MHB Tensor Products - Dummit and Foote Section 10.4 Corollary 9

[Math Amateur]

I am reading Dummit and Foote, Section 10.4: Tensor Products of Modules. I am studying Corollary 9 and attempting to fully understand the Corollary and it proof. (For details see the attachement page 362 in which Theorem 8 is stated and proved. This is followed by the statement and proof of Corollary 9.The proof of Corollary 9 reads as follows:

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

Proof: The quotient$$ N/ ker \ i $$ is mapped injectively (by i) into the S-module $$ S \oplus_R N$$.

Suppose now that $$ \phi $$ is an R-module homomorphism injecting the quotient $$ N/ker \ \phi $$ of N into an S-module L.

Then, by Theorem 8, ker i is mapped to 0 by $$ \phi $$, that is $$ ker \ i \subseteq ker \ \phi $$.

... ... etc etc

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

I do not fully understand how D&F reached the conclusion that $$ ker \ i \subseteq ker \ \phi $$

Can someone show me (formally and rigorously) why, as D&F assert, by Theorem 8, it follows that $$ ker \ i \subseteq ker \ \phi $$? (Edit : I suppose this reduces to the question of why, exactly, ker i is mapped to zero by $$ \phi $$.)

A simple diagram showing the maps of Corollary 9 is attached.Could someone also clarify the following issue for me:

In corollary 9 D&F refer to "the quotient $$ N/ker \ \phi $$ of $$N$$ ... ... does this actually mean the coset of the quotient module $$ N/ker \ \phi $$ or are they referring to the quotient module? (Hope I am making myself clear - I am a bit confused by the term ... )

Peter

 

[Math Amateur]

I am reading Dummit and Foote, Section 10.4: Tensor Products of Modules. I am studying Corollary 9 and attempting to fully understand the Corollary and it proof. (For details see the attachement page 362 in which Theorem 8 is stated and proved. This is followed by the statement and proof of Corollary 9.The proof of Corollary 9 reads as follows:

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

Proof: The quotient$$ N/ ker \ i $$ is mapped injectively (by i) into the S-module $$ S \oplus_R N$$.

Suppose now that $$ \phi $$ is an R-module homomorphism injecting the quotient $$ N/ker \ \phi $$ of N into an S-module L.

Then, by Theorem 8, ker i is mapped to 0 by $$ \phi $$, that is $$ ker \ i \subseteq ker \ \phi $$.

... ... etc etc

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

I do not fully understand how D&F reached the conclusion that $$ ker \ i \subseteq ker \ \phi $$

Can someone show me (formally and rigorously) why, as D&F assert, by Theorem 8, it follows that $$ ker \ i \subseteq ker \ \phi $$? (Edit : I suppose this reduces to the question of why, exactly, ker i is mapped to zero by $$ \phi $$.)

A simple diagram showing the maps of Corollary 9 is attached.Could someone also clarify the following issue for me:

In corollary 9 D&F refer to "the quotient $$ N/ker \ \phi $$ of $$N$$ ... ... does this actually mean the coset of the quotient module $$ N/ker \ \phi $$ or are they referring to the quotient module? (Hope I am making myself clear - I am a bit confused by the term ... )

Peter

Just been reflecting on my question - maybe we can se this by noting that in the notation of D&F $$ \phi = \Phi i $$

So if we take x belonging to the kernel of the map i i.e. I(x) = 0 then we have

$$ \phi (x) = \Phi \circ i (x) = \Phi (i(x)) $$

Now if i(x) = o we have

$$ \phi (x) = = \Phi (0) = \Phi(1 \oplus 0 ) = 1 \cdot \phi(0) $$Now if $$ \phi(0) = 0 $$ then all is well but does it? Is this a property of a homomorphism?

Can someone confirm that my reasoning is correct (or not)?

If it is correct then does $$ \phi(0) = 0 $$?

Peter

PS Now having difficulties with the rest of the proof of Corollary 9!

 

[ThePerfectHacker]

This is because by Theorem 8 we can factor $\varphi = \Phi i$. If $a\in \ker i$ then $\varphi(a) = \Phi(i(a))$ but $i(a) = 0$ since, by definition, $a\in \ker i$. Therefore, $\varphi(a) = \Phi(0) = 0$, and so $a\in \ker \Phi$. This shows that any element of $\ker i$ is in $\ker \Phi$.

 

[Math Amateur]

This is because by Theorem 8 we can factor $\varphi = \Phi i$. If $a\in \ker i$ then $\varphi(a) = \Phi(i(a))$ but $i(a) = 0$ since, by definition, $a\in \ker i$. Therefore, $\varphi(a) = \Phi(0) = 0$, and so $a\in \ker \Phi$. This shows that any element of $\ker i$ is in $\ker \Phi$.

Thanks ThePerfectHacker, I appreciate your help.

Just one point why does $$\Phi(0) = 0$$?

Peter

 

[Math Amateur]

Thanks ThePerfectHacker, I appreciate your help.

Just one point why does $$\Phi(0) = 0$$?

Peter

Thanks to ThePerfectHacker I am now confident that $$ ker \ i \subseteq ker \ \phi $$ in the proof of Corollary 9 (D&F page 362 - see attached for the proof)

After the statement that $$ ker \ i \subseteq ker \ \phi $$ D&F continue as follows:

"Hence $$ N/ker \ \phi $$ is a quotient of $$N/ker \ i $$ ( namely the quotient by the submodule $$ ker \ \phi / ker \ i $$. "

BUT ... what does the statement: "$$ N/ker \ \phi $$ is a quotient of $$N/ker \ i $$ " mean?

I can see that if $$ ker \ \phi $$ is a module (is it?), then since $$ ker \ i $$ is a subset of $$ ker \ \phi $$ we could form the quotient module $$ ker \ \phi / ker \ i $$ but how do we interpret the statement "$$ N/ker \ \phi $$ is a quotient of $$N/ker \ i $$ "?Next D&F write:

"It follows that $$N/ker \ i $$ is the unique largest quotient of N that can be embedded in any S-module. The last statement in the Corollary follows."

Can anyone help me by giving a clear explanation of why, exactly, it follows that $$N/ker \ i $$ is the unique largest quotient of N that can be embedded in any S-module, and then, indeed, why the last statement in the Corollary follows?

I would really appreciate some help.

Peter

 

[Deveno]

Recall that in groups in general, if $K$ is a normal subgroup of $G$, that is also a subgroup of a normal subgroup $H$ of $G$ (so that $K$ is likewise normal in $H$), we have:

$G/H \cong (G/K)/(H/K)$

and it is evident that $K$ (which is the kernel of the homomorphism $G \to G/K$) is a subgroup of the kernel of the homomorphism $G \to G/H$ (namely, $H$).

