4

I would like to know the following "local-global" principle holds (all the tensors are taken over $\Bbb Z$):

Let $A,B$ be two abelian groups. Assume that $f \otimes \text{id}_{\Bbb Q}$ and $f \otimes \text{id}_{\Bbb{F}_p}$ are isomorphisms (for all primes $p$). Is it true that $f:A\to B$ is a group isomorphism? Does it become true if we assume that $A$ and $B$ are finitely generated?

The converse is trivial, since $- \otimes M$ is a functor for every abelian group $M$. My question is somehow related to the notion of conservative functor or functors reflecting isomorphisms.

(Notice that in a general situation, if $f \otimes \text{id}_M : A \otimes_R M \to B \otimes_R M$ is an isomorphism for all $R$-modules $M$, then it follows that $f$ is an isomorphism because we just have to take $M=R$).

— I will assume here that $A,B$ are finitely generated. Then we can write $$A=\mathbb{Z}^{r} \oplus \bigoplus_{i=1}^{N} \mathbb{Z}/p_{i}^{a_{i}}\mathbb{Z}, \quad p_{i} \text{ prime numbers, } a_{i}, r \geq 0$$ $$B=\mathbb{Z}^{s} \oplus \bigoplus_{i=1}^{K} \mathbb{Z}/q_{i}^{b_{i}}\mathbb{Z}, \quad q_{i} \text{ prime numbers, } b_{i}, s \geq 0$$

Since $A \otimes \Bbb Q \cong B \otimes \Bbb Q$ and $A \otimes \Bbb F_p \cong B \otimes \Bbb F_p$ for all primes $p$, it follows that $A \cong B$ (because $r=s,K=N,p_i=q_i,a_i=b_i$). We know that there exists an isomorphism from $A$ to $B$, but I don't see why $f$ should be an iso.

— I believe that for the non-finitely generated case we could find sufficiently big abelian groups to make my statement wrong. A good way to start would be to find non isomorphic abelian groups $A,B$ such that $$A \otimes \Bbb Q \cong B \otimes \Bbb Q \text{ and }A \otimes \Bbb F_p \cong B \otimes \Bbb F_p \text{ for all primes } p.$$

— Here are some possibly related questions: Is the image of a tensor product equal to the tensor product of the images?, Tensor product and injective maps, What is the kernel of the tensor product of two maps?, Relation between $\operatorname{Coker}(f)$ and $\operatorname{Coker}(f \otimes 1_P)$.

Any suggestion will be appreciated!

Community
  • 1
Watson
  • 24,404

1 Answers1

7

Let $D$ be a divisible abelian group, and let $D'$ be a divisible torsion abelian group.

The inclusion $j:D\to D\oplus D'$ satisfies your requierements, because

  1. Since $D$ is divisible we know $\mathbb F_p\otimes D=D/pD=0$ for every prime $p$, and the same holds for $D'$.
  2. Since $D'$ is torsion, $j\otimes \mathbb Q$ is the map $1_D\otimes \mathbb Q$, which is an isomorphism.

For a concrete example, take the inclusion of $\mathbb Q$ in $\mathbb Q\oplus \mathbb Q/\mathbb Z$, or the inclusion $\mathbb Q/\mathbb Z$ in $\mathbb Q/\mathbb Z\oplus \mathbb Q/\mathbb Z$.

Observe that the functors $T_p=\mathbb F_p\,\otimes \;?$ "detects divisibility", in the sense an abelian group $A$ is divisible if and only if $T_pA=0$ for every prime $p$, while the functor $T=\mathbb Q\,\otimes \;?$ "detects torsion", in the sense an abelian group $A$ is torsion if and only if $TA=0$.

Pedro
  • 125,149
  • 19
  • 236
  • 403