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I almost feel embarrassed to ask this, but I am trying to learn about tensor products (for now over Abelian groups). Here is the definition given:

Let $A$ and $B$ be abelian groups. Their tensor product, denoted by $A \otimes B$, is the abelian group having the following presentation

Generators: $A \times B$ that is, all ordered pairs $(a,b)$

Relations: $(a+a',b)=(a,b)+(a',b)$ and $(a,b+b')=(a,b)+(a,b')$ for all $a,a' \in A$ and $b,b' \in B$

So from this, why is $a \otimes 0 = 0$? Looks to me like if $b$ is zero, then any $a,a' \in A$ will still satisfy the relations. I'm just after a simple explanation, then hopefully once that makes sense, it will all make sense!

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    This doesn't answer your question, but Tim Gowers has a very nice write up about tensor products that is worth a look! http://www.dpmms.cam.ac.uk/~wtg10/tensors3.html2011-04-11
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    @DJC - thanks, I already had it open!2011-04-11

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By the relation: $(a,0)+(a,0)=(a,0+0)=(a,0)$, so $(a,0)=0$.

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You are missing the relations $(na,b)=(a,nb)=n(a,b)$ for all $n\in\mathbb{Z}$ (recall that abelian groups are $\mathbb{Z}$-modules). Thus $$a\otimes 0_B=a\otimes (0_{\mathbb{Z}}\cdot 0_B)=(0_{\mathbb{Z}}\cdot a)\otimes 0_B=0_A\otimes 0_B=0.$$

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    thanks - I suppose that needs to be proved (indeed it is also an exercise!)2011-04-11
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    These rules follow from the definition as given above by induction.2011-04-11
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    @Thomas, ah, you're right of course. In fact this makes our answers equivalent, though yours is much simpler :) But still, it may still be useful to the OP to understand that the rules for scaling by elements of the ring have to be forced in general.2011-04-11
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    yes still useful. Without writing the induction out in full $(ma) \otimes b = \underbrace{(a+b) + \cdots + (a+b)}_{m \text{ times }}=m(a \otimes b)$ and similar for $a \otimes (mb)$2011-04-11