For abelian groups, ANY subgroup is normal. So we see if $K < H$, that $G/H$ is a quotient of $G/K$ (by $H/K$ of course).

Thus all one really has to do is verify that we get a well-defined $R$ action in the case of $R$-modules (and we do).

Also: ANY $R$-module homomorphism $f$ has to preserve 0:

$f(0) = f(m + -m) = f(m) + f(-m) = f(m) + f((-1)m)) = f(m) + (-1)f(m) = f(m) + -f(m) = 0$.

Recall that any $R$-module homomorphism is always a homomorphism of abelian groups, and abelian group homomorphisms preserve the group identity (as does ANY group homomorphism).

$R$-linear maps are always additive (as well as having other properties).

Every commutative ring with unity $R$ has an identity, and the additive subgroup generated by 1 is a quotient of $\Bbb Z$. This means we can consider any $R$ action as having a "sub-action" by $\Bbb Z$ (which is either an actual $\Bbb Z$ action if our ring has characteristic 0, or factors through a quotient map $\Bbb Z \to \Bbb Z/n\Bbb Z$).

The resulting $\Bbb Z$ action turns out just to give us $(M,+)$ as an abelian group, which shouldn't be surprising.

The thing to realize is that for $M \otimes N$ to "make sense", we have to have a "bi-action" in the middle: a RIGHT action on $M$, and a LEFT action on $N$.

If our ring is commutative, we can not worry about this too much, but in general it affects what tensor products are possible. In the general case, with an $R$-module $M$ and an $S$-module $N$, the structure on $M \otimes N$ is perhaps only an abelian group, without more information on the rings $R$ and $S$ (to have some OTHER ring action over the tensor product, we need some "compatibility" conditions, like some homomorphic relationship).

 

[ThePerfectHacker]

To understand the proof of the Theorem in Dummit and Foote make sure you understand the problem of extending scalars. Ignore the theorem for the time being and make sure you understand what problem they are trying to solve.

Suppose you have a ring $R$ and a module $N$ over this ring and say that $S$ is a subring of $R$. Since $N$ is a module over $R$ there is a binary operator $R\times N \to N$ which satisfies all the module properties. If we simply restrict this binary operator to $S\times N \to N$ then these module properties carry over and it allows us define a module structure of $N$ over $S$. This sort of simple construction is known as restricting scalars. A "scalar" is just another way of saying an element of the ring.

Now suppose we have a ring $R$ and a module $N$ over this ring but this time $R$ is a subring over $S$. The question now is can we "extend scalars", in other words, can we extend the binary operator $R\times N \to N$ to $S\times N\to N$ that still satisfies the module properties, and so this will produce $N$ as a module over $S$. This is not always possible and I think Dummit and Foote give an example of such.

Exercise: Find an example of rings $R\subset S$ and a module $N$ over $R$ such that scalars cannot be extended and so $N$ cannot be made into a module over $S$. Find an example of a rings $R\subset S$ and a module $N$ over $R$ such that scalars can be extended and so $N$ can be given a module structure over $S$.

The next natural question to explore is, okay it is not always possible to extend scalars in order to turn $N$ into a module over $S$. But perhaps we can enlarge $N$ itself into a bigger module $M$ such that $M$ is module over $S$. This question is exactly what Dummit and Foote are exploring in that section.

Before we go on you should understand that in mathematicians, especially in algebra, we deal with objects that often are not exactly sitting inside larger objects but in a sense they still are. For example, consider $\mathbb{Z}$ and $\mathbb{Z}\times \mathbb{Z}$ as groups. The group $\mathbb{Z}$ is technically not sitting inside $\mathbb{Z}\times \mathbb{Z}$ because it is not a subset, however, $\{ 1 \}\times \mathbb{Z}$ is essentially the same group and it is legit sitting inside $\mathbb{Z}\times \mathbb{Z}$. The way we make this notion precise is by saying that there is an embedding $j:\mathbb{Z}\to \mathbb{Z}\times \mathbb{Z}$ defined by $j(n) = (1,n)$, this map is a homomorphism and it is injective, so it identifies $\mathbb{Z}$ with a subset of $\mathbb{Z}\times \mathbb{Z}$ that is isomorphic to it.

Let us make the question of extending scalars to a larger module formal this time. The problem is given a ring $R$ and a ring $S$, with $R$ a subring of $S$, and a module $N$ over $R$. We say that we can extend scalars if there is an $S$ module, $M$, together with an embedding $j:N\to M$ as an $R$-module. [So in a sense $N$ is sitting inside a larger module $M$ though technically it does not need to be a legit subset of $M$.]

You have to understand this first. Otherwise that theorem is kinda pointless.

 

[Deveno]

One thing to point out:

If we have an extension ring $S$ of $R$, and an $R$-module $N$, then we have $S$ is an $R$-module (using the ring multiplication) so if we extend $N$ to an $S$-module $M$ "somehow" the $S$-module axioms dictate:

$(s+s')n = sn + s'n$
$s(n+n') = sn + sn'$

$s(rn) = (sr)n = (rs)n = r(sn)$
$(rs)n = r(sn)$

in other words: the map $B: S \times N \to M$ given by:

$(s,n) \mapsto sn$ is $R$-bilinear.

Well, what bilinear map will $B$ have to factor through (because EVERY $R$-bilinear map does)?

Once you answer THAT question, you get "some" $R$-module, and all that is necessary is to show you can actually get an $S$ action on it in a certain "natural" way.

********

I'd like to give an actual example, that may make things "more concrete". Take the ring $R = \Bbb R$, the real numbers. We have the extension ring $S = \Bbb C$ the complex field.

Now suppose we have a real vector space (an $\Bbb R$-module) like $\Bbb R^2$, the Euclidean plane. You should suspect that the object we want ought to be:

$\Bbb C \otimes_{\Bbb R} \Bbb R^2$

and that an elementary tensor is of the form:

$\alpha \otimes (x,y)$

Bilinearity let's us define a $\Bbb C$ action on this by:

$\beta(\alpha \otimes (x,y)) = (\beta\alpha)\otimes (x,y)$, and we can also write for

$\alpha = a + bi$:

$\alpha \otimes (x,y) = (a + bi)\otimes (x,y) = a\otimes (x,y) + (bi)\otimes(x,y)$

$= 1\otimes (ax,ay) + i\otimes(bx,by)$

Convince yourself that the map:

$\Bbb R^4 \to \Bbb C \otimes_{\Bbb R} \Bbb R^2$ given by:

$(a,b,x,y) \mapsto 1\otimes (ax,ay) + i\otimes (bx,by)$

is an $\Bbb R$-linear isomorphism, and that we can write the tensor product as a direct sum:

$\Bbb R^2 \oplus i\Bbb R^2$

This embeds the plane into a 4-dimensional space as:

$(x,y) \mapsto 1\otimes(x,y)$

Calculate the 4 coordinates of:

$(c + di)((1\otimes(x,y) + i\otimes(x',y'))$

seen as an element of $\Bbb R^4$.

