$M,N$ are $A$-modules. I don't see why the statement is true. Can you explain please?
Why is the image of $M\times N\rightarrow M\otimes N, (m,n)\mapsto m\otimes n$ not a submodule?
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modules
tensor-products
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0@KCd спасибо за ценную ссылку. У вас всё подробно и понятно объяснено. А то что вы русский знаете — вообще удивительно :) – 2012-04-23
1 Answers
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Edit: Suppose $M,N$ are two free modules of finite rank both 2 or greater. Take $m_1,m_2\in M$ (resp. $n_1,n_2\in N$) which are linearly independent and both are in a basis. Then $m_1\otimes n_1+m_2\otimes n_2$ cannot be realized as the image of any pair $(m,n)\in M\times N$.
As KCd points out, some care should probably be taken with the generality of the original statement.
Original: The set $\{m\otimes n| m\in M, n\in N\}\subset M\otimes N$ is not even closed under addition. Suppose that one can find two $A$-linearly independent elements $m_1,m_2\in M$ (resp. $n_1,n_2\in N$). Then $m_1\otimes n_1+m_2\otimes n_2$ cannot be realized as the image of any pair $(m,n)\in M\times N$.
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0Hmm. I think you're right to be alarmed here- I did not check up on this condition before posting this. I'll edit the post, think about the general case, and return when I figure it out. – 2012-04-23