Tensor Products - Dummit and Foote - Section 10-4, pages 359 - 362

[Math Amateur]

In Dummit and Foote, Section 10.4: Tensor Products of Modules, on pages 359 - 364 (see attachment) the authors deal with a process of 'extension of scalars' of a module, whereby we construct a left $$S$$-module $$ S \oplus_R N $$ from an $$R$$-module $$N$$. In this construction the unital ring $$R$$ is a subring of the unital ring $$S$$. (For a detailed description of this construction see the attachment pages 359 - 361 or see D&F Section 10.4)

To construct $$ S \oplus_R N $$ take the abelian group $$N$$ together with a map from $$ S \times N $$ to $$N$$, where the image of the pair (s,n) is denoted by sn.

D&F then argue that it is "natural" (but why is it natural?) to consider the free $$ \mathbb{Z} $$-module (the free abelian group) on the set $$ S \times N $$ - that is, the collection of all finite commuting sums of elements of the form $$ (s_i, n_i) $$ where $$ s_i \in S $$ and $$ n_i \in N $$.

To satisfy the relations necessary to attain an S-module structure, D&F argue that we must take the quotient of this abelian group by the subgroup H generated by all elements of the form:

$$ (s_1 + s_2, n) - (s_1, n) - (s_2, n) $$

$$ (s, n_1 + n_2) - (s, n_1) - (s, n_2)$$

$$ (sr,n) - (s, rn) $$

for $$ s, s_1, s_2 \in S, n, n_1, n_2 \in N $$ and $$ r \in R $$ where rn in the last element refers to the R-module structure already defined on N.

The resulting quotient group is denoted by $$ S \oplus_R N $$ and is called the tensor product of S and N over R.

If $$ s \oplus n $$ denotes the coset containing (s,n) then by definition of the quotient we have forced the relations:

$$ (s_1 + s_2) \oplus n = s_1 \oplus n + s_2 \oplus n $$

$$ s \oplus (n_1 + n_2) = s \oplus n_1 + s \oplus n_2 $$

$$ sr \oplus n = s \oplus rn $$

The elements of $$ S \oplus_R N $$ are called tensors and can be written (non-uniquely in general) as finite sums of "simple tensors" of the form $$ s \oplus n $$.

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Issues/Problems

Issue/Problem (1)

I am having real trouble understanding/visualizing the nature and character of the cosets of the quotient group defined above - I would really like to get a tangible and concrete view of the nature of the cosets. Can someone help in this matter either by general explanation and/or a concrete example.

(I can see in the case of a quotient group like $$ \mathbb{Z}/\mathbb{5Z} $$ that the cosets are clearly $$ 0 + 5 \mathbb{Z}, 1 + 5 \mathbb{Z}, 2 + 5 \mathbb{Z}, 3 + 5 \mathbb{Z}, 4 + 5 \mathbb{Z}$$, and that x and y are in the same coset if x - y is divisible by 5 - but I cannot get the same feeling for and understanding of the cosets of $$ s \oplus n $$)

I really hope someone can help make the nature of the cosets a little clearer. Certainly no texts or online notes attempt top make this clearer for the student/reader ... nor do they give helpful examples ...Issue/Problem 2

D&F state that:

"by definition of the quotient we have forced the relations:

$$ (s_1 + s_2) \oplus n = s_1 \oplus n + s_2 \oplus n $$

$$ s \oplus (n_1 + n_2) = s \oplus n_1 + s \oplus n_2)$$

$$ sr \oplus n) = s \oplus rn $$.

My question is, how exactly, does taking the quotient of the abelian group N by the subgroup H generated by all elements of the form:

$$ s_1 + s_2, n) - (s_1, n) - (s_2, n) $$

$$ (s, n_1 + n_2) - (s, n_1) - (s, n_2)$$

$$ (sr,n) - (s, rn) $$

guarantee or force the relations required?

I would be really grateful if someone can help. Again, as with issue/problem 1 no text or online notes have given a good explanation of this matter.

Peter

 

[Euge]

If you want to better understand cosets (and for that matter, quotient groups) consider reading through sections of Gallian's Contemporary Abstract Algebra. Note that the symbol for tensor product is $\otimes$, not $\oplus$.

If $s_1, s_2\in S$ and $n\in N$, then $(s_1 + s_2, n) - (s_1,n) - (s_2,n)$ belongs to $H$, and therefore the coset $(s_1 + s_2,n) + H$ equals the coset $(s_1,n) + (s_2,n) + H$. Now $(s_1,n) + (s_2,n) + H = [(s_1,n) + H] + [(s_2,n) + H]$, so that $$(s_1 + s_2,n) + H = [(s_1,n) + H] + [(s_2,n) + H]$$ In other words, $$(s_1 + s_2)\otimes n = s_1 \otimes n + s_2\otimes n$$ Similarly the other two tensor equations hold.