The generalized rank-nullity theorem

  • Thread starter quasar987
  • Start date

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
If one has a short exact sequence 0-->A-->B-->C-->0 of finitely generated abelian groups, how does one show that rank(B)=rank(A)+rank(C) ?

We have that A embeds in B and C is isomorphic to B/A. The natural thing to try to use I think is the uniqueness of the decomposition of a finitely generated abelian group into cyclic groups, but how?

Like, if [tex]\phi :B\cong \mathbb{Z}^r\oplus\mathbb{Z}_{n_1}\oplus\ldots\mathbb{Z}_{n_b}[/tex], even though [tex]A\cong \mathbb{Z}^s\oplus\mathbb{Z}_{m_1}\oplus\ldots\mathbb{Z}_{m_a}[/tex] for some isomorphism, there is no guarantee that the embedded A (in B) will map to [tex]A\cong \mathbb{Z}^s\oplus\mathbb{Z}_{m_1}\oplus\ldots\mathbb{Z}_{m_a}[/tex] under [itex]\phi[/itex]. Or is there?
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17
I'm confused by your notation.

Anyways, since the rank is a property of the torsion-free part, I would look for some way to get rid of the torsion subgroups. However, you could probably work out a contradiction directly by assuming the rank(B) < rank(A) + rank(C), and also for the other inequality.
 
316
0
Try tensoring all the groups with [tex]\mathbb{Q}[/tex] and proving that the resulting sequence is still exact (Tensoring with [tex]\mathbb{Q}[/tex] eliminates the torsion part, as Hurkyl suggested).
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17
That was the main idea I had in mind, and one of the easiest approaches -- if you know it. But, happily, it isn't the only way to get rid of the torsion. :smile:
 

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
It seems that I end up with the same kind of problem:

Tensoring a finitely generated abelian group A with Q gives, I believe, [tex]\mathbb{Q}^{\mbox{rank}(A)}[/tex] (as an abelian group) so we have a short exact sequence [tex]0\rightarrow \mathbb{Q}^{\mbox{rank}(A)} \stackrel{f}\rightarrow \mathbb{Q}^{\mbox{rank}(B)}\stackrel{g}\rightarrow \mathbb{Q}^{\mbox{rank}(C)} \rightarrow0[/tex] and I would like to say that [tex]f(\mathbb{Q}^{\mbox{rank}(A)})\cong (\mathbb{Q}^{\mbox{rank}(A)}\subset \mathbb{Q}^{\mbox{rank}(B)})[/tex], so that [tex]\mathbb{Q}^{\mbox{rank}(B)}/f(\mathbb{Q}^{\mbox{rank}(A)})\cong \mathbb{Q}^{\mbox{rank}(B)-\mbox{rank}(A)}[/tex].

But why must [tex]f(\mathbb{Q}^{\mbox{rank}(A)})\cong (\mathbb{Q}^{\mbox{rank}(A)}\subset \mathbb{Q}^{\mbox{rank}(B)})[/tex] hold?
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17
[tex]
f(\mathbb{Q}^{\mbox{rank}(A)})\cong (\mathbb{Q}^{\mbox{rank}(A)}\subset \mathbb{Q}^{\mbox{rank}(B)})
[/tex]
I don't get it -- the notation is somewhat confusing, and it looks like you're asking whether or not the image of f is isomorphic to the image of f.

Anyways, why can't you just apply the rank-nullity theorem to
[tex]
0\rightarrow \mathbb{Q}^{\mbox{rank}(A)} \stackrel{f}\rightarrow \mathbb{Q}^{\mbox{rank}(B)}\stackrel{g}\rightarrow \mathbb{Q}^{\mbox{rank}(C)} \rightarrow0
[/tex]
?
 

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
I would think because f and g are only homomorphisms of abelian groups and not of vector spaces.

I don't get it -- the notation is somewhat confusing, and it looks like you're asking whether or not the image of f is isomorphic to the image of f.
You know when sometimes we talk about R^k as being a subspace of R^n, and we write [tex]\mathbb{R}^k\subset \mathbb{R}^n[/tex] to mean the subspace [itex]\{(x_1,...,x_k,0,...,0) \in \mathbb{R}^n: x_i\in \mathbb{R}, \ 1\leq i\leq k\}[/tex] of R^n?

