Question regarding faithfully flat module Let $\phi: R \rightarrow S$ be a ring homomorphism. Prove that $S$ is faithfully flat $R$-module  iff $\phi$ is injective and $S/\phi(R)$ is flat R-module
Although I am able to prove faithfully flat implies the injectivity, I am unable to prove $S/\phi(R)$ is flat $R$-module. So, any help will be appreciated. Thanks 
 A: Consider the short exact sequence of $R$-modules 
$$0\rightarrow R\xrightarrow{\phi}S\rightarrow \frac{S}{\phi(R)}\rightarrow 0$$ Fix any $R$-module $M$. Then we get a long exact sequence in homology $$...\rightarrow Tor_2^R\left (\frac{S}{\phi(R)},M\right )\rightarrow Tor_1^R(R,M)\rightarrow Tor_1^R(S,M)\rightarrow Tor_1^R\left( \frac{S}{\phi(R)},M\right)\rightarrow M \rightarrow M\otimes S\rightarrow M\otimes \frac{S}{\phi(R)}\rightarrow 0 $$ We know $Tor_1^R(S,M)=Tor_1^R(R,M)=0$ . 
Since $S$ is faithfully flat we also know $M \xrightarrow{m\mapsto m\otimes 1} M\otimes S$ is an injective map. Thus the long exact sequence gives us $$Tor_1^R\left( \frac{S}{\phi(R)},M\right)=0$$ for any $R$-module $M \implies \frac{S}{\phi(R)} $ is a flat $R$-module. 
Edit:
Since $S$ is faithfully flat we know $N\otimes S=0\implies N=0$
Suppose $m\otimes 1=0$ in $M\otimes S\implies m\otimes s=0$ in $M\otimes S$ by considering the map $id\otimes \lambda_s :M\otimes S\rightarrow M\otimes S$ where $\lambda_s$ is the map multiplication by $s$.
Thus we can conclude $Rm\otimes S=0\implies Rm=0\implies m=0$
