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 $$0 \rightarrow A\otimes C \rightarrow A\otimes C' \rightarrow A\otimes C'' \rightarrow 0$$

  • $\begingroup$ What is your definition of a ‘flat’ $R$-module? $\endgroup$
    – Bernard
    Sep 24, 2016 at 12:30
  • $\begingroup$ A is a flat $R$-module if the functor $A\otimes - $ is exact which is the same as my last 2 sentences above. $\endgroup$
    – user53970
    Sep 24, 2016 at 13:59
  • $\begingroup$ Don't you know it is enough to have the exactness property for $I\hookrightarrow R$, where $I$ is an ideal of $R$ (not necessarily finitely generated)? $\endgroup$
    – Bernard
    Sep 24, 2016 at 14:29
  • $\begingroup$ I only know the injectivity test (Baer's lemma). But that is for the injective modules and not the flat ones. What do you mean exactly with "exactness property for $ I \hookrightarrow R$"? $\endgroup$
    – user53970
    Sep 24, 2016 at 16:48
  • $\begingroup$ That it is equivalent to say $A$ is flat and that, for every ideal $I\subset R $, $\;A \otimes_R I\rightarrow A \otimes_R R=A$ deduced from the canonical injection $I\hookrightarrow R$ is injective. $\endgroup$
    – Bernard
    Sep 24, 2016 at 18:10

1 Answer 1



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.
  • $\begingroup$ How does injectivity of the second arrow $1_A\otimes i_1$ on the bottom of your second diagram follow? $\endgroup$
    – gen
    Mar 2, 2022 at 14:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .