MHB Tensor Products - Dummit and Foote Section 10.4 Corollary 9

[Math Amateur]

Do you understand the three points first? This is just leading up to the punchline of the proof.

Well, yes, I believe I understand the three points ...

Whether, of course, I have as much insight into those points as you do, however, I very much doubt ... ...

I must say I still have the issues outlined in my last post ...

Maybe I have missed some of the implications of the three points ...

Peter

 

[ThePerfectHacker]

As before $N$ might not be scalar extendable to a module over $S$. But we do the map $i:N\to S\otimes N$ where $S\otimes N$ is a module over $S$. Since this map is not necessary injective this is not necessary an embedding of $N$ into a larger module over $S$. However, if we mod out $\ker i$ we are left with a new map $i_*: N/\ker i \to S\otimes N$ which is now injective (do you see why?). Thus, $N/\ker i$ can be enlarged (embedded into) a module over $S$, and so scalars can be extended.

Now suppose that we have a quotient $N/P$, for some submodule $P$ of $N$, for which scalars can be extended. Thus, there is a module $M$ over $S$ and an embedding $j:N/P \to M$. This map induces a map $j_*:N\to M$ defined by $j_*(n) = j(n+P)$ and with $\ker j_* = P$. Now by the universal property of the tensor product there is a map of $S$-modules $\varphi:S\otimes N\to M$ such that $j_* = \varphi i$. It follows from this that $\ker i \subseteq \ker j_*$, for reasons we explained before. But notice that $\ker j_* = P$ so we have shown that $\ker i \subseteq P$.

Therefore, every quotient of $N$, of the form $N/P$, for which scalars can be extended, must satisfy the condition that $\ker i \subseteq P$ and so $N/\ker i$ is "larger" than $N/P$. Thus, $N/\ker i$ is the largest of all quotients for which scalars can be extended.

 

[Deveno]

Basically, "the larger the quotient, the smaller the kernel", and vice-versa.

Since we know that at least for SOME $R$ and an $R$-module $N$, and some extension $S$ and an $S$-module $M$ we can't injectively map $N$ to $M$ (there have been several examples shown), it is a legitimate question to ask:

"What characterizes the largest possible quotient of $N$ that can possibly work (given $R,S,N, M$)"?

So we are looking for a $R$-submodule $P$ of $N$ such that we can embed:

$N/P \to M$.

Let's call such a possible embedding $f$. What is $f$? It is an $R$-module monomorphism.

So how and why do we involve the tensor product?

Well the module axioms force upon us the $R$-bilinearity of the scalar product in $M$ as a mapping: $S \times N \to M,\ (s,n) \mapsto sn$.

And tensor products have (as a DEFINING property) that any such bilinear map factors through the tensor product.

The tensor product is two things:

1) A bilinear map $\otimes: S \times N \to S \otimes_R N$

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

2) the $S$-module that is the "target" of this bilinear map.

For the time being, let's call $sn, B(s,n)$. By the defining property of tensor products, we are guaranteed that there is an $S$-module homomorphism: $\phi: S \otimes_R N \to M$, with:

$B = \phi \circ \otimes$

Now, I claim:

$f: N \to M,\ f(n) = B(1,n)$ is an $R$-module homomorphism.

Let's check this:

$f(n + n') = B(1,n+n') = (\phi \circ \otimes)(1,n+n')$

$=\phi(1\otimes(n+n')) = \phi(1\otimes n + 1\otimes n')$

$= \phi(1\otimes n) + \phi(1\otimes n')$ (since $\phi$ is a homomorphism)

$=\phi(\otimes(1,n)) + \phi(\otimes(1,n')) = B(1,n) + B(1,n')$

$= f(n) + f(n')$, so $f$ is additive (a shorter proof uses the bilinearity of $B$).

$f(rn) = B(1,rn) = (\phi \circ \otimes)(1,rn)$

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

(since $r \in R \subseteq S$ and $\phi$ is an $S$-module homomorphism)

$=r((\phi \circ \otimes)(1,n)) = r(B(1,n)) = rf(n)$

(this also follows from bilinearity of $B$, but I wanted to show the "nuts and bolts").

So $f$ is $R$-linear.

To turn this into an injection (embedding), we want to take the $R$ quotient module:

$N/\text{ker }f$ which gives us one possible embedding:

$f_{\ast}: N/\text{ker }f \to M$.

