Tensor Products - Dummit and Foote - Section 10-4, Theorem 8, page 362

[Math Amateur]

I am reading Dummit and Foote, Section 10 on tensor products of modules.

I am at present trying to understand the use of Theorem 6 (D&F, page 354 - see attachment) in Theorem 8 (D&F page 362, see attachment).

The proof of Theorem 8 in D&F Chapter 10 (see attachment) reads as follows:

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Proof: Suppose \(\displaystyle \phi : \ N \to L \) is an R-module homomorphism to the S-module L.

By the universal property of free modules (Theorem 6 in Section 3) there is a Z-module homomorphism from the free Z-module F on the set S x N to L that sends each generator \(\displaystyle (s,n) \) to \(\displaystyle s \phi (n) \). ... ...

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I am trying to see how the use of Theorem 6 applies and operates in the proof of Theorem 8 - so I am essentially trying to "tie up" or "tie together" the symbols and structures of Therem 6 with those of Theorem 8.

Now we know that in Theorem 8 " ... there is a Z-module homomorphism from the free Z-module F on the set S x N to L ..."

Thus for the map \(\displaystyle A \to F(A) \) in Theorem 6

we have the map

\(\displaystyle S \times N \to F(S \times N) \) in Theorem 8, but this map is not showing in the diagram for Theorem 8 on page 362 (see attachment). ?

To link up to \(\displaystyle S \oplus_R N = F(S \times N)/H \) we would need to extend our map above from \(\displaystyle F(S \times N) \) to \(\displaystyle F(S \times N)/H \) and then apply \(\displaystyle \Phi \) to map to L, thus:

\(\displaystyle S \times N \to F(S \times N) \to F(S \times N)/H = S \oplus_R \to L \)

But how does this fit with the figure for Theorem 8 on page 362 which shows the maps

\(\displaystyle N \to S \oplus_R N \to L \) and \(\displaystyle N \to L \)

So basically I am asking for guidance and help on how the use of Theorem 6 "works" in Theorem 8. In other words, how exactly does it apply? What elements does it apply to and how are its preconditions satisfied?

Hoping someone can help.

Peter

 

[Deveno]

Huh. I think I understand your confusion, so let's see where we can go with this.

What we have from Theorem 6 is this:

The universal property of free ($\Bbb Z$-) modules tells us if $f$ is ANY mapping:

$f:S\times N \to L$

that there is a UNIQUE $\Bbb Z$-module homomorphism $\Phi: F_{\Bbb Z}(S \times N) \to L$(that is to say, an abelian group homomorphism) with:

$f(s,n) = \Phi(s,n)$

Now since in this case (Theorem 8), the set $L$ is also an $S$-module, and we are given an $R$-module homomorphism $\phi:N \to L$ we can define the mapping:

$f(s,n) = s\phi(n)$.

Now, just to be clear, here: the inclusion mapping $S \times N \to F_{\Bbb Z}(S \times N)$ and $f$ are just plain ol' functions, the only homomorphism we have so far is $\Phi$ (which is just a homomorphism of abelian groups).

Can we use this to define a new $\Bbb Z$-module homomophism:

$\Phi':F_{\Bbb Z}(S \times N)/H \to L$?

Well, all we need is for $\Phi'$ to be well-defined on cosets of $H$. And this is going to be true if $H \subseteq \text{ker}(\Phi)$.

It clearly suffices to verify this for any generator of $H$.

Now $\Phi((s_1+s_2,n) - (s_1,n) - (s_2,n)) = \Phi(s_1+s_2,n) - \Phi(s_1,n) - \Phi(s_2,n)$ (since $\Phi$ is additive)

$= (s_1+s_2)\phi(n) - s_1\phi(n) - s_2\phi(n) = 0$ (since $L$ is an $S$-module).

Similarly:

$\Phi((s,n_1+n_2) - (s,n_1) - (s,n_2)) = \Phi(s,n_1+n_2) - \Phi(s,n_1) - \Phi(s,n_2)$

$= s\phi(n_1+n_2) - s\phi(n_1) - s\phi(n_2) = s\phi(n_1) + s\phi(n_2) - s\phi(n_1) - s\phi(n_2) = 0$ (since $\phi$, being an $R$-module homomorphism, is additive).