Same thing here.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17
I would think because f and g are only homomorphisms of abelian groups and not of vector spaces.
I bet they are homomorphisms of vector spaces! A direct proof should be uncomplicated... also, don't you already know they are homomorphisms of Q-modules?


Same thing here.
Your question still doesn't make sense. f is monic, so it's automatically true that [itex]f(X)[/itex] is isomorphic to X.



By the way... I don't know if you need to see this or not, but I'll put it up just in case. Consider these two special cases of the original question:

[tex]0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \to \mathbb{Z} / 2 \to 0[/tex]

[tex]0 \to \mathbb{Z} \xrightarrow{(0, 1)} \mathbb{Z}^2 \xrightarrow{(1,0)^T} \mathbb{Z} \to 0[/tex]

[tex]0 \to \mathbb{Z} \xrightarrow{(2, 3)} \mathbb{Z}^2 \xrightarrow{(3,-2)^T} \mathbb{Z} \to 0[/tex]


(If those sequences doesn't make sense... I'm treating elements of Zn as row vectors, so the maps are are "multiplication on the right by this matrix")
 
Last edited:
316
0
It seems that I end up with the same kind of problem:

Tensoring a finitely generated abelian group A with Q gives, I believe, [tex]\mathbb{Q}^{\mbox{rank}(A)}[/tex] (as an abelian group) so we have a short exact sequence [tex]0\rightarrow \mathbb{Q}^{\mbox{rank}(A)} \stackrel{f}\rightarrow \mathbb{Q}^{\mbox{rank}(B)}\stackrel{g}\rightarrow \mathbb{Q}^{\mbox{rank}(C)} \rightarrow0[/tex] and I would like to say that [tex]f(\mathbb{Q}^{\mbox{rank}(A)})\cong (\mathbb{Q}^{\mbox{rank}(A)}\subset \mathbb{Q}^{\mbox{rank}(B)})[/tex], so that [tex]\mathbb{Q}^{\mbox{rank}(B)}/f(\mathbb{Q}^{\mbox{rank}(A)})\cong \mathbb{Q}^{\mbox{rank}(B)-\mbox{rank}(A)}[/tex].

But why must [tex]f(\mathbb{Q}^{\mbox{rank}(A)})\cong (\mathbb{Q}^{\mbox{rank}(A)}\subset \mathbb{Q}^{\mbox{rank}(B)})[/tex] hold?
An abelian group is the same thing as a [tex]\mathbb{Z}[/tex]-module. This is why the tensor product, which only makes sense for modules, can be defined. When tensoring with [tex]\mathbb{Q}[/tex] you get a [tex]\mathbb{Q}[/tex]-module, which is of course just a vector space over the field [tex]\mathbb{Q}[/tex].
Moreover, the homorphisms f,g induce linear maps between the corresponding vector spaces, so the problem is now reduced to linear algebra.
 

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
An abelian group is the same thing as a [tex]\mathbb{Z}[/tex]-module. This is why the tensor product, which only makes sense for modules, can be defined. When tensoring with [tex]\mathbb{Q}[/tex] you get a [tex]\mathbb{Q}[/tex]-module, which is of course just a vector space over the field [tex]\mathbb{Q}[/tex].
How does this work? We have on the one hand Z which is at best a Z-module, and we have Q which is at best a Q-module. Happily Z is a subring of Q so we can define the tensor product wrt Q ([tex]\mathbb{Q}\otimes_{\mathbb{Q}}\mathbb{Z}[/tex]) as the free Q-module on Q x Z modulo the Q-module generated by all the elements of the form

(x+x'y)-(x,y)-(x',y),
(x,y+y')-(x,y)-(x,y'),
(rx,y)-r(x,y) for r rational,
n(x,y)-(nx,y)-(x,ny) for n interger.

Is this it?
 