We want to show that this embedding $f_{\ast}$ is "smaller" than the one given by the Corollary in D&F, that is, that the kernel of $\iota$ is contained in $\text{ker }f$.

Alright, let's go back to the $R$-module homomorphism $\iota$ now:

$\iota: N \to S \otimes_R N$ given by $\iota(n) = 1\otimes n$.

Suppose $n_0 \in \text{ker }\iota$. This means that:

$1 \otimes n_0 = 0$ (the 0-element of the tensor product).

Since $f(n_0) = B(1,n_0) = \phi(1\otimes n_0) = \phi(0) = 0$ (here the last 0 is in $M$), we see that the kernel of $\iota$ is indeed contained in $\text{ker }f$.

***********

It may not be clear that every embedding arises in this way. What *is* clear is that any such embedding defines an $R$-bilinear map from $S \times N \to M$ (namely the $S$-multiplication in $M$).

If $g:N \to M$ is a $R$-module homomorphism, the map:

$B'(s,n) = sg(n)$ has to be ($R$-) bilinear.

Since we have a homomorphism:

$\phi': S \otimes_R N \to M$ with:

$sg(n) = \phi'(s\otimes n) = \phi'(s(1\otimes n)) = s\phi'(1\otimes n)$

$= s\phi'(\iota(n))$, it follows that $\text{ker }\iota \subseteq \text{ker }g$.

If the kernel of $g$ is trivial (that is, if $g$ is injective), the kernel of $\iota$ has to be.

By the same token, if instead we have an $R$-module homomorphism:

$g_{\ast}: N/P \to M$ which is injective, we can define (extend this to) a homomorphism:

$g: N \to M$ by letting $g(n) = g_{\ast}(n + P)$

What is $\text{ker }g$?

(here is a good exercise for you to do:

Let $f: \Bbb Z_m \to \Bbb Z_n$ be an abelian group homomorphism. Define an abelian group homomorphism

$g:\Bbb Z \to \Bbb Z_n$ using $f$. If you want a specific $m$ and $n$ use $m = 8, n= 12$).

 

[Math Amateur]

Basically, "the larger the quotient, the smaller the kernel", and vice-versa.

Since we know that at least for SOME $R$ and an $R$-module $N$, and some extension $S$ and an $S$-module $M$ we can't injectively map $N$ to $M$ (there have been several examples shown), it is a legitimate question to ask:

"What characterizes the largest possible quotient of $N$ that can possibly work (given $R,S,N, M$)"?

So we are looking for a $R$-submodule $P$ of $N$ such that we can embed:

$N/P \to M$.

Let's call such a possible embedding $f$. What is $f$? It is an $R$-module monomorphism.

So how and why do we involve the tensor product?

Well the module axioms force upon us the $R$-bilinearity of the scalar product in $M$ as a mapping: $S \times N \to M,\ (s,n) \mapsto sn$.

And tensor products have (as a DEFINING property) that any such bilinear map factors through the tensor product.

The tensor product is two things:

1) A bilinear map $\otimes: S \times N \to S \otimes_R N$

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

2) the $S$-module that is the "target" of this bilinear map.

For the time being, let's call $sn, B(s,n)$. By the defining property of tensor products, we are guaranteed that there is an $S$-module homomorphism: $\phi: S \otimes_R N \to M$, with:

$B = \phi \circ \otimes$

Now, I claim:

$f: N \to M,\ f(n) = B(1,n)$ is an $R$-module homomorphism.

Let's check this:

$f(n + n') = B(1,n+n') = (\phi \circ \otimes)(1,n+n')$

$=\phi(1\otimes(n+n')) = \phi(1\otimes n + 1\otimes n')$

$= \phi(1\otimes n) + \phi(1\otimes n')$ (since $\phi$ is a homomorphism)

$=\phi(\otimes(1,n)) + \phi(\otimes(1,n')) = B(1,n) + B(1,n')$

$= f(n) + f(n')$, so $f$ is additive (a shorter proof uses the bilinearity of $B$).

$f(rn) = B(1,rn) = (\phi \circ \otimes)(1,rn)$

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

(since $r \in R \subseteq S$ and $\phi$ is an $S$-module homomorphism)

$=r((\phi \circ \otimes)(1,n)) = r(B(1,n)) = rf(n)$

(this also follows from bilinearity of $B$, but I wanted to show the "nuts and bolts").

