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Tensor Products - Keith Conrad - Theorem 3.3 - Tensor Products I

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
I am reading and trying to fully understand Keith Conrad's paper: Tensor Products I. These notes are available at Expository papers by K. Conrad or the specific paper at http://www.math.uconn.edu/~kconrad/blurbs/linmultialg/tensorprod.pdf.

Conrad's Theorem 3.3 (see attachment - page 10) is important since it, to an extent at least, begins to define the nature of elementary tensors \(\displaystyle m \oplus n \) in the tensor product \(\displaystyle M \oplus_R N \) where M and N are R-modules.

Theorem 3.3 and its proof read as follows:

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Theorem 3.3. Let M and N be R-modules with respective spanning sets with respective spanning sets \(\displaystyle \{ x_i \}_{i \in I} \) and \(\displaystyle \{ y_j \}_{j \in J} \).

The tensor product \(\displaystyle M \oplus_R N \) is spanned linearly by the elementary tensors \(\displaystyle x_i \oplus y_j \)



Proof: An elementary tensor in \(\displaystyle M \oplus_R N \) has the form \(\displaystyle m \oplus n \).

Write \(\displaystyle m = \sum_i a_ix_i \) and \(\displaystyle n = \sum_j b_jy_j \), where the \(\displaystyle a_i s \) and \(\displaystyle b_j s \) are 0 for all but finitely many i and j.

From the bilinearity of \(\displaystyle \oplus \) we have:

\(\displaystyle m \oplus n = \sum_i a_ix_i \oplus \sum_j b_jy_j = \sum_{i,j} a_ib_jx_i \oplus y_j \)

is a linear combination of the tensors \(\displaystyle x_i \oplus y_j \).

Since every elementary tensor is a sum of elementary tensors, the \(\displaystyle x_i \oplus y_j 's \) span \(\displaystyle M \oplus_R N \) as an R-module.

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Now my issue/problem ...

In the above Conrad assumes that the R-modules M and N have spanning sets \(\displaystyle \{ x_i \}_{i \in I} \) and \(\displaystyle \{ y_j \}_{j \in J} \).

Do all modules have spanning sets? (they do not necessarily have bases - which is a similar concept - unless they are free modules, of course.)

Is this simply a case of taking more and more elements in the spanning set ... even up to all the elements of the module? But then how do we guarantee that for elements m and n from M and N that we can write:

\(\displaystyle m = \sum_i a_ix_i \) and \(\displaystyle n = \sum_j b_jy_j \), where the \(\displaystyle a_i s \) and \(\displaystyle b_j s \) are 0 for all but finitely many i and j.

Can someone please clarify this issue for me. I would appreciate guidance on this matter.

Peter
 
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Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
The spanning sets need not be finite (in fact, for many important examples, this will NOT be true).

For example, with $R[x]$, considered as an $R$-module, we see we have the infinite spanning set:

$\{1,x,x^2,\dots\}$.

In some cases, the spanning set may be almost all of $M$ (we can always leave out 0). This is the case, for example, for $\Bbb Z_2$, where the spanning set is the sole non-zero element.

And yes, $M\setminus\{0\}$ certainly spans $M$. So if that was the case, you would have $|M-1|\ast|N-1|$ elementary tensors. Usually, though, we have a much more modest generating set for $M$ (often we are interested in finitely-generated modules). In such a case (and I have a hard time imagining how this would occur, since for every element $x$ is the spanning set, we might have as many as $|R|$ distinct elements spanned by $\{x\}$) we certainly have $m \in M$ as the finite linear combination:

$m = 1m$.
 

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
The spanning sets need not be finite (in fact, for many important examples, this will NOT be true).

For example, with $R[x]$, considered as an $R$-module, we see we have the infinite spanning set:

$\{1,x,x^2,\dots\}$.

In some cases, the spanning set may be almost all of $M$ (we can always leave out 0). This is the case, for example, for $\Bbb Z_2$, where the spanning set is the sole non-zero element.

And yes, $M\setminus\{0\}$ certainly spans $M$. So if that was the case, you would have $|M-1|\ast|N-1|$ elementary tensors. Usually, though, we have a much more modest generating set for $M$ (often we are interested in finitely-generated modules). In such a case (and I have a hard time imagining how this would occur, since for every element $x$ is the spanning set, we might have as many as $|R|$ distinct elements spanned by $\{x\}$) we certainly have $m \in M$ as the finite linear combination:

$m = 1m$.
Thanks Deveno ... that helps ... but I am still puzzling over the question:

How do we guarantee that for elements m and n from M and N that we can write:

\(\displaystyle m = \sum_i a_ix_i \) and \(\displaystyle n = \sum_j b_jy_j \), where the \(\displaystyle a_i s \) and \(\displaystyle b_j s \) are 0 for all but finitely many i and j.

As you point out the spanning sets need not be finite, so how do we guarantee that the number of non-zero elements in the above sums is finite?

Can you help?

Peter
 

Deveno

Well-known member
MHB Math Scholar
Feb 15, 2012
1,967
That is what spanning MEANS.

The idea is: a module consists of $R$-linear combinations of an abelian group. Linear combinations are by definition finite (an infinite sum doesn't mean anything unless you have some notion of convergence).

Now, to be fair, is IS possible to have such things as "formal power series" (where we consider "infinite linear combinations of $x^n$") which can be identified with the space of all functions:

$f: \Bbb N \to R$.

But we consider the free $R$-module on $R^{\infty}$ to be:

$\displaystyle \bigoplus_{\Bbb N} R$

instead of:

$\displaystyle \prod_{\Bbb N} R$

because the former is a lot SMALLER than the latter (we want "free objects" to be minimal, in some sense).

When $R = F$, a field, we want a minimal spanning set to agree with the notion of BASIS, and a basis consists of a $F$-linearly independent set such that any element of $V$ (an $F$-module) can be written as a FINITE $F$-linear combination of basis elements.

Recall that the $R$-module generated by a set $S$ is the SMALLEST $R$-submodule $N$ of a module $M$ such that $S \subseteq N$.

While it is certainly possible to consider "infinite $R$-linear combinations of $S$", the set of "finite $R$-linear combinations of $S$" has the closure properties:

finite + finite = finite

scalar(finite) = finite

so the set of finite $R$-linear combinations is clearly smaller, and also contains $S$.

For example, "finite power series" are polynomials, so $R[x]$ is the $R$-module generated by $\{x^n:n \in \Bbb N\}$, not $R[[x]]$.

This discussion can get rather tricky when considering "infinite copies of $R$" (as is the case, for example, when considering "infinite-dimensional vector spaces" such as the vector space of all functions:

$f: \Bbb R \to \Bbb R$

where finding a basis is clearly a difficult task).
 
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