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With M and N being R-modules, how does Z(M,N) have $M\times N$as a basis and therefore becomes a free abelian group?

Consider the element n(m,0) for an element$\,m\in M $ of order n. This is zero with coordinates not zero.

user123124
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    What is Z(M,N)? If it's defined to be the free abelian group on $M\times N$, then of course it has $M\times N$ as a basis. And $(m,0)+\cdots(m,0)$ does not stand for $(m+\cdots+m,0)$ so it's not zero. The addition operation in the free abelian group on a set $X$ is its own thing, it is not defined to be the addition already in place in $X$ - indeed, $X$ is simply treated as a barren set with no additional structure when forming the free abelian group. – anon Aug 10 '14 at 06:45
  • @blue so sums in Z(M,N) which is the free abelian group with $M\times N $ as basis is just a new random element in Z(M,N) without connection to the coordiatewise addition that could be imposed on $M\times N $ as outer product of modules? – user123124 Aug 10 '14 at 06:56
  • Yes, $M\times N$ is just a set when you form the free abelian group on it. The operation in this group has nothing at all to do with anything we might know about $M$ or $N$ beyond their being sets. – anon Aug 10 '14 at 08:26
  • You don't have to understand free modules in order to construct tensor products. See http://math.stackexchange.com/questions/291644/alternative-construction-of-the-tensor-product-or-pass-this-secret – Martin Brandenburg Aug 10 '14 at 10:51

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When you form the free Abelian group $F$ generated by the set $X$ (we will use the notation $Free(X)$ from here on, as introduced in Paulo Aluffi's text $Algebra:$ $Chapter$ $0$) is the set of formal combinations of "letters" $$ \sum\limits_{x \in X} a_x $$ where $a_x + b_x = (a+b)_x$. In this way the set $X$ acts as a basis for $Free(X)$; in fact this is easy to check. For finite sets $X$ (say of cardinality $n \in \mathbb{N}^\times$) we get that $$ \bigoplus\limits_{n \in \mathbb{N}} \mathbb{Z} =: \mathbb{Z}^n \cong Free(X)$$ by using the fact that every Abelian group is a $\mathbb{Z}$-module and every $\mathbb{Z}$-module is an Abelian group, combined with the fact that over a ring of unity $R$ the free $R$-modules are isomorphic to some direct sum of copies of $R$. Thus, for any ring $R$ (possibly noncommutative, possibly without identity) with modules $M, N$ we get that $$ Free(M \times N) = \left\lbrace \sum\limits_{(m,n) \in M \times N} a_{(m,n)} \right\rbrace$$ where the addition $a_{(m,n)} + b_{(m,n)} = (a+b)_{(m,n)}$ is isomorphic to integer addition in the $(m,n)$th position as above. This means that $M \times N$ acts as a basis on $Free(M \times N)$ by indexing the basis elements with the "letters" in $M \times N$, just as how the "letters" in the set $A = \lbrace 1, 2, 3, 4 \rbrace$ index the basis elements in the group $$Free(A) = \left\lbrace \sum\limits_{i = 1}^4 a_i \right\rbrace \cong \mathbb{Z}^4. $$

Geoff
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  • I dont understand what a "letter" is and where it belongs to? Is it a coefficient for some $x\in X$. Why dont we replace it with an integer in that case and remark that the important thing is the index and not the integer and the plus sign? – user123124 Aug 10 '14 at 19:43
  • @Johan A letter is just an element of the set (not the coefficient of that element in some sum in the free abelian group generated by the set). The word choice is supposed to be suggestive. When one forms multiplicative free abelian groups or free groups, the elements are always words and the original set is the alphabet. The ideas of words, letters, relations, and presentations are extremely important in group theory. I don't understand what your last sentence is asking. – anon Aug 10 '14 at 20:58
  • Also, Geoff, you don't want the subscript of $\bigoplus$ to be $n\in\Bbb N$... – anon Aug 10 '14 at 21:01
  • what we mean by $\sum_{x\in X} a_x$ is $a_x\in X$ and only one $a_x$ per $x \in X$? – user123124 Aug 11 '14 at 07:47