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Suppose E be a right R-module, and E is flat, then how can I prove that for any exact sequence of left R-modules $A\to B\to C$, the sequence $E\otimes A\to E\otimes B\to E\otimes C$ is exact? I am reaaly confused

E is flat means that Given any exact sequence $0\to A'\to A$, $E\otimes A'\to E\otimes A$ is injective (over here A' and A are just left R-modules)

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    What's your definition of flat? For me, preserving exactness under tensor product IS the definition of flatness. See https://en.wikipedia.org/wiki/Flat_module – Arkady Mar 07 '16 at 01:49
  • One can prove that the tensor product is right exact, i.e. for every module $E$ and every exact sequence $A\to B\to C\to 0$, the sequence $E\otimes A\to E\otimes B\to E\otimes C\to 0$ is exact. See http://math.stackexchange.com/questions/81640/proving-that-the-tensor-product-is-right-exact

    I think you don't even need $E$ to be flat, but I am not very sure. If $D$ is the image of the morphism ($B\to C$), your sequence $A\to B\to C$ is exact iff $A\to B\to D\to 0$ is exact. And also $E\otimes A\to E\otimes B\to E\otimes C$ is exact iff $E\otimes A\to E\otimes B\to E\otimes D\to 0$ is exact.

     – Marco Flores Mar 07 '16 at 02:46

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Let $\alpha: A \to B$ and $\beta: B \to C$ be the maps. Then the sequence $A \to B \to C$ is exact (at B) if and only if $$ 0 \to \alpha (A) \to B \to B/\ker \beta \to 0$$ is exact.

Similarly, $E \otimes A \to E \otimes B \to E \otimes C$ is exact if and only if

$$0 \to (\mathbf{1} \otimes \alpha )(A) \to E \otimes B \to E \otimes B/ \ker \mathbf{1} \otimes \beta \to 0 $$ is exact. Observe that this sequence is obtained by tensoring the first short exact sequence by $E$.

Now tensoring is a right-exact functor. Flatness, in the definition given in your question, gives left-exact.

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