Proving equivalent versions of faithfully flatness.

Question

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.

Answer 1

1. $Ax\simeq A/\operatorname{ann}(x)$ and $A/I\otimes_A M\simeq M/IM$, so $Ax\otimes_AM=0$ iff $M=\operatorname{ann}(x)M$.

2. $(U/V)\otimes M\simeq (U\otimes M)/(V\otimes M)$ and $(\ker g)\otimes M=\ker g_M$.

Answer 2

For your second question, I really like Exercise 1.6.H in Vakil's notes, which deals with the interaction between left/right exact functors and homology. I think the hint there is pretty generous and you'll learn three useful facts. The upshot is that tensoring by $M$ "preserves" kernels and cokernels, and one has $\operatorname{im} f = \ker(\operatorname{coker} f)$ so images are preserved as well. The exact sequences on the preceding page are fundamental.