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- Jun 22, 2012

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I am trying (struggling! ) to understand tensor products as developed by Dummit and Foote in Section 10.4 - specifically the early section devoted to the "extension of scalars".

I have been reflecting on my attempts to understand the material of Dummit and Foote, pages 359 -362 (see attachment) and wondering if I really fully understand the basic notions. Thus I am now examining my understanding of the (free) \(\displaystyle \mathbb{Z} \)-module on the set \(\displaystyle S \times N \) (see D&F page 360 - see attached) - as this concept is critical to the development.

My understanding of the nature of the (free) \(\displaystyle \mathbb{Z} \)-module on the set \(\displaystyle S \times N \) follows - I am carefully following D&F's definition of a module on page 337 (see attachment).

Let \(\displaystyle R = \mathbb{Z} \) and \(\displaystyle M = S \times N \).

Then ...

(1) MATH] M = S \times N [/MATH] is an abelian group under + where + operates as follows:

\(\displaystyle (s_1, n_1) + (s_2, n_2) = (s_1 + s_2, n_1 + n_2) \)

(2) we have an action of \(\displaystyle \mathbb{Z} \) on \(\displaystyle M\); that is a map

\(\displaystyle \mathbb{Z} \times M \to M \), that is

\(\displaystyle \mathbb{Z} \times (S \times N) \to (S \times N) \)

denoted by \(\displaystyle xm = x(s,n) \) where \(\displaystyle x \in \mathbb{Z}, (s,n) \in M = S \times N \)

which satisfies

(a) \(\displaystyle (x + y)(s,n) = x(s,n) + y(s,n) \) where \(\displaystyle x, y \in \mathbb{Z}, (s,n) \in M = S \times N \)

(b) \(\displaystyle (xy)(s,n) = x(y(s,n)) \) where \(\displaystyle x, y \in \mathbb{Z}, (s,n) \in M = S \times N \)

(c) \(\displaystyle x((s_1, n_1) + (s_2, n_2)) = x(s_1, n_1) + x(s_2, n_2) \) where \(\displaystyle x \in \mathbb{Z}, (s_1, n_1), (s_2, n_2) \in M = S \times N \)

and, since \(\displaystyle \mathbb{Z} \) has a 1, we impose the additional axiom:

(d) \(\displaystyle 1(s,n) = (s,n) \)

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Now looking at (2) more carefully we take

\(\displaystyle x(s,n) = (s,n) + (s,n) + ... \ ... (s,n) \) (x times)

So essentially the \(\displaystyle \mathbb{Z} \)-module is made up of sums of \(\displaystyle (s_i, n_i) \) of the form (for example)

\(\displaystyle 3(s_1, n_1) + 8(s_2, n_2) + 25(s_7, n_7) + 2(s_{10}, n_{10}) \)

I am uncertain of the construction because on pages 360 and 361 at various points D&F seem to indicate or imply at least that the elements of the free module are of the form:

\(\displaystyle \sum s_in_i = s_1n_1 + s_2n_2 + ... \ ... + s_kn_k \)

Another issue that I would like a comment n is the fact in the above that I have not explicitly discussed the free nature of the module being constructed - basically because I think that \(\displaystyle \mathbb{Z} \)-modules are free.

I would welcome some help.

Peter

I have been reflecting on my attempts to understand the material of Dummit and Foote, pages 359 -362 (see attachment) and wondering if I really fully understand the basic notions. Thus I am now examining my understanding of the (free) \(\displaystyle \mathbb{Z} \)-module on the set \(\displaystyle S \times N \) (see D&F page 360 - see attached) - as this concept is critical to the development.

**I hope that some MHB member will confirm the correctness of my construction.**My understanding of the nature of the (free) \(\displaystyle \mathbb{Z} \)-module on the set \(\displaystyle S \times N \) follows - I am carefully following D&F's definition of a module on page 337 (see attachment).

Let \(\displaystyle R = \mathbb{Z} \) and \(\displaystyle M = S \times N \).

Then ...

(1) MATH] M = S \times N [/MATH] is an abelian group under + where + operates as follows:

\(\displaystyle (s_1, n_1) + (s_2, n_2) = (s_1 + s_2, n_1 + n_2) \)

(2) we have an action of \(\displaystyle \mathbb{Z} \) on \(\displaystyle M\); that is a map

\(\displaystyle \mathbb{Z} \times M \to M \), that is

\(\displaystyle \mathbb{Z} \times (S \times N) \to (S \times N) \)

denoted by \(\displaystyle xm = x(s,n) \) where \(\displaystyle x \in \mathbb{Z}, (s,n) \in M = S \times N \)

which satisfies

(a) \(\displaystyle (x + y)(s,n) = x(s,n) + y(s,n) \) where \(\displaystyle x, y \in \mathbb{Z}, (s,n) \in M = S \times N \)

(b) \(\displaystyle (xy)(s,n) = x(y(s,n)) \) where \(\displaystyle x, y \in \mathbb{Z}, (s,n) \in M = S \times N \)

(c) \(\displaystyle x((s_1, n_1) + (s_2, n_2)) = x(s_1, n_1) + x(s_2, n_2) \) where \(\displaystyle x \in \mathbb{Z}, (s_1, n_1), (s_2, n_2) \in M = S \times N \)

and, since \(\displaystyle \mathbb{Z} \) has a 1, we impose the additional axiom:

(d) \(\displaystyle 1(s,n) = (s,n) \)

--------------------------------------------------------------------------

Now looking at (2) more carefully we take

\(\displaystyle x(s,n) = (s,n) + (s,n) + ... \ ... (s,n) \) (x times)

So essentially the \(\displaystyle \mathbb{Z} \)-module is made up of sums of \(\displaystyle (s_i, n_i) \) of the form (for example)

\(\displaystyle 3(s_1, n_1) + 8(s_2, n_2) + 25(s_7, n_7) + 2(s_{10}, n_{10}) \)

**Can someone please confirm that the above construction and reasoning is correct or indicate otherwise ...**I am uncertain of the construction because on pages 360 and 361 at various points D&F seem to indicate or imply at least that the elements of the free module are of the form:

\(\displaystyle \sum s_in_i = s_1n_1 + s_2n_2 + ... \ ... + s_kn_k \)

Another issue that I would like a comment n is the fact in the above that I have not explicitly discussed the free nature of the module being constructed - basically because I think that \(\displaystyle \mathbb{Z} \)-modules are free.

**Is that correct?**I would welcome some help.

Peter

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