I was reading a proof of the the following theorem from Matsumura (p.47)
There was something confusing about $(3) \implies (2)$ and $(2) \implies (1)$.
Question 1
Here, it says $M \not= \text{ann(x)}M \implies Ax \otimes M \not=0$. This is what I was thinking,
Since $M \not= \text{ann(x)}M$, we know that there exists $m \in M$ which cannot be written in the form $m=rn$ for some $r \in \text{ann}(x)$. So, we have a nonzero element $1+ \text{ann}(x) \otimes m$. Am I correct?
However, this leads me to ask another question. Suppose we have a ring $R$ and two $R$-modules $U$ and $V$. Let $u \in U, v \in V$, where $u$ is not a multiple of any element contained in the annihilator of $v$ and $v$ is not a multiple of any element contained in the annihilator of $u$. Is it possible to conclude that $u \otimes v$ is a nonzero element in $U \otimes V$? The answer must be wrong, since for $R = \Bbb{Q}$, we have $1 \otimes 1 \in \Bbb{Q} \otimes \Bbb{Z}_3$. But $1 \otimes 1 = (\frac{3}{3})(1 \otimes 1) = \frac{1}{3} \otimes 0 = 0$. So why are we assuming this in the proof (if I understood the proof correctly)?
Question 2
I was stuck on the last part, where it says $H \otimes M = Ker(g_M)/Im(f_M)$ by the flatness of $M$. So I probably need to find an exact sequence of $A$-modules which is going to be exact when you tensor it with $M$ (since $M$ is flat), which will imply the result...but I couldn't really see how to do that.