Equivalent definitions of faithful flatness I am currently studying the equivalent definitions of faithful flatness over (probably noncommutative) unital rings. In particular, there is a version that I have doubts about:

Let $R \subseteq S$ be unital rings, and regard $S$ as a right $R$-module. Then $S$ is faithfully flat over $R$ $\iff$ $S$ is flat over $R$, and for any left maximal ideal $I$ in $R$, $S\neq SI$.

My question is that can we replace part of the condition on the right by a somehow weaker condition, that it holds for all finitely generated left maximal ideal instead?
I have investigated some examples on it (of course assuming $R$ are not Noetherian), but they are not quite useful as I can only think of a few left maximal ideals which is not finitely generated. Is there any good example to think about? 
 A: If $R$ is a local commutative integral domain (not a field) with infinitely generated maximal ideal, and $S$ is its field of fractions, then $S$ is flat but not faithfully flat over $R$, but since $R$ has no finitely generated maximal ideals the "finitely generated" version of $S\neq SI$ is vacuously satisfied.
A: After some works, I found that the problem lies in the fact that finitely generated maximal ideal might not exist in every ring, as demonstrated by Jeremy. However, finitely generated proper ideal does exist in any nonzero ring (like the zero ideal). Then if we replace every "maximal" by "proper" in the statement (and specifying that the rings are nonzero, although it should be clear if we call them unital rings), it would not become a vacuous statement. More importantly, it can be shown by the definition of faithful flatness that the statement holds even with such replacement. That means

Let $R \subseteq S$ be unital rings, and regard $S$ as a right $R$-module. Then $S$ is faithfully flat over $R$ $\iff$ $S$ is flat over $R$, and for any left proper ideal $I$ in $R$, $S\neq SI$.

If so, then it seems that the assertion really holds (i.e. it suffices to consider all the finitely generated left proper ideals only). More precisely, the following statement holds:

Let $R \subseteq S$ be unital rings, and regard $S$ as a right $R$-module. Then $S\neq SI$ for any left proper ideal $I$ in $R$ $\iff$ $S\neq SI$ for any finitely generated left proper ideal $I$ in $R$.

$(\Rightarrow)$ is of course trivial. For $(\Leftarrow)$, if on the contrary $S = SI$ for some left proper ideal of $R$. Then $$\displaystyle\sum_{k=1}^{n}s_k t_k = 1$$ for some $s_k \in S, t_k \in I$. Then let $J$ be the left ideal of R generated by $t_k$'s. Then As $J \subseteq I$, it is proper and finitely generated by definition. As $1 \in SJ$, $S = SJ$, which contradicts with the assumption. Hence $S \neq SI$.
The same proof fails in the maximal case, as $J$ might not be a maximal ideal.
