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

#### Peter

##### Well-known member
MHB Site Helper
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 $$\displaystyle S$$-module $$\displaystyle S \oplus_R N$$ from an $$\displaystyle R$$-module $$\displaystyle N$$. In this construction the unital ring $$\displaystyle R$$ is a subring of the unital ring $$\displaystyle S$$. (For a detailed description of this construction see the attachment pages 359 - 361 or see D&F Section 10.4)

To construct $$\displaystyle S \oplus_R N$$ take the abelian group $$\displaystyle N$$ together with a map from $$\displaystyle S \times N$$ to $$\displaystyle N$$, 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 $$\displaystyle \mathbb{Z}$$-module (the free abelian group) on the set $$\displaystyle S \times N$$ - that is, the collection of all finite commuting sums of elements of the form $$\displaystyle (s_i, n_i)$$ where $$\displaystyle s_i \in S$$ and $$\displaystyle n_i \in N$$.

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:

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

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

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

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

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

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

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

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

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

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

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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 $$\displaystyle \mathbb{Z}/\mathbb{5Z}$$ that the cosets are clearly $$\displaystyle 0 + 5 \mathbb{Z}, 1 + 5 \mathbb{Z}, 2 + 5 \mathbb{Z}, 3 + 5 \mathbb{Z}, 4 + 5 \mathbb{Z}$$, 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 $$\displaystyle s \oplus n$$)

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:

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

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

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

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:

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

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

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

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

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