Suppose that $B$ is a flat $A$-algebra, where $A$ is a commutative ring. I would like to show that iv) $\Rightarrow$ v), where iv) and v) are as follows :
iv) If $M$ is any non-zero $A$-module, then $M_B \neq 0$.
v) For every $A$-module $M$, the mapping $x \mapsto 1 \otimes x$ of $M$ into $M_B$ is injective.
I've started a proof (shown below) that I'd like to finish :
We will show that iv) $\Rightarrow$ v) by the contrapositive. Suppose that v) is false. That is, suppose that there exists an $A$-module $M$ such that the mapping $x \mapsto 1 \otimes x$ of $M$ into $M_B$ is not injective. Then the kernel of this mapping is not trivial, so there exists a non-zero $x \in M$ such that $1 \otimes x = 0$ in $M_B$. Since the inclusion map $\iota:Ax \hookrightarrow M$ is injective and $B$ is a flat $A$-module, the map $\iota \otimes 1: Ax \otimes_A B \rightarrow M \otimes_A B = M_B$ is injective.
How can I finish this proof ? We need to show that iv) is false to finish the proof by contrapositive. To do this, my idea was to show that $(Ax)_B = Ax \otimes_A B = 0$, since $Ax$ is a non-zero $A$-module.
Edit : If I can show that $\text{Im}(\iota \otimes 1)$ is zero, then I believe that I am done.
Thanks !