316
0
How does this work? We have on the one hand Z which is at best a Z-module, and we have Q which is at best a Q-module. Happily Z is a subring of Q so we can define the tensor product wrt Q ([tex]\mathbb{Q}\otimes_{\mathbb{Q}}\mathbb{Z}[/tex]) as the free Q-module on Q x Z modulo the Q-module generated by all the elements of the form

(x+x'y)-(x,y)-(x',y),
(x,y+y')-(x,y)-(x,y'),
(rx,y)-r(x,y) for r rational,
n(x,y)-(nx,y)-(x,ny) for n interger.

Is this it?
The third one is really the definition of scalar multiplication, not a relation, and I think the last one should be just (nx,y)-(x,ny) for n an integer.

More generally, abelian groups and [tex]\mathbb{Q}[/tex] are both [tex]\mathbb{Z}[/tex]-modules, so one can form the http://en.wikipedia.org/wiki/Tensor_product_of_modules" [Broken] [tex]\mathbb{Q}\otimes_{\mathbb{Z}}A[/tex], which is not only a [tex]\mathbb{Z}[/tex]-module but also a [tex]\mathbb{Q}[/tex]-module (vector space), because [tex]\mathbb{Q}[/tex] is a ring. Scalar multiplication with [tex]\lambda\in\mathbb{Q}[/tex] is just defined by

[tex]\lambda(r\otimes a)=(\lambda r)\otimes a[/tex].
 
Last edited by a moderator:

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
The third one is really the definition of scalar multiplication, not a relation, and I think the last one should be just (nx,y)-(x,ny) for n an integer.
Yes, sorry about that.

More generally, abelian groups and [tex]\mathbb{Q}[/tex] are both [tex]\mathbb{Z}[/tex]-modules, so one can form the http://en.wikipedia.org/wiki/Tensor_product_of_modules" [Broken] [tex]\mathbb{Q}\otimes_{\mathbb{Z}}A[/tex], which is not only a [tex]\mathbb{Z}[/tex]-module but also a [tex]\mathbb{Q}[/tex]-module (vector space), because [tex]\mathbb{Q}[/tex] is a ring. Scalar multiplication with [tex]\lambda\in\mathbb{Q}[/tex] is just defined by

[tex]\lambda(r\otimes a)=(\lambda r)\otimes a[/tex].
Why do you write [tex]\mathbb{Q}\otimes_{\mathbb{Z}}A[/tex] and not [tex]\mathbb{Q}\otimes_{\mathbb{Q}}A[/tex]?

I would think [tex]\mathbb{Q}\otimes_{\mathbb{Z}}A[/tex] denotes the Z-module obtained from the smaller set of relations

(x+x'y)-(x,y)-(x',y),
(x,y+y')-(x,y)-(x,y'),
n(x,y)-(nx,y) for n integer
n(x,y)-(x,ny) for n interger.
 
Last edited by a moderator:

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17
I would think [tex]\mathbb{Q}\otimes_{\mathbb{Z}}A[/tex] denotes the Z-module obtained from the smaller set of relations

(x+x'y)-(x,y)-(x',y),
(x,y+y')-(x,y)-(x,y'),
n(x,y)-(nx,y) for n integer
n(x,y)-(x,ny) for n interger.
That's certainly right for its underlying group. However, canonically defined for this group are two Z-module structures and one Q-module structure.

The Q-module structure comes from the fact Q is a Q-module. The two Z-module structures come from the fact that Q is a Z-module, and Z is a Z-module.

The tensor product is over Z, because we want to force the two Z-module structures to be the same. It wouldn't make sense to take the tensor product over Q, because Z doesn't have a Q-module structure!
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17
If you pay attention to left-right modules and are stickler for details, this becomes more pronounced, because the tensor product "consumes" the module structure.

e.g. if M is a right R-module, and N is a left R-module, then the tensor product [itex]M \otimes_R N[/itex] is a plain Abelian group, and no longer has any sort of module structure. (Of course, we could give it a Z-module structure, but we didn't, because we're being a stickler for details!) However, if M was also a left S-module structure, that would be retained in the tensor product, and similarly if N had a right T-module structure. (more complicated things can happen too, but this is the most common case)

If we viewed Q as a left Q- right Z-module, and Z as a left Z- right Z-module, then the tensor product [itex]\mathbb{Q} \otimes_Z \mathbb{Z}[/itex] would be a left Q- right Z-module.

