Extension of Scalars: Dummit & Foote's Section 10.4 Q&A

[Math Amateur]

I am reading Dummit and Foote's book: Abstract Algebra ... ... and am currently focused on Section 10.4 Tensor Products of Modules ... ...

I have a basic question regarding the extension of the scalars ...

Dummit and Foote's exposition regarding extension of the scalars reads as follows:https://www.physicsforums.com/attachments/5491
View attachment 5492
View attachment 5493

Question 1

In the above text from D&F (towards the end of the quote) we read the following:

"... ... Suppose now that \(\displaystyle \sum s_i \otimes n_i = \sum s'_i \otimes n'_i\)

are two representations for the same element in \(\displaystyle S \otimes_R N\). Then \(\displaystyle \sum (s_i, n_i) - \sum (s'_i, n'_i)\) is an element of \(\displaystyle H\) ... ... ... " Can someone please explain exactly why \(\displaystyle \sum s_i \otimes n_i = \sum s'_i \otimes n'_i\) in \(\displaystyle S \otimes_R N\) implies that \(\displaystyle \sum (s_i, n_i) - \sum (s'_i, n'_i)\) is an element of \(\displaystyle H\) ... ...
[ ***Note*** I am a little unsure of the general nature of elements of \(\displaystyle H\) ... and even more unsure of the nature of elements of \(\displaystyle S \otimes_R N\) ... ... ]

Question 2

In the above text from D&F (towards the end of the quote) we read the following:

"... ... for any \(\displaystyle s \in S\) also \(\displaystyle \sum (ss_i, n_i) - \sum (ss'_i, n'_i)\) is an element of \(\displaystyle H\). But this means that \(\displaystyle \sum ss_i \otimes n_i - \sum ss'_i \otimes n'_i)\) in \(\displaystyle S \otimes_R N\) ... ... "Can someone please explain exactly why if \(\displaystyle \sum (ss_i, n_i) - \sum (ss'_i, n'_i)\) is an element of \(\displaystyle H\) ... ... that we then have \(\displaystyle \sum ss_i \otimes n_i - \sum ss'_i \otimes n'_i)\) in \(\displaystyle S \otimes_R N\) ... ...

... ... although the above seems right, why exactly is it the case ... ?

Hope someone can help ... I suspect my main problem is the general nature and characteristics of elements of \(\displaystyle H\) and elements of \(\displaystyle S \otimes_R N\) ... ... Peter

 

[Deveno]

This is pretty much the same as the vector space version, except now we are tensoring over a ring (considering $R$-modules, instead of vector spaces ($F$-modules)).

The idea is we want to turn $\Bbb Z$-multiplinear maps into $\Bbb Z$-linear ones.

The trouble with $FA(S \times N)$ is that it is "much too large", we have the abelian group structure we need, but not any reasonable $S$-module structure.

So we "quotient out" as many elements as we need to to FORCE $S$-multilinearity of the $S$-action.

This results in some pairs $(s,n)$ being equivalent, so just writing $s \otimes n$ only determines an equivalence class (and thus the pair $(s,n)$ is just a "representative").

$H$ is the subgroup *generated* by everything we quotiented out-there is no reason to suppose that these are closed under addition, so we may have to throw some extra stuff into get closure. That's fine, as long as all the stuff of the forms listed in (10.3) is included (that's the stuff we want to be "equivalent to 0", so we can put the stuff with the minus signs on the other side of an "equals" sign (actually an "equivalent to", which is why we work in the quotient, where "equivalent to" BECOMES "equal to")).

 

[Math Amateur]

This is pretty much the same as the vector space version, except now we are tensoring over a ring (considering $R$-modules, instead of vector spaces ($F$-modules)).

The idea is we want to turn $\Bbb Z$-multiplinear maps into $\Bbb Z$-linear ones.

The trouble with $FA(S \times N)$ is that it is "much too large", we have the abelian group structure we need, but not any reasonable $S$-module structure.

So we "quotient out" as many elements as we need to to FORCE $S$-multilinearity of the $S$-action.

This results in some pairs $(s,n)$ being equivalent, so just writing $s \otimes n$ only determines an equivalence class (and thus the pair $(s,n)$ is just a "representative").