This construction is called "the complexification of $\Bbb R^n$", and when applied to the one-dimensional vector space $\Bbb R$, gives the usual view of the complex numbers as a two-dimensional real vector space:

$\Bbb C \cong \Bbb R \oplus i\Bbb R$

 

[Math Amateur]

Recall that in groups in general, if $K$ is a normal subgroup of $G$, that is also a subgroup of a normal subgroup $H$ of $G$ (so that $K$ is likewise normal in $H$), we have:

$G/H \cong (G/K)/(H/K)$

and it is evident that $K$ (which is the kernel of the homomorphism $G \to G/K$) is a subgroup of the kernel of the homomorphism $G \to G/H$ (namely, $H$).

For abelian groups, ANY subgroup is normal. So we see if $K < H$, that $G/H$ is a quotient of $G/K$ (by $H/K$ of course).

Thus all one really has to do is verify that we get a well-defined $R$ action in the case of $R$-modules (and we do).

Also: ANY $R$-module homomorphism $f$ has to preserve 0:

$f(0) = f(m + -m) = f(m) + f(-m) = f(m) + f((-1)m)) = f(m) + (-1)f(m) = f(m) + -f(m) = 0$.

Recall that any $R$-module homomorphism is always a homomorphism of abelian groups, and abelian group homomorphisms preserve the group identity (as does ANY group homomorphism).

$R$-linear maps are always additive (as well as having other properties).

Every commutative ring with unity $R$ has an identity, and the additive subgroup generated by 1 is a quotient of $\Bbb Z$. This means we can consider any $R$ action as having a "sub-action" by $\Bbb Z$ (which is either an actual $\Bbb Z$ action if our ring has characteristic 0, or factors through a quotient map $\Bbb Z \to \Bbb Z/n\Bbb Z$).

The resulting $\Bbb Z$ action turns out just to give us $(M,+)$ as an abelian group, which shouldn't be surprising.

The thing to realize is that for $M \otimes N$ to "make sense", we have to have a "bi-action" in the middle: a RIGHT action on $M$, and a LEFT action on $N$.

If our ring is commutative, we can not worry about this too much, but in general it affects what tensor products are possible. In the general case, with an $R$-module $M$ and an $S$-module $N$, the structure on $M \otimes N$ is perhaps only an abelian group, without more information on the rings $R$ and $S$ (to have some OTHER ring action over the tensor product, we need some "compatibility" conditions, like some homomorphic relationship).

Hi Deveno,

Sorry, but I am going to have to ask for further clarification on your post that begins:

"Recall that in groups in general, if $K$ is a normal subgroup of $G$, that is also a subgroup of a normal subgroup $H$ of $G$ (so that $K$ is likewise normal in $H$), we have:

$G/H \cong (G/K)/(H/K)$

... ... etc "

I am assuming that you are answering my question as to how and why it follows that $$ ker \ i \subseteq ker \ \phi $$. Is that correct?

If that is the case then I have to confess to having some difficulty relating what you are saying to the entities of Theorem 8. Can you help?

I suspect that in what you have said $$G/H$$ is actually referring to $$ G/ker \ \phi $$ in the terminology and notation of D&F's Theorem 8, where G is the abelian group of the module N, so that $$ G \to G/H $$ is the homomorphism $$ \phi $$. So in the above notation, $$H = ker \ \phi$$.

Further, I suspect that that $$G/K$$ is referring to $$ G/ker \ i $$ in the terminology and notation of D&F's Theorem 8, and so $$ G \to G/K $$ is the mapping $$ i $$. So in the above notation, $$H = ker \ i $$.

If I am correct (and my interpretation of your post may be way off target!) then $$H/K $$ must refer to the mapping of $$ker \ i $$ to L under the unique S-module homomorphism $$ \Phi $$.Can you confirm that I am correct (or otherwise)?Now, proceeding on the assumption that I have understood/interpreted you correctly ...Above, you invoked the Third Isomorphism Theorem for Groups (Theorem 19, Dummit and Foote, page 98) which reads as follows:

------------------------------------------------------------------------------------------------
Theorem 19. (The Third Isomorphism Theorem for Groups)

Let $$G $$be a group and let $$K$$ and $$H$$ be normal subgroups of G with $$K \le H $$.

Then $$ H/K $$ is a normal subgroup of $$ G/K $$

and $$ G/H \cong (G/K) / (H/K)$$

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

BUT to apply this theorem we must have that K is a subgroup of H.

But surely then in order to apply the theorem we are assuming that $$K = ker \ i $$ is a subset of $$ H = ker \ \phi $$ i.e we are assuming what we need to show?

Are you able to clarify the situation further for me?
Further, are you able to shed some light on a previous question which related to the proof of Corollary 9 - as follows:

In the proof of Corollary 9, Next D&F write:

"It follows that $$ N/ ker \ i $$ is the unique largest quotient of $$N$$ that can be embedded in any $$S$$-module. The last statement in the Corollary follows."

Can anyone help me by giving a clear explanation of why, exactly, it follows that [FONT=MathJax_Math]N[FONT=MathJax_Main]/[FONT=MathJax_Math]k[FONT=MathJax_Math]e[FONT=MathJax_Math]r[FONT=MathJax_Math]i is the unique largest quotient of N that can be embedded in any S-module, and then, indeed, why the last statement in the Corollary follows?

Peter

 

[ThePerfectHacker]

Can anyone help me by giving a clear explanation of why, exactly, it follows that [FONT=MathJax_Math]N[FONT=MathJax_Main]/[FONT=MathJax_Math]k[FONT=MathJax_Math]e[FONT=MathJax_Math]r[FONT=MathJax_Math]i is the unique largest quotient of N that can be embedded in any S-module, and then, indeed, why the last statement in the Corollary follows?

First, it is false, $N/\ker i$ cannot be embedded into any $S$-module. For example, $N/\ker i$ will not embed into the $0$-module if $\ker i$ is not all of $N$, since an embedding must be injective.

