Several questions about direct limit of rings Direct limit in $CRing$ is driving me mad these days.
I have difficulty verifying the proof of some theorems in $EGA1 new$. 


*

*How to construct the direct limit of commutative rings? Is there any standard reference?

*$EGA1 new$ 0.6.1.1 Let $A=\varinjlim A_i =0$, then $\exists k$ s.t. $\forall i \ge k, A_i=0$. 

*$EGA1 new$ 0.6.3.3 Let $A=\varinjlim A_i =0$, $M$ be an $A$-module of finite presentation. Then $\exists i,\exists M_i$ be an $A_i$-module s.t. $M=A \otimes_{A_i}M_i$.
The proof in  $EGA1 new$ is incomplete to me.
 A: *

*If $(R_i, \nu_{ij}:R_i\rightarrow R_j))_{i, j}$ is a directed system of commutative rings, consider the set 
$$R=(\coprod_i R_i)/\sim\;, $$
where $\sim$ is the equivalence on $\coprod_i R_i$ given by $$r_i \sim r_j, \;\; r_i \in R_i,\; r_j \in R_j \text{  iff  } \nu_{ik}(r_i)=\nu_{jk}(r_j) \text{ for some }k\geq i, j $$
(note that here $\coprod$ denotes the disjoint union of sets, not coproduct of comm rings or anything else).


Define operations of $R$ as follows: Given $[r_i], [r_j] \in R,$ consider $k \geq i, j$ and put $[r_i]+[r_j]=[\nu_{ik}(r_i)+\nu_{jk}(r_j)]$ and $[r_i]\cdot[r_j]=[\nu_{ik}(r_i)\cdot \nu_{jk}(r_j)]$ (the correctness of this definition should be checked, but it works). Of course then the unit is given by the class $[1], \;1 \in R_i$ for arbitrary $i$, and define similarly $0=[0]$. The result, together with the co-cone of morphisms $R_i \rightarrow R, ; r_i \mapsto [r_i]$ gives the direct limit.
(Remark: although this description of direct limits seems kind of unnatural (i.e. does not come naturally as a coequalizer of a coproduct in the cat. of comm. rings), it has the advanage that this construction works in any variety (in the universal algebraic sense), i.e. one can describe in a similar manner direct limits of groups / modules / monoids/ lattices  ... abstractly, one could describe it by saying that the forgetful functor from a variety to sets preserves direct limits.)
This should take care of the statement 2. straightforwardly: If $R=0$, this means that $0=1$, which by the description of $\sim$ translates into $0=1$ in $R_k$ for some $k$. From this point on, of course $0=1$ in every $i \geq k$, since there is a homomorphism $R_k \rightarrow R_i$.
As for 3., if there is an exact sequence 
$$R^m \stackrel{\varphi}{\rightarrow} R^n \rightarrow M \rightarrow 0,$$
the map $\varphi$ is essentially given by an $m \times n$ matrix with entries in $R$. Call the entries $s^{a, b}, (a, b) \in \{0, 1, \dots, m\}\times \{0, 1, \dots, n\}$. Then we can take the representatives: $s^{a, b}=[r_{i_{a, b}}]$ for some $r_{i_{a, b}} \in R_{i_{a, b}}$. Taking $k$ such that $k\geq i_{a, b}$ for all $a, b$, we have
$$s^{a, b}=[r_{i_{a, b}}]=[\nu_{i_{a, b}k}(r_{i_{a, b}})], \; \nu_{i_{a, b}k}(r_{i_{a, b}}) \in R_k,$$
which is a complicated way of saying that all the representatives can be taken from the single ring $R_k$. Now take the module $M_k$ to be the module given by the presentation
$$(R_k)^m \stackrel{\tilde{\varphi}}\rightarrow (R_k)^n \rightarrow M_k \rightarrow 0, $$
where $\tilde{\varphi}$ is given by the matrix of representatives. Now apply the right exact functor $R\otimes_{R_k}-$ and observe that you obtain the original exact sequence.
