2
$\begingroup$

$M,N$ are $A$-modules. I don't see why the statement is true. Can you explain please?

  • 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 1

4

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 $(m,n)\in M\times N$.

  • 0
    Hmm. 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