$H$ is the subgroup *generated* by everything we quotiented out-there is no reason to suppose that these are closed under addition, so we may have to throw some extra stuff into get closure. That's fine, as long as all the stuff of the forms listed in (10.3) is included (that's the stuff we want to be "equivalent to 0", so we can put the stuff with the minus signs on the other side of an "equals" sign (actually an "equivalent to", which is why we work in the quotient, where "equivalent to" BECOMES "equal to")).

Thanks for the help Deveno ... helpful ... but still reflecting on what you have written ...

... however just a futher question you may be able to help with ...

In the above quote (near the start), D&F write the following:"... ... These axioms start with an abelian group \(\displaystyle N\) together with a map from \(\displaystyle S \times N\) to \(\displaystyle N\), where the image of the pair \(\displaystyle (s, n)\) is denoted by \(\displaystyle sn\). It is therefore natural to consider the free \(\displaystyle \mathbb{Z}\)-module (i.e. the free abelian group) on the set \(\displaystyle S \times N\) ... ... My question is as follows:

Why is it "natural"to consider the free \(\displaystyle \mathbb{Z}\)-module ... ... ? ... ... and what is the free \(\displaystyle \mathbb{Z}\)-module ... ... or the free abelian group, for that matter ...Hope you can help ... ...

Peter

 

[Deveno]

We want an abelian group to have an $S$-module structure ($S$-modules are basically abelian groups "with extra structure"-namely, an $S$-action).

We would also like this abelian group to"contain" $N$, or something like it, and the $S$-action is a map $\mu: S \times N \to N$ (we usually write $\mu(s,n)$ as $s\cdot n$, or simply $sn$).

We can create the free abelian group on the "generating set" $S \times N$ in one of several ways:

1. Create the free group $F(S \times N)$ and pass to the quotient modulo the commutator subgroup.

2. Take the direct sum of $|S \times N|$ copies of $\Bbb Z$: $FA(S \times N) \cong \bigoplus\limits_{\alpha \in S\times N} \Bbb Z$

3. Take every element of $S \times N$ to be a "basis" element, and take formal $\Bbb Z$-linear combinations (this is *most* like the construction of "the vector space based on $X$" Cooperstein writes about).

4. Create *any* object that satisfies the following universal mapping property: for any mapping $f: S \times N \to A$, where $A$ is an abelian group, there exists a injection: $\iota: S \times N \to FA(S \times N)$ and a UNIQUE group homomorphism $\phi: FA(S \times N) \to A$ such that:

$\phi \circ \iota = f$

(this is called "unique extension from generators" of $\phi$)

The size of any basis (generating set) for $FA(X)$ (typically $|X|$) is called the *rank* of the free abelian group. This is the analogue of dimension for free abelian groups (but be careful...there are subtleties to be considered for abelian groups that are *not* free, including any finite abelian group).

Anyway, what we are going to aim at, is the creation of a tensor product $S \otimes_R N$, in which $s \otimes n$ is going to be a "stand-in" for $sn$ (and we are going to realize $N$ as the elementary tensors $1_S \otimes n$). By construction, we wind up with an abelian group where the map:

$\otimes: S \times N \to S\otimes_R N$

is ($\Bbb Z$-)bilinear, or "bi-additive". This mimics the first two rules of $\mu$ (the usual $S$-action, where defined):

1. $(s_1 + s_2)\cdot n = s_1\cdot n + s_2\cdot n$
2. $s\cdot(n_1+n_2) = s\cdot n_1 + s\cdot n_2$

so we just need to define an $S$-action that gives us the other $S$-module axioms. This is what defining:

$s\cdot \left(\sum\limits_i s_i \otimes n_i\right) = \sum\limits_i (ss_i)\otimes n$

does for us.

 

[Math Amateur]

We want an abelian group to have an $S$-module structure ($S$-modules are basically abelian groups "with extra structure"-namely, an $S$-action).

We would also like this abelian group to"contain" $N$, or something like it, and the $S$-action is a map $\mu: S \times N \to N$ (we usually write $\mu(s,n)$ as $s\cdot n$, or simply $sn$).

