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So I'm trying to prove that if $G$ is an Abelian Group of rank $r$ (As $\mathbb{Z}$-module) then $G\otimes_\mathbb{Z}\mathbb{Q}$ is isomorphic to $\mathbb{Q}^r.$

Using the results I know about factorization of modules over PIDs, and the fact that the tensor product is distributive in respect to the direct sum, I restricted the problem to prove that $\frac{\mathbb{Z}}{(p^i)}\otimes_\mathbb{Z}\mathbb{Q}$ is isomorphic to $\mathbb{Q}$; where $p$ is a prime number.

Now, I can see how that tensor product has a Surjective morphism in $\mathbb{Q}$, and by writing the following short exact sequence

$$0 \longrightarrow (p^i) \longrightarrow \mathbb{Z} \longrightarrow \frac{\mathbb{Z}}{(p^i)} \longrightarrow 0$$

And tensorising with $\mathbb{Q}$ (which is flat, even though I wouldn't need the info since I only care about the end of the sequence) I get

$$0 \longrightarrow (p^i)\otimes_\mathbb{Z}\mathbb{Q} \longrightarrow \mathbb{Z}\otimes_\mathbb{Z}\mathbb{Q} \longrightarrow \frac{\mathbb{Z}}{(p^i)}\otimes_\mathbb{Z}\mathbb{Q} \longrightarrow 0$$

which is isomorphic to

$$0 \longrightarrow (p^i)\otimes_\mathbb{Z}\mathbb{Q} \longrightarrow \mathbb{Q} \longrightarrow \frac{\mathbb{Z}}{(p^i)}\otimes_\mathbb{Z}\mathbb{Q} \longrightarrow 0$$

So I have two surjections in $\mathbb{Z}$-$\mathsf{Mod}$:

$$\mathbb{Q} \rightarrow \frac{\mathbb{Z}}{(p^i)}\otimes_\mathbb{Z}\mathbb{Q}$$ $$\frac{\mathbb{Z}}{(p^i)}\otimes_\mathbb{Z}\mathbb{Q} \rightarrow \mathbb{Q}$$

My question is whether this is enough to say that the two are isomorphic or not. I know this doesn't work on $\mathsf{Set}$ and intuitively I can see this not being enough on generic modules if the operations were made to fit in the same counterexamples that we can make on $\mathsf{Set}$, but I might be wrong and the operation might be too restrictive.

(EDIT: It does work on $\mathsf{Set}$, sorry, I forgot about Schröder–Bernstein Theorem and I thought about it shallowly. This said I also know that having an inverse in $\mathsf{Set}$ of a Module homomorphism is enough to say that the inverse is an Homomorphism itself, thus saying that they both are isomorphisms. The problem is that I don't see a way to build an inverse in set even using the surjective version of Schröder–Bernstein. Also as some kind soul already made me notice any counterexemple shouldn't be finitely generated.)

Also, whether this works or not may I ask for a tip or two on how would you solve this problem?

EDIT: So, the person who wrote the exercise forgot to specify that $G$ was free, so the particular problem is solved and the morphism $$\frac{\mathbb{Z}}{(p^i)}\otimes_\mathbb{Z}\mathbb{Q} \rightarrow \mathbb{Q}$$ cannot be the canonic projection over $\mathbb{Q}$, so the abstract question doesn't apply in this case. So the only thing left is to know if, in general, given:

$$f: M \rightarrow N \\ g: N \rightarrow M$$ both $R$-Module surjective homomorphism ($R$ unitary commutative ring) there exist an isomorphism between the two or not necessarily.

Gnocchi
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  • $\frac{\mathbb{Z}}{(p^i)}\otimes_\mathbb{Z}\mathbb{Q}\cong 0$ since $m\otimes n=m\otimes p^i\frac{n}{p^i}=p^im\otimes \frac{n}{p^i}=0$, right? Am I missing something? – cqfd May 26 '20 at 14:31
  • That sounds correct actually, but this make things even more confusing, since $G\cong \mathbb{Z}^s \oplus tG$ where $tG$ is the torsion part of $G$, and $s \leq r$. but $tG$ is direct sum of stuff in the form of $\frac{\mathbb{Z}}{(p^i)}$. I mean, if the tensor product distribute over the direct sum, it should eat away all the torsion and I have no bound over G to not have it have any torsion: $s$ can be strictly less than $r$, am I wrong? – Gnocchi May 26 '20 at 14:43

1 Answers1

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Since $G$ is of free rank $r$, we have $G\cong \mathbb{Z}^{r}\oplus\bigoplus_{k=1}^{m}\dfrac{\mathbb{Z}}{(q_k)}$. Thus, \begin{align} G\otimes_{\mathbb Z}\mathbb Q &\cong\left(\mathbb{Z}^{r}\oplus\bigoplus_{k=1}^{m}\dfrac{\mathbb{Z}}{(q_k)}\right)\otimes_{\mathbb Z}\mathbb Q\\ &\cong (\mathbb{Z}^{r}\otimes_{\mathbb Z}\mathbb Q)\oplus\bigoplus_{k=1}^{m}\dfrac{\mathbb{Z}}{(q_k)}\otimes_{\mathbb Z}\mathbb Q\\ &\cong \mathbb Q^r\oplus 0\cong \mathbb Q^r. \end{align} I have used the isomorphisms $\mathbb{Z}\otimes_{\mathbb Z}\mathbb Q=\mathbb Q$ and $\dfrac{\mathbb{Z}}{(q_k)}\otimes_{\mathbb Z}\mathbb Q=0$.

A partial answer to your second question. Recall(?) that a surjective $R$-linear endomorphism $T: M\to M$ of a finitely generated $R$-module $M$ is an isomorphism(click here for a proof).

Suppose that $M$ and $N$ are finitely generated $R$-modules and $f:M \to N$ and $g:N \to M$ are surjective $R$-linear maps. Then $f\circ g: N\to N$ and $g\circ f: M\to M$ are surjective $R$-linear maps and hence isomorphisms. Thus we see that $f$ and $g$ are injective as well. So $M$ and $N$ are isomorphic. Now you know where to look for counterexamples.

cqfd
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    Unfortunately I see that there is a lot of different definition of rank of a non-free module. We used the term referred to the minimum number of generators, not of free generators, but since it's possible that the exercise don't follow the same notation we used during the rest of the course I contacted the professor in regard. If the definition is the one you used this way definitely works and was what I obtained after your first answer. – Gnocchi May 26 '20 at 15:19
  • Hmm, I see. My answer might be wrong because I got the rank and the free rank(see Dummit and Foote for the definition) mixed up. Anyway, I will wait until your professor clarifies the question. – cqfd May 26 '20 at 16:09
  • Nevermind, they forgot to say that G was free. Let's leave the thread only for the abstract question about surjective maps – Gnocchi May 26 '20 at 17:36