If $A\otimes I \mapsto AI$ is injective for every every finitely generated ideal of $R$ then $A$ is flat.

Question

If $A\otimes I \mapsto AI$ is injective for every every finitely generated ideal of $R$ then $A$ is flat.

I can prove that $$0 \rightarrow A\otimes I \rightarrow A\otimes R \rightarrow A\otimes R/I \rightarrow 0$$ is exact.

I don't know if this can help to prove that every short exact sequence $$0 \rightarrow C \rightarrow C'\rightarrow C''\rightarrow 0$$ goes to a short exact sequence $$0 \rightarrow A\otimes C \rightarrow A\otimes C' \rightarrow A\otimes C'' \rightarrow 0$$

Answer 1

Hints:

You have to prove the exactness property for submodules of free modules $F$ of finite rank $r$ first. Here is a sketch:

• Suppose first $r=2$. We may as well suppose $F=R^2$. Explicitly, we'll denote $F=R_1\oplus R_2$ to distinguish the two copies of $R$, $i_1,\, i_2$ the canonical injections, $p_1,\, p_2$ the canonical projections.

Let $G$ be a submodule of $F$. We'll denote $I_1$ the ideal $M\cap R_1$, $I_2=p_2(G)$. Consider the following commutative diagram of short exact sequences: $$\require{AMScd} \begin{CD} 0 @>>> I_1 @>{i_1'}>> G @>{p_i'}>> I_2 @>>> 0 \\ @. @V{j_1}VV @V{j}VV @V{j_2}VV \\ 0 @>>> R_1 @>{i_1}>> F @>{p_2}>> R_2 @>>> 0 \end{CD} $$ and tensor it by $A$ to obtain the commutative diagram of exact sequences: $$ \begin{CD} {} @. A\otimes_R I_1 @>{1_A\otimes i'_1}>> A\otimes_R G @>{1_A\otimes p'_2}>> A\otimes_R I_2@>>> 0 \\ @. @V{1_A\otimes j_1}VV @V{1_A\otimes j}VV @V{1_A\otimes j_2}VV \\ 0@>>> A\otimes_R R_1 @>{1_A\otimes i_1}>> A\otimes_R F @>{1_A\otimes p_2}>> A\otimes_R R_2 @>>> 0 \end{CD} $$ Note the left and right vertical arrows are injective, and some diagram a-hunting shows the middle map is also injective.

• An easy induction shows it is true for any rank $r$.
• For a finitely generated $R$ module $M$ and a submodule $N$, write $M$ as a quotient of a free module $F$ of finite rank, let $G$ be the inverse image of $N$ in $F$, and consider this commutative diagram, where $K$ denotes the kernel of the morphism $F\longrightarrow M$: $$ \begin{CD} 0 @>>> K @>{i'}>> G @>{p'}>> N @>>> 0 \\ @. @| @V{\ell}VV @V{j}VV \\ 0 @>>> K @>{i}>> F @>{p}>> M @>>> 0 \end{CD} $$ Tensor again with $A$ and do some diagram a-hunting to show $1_A\otimes j$ is injective from the fact that $1_A\otimes \ell$ is injective.