Finally:

$\Phi((sr,n) - (s,rn)) = \Phi(sr,n) - \Phi(s,rn) = (sr)\phi(n) - s\phi(rn)$

$= (sr)\phi(n) - s(r\phi(n)) = 0$ (since $\phi$ is $R$-linear, and $L$ is an $S$-module).

This means that $\Phi'$ is indeed well-defined on cosets of $H$ as:

$\displaystyle \Phi'\left(\sum_k a_k(s_k,n_k) + H\right) = \Phi\left(\sum_k a_k(s_k,n_k)\right)$

Now, we have that $F_{\Bbb Z}(S \times N)/H$ *is* $S \otimes_R N$, so we have established that we do have a $\Bbb Z$-module homomorphism:

$S \otimes_R N \to L$.

That's half the battle right there. Now we only need to show that $\Phi'$ is, in fact an $S$-module homomorphism, which the next part of the proof in D&F does quite well.

*********

That doesn't quite address your question, because we have shown we have the right kind of $S$-module homomorphism, but our mapping just goes from $S \times N \to L$ and not $N \to L$.

However, we have the subset $\{1\} \times N$ of $S \times N$, and surely the map:

$S \times N \to S \otimes_R N$ given by:

$(s,n) \mapsto s\otimes n$

when restricted to $\{1\} \times N$ can be thought of as a map $N \to S \otimes_R N$.

So really, the only piece missing is to show that:

$\iota:N \to S \otimes_R N$ given by $\iota(n) = 1\otimes n$ is an $R$-module homomorphism.

This boils down to:

$\iota(n_1+n_2) = 1\otimes(n_1+n_2) = 1\otimes n_1 + 1\otimes n_2 = \iota(n_1) + \iota(n_2)$

$\iota(rn) = 1\otimes(rn) = r(1\otimes n) = r\iota(n)$

which follows from the bilinearity of $\otimes$, so that $\iota$ is indeed an $R$-module homomorphism.

Put another way: the missing presence of $S$ in the commutative diagram is due to the fact that, as an $S$-module, $S$ is generated by a singleton set, {1}.

You may wish to convince yourself that, as SETS, the set:

$\{x\} \times Y$ and the set $Y$ are isomorphic

(that is, that $f:\{x\} \times Y \to Y$ given by: $f(x,y) = y$ is a bijection).

It should not be surprising that this is so:

$|\{x\}\times Y| = |\{x\}|\ast |Y| = 1\ast|Y| = |Y|$.

(visually, if you imagine embedding the real line as the $y$-axis in the plane and "translating it" by $x$, the resulting line is in one-to-one correspondence with the real line, a similar construction can be done with higher dimensions).

Heuristically, we think of $\otimes$ as a "product" and so when we want to pick a single element from a ring as "representative" we want to pick a UNIT. When we extend by scalars, we need a "ring in the middle" (something that works as a LEFT ring for the right factor of the tensor product, and a RIGHT ring for the left factor of the tensor product), because what we are doing is essentially constructing a "bi-module".

Working with rings with unity gives us a "default ring element to tensor by", and working with commutative rings let's us not worry about the left-versus-right issue. Perhaps you can see the the default "additive element" 0, of a ring, would not work out very well in the tensor construction: bilinearity forces $0\otimes n = 0(1\otimes n) = 0$ (since this is true of any $S$-module).

 

[Math Amateur]

Huh. I think I understand your confusion, so let's see where we can go with this.

What we have from Theorem 6 is this:

The universal property of free ($\Bbb Z$-) modules tells us if $f$ is ANY mapping:

$f:S\times N \to L$

that there is a UNIQUE $\Bbb Z$-module homomorphism $\Phi: F_{\Bbb Z}(S \times N) \to L$(that is to say, an abelian group homomorphism) with:

$f(s,n) = \Phi(s,n)$

Now since in this case (Theorem 8), the set $L$ is also an $S$-module, and we are given an $R$-module homomorphism $\phi:N \to L$ we can define the mapping:

$f(s,n) = s\phi(n)$.

Now, just to be clear, here: the inclusion mapping $S \times N \to F_{\Bbb Z}(S \times N)$ and $f$ are just plain ol' functions, the only homomorphism we have so far is $\Phi$ (which is just a homomorphism of abelian groups).

Can we use this to define a new $\Bbb Z$-module homomophism:

$\Phi':F_{\Bbb Z}(S \times N)/H \to L$?

Well, all we need is for $\Phi'$ to be well-defined on cosets of $H$. And this is going to be true if $H \subseteq \text{ker}(\Phi)$.