This is what Dummit and Foote are trying to say.

You have a module $N$ over a ring $R$, and $R$ is a subring of $S$. It is not always possible to find a module $M$ over $S$ that embeds $N$ into $M$ as an $R$-module.

But a quotient of $N$ might have this extension of scalars property. What is a quotient of $N$? It just means $N/P$ were $P$ is some submodule, this quotient $N/P$ is an $R$-module in a natural way by defining $r(n+P) = rn+P$.

For example, if we quotient by $N$ and get $N/N$, this is just the $0$ module, and the $0$ module can be extended as a module over $S$ in a trivial manner.

 

[Math Amateur]

First, it is false, $N/\ker i$ cannot be embedded into any $S$-module. For example, $N/\ker i$ will not embed into the $0$-module if $\ker i$ is not all of $N$, since an embedding must be injective.

This is what Dummit and Foote are trying to say.

You have a module $N$ over a ring $R$, and $R$ is a subring of $S$. It is not always possible to find a module $M$ over $S$ that embeds $N$ into $M$ as an $R$-module.

But a quotient of $N$ might have this extension of scalars property. What is a quotient of $N$? It just means $N/P$ were $P$ is some submodule, this quotient $N/P$ is an $R$-module in a natural way by defining $r(n+P) = rn+P$.

For example, if we quotient by $N$ and get $N/N$, this is just the $0$ module, and the $0$ module can be extended as a module over $S$ in a trivial manner.

Thanks so much ThePerfectHacker ... will work through your post in a minute.

Peter

 

[Math Amateur]

First, it is false, $N/\ker i$ cannot be embedded into any $S$-module. For example, $N/\ker i$ will not embed into the $0$-module if $\ker i$ is not all of $N$, since an embedding must be injective.

This is what Dummit and Foote are trying to say.

You have a module $N$ over a ring $R$, and $R$ is a subring of $S$. It is not always possible to find a module $M$ over $S$ that embeds $N$ into $M$ as an $R$-module.

But a quotient of $N$ might have this extension of scalars property. What is a quotient of $N$? It just means $N/P$ were $P$ is some submodule, this quotient $N/P$ is an $R$-module in a natural way by defining $r(n+P) = rn+P$.

For example, if we quotient by $N$ and get $N/N$, this is just the $0$ module, and the $0$ module can be extended as a module over $S$ in a trivial manner.

Hi ThePerfectHacker,

I am reflecting on your post, D&F's exposition of the extension of scalars, and Corollary 9.

When D&F refer to "the quotient $$N/ker \ \phi $$ of N", what precisely and exactly do they mean?

A quotient module is a set of cosets and so, if we think of it as a set then it is one point for each coset - and then we define operations between cosets. Is ""the quotient $$N/ker \ \phi $$ of N" simply the quotient module - the set of cosets plus coset operations?

Or is "the quotient $$N/ker \ \phi $$ of N" simply the coset $$N/ker \ \phi $$, that is the zero coset?

I suspect it must be the former as the latter does not make much sense?

Basically, another way of stating my issue with understanding is this - when we map the quotient $$N/ker \ i$$ into the S- module are we mapping each coset to to a point in S?

Further if when we refer to $$N/ker \ i$$ we are referring to the set of cosets (plus the relevant operations) then the largest quotient of N would be the one with the most cosets and hence the one with the fewest elements in its kernel? Is that correct?

I am also struggling with the meaning of "$$ N/ker \ \phi $$ is a quotient of $$N/ker \ i $$?

Can you clarify

Peter

 

[ThePerfectHacker]

When D&F refer to "the quotient $$N/ker \ \phi $$ of N", what precisely and exactly do they mean?

Let $N$ be a module over a ring $R$ with $M$ a submodule. We can form the quotient module $N/M$ we define addition and multiplication in the same usual way as with groups and scalar multiplication by $r(n+M) = (rm)+M$.

A quotient of $N$ is simply the module $N/M$ for some submodule $M$ of $N$.

You are probably confused about why they have a $\ker \varphi$. This has to do with the fact that whenever you have a quotient of $N$, say $N/M$, we can define the map between modules as $\varphi:N\to N/M$ by $\varphi(x) = x+M$, it is easy to see that $\ker \varphi = M$. So every quotient of $N$ has the form $N/\ker \varphi$ for some map $\varphi$ from $N$ into some other module.

I am also struggling with the meaning of "$$ N/ker \ \phi $$ is a quotient of $$N/ker \ i $$?

It be best to ignore their confusing terminology.

Think of it this way. You have a module $N$ over a ring $R$, and $M_1,M_2$ are submodules of $N$, with $M_1\subseteq M_2$.

The idea is that $N/M_2$ is "smaller" than $N/M_1$, in other words, modding out by a larger module results in a "smaller" module. You can make this a little more precise by noticing the function $j:N/M_1\to N/M_2$ defined by $j(n+M_1) = n+M_2$ is well-defined (do you see why it is well-defined?) and it is surjective. So we are tempted to say $N/M_1$ is larger than $N/M_2$ (because surjective maps only exist when the domain is "larger" than the codomain).

When Dummit and Foote say $N/\ker i$ is the largest quotient of $N$ for which scalars can be extended, they basically mean that out of all quotient of $N$ for which scalars are extendable the largest one is $N/\ker i$ in the way we explained above.

 

[Math Amateur]

Let $N$ be a module over a ring $R$ with $M$ a submodule. We can form the quotient module $N/M$ we define addition and multiplication in the same usual way as with groups and scalar multiplication by $r(n+M) = (rm)+M$.

A quotient of $N$ is simply the module $N/M$ for some submodule $M$ of $N$.

You are probably confused about why they have a $\ker \varphi$. This has to do with the fact that whenever you have a quotient of $N$, say $N/M$, we can define the map between modules as $\varphi:N\to N/M$ by $\varphi(x) = x+M$, it is easy to see that $\ker \varphi = M$. So every quotient of $N$ has the form $N/\ker \varphi$ for some map $\varphi$ from $N$ into some other module.
It be best to ignore their confusing terminology.

Think of it this way. You have a module $N$ over a ring $R$, and $M_1,M_2$ are submodules of $N$, with $M_1\subseteq M_2$.

The idea is that $N/M_2$ is "smaller" than $N/M_1$, in other words, modding out by a larger module results in a "smaller" module. You can make this a little more precise by noticing the function $j:N/M_1\to N/M_2$ defined by $j(n+M_1) = n+M_2$ is well-defined (do you see why it is well-defined?) and it is surjective. So we are tempted to say $N/M_1$ is larger than $N/M_2$ (because surjective maps only exist when the domain is "larger" than the codomain).

