Let $A$ be a ring (commutative with unity). Prove that if every finitely generated ideal is a direct summand of $A$ then $A$ is absolutely flat.
I have read a couple of solutions that use the $\text{Tor}$ functor, which I am unfamiliar with. I was wondering if there was a solution that does not involve said functor. Something I am familiar with is the following result: $M$ is flat over $A$ iff for every finitely generated ideal $I \subset A$ the map $I \otimes_A M \to A \otimes_A M=M$ is injective
Now here's what I tried: Let $\mathfrak{a}$ be a finitely generated ideal of $A$. By hypothesis there is an ideal $\mathfrak{b}$ of $A$ such that $A=\mathfrak{a} \oplus \mathfrak{b}$ (as an $A$-module). Then for any $A$-module $M$ the map $\mathfrak{a} \otimes_A M \to A \otimes_A M = (\mathfrak{a} \oplus \mathfrak{b}) \otimes_A M = (\mathfrak{a} \otimes_A M) \oplus (\mathfrak{b} \otimes_A M)$ is clearly injective.
Is this correct? I am sorry if it is really obvious or if there is a fatal flaw in my argument, as I tend to struggle with everything tensor related. As I said, I would also like to know if anyone has a different solution not involving the $\text{Tor}$ functor.
Thanks in advance!