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Let $A$ be a local ring, $M$ and $N$ finitely generated $A$-modules. Show that if $M \otimes_A N = 0$ then $M=0$ or $N=0$

I read one proof from my book and it goes as follow:
First, show that $(A/I)\otimes M \simeq M/IM$ by tensor the exact sequence $0\to I \to A \to A/I \to 0 $ with $M$. Then, let $m$ be the maximal ideal, $k=A/m$ the residue field. Let $M_k := k\otimes_A M \simeq M/mM$. By Nakayama's lemma, $M_k = 0 \Rightarrow M=0$. But $M\otimes_A N = 0 \Rightarrow (M\otimes_A N)_k = 0 \Rightarrow M_k \otimes_k N_k =0 \Rightarrow M_k = 0$ or $N_k =0$.

I could not understand 2 parts of above proof. First, how could we deduce $(A/I)\otimes M \simeq M/IM$ by tensor the exact sequence $0\to I \to A \to A/I \to 0 $ with $M$? Is there another easy argument to show $(A/I)\otimes M \simeq M/IM$ ? Secondly, how is it possible that $(M\otimes_A N)_k = 0 \Rightarrow M_k \otimes_k N_k = 0$?

Can anyone give me a clarification? Thank you in advance!

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marked as duplicate by user26857 abstract-algebra Jan 30 '16 at 12:08

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1) The functor $- \otimes_A M$ is right exact. Therefore, the exact sequence $I \to A \to A/I \to 0$ yields the exact sequence $I \otimes_A M \to A \otimes_A M \to A/I \otimes_A M \to 0$. Now use that $A \otimes_A M \cong M$ and that the image of $I \otimes_A M \to M$ is exactly $IM$. This shows $A/I \otimes_A M \cong M/IM$.

You can also show directly that $M/IM$ satisfies the universal property of $A/I \otimes_A M$: Check that $([a],m) \mapsto am$ is a universal bilinear map $|A/I| \times |M| \to |M/IM|$.

2) There is always an isomorphism $(M \otimes_A N) \otimes_A B \cong (M \otimes_A B) \otimes_B (N \otimes_A B)$ for commutative $A$-algebras $B$. Just use the universal properties to show this.

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  • $\begingroup$ What means $|M|$? $\endgroup$ – user26857 Jan 30 '16 at 11:53
  • $\begingroup$ $|M|$ denotes the underlying set of $M$. Notice that bilinear maps are not defined between modules, but rather their underlying sets: They are not linear. $\endgroup$ – Martin Brandenburg Jan 30 '16 at 11:53
  • $\begingroup$ But when define a bilinear map don't need the module (abelian group) structure? (Denoting the objects this way seems to suggest that you don't.) $\endgroup$ – user26857 Jan 30 '16 at 11:55
  • $\begingroup$ The map is just a map of sets. To check for bilinearity, of course we need the module structure. $\endgroup$ – Martin Brandenburg Jan 30 '16 at 12:43
  • $\begingroup$ I'm looking for a counterexample (with $M$ or $N$ not finitely generated) where the implication $M = 0$ or $N = 0$ does not hold. Do you have any ideas? $\endgroup$ – mathcourse Jul 2 '18 at 21:28

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