When Dummit and Foote say $N/\ker i$ is the largest quotient of $N$ for which scalars can be extended, they basically mean that out of all quotient of $N$ for which scalars are extendable the largest one is $N/\ker i$ in the way we explained above.

Still working through your post and reflecting on it but I must say that your post is extremely helpful to me ... Thank you so much ... Really appreciate your help ...

Peter

 

[Deveno]

In many kind of algebraic structures we have a 3-fold relationship between 3 things:

a) A homomorphism
b) A quotient object (which is the image of a homomorphism)
c) A kernel (which is what the homomorphism annihilates)

All of this boils down to the fact that homomorphisms shrink structures in a "regular" way.

One map we are particularly interested in is:

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

$\iota(n) = 1\otimes n$.

You might well ask, "how could $\iota$ not be injective"? Let's see if we can create an example that forces this. So suppose our "small ring" $R$ is the integers, and our "extension ring" $S$ is the rational numbers. We need to pick a $\Bbb Z$-module that is "too small" to faithfully embed in a rational vector space (a $\Bbb Q$-module), for our "$N$".

So let's pick a finite abelian group, say $\Bbb Z_p$.

In the tensor product: $\Bbb Q \otimes_{\Bbb Z} \Bbb Z_p$ we have:

$1 \otimes a = p\left(\dfrac{1}{p}\right) \otimes a$

$= \dfrac{1}{p} \otimes pa = \dfrac{1}{p} \otimes 0$

$= 0\left(\dfrac{1}{p} \otimes 1\right) = 0$

This should make sense, the only element of finite additive order in a rational vector space is the 0-vector.

 

[Math Amateur]

In many kind of algebraic structures we have a 3-fold relationship between 3 things:

a) A homomorphism
b) A quotient object (which is the image of a homomorphism)
c) A kernel (which is what the homomorphism annihilates)

All of this boils down to the fact that homomorphisms shrink structures in a "regular" way.

One map we are particularly interested in is:

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

$\iota(n) = 1\otimes n$.

You might well ask, "how could $\iota$ not be injective"? Let's see if we can create an example that forces this. So suppose our "small ring" $R$ is the integers, and our "extension ring" $S$ is the rational numbers. We need to pick a $\Bbb Z$-module that is "too small" to faithfully embed in a rational vector space (a $\Bbb Q$-module), for our "$N$".

So let's pick a finite abelian group, say $\Bbb Z_p$.

In the tensor product: $\Bbb Q \otimes_{\Bbb Z} \Bbb Z_p$ we have:

$1 \otimes a = p\left(\dfrac{1}{p}\right) \otimes a$

$= \dfrac{1}{p} \otimes pa = \dfrac{1}{p} \otimes 0$

$= 0\left(\dfrac{1}{p} \otimes 1\right) = 0$

This should make sense, the only element of finite additive order in a rational vector space is the 0-vector.

First, thanks again to ThePerfectHacker and Deveno for their help.

I still, however, have a very basic issue with the proof of Corollary 9 - as follows:

Now in D&F Section 10.4, the proof of Corollary 9 reads as follows:

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

Proof: The quotient $$ N/ ker \ i $$ is mapped injectively (by $$i$$) into the $$S$$-module $$ S \oplus_R N $$. ... ... etc etc

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

How can $$i$$ map $$ N/ ker \ i $$ into S when i is defined by:

$$ i (n) = 1 \oplus n $$

- the domain of i is the elements of N, not the cosets of $$ N/ ker \ i $$

Can someone please clarify this for me?

Peter[... thinking ... presumably, we define a map

$$ i(n + P) = (1 \oplus n + P) $$ where P is a sub-module (?) of the module N?

Is that right?]

 

[Deveno]

(you should use \otimes instead of \oplus to indicate tensor products)

In general, if we have a module homomorphism:

$f:M \to M'$

we can construct another module homomorphism:

$[f]: M/\text{ker}f \to M'$ by:

$[f](m + \text{ker}f) = f(m)$.

because if $m + \text{ker}f = m' + \text{ker}f$ this means:

$m - m' \in \text{ker}f$ so that:

$f(m - m') = 0$ and since $f$ is additive (because all module homomorphisms are), we have:

$f(m) - f(m') = 0 \implies f(m) = f(m')$.

Again, this is just a basic consequence of the Fundamental Isomorphism Theorem, which identifies the image of any homomorphism with a quotient of the domain:

$M \stackrel{f}{\rightarrow} f(M)$ <---surjective becomes:

$M/\text{ker}f \stackrel{[f]}{\rightarrow} f(M)$<---bijective

Technically, D&F should probably write $[\iota]$ or "the mapping induced by $\iota$" but this is a common abuse of notation, to call the induced mapping by the same name (in much that same way as we often denoted [k] in the integers mod n as simply k, if the context is understood).

So, $\iota(n+P) = 1\otimes n$ (the images of the induced map are the same as the images under $\iota$, since $P = \text{ker}\iota$ so that $\iota(P) = \{0\}$.

This is the whole point of a quotient, to "shrink" something we'd rather not bother with down to nothing. For example, when we map:

$\Bbb Z \to \Bbb Z_n$

the entire subgroup $n\Bbb Z$ "shrinks to 0" pulling the "straight line" of the integers into a circular "coil", like when you pull the drawstrings of a purse.

(This very idea is the basic concept of "wrapping the real numbers around the circle" which lies at the heart of the concept of "looped paths" and "winding numbers").

Here, what we find is $\iota$ tells us how faithfully we can reproduce the $R$-module $N$ inside an $S$-module...the "best-case scenario" (and the case we are most often interested in) is when the kernel is trivial, so that our $S$-module $M$ is "big enough" (with respect to $R$) to accommodate $N$ unharmed.

 

[ThePerfectHacker]

How can $$i$$ map $$ N/ ker \ i $$ into S when i is defined by:

$$ i (n) = 1 \oplus n $$

- the domain of i is the elements of N, not the cosets of $$ N/ ker \ i $$

Can someone please clarify this for me?

Peter

It is just math sloppiness. When you get comfortable with algebra you get sloppy for the sake of being brief. Of course, as you said, $i$ does not map $N/\ker I$ into $S$ because $i$ is defined on a different set. It is what you meant, rather $i$ induces a map $i_*: N/\ker i \to S$ by $i(n+\ker i) = 1\otimes n$.

