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Where can I find a canonical proof of the following statement?

If $M$ is a flat $A$-module and $N$ is an $A$-module with submodules $N_1, N_2$, then $$(N_1\cap N_2)\otimes_A M = (N_1\otimes_A M)\cap (N_2\otimes_A M)$$ inside $N \otimes_A M$.

This is a standard proposition in commutative algebra, so I expect it can be found in a textbook. I would like to know so that I can study this proof in detail until I understand it.

user26857
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going flatinum
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1 Answers1

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This is Theorem 7.4(i) in Matsumura's Commutative Ring Theory (on page 48).

The proof is simple enough that I may as well include it here.

Take $$\begin{array}{rcrcc} \varphi&:&N&\rightarrow &N/N_1\oplus N/N_2 \\ &&x&\mapsto &(x+N_1,x+N_2)\end{array}$$ and the sequence $$0\rightarrow N_1\cap N_2 \rightarrow N \overset{\varphi}{\rightarrow} \frac{N}{N_1}\oplus \frac{N}{N_2}\rightarrow 0$$ is exact. Since $M$ is flat, we then have that $$0\rightarrow (N_1\cap N_2)\otimes_A M \rightarrow N\otimes_A M \overset{\varphi\otimes 1}{\rightarrow} \frac{N\otimes_A M}{N_1\otimes_A M} \oplus \frac{N\otimes_A M}{N_2\otimes_A M} \rightarrow 0$$ is exact, and that's what we wanted to show.

Alexander Gruber
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  • What is your $N$? The map your defined in line 2 is not neccessarily surjective, so you cannot conclude that the induced sequence is exact. Instead, the sequence is exact at $N$ and then we have the second sequence is exact at $N\otimes M$. Then we win. – user119882 Mar 13 '14 at 17:42
  • @user119882 Can you provide a counterexample? I copied the proof (nearly) verbatim from Matsumura. They're his maps. – Alexander Gruber Mar 14 '14 at 00:52
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    For example $N_1=N_2=0,N=A$. We don't need the map $\phi$ being surjective. What we need is the sequence $0\rightarrow N_1\cap N_2 \rightarrow N \overset{\varphi}{\rightarrow} \frac{N}{N_1}\oplus \frac{N}{N_2}$ is exact. (Which was same as Matsumura's book). That is the difference between your answer and the proof in the book. – user119882 Mar 14 '14 at 16:27
  • @user119882 Ohhh, I see, the issue is with my use of "induced." I should read more carefully, thanks. – Alexander Gruber Mar 14 '14 at 17:14