Things are a lot simpler in the commutative case, because you don't have to worry about keeping track of left and right and stuff.
 

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
The tensor product is over Z, because we want to force the two Z-module structures to be the same. It wouldn't make sense to take the tensor product over Q, because Z doesn't have a Q-module structure!
What about the Q-module I defined in post #10 as the quotient Q<M x N>/R where R is the sub-module of Q<M x N> generated by the elements of the type

(x+x'y)-(x,y)-(x',y),
(x,y+y')-(x,y)-(x,y'),
r(x,y)-(rx,y) for r rational,
n(x,y)-(x,ny) for n interger

? How do we denote that creature in order to distinguish it from Q<M x N>/R' where R' is the sub-module of Q<M x N> generated by the elements of the type

(x+x'y)-(x,y)-(x',y),
(x,y+y')-(x,y)-(x,y'),
n(x,y)-(nx,y) for n interger,
n(x,y)-(x,ny) for n interger

?
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17
What about the Q-module I defined in post #10 as the quotient Q<M x N>/R where R is the sub-module of Q<M x N> generated by the elements of the type

(x+x'y)-(x,y)-(x',y),
(x,y+y')-(x,y)-(x,y'),
r(x,y)-(rx,y) for r rational,
n(x,y)-(x,ny) for n interger
What are M and N? If they're just Z-modules, then what you've written is not well-defined, because rx isn't defined.

If M is a Q-module being viewed as a Z-module and N is a Z-module, then what you wrote is isomorphic to the Q-module [itex]M \otimes_\mathbb{Z} N[/itex].

? How do we denote that creature in order to distinguish it from Q<M x N>/R' where R' is the sub-module of Q<M x N> generated by the elements of the type

(x+x'y)-(x,y)-(x',y),
(x,y+y')-(x,y)-(x,y'),
n(x,y)-(nx,y) for n interger,
n(x,y)-(x,ny) for n interger

?
M and N are Z-modules, I assume? This should be isomorphic to [itex]\mathbb{Q} \otimes_\mathbb{Z} M \otimes_\mathbb{Z} N[/itex] -- tensoring with Q accounts for the fact that you replaced Z with Q in the usual presentation of [itex]M \otimes_\mathbb{Z} N[/itex].



Aside: we can get away with a little bit of sloppiness in this case, because [itex]\mathbb{Q} \otimes_\mathbb{Z} M \cong M[/itex] for every Q-module M.
 
Last edited:

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
What are M and N? If they're just Z-modules, then what you've written is not well-defined, because rx isn't defined.

If M is a Q-module being viewed as a Z-module and N is a Z-module, then what you wrote is isomorphic to the Q-module [itex]M \otimes_\mathbb{Z} N[/itex].
Gahhh! Yes M is a Q-module and N a Z-module. I really don't get where the Z comes from in the notation [tex]M\otimes_{\mathbb{Z}}N[/tex]. We're taking a quotient of the free Q-module over MxN, so wouldn't [tex]M\otimes_{\mathbb{Q}}N[/tex] be infinitely more appropriate in order to distinguish it from Z<M x N>/R' ?


M and N are Z-modules, I assume? This should be isomorphic to [itex]\mathbb{Q} \otimes_\mathbb{Z} M \otimes_\mathbb{Z} N[/itex] -- tensoring with Q accounts for the fact that you replaced Z with Q in the usual presentation of [itex]M \otimes_\mathbb{Z} N[/itex].
Sorry, I meant Z<M x N>/R' here and not Q<M x N>/R' where again M is a Q-module and N a Z-module! (double gahhh!)
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17
Gahhh! Yes M is a Q-module and N a Z-module. I really don't get where the Z comes from in the notation [tex]M\otimes_{\mathbb{Z}}N[/tex].
The main property of the tensor product over R of two R-modules is that there is a bijective correspondence between bilinear R-module transformations [itex]M \times N \to P[/itex] and R-module homomorphisms [itex]M \otimes N \to P[/itex].