Very often in mathematics maps induces new maps, and quite often mathematicians get lazy and just use the same letter even though formally speaking those are different functions.

 

[Math Amateur]

It is just math sloppiness. When you get comfortable with algebra you get sloppy for the sake of being brief. Of course, as you said, $i$ does not map $N/\ker I$ into $S$ because $i$ is defined on a different set. It is what you meant, rather $i$ induces a map $i_*: N/\ker i \to S$ by $i(n+\ker i) = 1\otimes n$.

Very often in mathematics maps induces new maps, and quite often mathematicians get lazy and just the same letter even though formally speaking those are different functions.

Thanks ThePerfectHacker ... appreciate your help ...

Peter

 

[Math Amateur]

Thanks ThePerfectHacker ... appreciate your help ...

Peter

Thanks to significant help from ThePerfectHacker and Deveno, I now have a much better idea of the context of Corollary 9 (and its proof) from Dummit and Foote, Section 10.4.

However I still lack a complete understanding of the "punch line" of the proof (last two sentences) of the Corollary.

My understanding to this point is summarised below ... ...The first two lines of the proof (Dammit and Foote, page 362) read as follows:

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

"Proof: The quotient $$ N/ ker \ i $$ is mapped injectively (by i - in my notation by ) into the S-module $$ S \otimes_R N $$. Suppose now that $$ \phi $$ is an R-module homomorphism injecting the quotient $$ n / ker \ \phi $$ of N into an S-module L."

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

The mappings corresponding to these two statements, namely $$ i, \phi , \Phi , and [ \phi ] $$ are depicted in Figures 1 and 2 (see attachment).

In Figures 1 and 2, note the following maps:

$$ \ : \ N/ ker \ i \to S \otimes_R N $$

defined by $$ \ (n + P ) = 1 \otimes n $$ where P = ker i and (P) = 0.$$ [ \phi ] \ : \ \ N/ ker \ \phi \to L $$

Now by the First Isomorphism Theorem $$ \text{ and } [ \phi ] $$ are isomorphisms and hence injective.

The above facts cover the first two lines of the proof.The proof now goes on to state:

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

" Then, by Theorem 8, $$ ker \ i $$ is mapped to $$ 0 \text{ by } \phi \text{ i.e., } ker \ i \subseteq ker \ \phi $$. Hence $$ N / ker \ \phi $$ is a quotient of $$ N / ker \ i $$ (namely the quotient of the submodule $$ ker \ \phi / ker \ i $$ )."

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

Now $$ ker \ i \subseteq ker \ \phi $$ has been shown to be the case in a previous post, and kernels of R-module homomorphisms are sub-modules (i.e. modules) and so we can form the quotient module $$ ker \ \phi / ker \ i $$.

BUT ... ... the proof now states the following:

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

"It follows that $$ N / ker \ i $$ is the unique, largest quotient of N that can be embedded in any S-module. The last statement in the Corollary follws immediately."

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

Can someone please help me to understand how these two sentences follow from the previous sentences of the proof?

I would really appreciate the help.

(Apologies to ThePerfectHacker and Deveno if you have actually explained this before ... but I am still unclear on the above ... maybe the issue is quite simple, but I cannot see it ...)

Peter

 

[ThePerfectHacker]

I am going to explain this a little differently from how Dummit and Foote explain this in order to see the punchline of the proof. I am going to do it in order and make sure you understand each step.

1) Let me being by reminding you of our main problem: Given a module $N$ over a ring $R$, with $R$ a subring of $S$, can we find a module $M$ over $S$ together with an embedding $j:N\to M$ of $R$-modules. This is the problem of extending scalars that we said a few times, I hope it is clear why $M$ enlarges $N$.

2) It is not always possible to extend scalars to a larger module as in our main problem. Here is an example, $R = \mathbb{Z}$, $N=\mathbb{Z}_2$, the module structure of $N$ over $\mathbb{Z}$ is as you expect, namely view $N$ as a abelian group endowned with the standard module structure over $\mathbb{Z}$. Let $S = \mathbb{Q}$, so that $R$ is a subring of $S$. The question is can we find a module $M$ over $S$ such that we can an embedding of modules $j:N\to S$. The answer is NO! Let us see why.

Since $j:N\to S$ is an embedding (injective) it means $j(1) \not = 0$. Of course, "$1$" being a shorthand for $[1]_2\in N = \mathbb{Z}_2$ and "$0$" is shorthand for the identity element of $M$. Now $j(1) \in M $ which is a module over $S=\mathbb{Q}$ so we can write $j(1) = 1\cdot j(1) = (\tfrac{1}{2}\cdot 2)j(1) = \tfrac{1}{2}\cdot( 2\cdot j(1))$. Now since $j$ is an embedding of $R$-modules it follows that $2\cdot j(1) = j(2\cdot 1) = j(0) = 0$. And so $j(1) = 0$ which is a contradiction!

A small warning. There are are lot of $1$'s used above, and each one might technically be different in different sets .. but we use the same $1$ for brevity and easeness of notation. It is important to carefully understand that.

3) In general the problem in 1 may not be solvable as 2 shows. However, we may ask a question of whether there is a quotient of $N$ (this is again a module of the form $N/P$ where $P$ is a submodule of $N$) for which scalar extension is possible. The answer is yes. We can always form the quotient $N/N$, this is the zero-module. And for the zero-module scalar extension is always possible regardless of what the ring and subring is by just defining all scalar products to be equal to zero (do you see why?).

Let us return back to the business of "smaller" and "larger" quotients. Recall that if $N$ is a module and $M_1,M_2$ are submodules with $M_1\subseteq M_2$ then $N/M_1$ is "larger" than $N/M_2$.

Now the module $N/N$, the $0$-module, is in a way the smallest of all modules in the way explained above and for that module scalar extension is possible. This is the smallest quotient of $N$ that can be embedded into a module over $S$.

Basically what the theorem of Dummit and Foote is finding is the largest of all quotients of $N$ for which scalar extension is possible. Furthermore, it says that any quotient of $N$ for which scalar extension is possible must be "smaller" then $N/\ker i$. This is the largest quotient of $N$ that can be embedded into a module over $S$.

 

[Math Amateur]

I am going to explain this a little differently from how Dummit and Foote explain this in order to see the punchline of the proof. I am going to do it in order and make sure you understand each step.

1) Let me being by reminding you of our main problem: Given a module $N$ over a ring $R$, with $R$ a subring of $S$, can we find a module $M$ over $S$ together with an embedding $j:N\to M$ of $R$-modules. This is the problem of extending scalars that we said a few times, I hope it is clear why $M$ enlarges $N$.