So $f$ is $R$-linear.

To turn this into an injection (embedding), we want to take the $R$ quotient module:

$N/\text{ker }f$ which gives us one possible embedding:

$f_{\ast}: N/\text{ker }f \to M$.

We want to show that this embedding $f_{\ast}$ is "smaller" than the one given by the Corollary in D&F, that is, that the kernel of $\iota$ is contained in $\text{ker }f$.

Alright, let's go back to the $R$-module homomorphism $\iota$ now:

$\iota: N \to S \otimes_R N$ given by $\iota(n) = 1\otimes n$.

Suppose $n_0 \in \text{ker }\iota$. This means that:

$1 \otimes n_0 = 0$ (the 0-element of the tensor product).

Since $f(n_0) = B(1,n_0) = \phi(1\otimes n_0) = \phi(0) = 0$ (here the last 0 is in $M$), we see that the kernel of $\iota$ is indeed contained in $\text{ker }f$.

***********

It may not be clear that every embedding arises in this way. What *is* clear is that any such embedding defines an $R$-bilinear map from $S \times N \to M$ (namely the $S$-multiplication in $M$).

If $g:N \to M$ is a $R$-module homomorphism, the map:

$B'(s,n) = sg(n)$ has to be ($R$-) bilinear.

Since we have a homomorphism:

$\phi': S \otimes_R N \to M$ with:

$sg(n) = \phi'(s\otimes n) = \phi'(s(1\otimes n)) = s\phi'(1\otimes n)$

$= s\phi'(\iota(n))$, it follows that $\text{ker }\iota \subseteq \text{ker }g$.

If the kernel of $g$ is trivial (that is, if $g$ is injective), the kernel of $\iota$ has to be.

By the same token, if instead we have an $R$-module homomorphism:

$g_{\ast}: N/P \to M$ which is injective, we can define (extend this to) a homomorphism:

$g: N \to M$ by letting $g(n) = g_{\ast}(n + P)$

What is $\text{ker }g$?

(here is a good exercise for you to do:

Let $f: \Bbb Z_m \to \Bbb Z_n$ be an abelian group homomorphism. Define an abelian group homomorphism

$g:\Bbb Z \to \Bbb Z_n$ using $f$. If you want a specific $m$ and $n$ use $m = 8, n= 12$).

Thanks Deveno.

I am now working through, and reflecting upon, your post.

Just a quick clarification:

You write:
--------------------------------------------------------------------------

"Since we know that at least for SOME $R$ and an $R$-module $N$, and some extension $S$ and an $S$-module $M$ we can't injectively map $N$ to $M$ (there have been several examples shown), it is a legitimate question to ask:

"What characterizes the largest possible quotient of $N$ that can possibly work (given $R,S,N, M$)"? '

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

In what you are saying, is M actually the L of Dummit and Foote in this set up.

So, in Corollary 9, we are actually seeking to map a quotient of N into L? ... and further, the tensor product is only a step on the path to do this ...? Further, is the S-module where we have actually succeeded in "extending the scalars" actually L?Also, when you write:

------------------------------------------------------------------------------
"The tensor product is two things:

1) A bilinear map $\otimes: S \times N \to S \otimes_R N$

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

2) the $S$-module that is the "target" of this bilinear map."

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

In this case, the S-module that is the "target" is $$ S \otimes_R N $$ - or is it L? (I think it is $$S \otimes_R N$$ ... but can you confirm?

Peter

PS Just thinking that some of my confusion over Theorem 8 and Corollary 9 is due to D&F introducing extension of the scalars, in a sense, before tensor products (or at least at the same time as) - instead of introducing the extension of the scalars after tensor products - some of the confusion anyway :)

 

[Math Amateur]

As before $N$ might not be scalar extendable to a module over $S$. But we do the map $i:N\to S\otimes N$ where $S\otimes N$ is a module over $S$. Since this map is not necessary injective this is not necessary an embedding of $N$ into a larger module over $S$. However, if we mod out $\ker i$ we are left with a new map $i_*: N/\ker i \to S\otimes N$ which is now injective (do you see why?). Thus, $N/\ker i$ can be enlarged (embedded into) a module over $S$, and so scalars can be extended.

