Absolutely Flat Ring A commutative ring $\mathcal R$ with identity is said to be absolutely flat if every $\mathcal R$-module is flat. I know that for an absolutely flat ring, every principal ideal is idempotent. Also, know a proof of converse using Tor functor. Is there any proof of converse without using properties of $\text{Tor}$ functor?
 A: Method 1:  We just need to know that being flat is a local property, and that every module over a field is free (that's a vector space!) and a fortiori flat.  First show that principal ideals being generated by idempotents implies that $R$ is locally a field, and then the conclusion follows immediately.  
Method 2: At first glance this might seem unnecessarily complicated, but I prefer this argument because it generalizes well to characterizations of rings of fixed weak global dimension.  
In the following approach we only need to know a few things about flatness: (1) its definition, i.e. tensoring preserves injectivity (2) that colimits of directed systems of flat modules are flat (3)$^{1}$ that if $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is exact with $A,C$ flat, then $B$ is flat.  
First we reduce the problem:

Let $R$ be a ring.  Every module of $R$ is flat iff every cyclic module $R/I$ is flat.  

Proof: First reduce to finitely generated modules, since every module is a colimit of its finitely generated submodules, and since a colimit of a directed system of flat modules is flat.   
To reduce to cyclic modules, recall that any finitely generated module $M$ has a filtration $0 \subseteq M_1 \subseteq \cdots \subseteq M_n = M$ such that for any $i>0$ we have $M_{i}/M_{i-1} \cong R/I_{i}$ for some ideal $I_i$.  (Note that any cyclic module $N$ is isomorphic to $R/\operatorname{Ann}_R(N)$, and take $M_i$ to be the module generated by the first $i$ generators of $M$).    
Thus we have $M_1 \cong R/I_1$ flat, and inductively we show that $M_i$ is flat by considering the short exact sequence $0 \rightarrow M_{i-1} \rightarrow M_{i} \rightarrow R/I_{i} \rightarrow 0$.  $\square$ 

If every principal ideal of $R$ is idempotent, then every finitely generated ideal is principally generated by an idempotent.

Proof: This follows from the observation that if $e,f$ are idempotents then $g = e + f - ef$ is an idempotent and $ge = e, gf = f$, hence $(e,f) = (g)$.  $\square$
All that remains is to prove that

If every principal ideal of $R$ is generated by an idempotent, then every cyclic module $R/I$ must be flat.  

Proof
Firstly, it is easy to see that the module $R/eR$ is flat for any idempotent $e$.  (Check this locally -- since the only idempotents of local rings are $0$ and $1$, the ideal $eR$ becomes free of rank either $1$ or $0$, and likewise for $R/eR$). Thus in our ring $R/I$ is flat for any f.g. ideal $I$. 
Now for any ideal $I$, $R/I$ is the colimit of the directed system of $R/I_\alpha$ where $I_\alpha$ are the finitely generated sub-ideals of $I$ and the maps of the directed system are the canonical projections $R/I_\alpha \twoheadrightarrow R/I_\beta$ for $I_\alpha \subseteq I_\beta$.  $\square$
$^{1}$ (A note about (3).  Sometimes this is done with the long exact sequence for Tor, or by developing more sophisticated tests for flatness, but it can also be shown in two breezy diagram chases.  The first diagram chase is in showing that if $\mathcal{C} := 0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ is exact and $C$ is flat, then tensoring with any module $M$ is exact.  An easy approach comes from embedding $M$ in an exact sequence $\mathcal{C}' := 0 \rightarrow K \rightarrow F \rightarrow M \rightarrow 0$ with $F$ free and chasing the diagram $\mathcal{C} \otimes \mathcal{C'}$. To get the desired result (3), we then chase the diagram obtained by tensoring $0 \rightarrow N \rightarrow M$ with $0 \rightarrow A \rightarrow B \rightarrow C \rightarrow 0$ (where $A, C$ flat) and utilizing the previous fact.)