2) It is not always possible to extend scalars to a larger module as in our main problem. Here is an example, $R = \mathbb{Z}$, $N=\mathbb{Z}_2$, the module structure of $N$ over $\mathbb{Z}$ is as you expect, namely view $N$ as a abelian group endowned with the standard module structure over $\mathbb{Z}$. Let $S = \mathbb{Q}$, so that $R$ is a subring of $S$. The question is can we find a module $M$ over $S$ such that we can an embedding of modules $j:N\to S$. The answer is NO! Let us see why.

Since $j:N\to S$ is an embedding (injective) it means $j(1) \not = 0$. Of course, "$1$" being a shorthand for $[1]_2\in N = \mathbb{Z}_2$ and "$0$" is shorthand for the identity element of $M$. Now $j(1) \in M $ which is a module over $S=\mathbb{Q}$ so we can write $j(1) = 1\cdot j(1) = (\tfrac{1}{2}\cdot 2)j(1) = \tfrac{1}{2}\cdot( 2\cdot j(1))$. Now since $j$ is an embedding of $R$-modules it follows that $2\cdot j(1) = j(2\cdot 1) = j(0) = 0$. And so $j(1) = 0$ which is a contradiction!

A small warning. There are are lot of $1$'s used above, and each one might technically be different in different sets .. but we use the same $1$ for brevity and easeness of notation. It is important to carefully understand that.

3) In general the problem in 1 may not be solvable as 2 shows. However, we may ask a question of whether there is a quotient of $N$ (this is again a module of the form $N/P$ where $P$ is a submodule of $N$) for which scalar extension is possible. The answer is yes. We can always form the quotient $N/N$, this is the zero-module. And for the zero-module scalar extension is always possible regardless of what the ring and subring is by just defining all scalar products to be equal to zero (do you see why?).

Let us return back to the business of "smaller" and "larger" quotients. Recall that if $N$ is a module and $M_1,M_2$ are submodules with $M_1\subseteq M_2$ then $N/M_1$ is "larger" than $N/M_2$.

Now the module $N/N$, the $0$-module, is in a way the smallest of all modules in the way explained above and for that module scalar extension is possible. This is the smallest quotient of $N$ that can be embedded into a module over $S$.

Basically what the theorem of Dummit and Foote is finding is the largest of all quotients of $N$ for which scalar extension is possible. Furthermore, it says that any quotient of $N$ for which scalar extension is possible must be "smaller" then $N/\ker i$. This is the largest quotient of $N$ that can be embedded into a module over $S$.

Thanks so much, Theperfecthacker.

Will now work through your post carefully and reflect on what you have said.

Thanks again,

Peter

 

[Math Amateur]

I am going to explain this a little differently from how Dummit and Foote explain this in order to see the punchline of the proof. I am going to do it in order and make sure you understand each step.

1) Let me being by reminding you of our main problem: Given a module $N$ over a ring $R$, with $R$ a subring of $S$, can we find a module $M$ over $S$ together with an embedding $j:N\to M$ of $R$-modules. This is the problem of extending scalars that we said a few times, I hope it is clear why $M$ enlarges $N$.

2) It is not always possible to extend scalars to a larger module as in our main problem. Here is an example, $R = \mathbb{Z}$, $N=\mathbb{Z}_2$, the module structure of $N$ over $\mathbb{Z}$ is as you expect, namely view $N$ as a abelian group endowned with the standard module structure over $\mathbb{Z}$. Let $S = \mathbb{Q}$, so that $R$ is a subring of $S$. The question is can we find a module $M$ over $S$ such that we can an embedding of modules $j:N\to S$. The answer is NO! Let us see why.

Since $j:N\to S$ is an embedding (injective) it means $j(1) \not = 0$. Of course, "$1$" being a shorthand for $[1]_2\in N = \mathbb{Z}_2$ and "$0$" is shorthand for the identity element of $M$. Now $j(1) \in M $ which is a module over $S=\mathbb{Q}$ so we can write $j(1) = 1\cdot j(1) = (\tfrac{1}{2}\cdot 2)j(1) = \tfrac{1}{2}\cdot( 2\cdot j(1))$. Now since $j$ is an embedding of $R$-modules it follows that $2\cdot j(1) = j(2\cdot 1) = j(0) = 0$. And so $j(1) = 0$ which is a contradiction!

A small warning. There are are lot of $1$'s used above, and each one might technically be different in different sets .. but we use the same $1$ for brevity and easeness of notation. It is important to carefully understand that.

3) In general the problem in 1 may not be solvable as 2 shows. However, we may ask a question of whether there is a quotient of $N$ (this is again a module of the form $N/P$ where $P$ is a submodule of $N$) for which scalar extension is possible. The answer is yes. We can always form the quotient $N/N$, this is the zero-module. And for the zero-module scalar extension is always possible regardless of what the ring and subring is by just defining all scalar products to be equal to zero (do you see why?).

Let us return back to the business of "smaller" and "larger" quotients. Recall that if $N$ is a module and $M_1,M_2$ are submodules with $M_1\subseteq M_2$ then $N/M_1$ is "larger" than $N/M_2$.

Now the module $N/N$, the $0$-module, is in a way the smallest of all modules in the way explained above and for that module scalar extension is possible. This is the smallest quotient of $N$ that can be embedded into a module over $S$.

Basically what the theorem of Dummit and Foote is finding is the largest of all quotients of $N$ for which scalar extension is possible. Furthermore, it says that any quotient of $N$ for which scalar extension is possible must be "smaller" then $N/\ker i$. This is the largest quotient of $N$ that can be embedded into a module over $S$.

Hi ThePerfectHacker,

I just need to clarify my understanding of your outline of the problem in step 1, including your notation.

In your statement of the problem it seems that we are extending the scalars from the left R-module N to the left S-module $$ M = S \otimes_R N $$? (Is that right? Just trying to line up your notation and problem statement with D&F)

So given the above j is the map:

$$ j \ : \ N \to S \otimes_R N = M $$

defined by $$ j(n) = 1 \otimes n $$.

Thus we are looking for a module M over S (denoted $$ M = S \otimes_R N $$) such that N is embedded in M via $$ j \ : N \to S $$?

BUT ... should this last definition of j as $$ j \ : N \to S $$ actually be $$ j \ : \ N \to S \otimes_R N = M $$?

OR

have you (slightly) "redefined" j to be the embedding

$$ j' \ : \ N \to S $$ where $$ j'(n) = n $$?

OR

Is the second j, that is $$ j \ : N \to S $$ a different map?