We can create the free abelian group on the "generating set" $S \times N$ in one of several ways:

1. Create the free group $F(S \times N)$ and pass to the quotient modulo the commutator subgroup.

2. Take the direct sum of $|S \times N|$ copies of $\Bbb Z$: $FA(S \times N) \cong \bigoplus\limits_{\alpha \in S\times N} \Bbb Z$

3. Take every element of $S \times N$ to be a "basis" element, and take formal $\Bbb Z$-linear combinations (this is *most* like the construction of "the vector space based on $X$" Cooperstein writes about).

4. Create *any* object that satisfies the following universal mapping property: for any mapping $f: S \times N \to A$, where $A$ is an abelian group, there exists a injection: $\iota: S \times N \to FA(S \times N)$ and a UNIQUE group homomorphism $\phi: FA(S \times N) \to A$ such that:

$\phi \circ \iota = f$

(this is called "unique extension from generators" of $\phi$)

The size of any basis (generating set) for $FA(X)$ (typically $|X|$) is called the *rank* of the free abelian group. This is the analogue of dimension for free abelian groups (but be careful...there are subtleties to be considered for abelian groups that are *not* free, including any finite abelian group).

Anyway, what we are going to aim at, is the creation of a tensor product $S \otimes_R N$, in which $s \otimes n$ is going to be a "stand-in" for $sn$ (and we are going to realize $N$ as the elementary tensors $1_S \otimes n$). By construction, we wind up with an abelian group where the map:

$\otimes: S \times N \to S\otimes_R N$

is ($\Bbb Z$-)bilinear, or "bi-additive". This mimics the first two rules of $\mu$ (the usual $S$-action, where defined):

1. $(s_1 + s_2)\cdot n = s_1\cdot n + s_2\cdot n$
2. $s\cdot(n_1+n_2) = s\cdot n_1 + s\cdot n_2$

so we just need to define an $S$-action that gives us the other $S$-module axioms. This is what defining:

$s\cdot \left(\sum\limits_i s_i \otimes n_i\right) = \sum\limits_i (ss_i)\otimes n$

does for us.

Thanks for the help, Deveno ...

Just working through your post now ...

You have certainly given me a lot to think about ...

Peter

 

[Math Amateur]

We want an abelian group to have an $S$-module structure ($S$-modules are basically abelian groups "with extra structure"-namely, an $S$-action).

We would also like this abelian group to"contain" $N$, or something like it, and the $S$-action is a map $\mu: S \times N \to N$ (we usually write $\mu(s,n)$ as $s\cdot n$, or simply $sn$).

We can create the free abelian group on the "generating set" $S \times N$ in one of several ways:

1. Create the free group $F(S \times N)$ and pass to the quotient modulo the commutator subgroup.

2. Take the direct sum of $|S \times N|$ copies of $\Bbb Z$: $FA(S \times N) \cong \bigoplus\limits_{\alpha \in S\times N} \Bbb Z$

3. Take every element of $S \times N$ to be a "basis" element, and take formal $\Bbb Z$-linear combinations (this is *most* like the construction of "the vector space based on $X$" Cooperstein writes about).

4. Create *any* object that satisfies the following universal mapping property: for any mapping $f: S \times N \to A$, where $A$ is an abelian group, there exists a injection: $\iota: S \times N \to FA(S \times N)$ and a UNIQUE group homomorphism $\phi: FA(S \times N) \to A$ such that:

$\phi \circ \iota = f$

(this is called "unique extension from generators" of $\phi$)

The size of any basis (generating set) for $FA(X)$ (typically $|X|$) is called the *rank* of the free abelian group. This is the analogue of dimension for free abelian groups (but be careful...there are subtleties to be considered for abelian groups that are *not* free, including any finite abelian group).

Anyway, what we are going to aim at, is the creation of a tensor product $S \otimes_R N$, in which $s \otimes n$ is going to be a "stand-in" for $sn$ (and we are going to realize $N$ as the elementary tensors $1_S \otimes n$). By construction, we wind up with an abelian group where the map:

$\otimes: S \times N \to S\otimes_R N$

is ($\Bbb Z$-)bilinear, or "bi-additive". This mimics the first two rules of $\mu$ (the usual $S$-action, where defined):

1. $(s_1 + s_2)\cdot n = s_1\cdot n + s_2\cdot n$
2. $s\cdot(n_1+n_2) = s\cdot n_1 + s\cdot n_2$

so we just need to define an $S$-action that gives us the other $S$-module axioms. This is what defining:

$s\cdot \left(\sum\limits_i s_i \otimes n_i\right) = \sum\limits_i (ss_i)\otimes n$

does for us.

