The generalized rank-nullity theorem

[quasar987]

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

 

[Hurkyl]

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.

 

[yyat]

Try tensoring all the groups with \mathbb{Q} and proving that the resulting sequence is still exact (Tensoring with \mathbb{Q} eliminates the torsion part, as Hurkyl suggested).

 

[Hurkyl]

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.

 

[quasar987]

It seems that I end up with the same kind of problem:

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

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

 

[Hurkyl]

<br /> f(\mathbb{Q}^{\mbox{rank}(A)})\cong (\mathbb{Q}^{\mbox{rank}(A)}\subset \mathbb{Q}^{\mbox{rank}(B)})<br />

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
<br /> 0\rightarrow \mathbb{Q}^{\mbox{rank}(A)} \stackrel{f}\rightarrow \mathbb{Q}^{\mbox{rank}(B)}\stackrel{g}\rightarrow \mathbb{Q}^{\mbox{rank}(C)} \rightarrow0<br />
?

 

[quasar987]

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 \mathbb{R}^k\subset \mathbb{R}^n to mean the subspace \{(x_1,...,x_k,0,...,0) \in \mathbb{R}^n: x_i\in \mathbb{R}, \ 1\leq i\leq k\}[/tex] of <b>R</b>^n?<br /> <br /> Same thing here.

 

[Hurkyl]

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 f(X) 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:

0 \to \mathbb{Z} \xrightarrow{2} \mathbb{Z} \to \mathbb{Z} / 2 \to 0

0 \to \mathbb{Z} \xrightarrow{(0, 1)} \mathbb{Z}^2 \xrightarrow{(1,0)^T} \mathbb{Z} \to 0

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


(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")

 

[yyat]

It seems that I end up with the same kind of problem:

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

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

An abelian group is the same thing as a \mathbb{Z}-module. This is why the tensor product, which only makes sense for modules, can be defined. When tensoring with \mathbb{Q} you get a \mathbb{Q}-module, which is of course just a vector space over the field \mathbb{Q}.
Moreover, the homorphisms f,g induce linear maps between the corresponding vector spaces, so the problem is now reduced to linear algebra.

 

[quasar987]

An abelian group is the same thing as a \mathbb{Z}-module. This is why the tensor product, which only makes sense for modules, can be defined. When tensoring with \mathbb{Q} you get a \mathbb{Q}-module, which is of course just a vector space over the field \mathbb{Q}.

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 (\mathbb{Q}\otimes_{\mathbb{Q}}\mathbb{Z}) 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?

 

[yyat]

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

\lambda(r\otimes a)=(\lambda r)\otimes a.

 

[quasar987]

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

\lambda(r\otimes a)=(\lambda r)\otimes a.

Why do you write \mathbb{Q}\otimes_{\mathbb{Z}}A and not \mathbb{Q}\otimes_{\mathbb{Q}}A?

I would think \mathbb{Q}\otimes_{\mathbb{Z}}A 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.

 

[Hurkyl]

I would think \mathbb{Q}\otimes_{\mathbb{Z}}A 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]

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 M \otimes_R N 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 \mathbb{Q} \otimes_Z \mathbb{Z} 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]

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]

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 M \otimes_\mathbb{Z} N.

? 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 \mathbb{Q} \otimes_\mathbb{Z} M \otimes_\mathbb{Z} N -- tensoring with Q accounts for the fact that you replaced Z with Q in the usual presentation of M \otimes_\mathbb{Z} N.



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

 

[quasar987]

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 M \otimes_\mathbb{Z} N.

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 M\otimes_{\mathbb{Z}}N. We're taking a quotient of the free Q-module over MxN, so wouldn't M\otimes_{\mathbb{Q}}N 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 \mathbb{Q} \otimes_\mathbb{Z} M \otimes_\mathbb{Z} N -- tensoring with Q accounts for the fact that you replaced Z with Q in the usual presentation of M \otimes_\mathbb{Z} N.

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]

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 M\otimes_{\mathbb{Z}}N.

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 M \times N \to P and R-module homomorphisms M \otimes N \to P.

