_p$, the first obtained by tensoring the inclusion map with $A$ on the left and the second obtained by tensoring the inclusion map wiht $\mathbb{Z}_p$ on the right). But now, since $1$ and $a$ are linearly independent... – Eric Wofsey Oct 28 '18 at 3:38
• ...$A$ is freely generated by $1$ and $a$. It follows that $A\otimes A$ is freely generated by $1\otimes 1,a\otimes 1,1\otimes a,$ and $a\otimes a$. In particular, $1\otimes a-a\otimes 1$ is nonzero in $A\otimes A$, and hence also in $\mathbb{Z}_p\otimes\mathbb{Z}_p$. – Eric Wofsey Oct 28 '18 at 3:39