$0$ element of $A\otimes \mathbb{Q}$ I am studying tensor product and I feel that it is not easy to distinguish $a\otimes b=0$ or $\neq 0$.
I'm thinking the following special case and it should be true.
Let $A$ be an abelian group, in other word a $\mathbb{Z}$ module, and  $a\otimes 1 \in A\bigotimes_{\mathbb{Z}} \mathbb{Q}$.
Then, $a\otimes 1 =0$ if and only if there exists some $0\neq n\in \mathbb{Z}$ such that $na=0$.
I think "if part" is trivial, but the other part is difficult. I want a proof of this claim or a counter examaple.
 A: Define the abelian group $A_0$ as follows:
The underlying set is $(A \times (\mathbb{Z} \setminus \{0\})) / \sim$, where $\sim$ is the equivalence relation defined by $(a, n) \sim (b, m) \iff \exists u \in \mathbb{Z} \setminus \{0\} (u(ma - nb) = 0)$. The group operation is defined by
$$[(a,n)] + [(b,m)] = [(ma+nb,nm)]$$
(check this is well-defined and gives an abelian group!)
In other words, $A$ is the abelian group of formal quotients $a/n$ where $a \in A$ in $n \in \mathbb{Z} \setminus \{0\}$. In even more different words, we have just constructed the localization $A_0$. By basic facts of commutative algebra, $A_0 \cong A \otimes_{\mathbb{Z}} \mathbb{Z}_0 = A \otimes_{\mathbb{Z}} \mathbb{Q}$. I'll explain this isomorphism, then say how it solves the problem.
Claim $A_0 \cong A \otimes_{\mathbb{Z}} \mathbb{Q}$.
Proof Sketch. First, we define $[(a,n)] \mapsto a \otimes 1/n : A_0 \to A \otimes_{\mathbb{Z}} \mathbb{Q}$ -- check this is well-defined and a homomorphism. Next, we construct the inverse by the universal property of tensor products. Define the function $(a,p/q) \mapsto [(pa,q)] : A \times \mathbb{Q} \to A_0$. Check that this is $\mathbb{Z}$-bilinear, so it yields a homomorphism $A \otimes_{\mathbb{Z}} \mathbb{Q} \to A_0$ determined by $a \otimes p/q \mapsto [(pa,q)]$. Check that this is inverse to the homomorphism we constructed previously.
Corollary $a \otimes 1 = 0$ in $A \otimes_{\mathbb{Z}} \mathbb{Q}$ if and only if $na = 0$ for some $n \in \mathbb{Z} \setminus \{0\}$.
Proof. The element $a \otimes 1 \in A \otimes_{\mathbb{Z}} \mathbb{Q}$ corresponds to $[(a,1)] \in A_0$ via the aforementioned isomorphism, so $a \otimes 1 = 0$ if and only if $[(a,1)] = 0 = [(0,1)]$. By definition, this is equivalent to the existence of some $n \in \mathbb{Z} \setminus \{0\}$ such that $na = n(1 \cdot a - 1 \cdot 0) = 0$, as desired.
A: When you tensor over $\mathbb{Z}$, you can pass integers over the tensor. As you seem to be alluding to, this allows you to annihilate torsion elements from $A$. Specifically, if $an = 0 \in A$, then
$$
a \otimes 1 
= a \otimes \bigl( n \cdot \tfrac{1}{n} \bigr) 
= (an) \otimes \tfrac{1}{n} 
= 0 \otimes \tfrac{1}{n} 
= 0 \otimes 0 \in A \otimes \mathbb{Q}
$$
How do you characterize a nonzero element $a \in A$ that is not torsion? If $na = 0 \in A$ then $n=0$ (linear independence of the singleton set $\{a\}$). So every pure tensor looks like
$$
a \otimes 1 = (an) \otimes \tfrac{1}{n}, 
$$
but $an \neq 0 \in A$ and $\tfrac{1}{n} \neq 0 \in \mathbb{Q}$, so this is not the zero element in $A \otimes \mathbb{Q}$.
A: The proof is to explicitly construct the localization $M = (\mathbf{Z} - \{0\})^{-1}A$. This carries a $\mathbf{Q}$-vector space structure. Then there is an obvious $\mathbf{Z}$-bilinear mapping $A \times \mathbf{Q} \to M$ carrying $(a,1)$ to a nonzero element.
A: Here is another solution: Tensor the exact sequence $0 \rightarrow (a) \rightarrow A \rightarrow A / (a) \rightarrow 0$ with $\mathbb{Q}$,  which is flat over $\mathbb{Z}$. $(a)$ is the subgroup generated by $a$.
If such an $n$ exists, then $a \otimes_{\mathbb{Z}} \mathbb{Q} \sim (\mathbb{Z}/n\mathbb{Z}) \otimes_{\mathbb{Z}} \mathbb{Q} = 0$. Therefore, $a \otimes 1$ (which is in the image of $(a) \otimes_{\mathbb{Z}} \mathbb{Q} \rightarrow A \otimes_{\mathbb{Z}} \mathbb{Q}$) must be 0.
If no such $n$ exists, then $(a) \sim \mathbb{Z}$ , so $(a) \otimes_{\mathbb{Z}} \mathbb{Q} \sim \mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Q} \sim \mathbb{Q}$, with the isomorphism function being $na \otimes p/q = np /q$. Under this isomorphism, $1 \in \mathbb{Q}$ corresponds to $a \otimes 1$, so $a \otimes 1$ is not 0 in $(a) \otimes_{\mathbb{Z}} \mathbb{Q}$. Therefore, by injectivity of $(a) \otimes_{\mathbb{Z}} \mathbb{Q} \rightarrow A \otimes_{\mathbb{Z}} \mathbb{Q}$, its image in $A \otimes_{\mathbb{Z}} \mathbb{Q}$ is not 0 either.