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

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 M \otimes_R N. You can construct your presentation in three stages:
(1) Take the usual presentation for M \otimes_R N
(2) Replace "free R-module" with "free S-module"
(3) Add the remaining relations s(m \otimes n) = (sm \otimes n)

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

 

[quasar987]

Ok, I understand now, thank you both.

 

[quasar987]

Try tensoring all the groups with \mathbb{Q} and proving that the resulting sequence is still exact (Tensoring with \mathbb{Q} 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
0\rightarrow A \stackrel{f}\rightarrow B\stackrel{g}\rightarrow C \rightarrow 0, then knowing that the functor \mathbb{Q}\otimes\underline{ \ \ } is right exact, it suffices to prove that \mathbb{Q}\otimes A \stackrel{id\otimes f}\rightarrow \mathbb{Q}\otimes B is injective.

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

 

[Hurkyl]

What is your counterexample?

 

[quasar987]

For simplicity, I assumed that A=\mathbb{Z}^2\oplus \mbox{torsion terms} and B=\mathbb{Z}^2\oplus \mbox{torsion terms} so f is of the "form" f((x,y),z)=((f_{11}((x,y),z),f_{12}((x,y),z)),f_{2}((x,y),z)).

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

(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)).

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

 

[Hurkyl]

You should make your examples more specific.

with f_2 injective

f₂ isn't injective.

 

[quasar987]

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 f((x,y),z):=(0,f_{2}((x,y),z)).

 

[quasar987]

Oh right, f_2 cannot be injective because the codomain is finite and the domain not...mmh

 

[quasar987]

Ok, so f_2 cannot be injective but f is injective, so it must be that f_1 is injective.

Proof: Suppose the torsion part of B is \mathbb{Z}_{m_1}\oplus\ldots\oplus\mathbb{Z}_{m_s} with m_s|m_{s-1}|\ldots|m_1. If f_1((x,y),z)=((0,0),0) for some ((x,y),z)\neq ((0,0),0), then by injectivity of f, it must be that f_2((x,y),z)=(\bar{k_1},\ldots,\bar{k_s})\neq 0. However, f_1(m_1((x,y),z))=m_1f_1((x,y),z)=0 and f_2(m_1((x,y),z))=m_1f_2((x,y),z)=m_1(\bar{k_1},\ldots,\bar{k_s})=(m_1\bar{k_1},\ldots,m_1\bar{k_s})=(k_1\bar{m_1},\ldots,k_s\bar{m_1})=(k_1\bar{0},\ldots,k_s\bar{0})=0, thus violating the injectivity of f. Q.E.D.

Mmh, there must be something wrong with the above proof because then f_1(A) would be an isomorphic copy of A in Z^2. But as a subgroup of Z^2, f_1(A) must be free, which need not be the case in general.

Most likely is that only the restriction of f_1 to Z^2 is injective I think... but what is wrong with the above proof?!

 

[quasar987]

Ok, I found what is wrong with the above proof and this out of the way, I was able to complete the proof of the generalized rank-nullity theorem.

Hurkyl, may I know what is the other proof you had in mind? I suppose it is more elementary.

 

[Hurkyl]

Partial sketch -- A \otimes \mathbb{Q} simply amounts to including denominators: any element can be written in the form a \otimes \frac{1}{n} for some a \in A, or more briefly, as a/n. (You add terms by finding a common denominator) Also, a/n = 0 iff a has finite order. If f:A->B, then the function f \otimes \mathbb{Q} sends a/n to f(a)/n.


Even shorter answer -- simply remember that Q is a flat Z-module. (Or more generally, if S is a localization of the ring R, then S is a flat R-module) I suppose that's hard to remember if you haven't learned it yet, though.

 

[dvs]

quasar987: It's sometimes helpful to think about tensoring as a "change of base" type operation, and this is one of these times. For more info, see the section of Dummit & Foote on tensor products of modules. This type of thinking will also be helpful, e.g., when you try to construct homology groups with arbitrary (abelian) coefficient groups. (I mention this because you've posted a few homology related questions in the topology forum.)