What does a complex of modules mean? I try to understand from Qing Liu's book Algebraic Geometry and Arithmetic Curves the problem 1.2.16. It goes as follows:

Let $(A,\mathfrak m)$ be a Noetherian local ring, and 
  $$C^\bullet:0\to M'\to M\to M''\to 0$$
  a complex of finitely generated flat $A$-modules. Show that if there exists an ideal $I\subset \mathfrak m$ such that $C^\bullet\otimes_AA/I$ is exact, then $C^\bullet$ is exact.

The notation $C^\bullet$ is new for me so does it mean only that $C^\bullet$ is a name of an exact sequence, and $C^\bullet\otimes_AA/I$ is the  sequence $0\to M'\otimes_AA/I\to M\otimes_AA/I\to M''\otimes_AA/I\to 0$? I am unable to solve the problem.
 A: I don't know if you solved the problem, but this follows easily from the following remark: if $0\to M'\otimes_AA/I\to M\otimes_AA/I$ is exact, then $0\to M'\to M$ is also exact, and moreover the image of $M'$ is a direct summand of $M$ (the argument is the same as the one given in this answer).
A: This already has an answer. But the following one maybe more adaptable to people who read the text book.

*

*By Thm 1.2.16 from the book, $M', M, M''$ are all free $A$-module. So after tensoring $A/I$, they become free $A/I$-module.

*As $M''\otimes_A A/I$ is free the complex $C^{\bullet}\otimes_A A/I$ is split, so is $C^{\bullet}\otimes_A A/\mathfrak m=C^{\bullet}\otimes_A A/I\otimes_{A/I} A/\mathfrak m$. (additive functor preserve split exact sequence.)

*By the proof of Thm 1.2.16, $\{x_1,\cdots,x_n\}$ is a basis of $M$ provided its image is a $A/\mathfrak m$-basis of  $M/\mathfrak mM$. So we can lift the basis of $C^{\bullet}\otimes_A A/\mathfrak m$ to $C^{\bullet}$ and deduce that $C^{\bullet}$ is split exact.

Also the Noetherian condition seems unnecessary.
