Homomorphisms and Tensor Products

Question

Suppose $A$ and $B$ are finitely generated $\mathbb{Z}$-modules. Then if $ \operatorname{Hom}_\mathbb{Z}(A,B) \neq 0$ and $ \operatorname{Hom}_\mathbb{Z}(B,A)=0$ then $B \otimes \mathbb{Q} =0 $ and $A\otimes \mathbb{Q} \neq 0$. Can u guys help me out. I've tried using the fact that any element in the tensor product is of the form $\sum b\otimes q$ and the properties of the module homomorphisms but I'm kinda stuck , should I use the fact they $A,B \cong \mathbb{Z}/(d_1)\oplus\cdots \oplus \mathbb{Z}/(d_k)$ ?

Answer 1

Hint: You may write $$ A \cong \bigoplus_{j=1}^m A_j, \qquad B \cong \bigoplus_{k=1}^n B_k$$ where $A_j, B_j$ are either isomorphic to $\mathbb{Z}^{d_j}$ or $\mathbb{Z}/p_j^{r_j}\mathbb{Z}$. Using $$\operatorname{Hom}_{\mathbb{Z}} \left(\bigoplus_{j=1}^m A_j, \bigoplus_{k=1}^n B_k\right) \cong \bigoplus_{1\leq j \leq m,\ 1\leq k \leq n} \operatorname{Hom}_{\mathbb{Z}}(A_j, B_k)$$ it suffices to check when $\operatorname{Hom}_{\mathbb{Z}}(A_j, B_k) $ is trivial (or not).

After that we can use $$ A \otimes \mathbb{Q} \cong \bigoplus_{j=1}^m (A_j \otimes \mathbb{Q}), \qquad B \otimes \mathbb{Q} \cong \bigoplus_{k=1}^n (B_k \otimes \mathbb{Q}) $$ Furthermore, we have for $d\neq 0$ (just use the fact that $[a]\otimes c= [da]\otimes (c/d)=0$) $$ (\mathbb{Z}/d \mathbb{Z}) \otimes \mathbb{Q} \cong 0$$ and (using $z \otimes q = 1\otimes (zg)$) $$ \mathbb{Z} \otimes \mathbb{Q} \cong \mathbb{Q} \qquad \text{and thus} \qquad \mathbb{Z}^d \otimes \mathbb{Q} \cong \mathbb{Q}^d.$$