# The generalized rank-nullity theorem

1. Mar 20, 2009

### 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?

2. Mar 20, 2009

### Hurkyl

Staff Emeritus

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.

3. Mar 20, 2009

### 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).

4. Mar 20, 2009

### Hurkyl

Staff Emeritus
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.

5. Mar 26, 2009

### 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?

6. Mar 26, 2009

### Hurkyl

Staff Emeritus
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
$$0\rightarrow \mathbb{Q}^{\mbox{rank}(A)} \stackrel{f}\rightarrow \mathbb{Q}^{\mbox{rank}(B)}\stackrel{g}\rightarrow \mathbb{Q}^{\mbox{rank}(C)} \rightarrow0$$
?

7. Mar 26, 2009

### quasar987

I would think because f and g are only homomorphisms of abelian groups and not of vector spaces.

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 $\{(x_1,...,x_k,0,...,0) \in \mathbb{R}^n: x_i\in \mathbb{R}, \ 1\leq i\leq k\}[/tex] of R^n? Same thing here. 8. Mar 27, 2009 ### Hurkyl Staff Emeritus I bet they are homomorphisms of vector spaces! A direct proof should be uncomplicated... also, don't you already know they are homomorphisms of Q-modules? Your question still doesn't make sense. f is monic, so it's automatically true that [itex]f(X)$ 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")

Last edited: Mar 27, 2009
9. Mar 27, 2009

### yyat

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.

10. Mar 27, 2009

### quasar987

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?

11. Mar 27, 2009

### yyat

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" [Broken] $$\mathbb{Q}\otimes_{\mathbb{Z}}A$$, 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

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

Last edited by a moderator: May 4, 2017
12. Mar 27, 2009

### quasar987

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.

Last edited by a moderator: May 4, 2017
13. Mar 27, 2009

### Hurkyl

Staff Emeritus
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!

14. Mar 27, 2009

### Hurkyl

Staff Emeritus
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.

15. Mar 27, 2009

### quasar987

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

?

16. Mar 27, 2009

### Hurkyl

Staff Emeritus
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 $M \otimes_\mathbb{Z} N$.

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.

Last edited: Mar 27, 2009
17. Mar 28, 2009

### quasar987

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' ?

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!)

18. Mar 28, 2009

### Hurkyl

Staff Emeritus
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 $M \otimes_R N$.

19. Mar 30, 2009

### quasar987

Ok, I understand now, thank you both.

20. Apr 7, 2009

### quasar987

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.