It clearly suffices to verify this for any generator of $H$.

Now $\Phi((s_1+s_2,n) - (s_1,n) - (s_2,n)) = \Phi(s_1+s_2,n) - \Phi(s_1,n) - \Phi(s_2,n)$ (since $\Phi$ is additive)

$= (s_1+s_2)\phi(n) - s_1\phi(n) - s_2\phi(n) = 0$ (since $L$ is an $S$-module).

Similarly:

$\Phi((s,n_1+n_2) - (s,n_1) - (s,n_2)) = \Phi(s,n_1+n_2) - \Phi(s,n_1) - \Phi(s,n_2)$

$= s\phi(n_1+n_2) - s\phi(n_1) - s\phi(n_2) = s\phi(n_1) + s\phi(n_2) - s\phi(n_1) - s\phi(n_2) = 0$ (since $\phi$, being an $R$-module homomorphism, is additive).

Finally:

$\Phi((sr,n) - (s,rn)) = \Phi(sr,n) - \Phi(s,rn) = (sr)\phi(n) - s\phi(rn)$

$= (sr)\phi(n) - s(r\phi(n)) = 0$ (since $\phi$ is $R$-linear, and $L$ is an $S$-module).

This means that $\Phi'$ is indeed well-defined on cosets of $H$ as:

$\displaystyle \Phi'\left(\sum_k a_k(s_k,n_k) + H\right) = \Phi\left(\sum_k a_k(s_k,n_k)\right)$

Now, we have that $F_{\Bbb Z}(S \times N)/H$ *is* $S \otimes_R N$, so we have established that we do have a $\Bbb Z$-module homomorphism:

$S \otimes_R N \to L$.

That's half the battle right there. Now we only need to show that $\Phi'$ is, in fact an $S$-module homomorphism, which the next part of the proof in D&F does quite well.

*********

That doesn't quite address your question, because we have shown we have the right kind of $S$-module homomorphism, but our mapping just goes from $S \times N \to L$ and not $N \to L$.

However, we have the subset $\{1\} \times N$ of $S \times N$, and surely the map:

$S \times N \to S \otimes_R N$ given by:

$(s,n) \mapsto s\otimes n$

when restricted to $\{1\} \times N$ can be thought of as a map $N \to S \otimes_R N$.

So really, the only piece missing is to show that:

$\iota:N \to S \otimes_R N$ given by $\iota(n) = 1\otimes n$ is an $R$-module homomorphism.

This boils down to:

$\iota(n_1+n_2) = 1\otimes(n_1+n_2) = 1\otimes n_1 + 1\otimes n_2 = \iota(n_1) + \iota(n_2)$

$\iota(rn) = 1\otimes(rn) = r(1\otimes n) = r\iota(n)$

which follows from the bilinearity of $\otimes$, so that $\iota$ is indeed an $R$-module homomorphism.

Put another way: the missing presence of $S$ in the commutative diagram is due to the fact that, as an $S$-module, $S$ is generated by a singleton set, {1}.

You may wish to convince yourself that, as SETS, the set:

$\{x\} \times Y$ and the set $Y$ are isomorphic

(that is, that $f:\{x\} \times Y \to Y$ given by: $f(x,y) = y$ is a bijection).

It should not be surprising that this is so:

$|\{x\}\times Y| = |\{x\}|\ast |Y| = 1\ast|Y| = |Y|$.

(visually, if you imagine embedding the real line as the $y$-axis in the plane and "translating it" by $x$, the resulting line is in one-to-one correspondence with the real line, a similar construction can be done with higher dimensions).

Heuristically, we think of $\otimes$ as a "product" and so when we want to pick a single element from a ring as "representative" we want to pick a UNIT. When we extend by scalars, we need a "ring in the middle" (something that works as a LEFT ring for the right factor of the tensor product, and a RIGHT ring for the left factor of the tensor product), because what we are doing is essentially constructing a "bi-module".

Working with rings with unity gives us a "default ring element to tensor by", and working with commutative rings let's us not worry about the left-versus-right issue. Perhaps you can see the the default "additive element" 0, of a ring, would not work out very well in the tensor construction: bilinearity forces $0\otimes n = 0(1\otimes n) = 0$ (since this is true of any $S$-module).

Thank you so much for the extensive help ... I am indeed confused and struggling ... I will now work through you post very carefully ...

Thank you again ...

Peter