Hi Deveno,Just a minor issue that I hope you can help with ...

You write:"We can create the free abelian group on the "generating set" $S \times N$ in one of several ways:

1. Create the free group $F(S \times N)$ and pass to the quotient modulo the commutator subgroup. ... ...
BUT ... how do you create the free group $F(S \times N)$ ... do you take all the words based on the set \(\displaystyle S \times N\) ... so there would be words like :

\(\displaystyle (n_1, s_1) (n_6, s_6)^{-1} (n_{31}, s_{31}) (n_{31}, s_{31})^{-1}\) ...and then reduce the words ...Is that right ...?


... ... if it is correct, then ...How would you take the required quotient exactly ...?


A further question regarding the meaning of the notation \(\displaystyle FA(S \times N)\) is as follows ...... now \(\displaystyle F(S \times N)\) is the free group generated by the set \(\displaystyle S \times N\) ... ... BUT ... ... what is meant by \(\displaystyle FA(S \times N)\) ... ?
Hope you can help ...

Peter

 

[Deveno]

Hi Deveno,Just a minor issue that I hope you can help with ...

You write:"We can create the free abelian group on the "generating set" $S \times N$ in one of several ways:

1. Create the free group $F(S \times N)$ and pass to the quotient modulo the commutator subgroup. ... ...
BUT ... how do you create the free group $F(S \times N)$ ... do you take all the words based on the set \(\displaystyle S \times N\) ... so there would be words like :

\(\displaystyle (n_1, s_1) (n_6, s_6)^{-1} (n_{31}, s_{31}) (n_{31}, s_{31})^{-1}\) ...and then reduce the words ...Is that right ...?

Yes (note you can reduce your "example word" since it has an inverse pair that are adjacent).

... ... if it is correct, then ...How would you take the required quotient exactly ...?

In any group (free, or otherwise), the commutator subgroup is defined as:

$[G,G] = \langle aba^{-1}b^{-1}: a,b \in G\rangle$.

Elements of the form $aba^{-1}b^{-1}$ are called *commutators* since:

$ab = ba \iff aba^{-1}b^{-1} = e$.

The SET of all commutators usually isn't a subgroup, so we have to take all possible *products* of commutators to ensure we have a subgroup.

The *abelianization* of a group $G$ is DEFINED to be $G/[G,G]$.

Now, it may not immediately be apparent that $[G,G]$ is normal in $G$. But look, the conjugate of a product is the product of the conjugates:

$g(ab)g^{-1} = (gag^{-1})(gbg^{-1})$, so to show that $[G,G]$ is normal it suffices to show the conjugate of a commutator is a commutator:

$g(aba^{-1}b^{-1})g^{-1} = (gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1}) = (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}$.

If one adopts the notation $aba^{-1}b^{-1} = [a,b]$, and $gag^{-1} =\ ^ga$, then this becomes:

$^g[a,b] = [^ga,^gb]$

The abelianization of $G$ satisfies the following universal mapping property:

If $\pi: G \to G/[G,G]$ is the canonical quotient map, and $\phi:G \to A$ is any homomorphism from $G$ to an abelian group $A$, then we say $\phi$ factors through $\pi$: that is, there exists a unique homomorphism:

$\psi: G/[G,G] \to A$ such that $\phi = \psi\circ \pi$.

Namely, we set $\psi(g[G,G]) = \phi(g)$. To show this is well-defined, suppose:

$g[G,G] = h[G,G]$, so that $gh^{-1} \in [G,G]$.

Note that since $A$ is abelian, for any commutator $[x,y]$ we have:

$\phi([x,y]) = \phi(xyx^{-1}y^{-1}) = \phi(x)\phi(y)\phi(x)^{-1}\phi(y)^{-1} = \phi(x)\phi(x)^{-1}\phi(y)\phi(y)^{-1} = e$.

