# Every $A$-mod is flat iff every mod-$A$ is flat.

$\require{AMScd}$ Suppose $A$ is a ring, and we have a SES of left $A$-modules $$0\longrightarrow M'\mathop\longrightarrow\limits^f M \mathop\longrightarrow \limits^g M''\longrightarrow 0$$ where $M''$ is flat. Now let $N$ be any right $A$-module. Show that $$0\longrightarrow N\otimes M'\longrightarrow N \otimes M \longrightarrow N\otimes M''\longrightarrow 0$$ is exact. Still assuming $M''$ is flat, show also that if $M'$ is flat, then so is $M$, and conversely.

I am sure the second claim follows easily from the nine lemma, but I am stuck with the first one. The hint is to consider a SES $0 \longrightarrow K\longrightarrow F \longrightarrow N \longrightarrow 0$ where $F$ is free and analyze the resulting diagram obtained from tensoring term by term. One gets the following $$\begin{CD} {}&{}&0&{}&0&{}&0\\ {}&@VV(*)V@VV(*)V@VVV \\ 0@>(?)>> K\otimes M^\prime @>>> K\otimes M @>>> K\otimes M^{\prime\prime} @>>> 0 \\ {}& @VVV @VVV @VVV \\ 0 @>>> F\otimes M^\prime @>>> F\otimes M @>>> F\otimes M^{\prime\prime} @>>> 0 \\ {}&@VVV @VVV @VVV \\ 0 @>\rm prove>> N\otimes M^\prime @>>> N\otimes M @>>> N\otimes M^{\prime\prime} @>>> 0 \\ {}& @VVV @VVV @VVV \\ {}&{}&0&{}&0&{}&0 \end{CD}$$

The second row is exact since $F$ is free hence flat, the third column is exact since $M''$ is flat. I want to prove the last row is exact, i.e. that $1_N\otimes f$ is injective, but I am not sure how to proceed. I tried some diagram chasing, but no luck. I am not sure $(?)$ the first row is exact i.e. that $1_K\otimes f$ is injective, since submodules of flat modules needn't be flat. Even assuming $(*)$ $M',M$ are both flat I do not get a diagram that allows the application of the nine lemma. At any rate, I don't want to make this assumption.

• The title seems to be unrelated to the question. Nowhere is that equivalence mentioned. – RghtHndSd Jun 16 '14 at 4:22
• @rghthndsd The first claim proves the equivalence stated in the title. If you can think of a better title, please let me know. – Pedro Tamaroff Jun 16 '14 at 4:23
• Dear Pedro, I went to look for you in chat, but then saw you'd posted your question here. I will post an answer below. Cheers, – Matt E Jun 16 '14 at 4:32
• @MattE Thank you very much! I wouldn't mind to get a hint. =) – Pedro Tamaroff Jun 16 '14 at 4:32
• Yeah, I've never used snake / 9 lemmas or proved flatness. Can't help ya here. – blue Jun 17 '14 at 4:05

Try using the snake lemma.

The second row is short exact, and the first row is almost short exact; you can make it short exact on the nose just by replacing $K \otimes M'$ with its image in $K \otimes M$. (Note that by short exactness of the second row, the map from $K \otimes M'$ to $F \otimes M'$ factors through this image.)

Now you have a map between two short exact sequences, and the snake lemma gives a six term exact sequence. The first two terms will be irrelevant, but the rest of them will give what you want.

[One way to think of this approach is that your (?) , while it generally has a negative answer --- you are right that submodules of flat modules need not be flat, and it's not hard to find examples where this map won't be injective --- this doesn't matter for the problem at hand!]

[And a general philosophical remark: when you have a diagram like the one you wrote, and you are trying to show that first arrow in a putative s.e.s. is injective, the snake lemma is a natural thing to use, so you want to try to set things up so that you putative s.e.s. is the last three terms of a snake lemma 6-term exact sequence. This is made systematic by the theory of derived functors, and the theory of Tor is underlying this question. More precisely, this question is a special case of the general fact that Tor${}^R_i(M,N)$ can be computed using a flat resolution of either $M$ or $N$.]

• OK, I will try this and get back at you. It is almost 2 a.m. here, so I'll be doing this tomorrow. =) – Pedro Tamaroff Jun 16 '14 at 4:38
• Dear Matt, I have added what I think is a correct solution. If you have any criticism, let me know. – Pedro Tamaroff Jun 17 '14 at 4:20

I will add the solution here. I am hoping all is correct. We have the diagram (I've removed some irrelevant arrows) $$\begin{CD} {}&{}&{}&{}&{}&{}&0\\ {}&{}&{}&{}&{}&@VVV \\ &{}& K\otimes M^\prime @>>> K\otimes M @>>> K\otimes M^{\prime\prime} @>>> 0 \\ {}& @VV\iota \otimes 1 V @VVV @VVV \\ 0 @>>> F\otimes M^\prime @>>> F\otimes M @>>> F\otimes M^{\prime\prime} \\ {}&@VVV @VVV @VVV \\ &{}& N\otimes M^\prime @>>> N\otimes M @>>> N\otimes M^{\prime\prime} \\ {}& @VVV @VVV @VVV \\ {}&{}&0&{}&0&{}&0 \end{CD}$$

Since the columns are exact, the various $\eta\otimes 1:F\otimes M'\longrightarrow N\otimes M'$ are cokernels for various $\iota\otimes 1$, and the uppermost $0$ is the kernel of the third $\iota\otimes 1$. Thus the snake lemma gives a connecting morphism $\partial$ making the sequence $0\mathop\longrightarrow\limits^{\partial}N\otimes M'\longrightarrow N\otimes M$ exact, as we wanted.