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Tensor Products - D&F - Extension of the scalars

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am attempting to understand Dummit and Foote exposition on 'extending the scalars' in Section 10.4 Tensor Products of scalars - see attachment - particularly page 360)

[I apologise in advance to MHB members if my analysis and questions are not clear - I am struggling with tensor products! - anyway, I hope readers can divine what I am on about :) ]

On page 360, D&F write the following:

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If the R-module N were already a an S-module then there is no difficulty in "extending" the scalars from R to S, so we begin the construction by returning to the basic module axioms in order to examine whether we can define "products" of the form sn, for \(\displaystyle s \in S \) and \(\displaystyle n \in N \). These axioms start with an abelian group N together with a map from S x N to N, where the image of the pair (s,n) is denoted sn. It is therefore natural to consider the free \(\displaystyle \mathbb{Z} \)-module (i.e. the free abelian group) on the set S x N i.e. the collection of all finite commuting sums of elements of the form \(\displaystyle (s_i, n_i) \) where \(\displaystyle s \in S \) and \(\displaystyle n \in N \). ... ... etc etc

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I would like someone to confirm (or correct) my statements and reasoning in the following ...

An abelian group on the set S x N would be an additive group, with an operation + defined componentwise, visually:

\(\displaystyle (s_1, n_1)+ (s_2, n_2) = (s_1 + s_2, n_1 + n_2) \) ... ... (1)

which seems OK since S is a ring and N is a module and so the two compnent + operations on the right of (1) above are OK. The operations + are associative, identity is (0,0), inverse of (s, n) is (-s, -n). Is that correct?

Now D&F write (see attachment or above)

"t is therefore natural to consider the free \(\displaystyle \mathbb{Z} \)-module (i.e. the free abelian group) on the set S x N ... ..."

Now I am uncertain about how to form the free \(\displaystyle \mathbb{Z} \)-module on the set S x N ... but anyway, I followed D&F's example on page 339 ... ... (see attachment)

Following D&F's example (where \(\displaystyle R = \mathbb{Z} \) ??? ... in our case \(\displaystyle R \ne \mathbb{Z} \) , but continuing anyway ...) ... ...

Make the abelian group into a \(\displaystyle \mathbb{Z} \)-module on the set S x N as follows:

For \(\displaystyle m \in \mathbb{Z} \) and (s,n) in S x N we define:

m(s,n) = (s,n) + ... ... + (s,n) (m times if \(\displaystyle m \gt 0 \))

and

m(s,n) =0 if m = 0

and

m(s,n) = - (s,n) - ... ... - (s,n) (m times if \(\displaystyle m \lt 0 \))

(and I take it that -(s,n) = (-s, -n) ...

But then, how do we (following what D&F say above) end up with elements of the free module being finite commuting sums of elements of the form \(\displaystyle (s_i, n_i) \) (i.e. elements of the form \(\displaystyle \sum_i (s_i, n_i) \) where the sum is finite?)

Do we actually "forbid" operations of addition of elements such as \(\displaystyle (s_1, n_1)+ (s_2, n_2) = (s_1 + s_2, n_1 + n_2) \)? Can someone please clarify this matter?

Further, how does the map from S x N to N with the images sn come into the above construction ...

I would appreciate some help.

Peter
 
Last edited:

ThePerfectHacker

Well-known member
Jan 26, 2012
236
No.

You start with the set $S\times N$. But you do not define $(s_1,n_1) + (s_2,n_2) = (s_1+s_2,n_1+n_2)$. If you are doing that, you are doing it incorrectly. So how do you define addition? You basically just defined formal sums. In other words, you just keep $(s_1,n_1) + (s_2,n_2)$ without adding anything together. There is no component addition. It is just formal addition.

You start with $S\times N$ and then form $F(S\times N)$ this is the free abelian group generated by the set $S\times N$. So it consists of all formal sums.

Any abelian group can be regarded as a $\mathbb{Z}$-module by defining $ng = g+g+...+g$ and with negatives if $n$ is a negative integer. Now since $F(S\times N)$ is an abelian group it can now be regarded as a $\mathbb{Z}$-module in the sense just explained above.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
No.

You start with the set $S\times N$. But you do not define $(s_1,n_1) + (s_2,n_2) = (s_1+s_2,n_1+n_2)$. If you are doing that, you are doing it incorrectly. So how do you define addition? You basically just defined formal sums. In other words, you just keep $(s_1,n_1) + (s_2,n_2)$ without adding anything together. There is no component addition. It is just formal addition.

You start with $S\times N$ and then form $F(S\times N)$ this is the free abelian group generated by the set $S\times N$. So it consists of all formal sums.

Any abelian group can be regarded as a $\mathbb{Z}$-module by defining $ng = g+g+...+g$ and with negatives if $n$ is a negative integer. Now since $F(S\times N)$ is an abelian group it can now be regarded as a $\mathbb{Z}$-module in the sense just explained above.
Thanks, your post is most helpful!

Peter