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

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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!

marked as duplicate by user26857 abstract-algebra StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 30 '16 at 12:08

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.
• What means $|M|$? – user26857 Jan 30 '16 at 11:53
• $|M|$ denotes the underlying set of $M$. Notice that bilinear maps are not defined between modules, but rather their underlying sets: They are not linear. – Martin Brandenburg Jan 30 '16 at 11:53
• 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? – mathcourse Jul 2 '18 at 21:28