# construction of the tensor product, generating submodule which is factored out

The tensor product of modules $M_0, M_1$ is a quotient of a free module $F$, …, by a submodule $F'$. I found 2 definitions of this $F'$, and the difference is in this generating rules:

• Wikipedia: $(r\cdot m_0)\otimes m_1 - m_0\otimes (r\cdot m_1)$;
• Lang's “Algebra”: $r\cdot(m_0\otimes m_1) - (r\cdot m_0)\otimes m_1, r\cdot(m_0\otimes m_1) - m_0\otimes (r\cdot m_1)$.

IMO Lang's “Algebra” $\to$ Wikipedia, because

$(r\cdot(m_0\otimes m_1) - m_0\otimes (r\cdot m_1)) - (r\cdot(m_0\otimes m_1) - (r\cdot m_0)\otimes m_1)$ $= (r\cdot m_0)\otimes m_1 - m_0\otimes (r\cdot m_1)$.

Does the backward implication hold? If not, then it is an error in Wikipedia?

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I believe the problem is that a) these are dealing with two different situations and b) the Wikipedia article is inconsistent because it speaks of a free module over the symbols $m_0 \otimes m_1$ whereas I think it means a free abelian group.
The Wikipedia article is dealing with left and right modules, and in other parts of the article says that the tensor product is just an abelian group, not a module. (You misquoted the rule; in the article it's correctly written with $r$ to the right of $m_0$.) The rules given there are the appropriate ones for an abelian group (it makes no sense to multiply by $r$ in that case), but the article incorrectly states that the quotient of a free module is being taken.
Here's the link to the page with Lang's construction: books.google.de/… Unfortunately the page before it where it says "Let $R$ be a commutative ring" is missing in the preview, but you can see from the construction that he's not differentiating between left and right modules. –  joriki Apr 15 '11 at 19:09