Can you please clarify?A further point (aside from the definition of J) is the statement:

"Since $j:N\to S$ is an embedding (injective) it means $j(1) \not = 0$"

My question: how does j being an embedding (ie injective) imply that $j(1) \not = 0$?Yet another point in passing that you (or someone) may be able to help with - what is the definition of an embedding? I know it is an injective map but what are the other defining properties?

Peter

 

[ThePerfectHacker]

BUT ... should this last definition of j as $$ j \ : N \to S $$ actually be $$ j \ : \ N \to S \otimes_R N = M $$?

Yes, that was a mistake. It should be $j:N\to M$! Another thing. I am not using any tensor products so far. When I say $M$, I mean some module into which $N$ can be embedded. That is all. I am not using $M$ in place of $S\otimes N$.

Since $j:N\to M$ is an embedding it means distinct elements of $N$ get send to distinct elements of $M$. Because by definition embeddings are injective maps that preserve the algebraic structure. Since $j(0) = 0$ it means $j(1) \not = 0$, or else otherwise $j$ will map $0,1$ both into the same element. Is it clear why $j(0) = 0$?

 

[Math Amateur]

Yes, that was a mistake. It should be $j:N\to M$! Another thing. I am not using any tensor products so far. When I say $M$, I mean some module into which $N$ can be embedded. That is all. I am not using $M$ in place of $S\otimes N$.

Since $j:N\to M$ is an embedding it means distinct elements of $N$ get send to distinct elements of $M$. Because by definition embeddings are injective maps that preserve the algebraic structure. Since $j(0) = 0$ it means $j(1) \not = 0$, or else otherwise $j$ will map $0,1$ both into the same element. Is it clear why $j(0) = 0$?

Well, in the case of D&F's notation i was an R-module homomorphism and so i(0) = 0

But in your case where you are not invoking tensor products ... I am not sure ... but if j is defined to be a homomorphism then j(0) = 0 ...

Now you define j as an embedding ... then if an embedding is a homomorphism then j(0) = 0 as this is the case with homomorphisms

Can you help, as clearly I am not sure ...

Peter

 

[ThePerfectHacker]

Well, in the case of D&F's notation i was an R-module homomorphism and so i(0) = 0

But in your case where you are not invoking tensor products ... I am not sure ... but if j is defined to be a homomorphism then j(0) = 0 ...

Can you help, as clearly I am not sure ...

Peter

This is a general fact. A group homomorphism maps the identity element to the identity element. A linear map between vector spaces maps the zero vector to the zero vector. And a map between modules maps $0\in N$ to $0\in M$, where "$0$", is the notation for the identity element of each module.

The reason for this is that suppose $f:N\to M$ is any map of modules over a ring $R$. Let $\alpha = f(0)$. Then $f(0) = f(0+0) = f(0) + f(0)$, so $\alpha = \alpha + \alpha$. Since $M$ is a module there is a $\beta$ such that $\alpha + \beta = 0$ (this $0$ is idenity element of $M$). Now add $\beta$ to both sides of $\alpha = \alpha + \alpha$ to get $\alpha + \beta = \alpha + \alpha + \beta$ this simplifies to $0 = \alpha + 0 = \alpha = f(0)$.

Basically, $f(0) = f(0) + f(0)$, now subtract $f(0)$ from both sides.

 

[Math Amateur]

This is a general fact. A group homomorphism maps the identity element to the identity element. A linear map between vector spaces maps the zero vector to the zero vector. And a map between modules maps $0\in N$ to $0\in M$, where "$0$", is the notation for the identity element of each module.

The reason for this is that suppose $f:N\to M$ is any map of modules over a ring $R$. Let $\alpha = f(0)$. Then $f(0) = f(0+0) = f(0) + f(0)$, so $\alpha = \alpha + \alpha$. Since $M$ is a module there is a $\beta$ such that $\alpha + \beta = 0$ (this $0$ is idenity element of $M$). Now add $\beta$ to both sides of $\alpha = \alpha + \alpha$ to get $\alpha + \beta = \alpha + \alpha + \beta$ this simplifies to $0 = \alpha + 0 = \alpha = f(0)$.

Basically, $f(0) = f(0) + f(0)$, now subtract $f(0)$ from both sides.

Just checking one point - you say that f(0) = 0 for any map between modules ... do you mean any linear map? or any map at all?

Does the fact that your proof uses f(0 + 0) = f(0) + f(0) mean that we are talking only about linear maps?

Just checking ...

Peter

 

[ThePerfectHacker]

Just checking one point - you say that f(0) = 0 for any map between modules ... do you mean any linear map? or any map at all?

It means a linear map. This is what I meant between a map between modules. A map that respects the module structure.

In topology you often say map between topological spaces instead of continous map.

 

[Math Amateur]

It means a linear map. This is what I meant between a map between modules. A map that respects the module structure.

In topology you often say map between topological spaces instead of continous map.

I have to say that despite extensive help from ThePerfectHacker and Deveno, I still have not achieved complete understanding of the 'punch lines' (final lines) of the proof.

The 'punch lines' as I see them (last two lines of the proof) are as follows:

"It follows that $$ N / ker \ i $$ is the unique, largest quotient of N that can be embedded in any S-module. The last statement in the Corollary follows immediately."

Now since $$ ker \ i \subseteq ker \ \phi $$ (see attachment illustrating this!) we have that $$ N/ ker \ i $$ is larger than (i.e. has more cosets than $$ N / ker \ \phi $$)

But $$ \phi $$ is pretty general, from my reading of the Corollary - just a (any?) R-module homomorphism from N to L.

So then, I imagine, the argument goes, that for any such $$ \phi $$, $$ N/ ker \ i $$ is larger than (i.e. has more cosets than $$ N / ker \ \phi $$.

So then $$ N/ ker \ i $$ is the largest quotient module that can be embedded in, I presume, $$ S \otimes_R N $$ (- or is it the largest that can be embedded in L?) [also have issue of uniqueness]

My problem is that $$ ker \ \phi $$ comes from a map from N to L and not N to $$ S \otimes_R N $$, our constructed left S-module. So how is $$ N / ker \ \phi $$ embedded in $$ S \otimes_R N $$.

Indeed I fail to understand the role of L at all!

I am also very uncertain of the quality of my logic in concluding that $$ N/ ker \ i $$ is the largest quotient module that can be embedded in, I presume, $$ S \otimes_R N $$.

Can someone please help with the above issues? I would really appreciate some clarity.

Peter

 

[ThePerfectHacker]

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

 

[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.