The group you defined satisfies that property for Z, thus we call it [itex]M\otimes_{\mathbb{Z}}N[/itex]. It doesn't satisfy that property for Q, so we can't call it [itex]M\otimes_{\mathbb{Q}}N[/itex].

I'm pretty sure that you can prove that for any morphism of rings f : R -> S, an S-module M and an R-module N, you can prove the method you used to present an S-module gives something isomorphic to [itex]M \otimes_R N[/itex]. You can construct your presentation in three stages:
(1) Take the usual presentation for [itex]M \otimes_R N[/itex]
(2) Replace "free R-module" with "free S-module"
(3) Add the remaining relations [itex]s(m \otimes n) = (sm \otimes n)[/itex]

Step 1 gives you [itex]M \otimes_R N[/itex]. Step 2 gives you [itex]S \otimes_R M \otimes_R N[/itex]. Step three knocks it back down to [itex]M \otimes_R N[/itex].
 

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
Ok, I understand now, thank you both.
 

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
Try tensoring all the groups with [tex]\mathbb{Q}[/tex] and proving that the resulting sequence is still exact (Tensoring with [tex]\mathbb{Q}[/tex] eliminates the torsion part, as Hurkyl suggested).
Are you guys sure this works? If I start with a short exact sequence of finitely generated abelian groups
[tex]0\rightarrow A \stackrel{f}\rightarrow B\stackrel{g}\rightarrow C \rightarrow 0[/tex], then knowing that the functor [tex]\mathbb{Q}\otimes\underline{ \ \ }[/tex] is right exact, it suffices to prove that [tex]\mathbb{Q}\otimes A \stackrel{id\otimes f}\rightarrow \mathbb{Q}\otimes B[/tex] is injective.

I figured out what [tex]id\otimes f[/tex] becomes under the natural isomorphisms [tex]\mathbb{Q}\otimes A\cong \mathbb{Q}^{\mbox{rank}(A)}[/tex] and [tex]\mathbb{Q}\otimes B\cong \mathbb{Q}^{\mbox{rank}(B)}[/tex] and it is definitely not injective in general. I verified each step several times so I'm kinda stumped.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17
What is your counterexample?
 

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
For simplicity, I assumed that [tex]A=\mathbb{Z}^2\oplus \mbox{torsion terms}[/tex] and [tex]B=\mathbb{Z}^2\oplus \mbox{torsion terms}[/tex] so f is of the "form" [tex]f((x,y),z)=((f_{11}((x,y),z),f_{12}((x,y),z)),f_{2}((x,y),z))[/tex].

And so, under the natural isomorphisms [tex]\mathbb{Q}\otimes A\cong \mathbb{Q}^2[/tex] and [tex]\mathbb{Q}\otimes B\cong \mathbb{Q}^2[/tex], I found that [tex]id\otimes f[/tex] becomes the map

[tex](r_1,r_2)\mapsto (r_1f_{11}((1,0),0)+r_2f_{11}((0,1),0),r_1f_{12}((1,0),0)+r_2f_{12}((0,1),0))[/tex].

So for example, if f is of the form [tex]f((x,y),z)=(0,f_{2}((x,y),z))[/tex] with f_2 injective, then f is injective (as per the hypothesis that the original short exact sequence is exact at A), but [tex]id\otimes f[/tex] is identically zero.
 

Hurkyl

Staff Emeritus
Science Advisor
Gold Member
14,847
17

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
What do you mean "f_2 isn't injective"? I'm saying that one can take any injective f_2 and construct f by setting [tex]f((x,y),z):=(0,f_{2}((x,y),z))[/tex].
 

quasar987

Science Advisor
Homework Helper
Gold Member
4,771
5
Oh right, f_2 cannot be injective because the codomain is finite and the domain not...mmh
 

Related Threads for: The generalized rank-nullity theorem

  • Posted
Replies
3
Views
4K
  • Posted
Replies
2
Views
1K
Replies
6
Views
3K
  • Posted
Replies
2
Views
6K
  • Posted
Replies
2
Views
9K

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving
Top