$M,N$ 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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2Not every element of the tensor product is an elementary tensor. – 2012-04-22
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0@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? – 2012-04-22
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2If 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. – 2012-04-22
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0@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! – 2012-04-22
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0@Sergey: The term "elementary tensor" means any tensor of the form $m \otimes n$. What you wrote down in your comment, $1 \otimes 0$ and $0 \otimes 1$ (which, strictly, don't make sense generally since "1" isn't an element of $M$ or $N$) would both be $0$ in $M \otimes_A N$. We usually have more interesting elementary tensors in mind than $0$. – 2012-04-22
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0@Kcd I don't understand why $a\otimes0,0\otimes a,0\otimes0$ are all the same element. Could you elaborate on that please? – 2012-04-22
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0@SergeyFilkin: They are all the zero element. $a\otimes 0=a \otimes (0\cdot 0)=a\cdot 0\otimes 0=0\otimes0 = 0\otimes (0\cdot a)=0\cdot 0 \otimes a=0\otimes a$. (Here, a 0 on the right in the left argument or on the left in the right argument is $0\in A$ and all other 0s are $0\in M$ or $0\in N$.) – 2012-04-22
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2@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 www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf, особенно вопросы и ответы на стр. 10 и 11. – 2012-04-22
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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
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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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1Do 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. – 2012-04-23
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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