Hence $\phi([G,G]) = \{e\}$, that is $[G,G] \subseteq \text{ker }\phi$.

So since $gh^{-1} \in [G,G]$, we have $\phi(g)\phi(h)^{-1} = e$, and thus $\phi(g) = \phi(h)$, and $\psi$ is well-defined (constant on cosets of $[G,G]$).

What this means "in lay terms" is $[G,G]$ is the "smallest normal subgroup $N$" such that $G/N$ is abelian (note if $G$ is *already* abelian that $[G,G] = \{e\}$).

perhaps it isn't obvious that $G/[G,G]$ is indeed abelian:

$(x[G,G])(y[G,G]) = xy[G,G] = xy[x^{-1},y^{-1}][G,G]$ (since $[x^{-1},y^{-1}] \in [G,G]$)

$= (xy)(y^{-1}x^{-1}yx)[G,G] = yx[G,G] = (y[G,G])(x[G,G])$.

So to "abelianize" the free group $F(S \times N)$ one creates:

$F(S \times N)/[F(S \times N),F(S \times N)]$

which basically adds the "reduction rule" $(s,n)(s',n')(s,n)^{-1}(s',n')^{-1} = \ \ $ (the empty word).

A further question regarding the meaning of the notation \(\displaystyle FA(S \times N)\) is as follows ...... now \(\displaystyle F(S \times N)\) is the free group generated by the set \(\displaystyle S \times N\) ... ... BUT ... ... what is meant by \(\displaystyle FA(S \times N)\) ... ?
Hope you can help ...

Peter

$F$ for "free", $FA$ for "free abelian".

 

[Math Amateur]

Yes (note you can reduce your "example word" since it has an inverse pair that are adjacent).

In any group (free, or otherwise), the commutator subgroup is defined as:

$[G,G] = \langle aba^{-1}b^{-1}: a,b \in G\rangle$.

Elements of the form $aba^{-1}b^{-1}$ are called *commutators* since:

$ab = ba \iff aba^{-1}b^{-1} = e$.

The SET of all commutators usually isn't a subgroup, so we have to take all possible *products* of commutators to ensure we have a subgroup.

The *abelianization* of a group $G$ is DEFINED to be $G/[G,G]$.

Now, it may not immediately be apparent that $[G,G]$ is normal in $G$. But look, the conjugate of a product is the product of the conjugates:

$g(ab)g^{-1} = (gag^{-1})(gbg^{-1})$, so to show that $[G,G]$ is normal it suffices to show the conjugate of a commutator is a commutator:

$g(aba^{-1}b^{-1})g^{-1} = (gag^{-1})(gbg^{-1})(ga^{-1}g^{-1})(gb^{-1}g^{-1}) = (gag^{-1})(gbg^{-1})(gag^{-1})^{-1}(gbg^{-1})^{-1}$.

If one adopts the notation $aba^{-1}b^{-1} = [a,b]$, and $gag^{-1} =\ ^ga$, then this becomes:

$^g[a,b] = [^ga,^gb]$

The abelianization of $G$ satisfies the following universal mapping property:

If $\pi: G \to G/[G,G]$ is the canonical quotient map, and $\phi:G \to A$ is any homomorphism from $G$ to an abelian group $A$, then we say $\phi$ factors through $\pi$: that is, there exists a unique homomorphism:

$\psi: G/[G,G] \to A$ such that $\phi = \psi\circ \pi$.

Namely, we set $\psi(g[G,G]) = \phi(g)$. To show this is well-defined, suppose:

$g[G,G] = h[G,G]$, so that $gh^{-1} \in [G,G]$.

Note that since $A$ is abelian, for any commutator $[x,y]$ we have:

$\phi([x,y]) = \phi(xyx^{-1}y^{-1}) = \phi(x)\phi(y)\phi(x)^{-1}\phi(y)^{-1} = \phi(x)\phi(x)^{-1}\phi(y)\phi(y)^{-1} = e$.

Hence $\phi([G,G]) = \{e\}$, that is $[G,G] \subseteq \text{ker }\phi$.

So since $gh^{-1} \in [G,G]$, we have $\phi(g)\phi(h)^{-1} = e$, and thus $\phi(g) = \phi(h)$, and $\psi$ is well-defined (constant on cosets of $[G,G]$).

