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

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Not every element of the tensor product is an elementary tensor. – user38268 Apr 22 '12 at 9:52
@BenjaminLim excuse my stupid, but I don't understand your argument. Elementary tensors are $1\otimes0$ and $0\otimes1$. But how does it relate to the question? – Sergey Filkin Apr 22 '12 at 10:31
If it is a submodule, we have closure under addition. That is to say, given $m \otimes n$ and $p \otimes q$, we have that $m \otimes n + p \otimes q = x \otimes y$ where $x \in M$ and $y \in N$. That is to say that the sum of two elementary tensors is an elementary tensor. But this is not true in general. – user38268 Apr 22 '12 at 10:47
@BenjaminLim because $(m_1+m_2)\otimes(n_1+n_2)=m_1\otimes n_1 + m_1 \otimes n_2 + m_2\otimes n_1 + m_2\otimes n_2$. Now it's clear. Thank you! – Sergey Filkin Apr 22 '12 at 10:56
@Sergey: Это как умножение в школе: $a \otimes 0 = a \otimes (0+0) = a \otimes 0 + a \otimes 0 \Rightarrow a \otimes 0 = 0$. Ясно, что Вы впервые пытаетесь понимать тензорные произведения. Смотрите файл, особенно вопросы и ответы на стр. 10 и 11. – KCd Apr 22 '12 at 19:46
up vote 4 down vote accepted

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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Do you really mean this for all $A$-modules $M$ and $N$ that each admit linearly independent pairs? If one uses pairs taken from a basis in two free modules $M$ and $N$ (each with rank greater than 1), that's one thing, but you are making no constraints on $M$ and $N$ whatsoever other than that each has a pair of linearly independent elements. – KCd Apr 23 '12 at 2:35
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. – KReiser Apr 23 '12 at 4:26

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