Now suppose that we have a quotient $N/P$, for some submodule $P$ of $N$, for which scalars can be extended. Thus, there is a module $M$ over $S$ and an embedding $j:N/P \to M$. This map induces a map $j_*:N\to M$ defined by $j_*(n) = j(n+P)$ and with $\ker j_* = P$. Now by the universal property of the tensor product there is a map of $S$-modules $\varphi:S\otimes N\to M$ such that $j_* = \varphi i$. It follows from this that $\ker i \subseteq \ker j_*$, for reasons we explained before. But notice that $\ker j_* = P$ so we have shown that $\ker i \subseteq P$.

Therefore, every quotient of $N$, of the form $N/P$, for which scalars can be extended, must satisfy the condition that $\ker i \subseteq P$ and so $N/\ker i$ is "larger" than $N/P$. Thus, $N/\ker i$ is the largest of all quotients for which scalars can be extended.

Thanks ThePerfectHacker.

Now working through your post and reflecting on what you have said.

Thanks again for your extensive help.

Peter

 

[Deveno]

Thanks Deveno.

I am now working through, and reflecting upon, your post.

Just a quick clarification:

You write:
--------------------------------------------------------------------------

"Since we know that at least for SOME $R$ and an $R$-module $N$, and some extension $S$ and an $S$-module $M$ we can't injectively map $N$ to $M$ (there have been several examples shown), it is a legitimate question to ask:

"What characterizes the largest possible quotient of $N$ that can possibly work (given $R,S,N, M$)"? '

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

In what you are saying, is M actually the L of Dummit and Foote in this set up.

So, in Corollary 9, we are actually seeking to map a quotient of N into L? ... and further, the tensor product is only a step on the path to do this ...? Further, is the S-module where we have actually succeeded in "extending the scalars" actually L?

Yes.

Also, when you write:

------------------------------------------------------------------------------
"The tensor product is two things:

1) A bilinear map $\otimes: S \times N \to S \otimes_R N$

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

2) the $S$-module that is the "target" of this bilinear map."

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

In this case, the S-module that is the "target" is $$ S \otimes_R N $$ - or is it L? (I think it is $$S \otimes_R N$$ ... but can you confirm?

Peter

by target, I mean the co-domain of the tensor product bilinear mapping.

PS Just thinking that some of my confusion over Theorem 8 and Corollary 9 is due to D&F introducing extension of the scalars, in a sense, before tensor products (or at least at the same time as) - instead of introducing the extension of the scalars after tensor products - some of the confusion anyway :)

My thought is that they were trying to introduce the tensor product in a limited setting first, as an abelian group construction ($\Bbb Z$-module) before tensoring over a ring. Other structures (most notably vector spaces and their duals) have tensor products as well.

I think Dummit and Foote assume one has some familiarity already with linear algebra (at least with the notions of basis, linear combination and matrix representation), and therefore some of their exposition on various topics is rather brief. Don't get me wrong, it's a darn good algebra book, but it's also known for being a challenge to work through.

 

[Math Amateur]

Yes.
by target, I mean the co-domain of the tensor product bilinear mapping.
My thought is that they were trying to introduce the tensor product in a limited setting first, as an abelian group construction ($\Bbb Z$-module) before tensoring over a ring. Other structures (most notably vector spaces and their duals) have tensor products as well.

I think Dummit and Foote assume one has some familiarity already with linear algebra (at least with the notions of basis, linear combination and matrix representation), and therefore some of their exposition on various topics is rather brief. Don't get me wrong, it's a darn good algebra book, but it's also known for being a challenge to work through.

Thanks Deveno.

You write:

"by target, I mean the co-domain of the tensor product bilinear mapping."

... since as you write: "bilinear map $\otimes: S \times N \to S \otimes_R N$ which takes $(s,n) \mapsto s\otimes n$ ...

that co-domain is $$ S \otimes_R N $$. Is that right? Can you confirm?

Peter

 

[Deveno]

Yes, that is indeed what I intended. The module $S \otimes_R N$ doesn't mean anything without the bilinear map $\otimes$.

In fact, most of the time, one can just think of $\otimes$ as "the most general bilinear map possible." Most of the theorems concerning the tensor product are to reinforce the utility of this view.