What this means "in lay terms" is $[G,G]$ is the "smallest normal subgroup $N$" such that $G/N$ is abelian (note if $G$ is *already* abelian that $[G,G] = \{e\}$).

perhaps it isn't obvious that $G/[G,G]$ is indeed abelian:

$(x[G,G])(y[G,G]) = xy[G,G] = xy[x^{-1},y^{-1}][G,G]$ (since $[x^{-1},y^{-1}] \in [G,G]$)

$= (xy)(y^{-1}x^{-1}yx)[G,G] = yx[G,G] = (y[G,G])(x[G,G])$.

So to "abelianize" the free group $F(S \times N)$ one creates:

$F(S \times N)/[F(S \times N),F(S \times N)]$

which basically adds the "reduction rule" $(s,n)(s',n')(s,n)^{-1}(s',n')^{-1} = \ \ $ (the empty word).

$F$ for "free", $FA$ for "free abelian".

Thanks Deveno ... really help ... good to see what's involved ... quite a lesson for me ... indeed I sometimes find your grasp of abstract algebra quite unnerving :) ...

Still reflecting on what you have written ... indeed, thinking of revising some relevant group theory ...Just one minor point ...

The free group for \(\displaystyle S \times N\) has elements of the form

\(\displaystyle (s_1, n_1) (s_6, n_6)^{-1} (s_{31}, n_{31}) \) ...How do we get to elements like

\(\displaystyle (s_1, n_1) + (s_6, n_6)^{-1} + (s_{31}, n_{31}) \) ... Is this just a matter of choosing the notation for the group operation or is their more in moving from concatenation to 'plus'Peter

 

[Deveno]

Thanks Deveno ... really help ... good to see what's involved ... quite a lesson for me ... indeed I sometimes find your grasp of abstract algebra quite unnerving :) ...

Still reflecting on what you have written ... indeed, thinking of revising some relevant group theory ...Just one minor point ...

The free group for \(\displaystyle S \times N\) has elements of the form

\(\displaystyle (s_1, n_1) (s_6, n_6)^{-1} (s_{31}, n_{31}) \) ...How do we get to elements like

\(\displaystyle (s_1, n_1) + (s_6, n_6)^{-1} + (s_{31}, n_{31}) \) ... Is this just a matter of choosing the notation for the group operation or is their more in moving from concatenation to 'plus'Peter

Well, yes, and no. Since we're "forcing" + to be commutative, we get extra reductions. For example, let's say our generating set has two elements $X = \{a,b\}$.

The word:

$aba'bab$ is reduced in the free group, since we don't have any "adjacent-inverse-pairs".

However, when we write it "additively", we have:

$a + b - a + b + a + b$, and since we're in an abelian group, now, we have:

$= a - a + b + b + a + b = 0 + 2b + a + b = 2b + a + b = a + 2b + b = a + 3b$.

Let's derive this in "multiplicative form", using the rule $xyx'y' = \ $ (for any words $x,y$). So we can insert, and remove such strings at will.

So let's put one such string of the form $ab'a'b$ after the first occurence of $a'$:

$aba'(ab'a'b)bab$
$ab(a'a)b'a'bbab$
$a(bb')a'bbab$
$(aa')bbab$
$bbab$

Now let's insert a string of the form $aba'b'$ after the first $b$:

$b(aba'b')bab$
$baba'(b'b)ab$
$bab(a'a)b$
$babb$

Now we'll insert $a'b'ab$ after the $a$:

$ba(a'b'ab)bb$
$b(aa')b'abbb$
$(bb')abbb$
$abbb = ab^3$, and it is clear that $ab^3 \mapsto a + 3b$ when we switch from "multiplicative" to "additive" notation.

In other words, being able to "reduce" commutators, allows us to switch "runs of $a$'s" and "runs of $b$'s" at will. So our "reduced words" can be put in the form:

$a^kb^m: k,m \in \Bbb Z$

It's "fairly obvious" that now the $a$'s and $b$'s are just "placeholders", we may as well replace them with:

$(k,m)$ (which we can think of as "representing" $ka + mb$).