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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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up vote 4 down vote accepted

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.

Lang is dealing with modules over commutative rings, and constructs the tensor product as you say as the quotient of a free module over the module generated by the given rules.

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.

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Here's the link to the page with Lang's construction:… 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
Actually, you can see in the section Additional Structure of the Wikipedia article (…) that they define a multiplication on the group to turn it into a module in the cases where it makes sense, and it's precisely the rule that you were missing. I presume the effect of introducing the module structure in the commutative case in this way is the same as starting from the free module and quotienting by the additional rules that you pointed out. – joriki Apr 15 '11 at 19:12
+1: this is all correct. Given a left R-module and a right R-module, their tensor product is just an abelian group. – wildildildlife Apr 15 '11 at 19:18
Ah, I see. So for commutative rings there is another way of constructing the tensor product which does not work for non-commutative ones. That's interesting. I will delete my answer because it might be misleading. Someone should correct the wikipedia article. ;) – Rasmus Apr 15 '11 at 20:15
(Also your answer perfectly summarises the situation.) – Rasmus Apr 15 '11 at 20:21

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