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I'm working on the following problem (Problem 3.17, to be exact) in Atiyah & MacDonald's "Introduction to Commutative Algebra" :

Let $A \xrightarrow{f} B \xrightarrow{g} C$ be ring homomorphisms. If $g \circ f$ is flat and $g$ is faithfully flat, then $f$ is flat.

This question has already been solved on this site (see here), but the proposed solution below uses a certain commutative diagram.

Let $h:M \rightarrow N$ be an injective map of $A$-modules, inducing the map $h_B:M \otimes_A B = M_B \rightarrow N \otimes_A B = N_B$. Let $K = \ker(h_B)$. Since $M_B$ and $N_B$ are $B$-modules, $K$ is a submodule of $M_B$ (and hence a $B$-module), and $g$ is faithfully flat (hence flat), the exact sequence $0 \rightarrow K \rightarrow M_B \rightarrow N_B$ induces the exact sequence $0 \rightarrow K \otimes_B C \rightarrow M_B \otimes_B C \rightarrow N_B \otimes_B C$. Further, since $g \circ f$ is flat, the exact sequence $0 \rightarrow M \rightarrow N$ induces the exact sequence $0 \rightarrow M \otimes_A C \rightarrow N \otimes_A C$. Also, we have isomorphisms $M_B \otimes_B C \cong M \otimes_A (B \otimes_B C) \cong M \otimes_A C$ and $N_B \otimes_B C \cong N \otimes_A (B \otimes_B C) \cong N \otimes_A C$. Even more, one obtains the commutative diagram enter image description here .Thus, $K \otimes_B C = 0$. Since $g$ is faithfully flat, it follows that $K = 0$. Thus, $h_B$ is injective, and $f$ is flat.

I have two questions about the diagram in the solution above :

  1. Can we really say that the provided diagram is commutative ?

  2. If we know the diagram is commutative, how can we conclude that $K \otimes_B C = 0$ ? Is this a consequence of a more general result on commutative diagrams ?

To try and answer 2., I noted Lemma 1.1 in Hilton & Stammbach's Homological Algebra text, shown below.

enter image description here

However, as we note, both the top and bottom rows of the commutative diagram in Lemma 1.1 involve an injection and a surjection, which I'm not sure we have with the maps in our situation.

Thanks !

michiganbiker898
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1 Answers1

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The isomorphisms you mention are natural in $M,N$ (because they are canonical ie induced by universal mapping properties) making the square on the right commute. The left square commutes by exactness of the upper row. The lower row can be extended by another 0 on the left giving a diagram of left exact sequences. They can be turned into a short 5-lemma diagram (the lemma you cite) via epi-mono factorization of the right horizontal maps. Hence the left vertical map is an isomorphism by 5-lemma proving $K\otimes_B C$ to be zero. By faithful flatness $K$ has to be zero.

Jonas Linssen
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