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The generalized rank-nullity theorem

  1. Mar 20, 2009 #1

    quasar987

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    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?
     
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  3. Mar 20, 2009 #2

    Hurkyl

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    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.
     
  4. Mar 20, 2009 #3
    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).
     
  5. Mar 20, 2009 #4

    Hurkyl

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    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:
     
  6. Mar 26, 2009 #5

    quasar987

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    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?
     
  7. Mar 26, 2009 #6

    Hurkyl

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    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]
    ?
     
  8. Mar 26, 2009 #7

    quasar987

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    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 [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.
     
  9. Mar 27, 2009 #8

    Hurkyl

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    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)[/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: Mar 27, 2009
  10. Mar 27, 2009 #9
    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.
     
  11. Mar 27, 2009 #10

    quasar987

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    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?
     
  12. Mar 27, 2009 #11
    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: May 4, 2017
  13. Mar 27, 2009 #12

    quasar987

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    Yes, sorry about that.

    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: May 4, 2017
  14. Mar 27, 2009 #13

    Hurkyl

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    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!
     
  15. Mar 27, 2009 #14

    Hurkyl

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    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.
     
  16. Mar 27, 2009 #15

    quasar987

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    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

    ?
     
  17. Mar 27, 2009 #16

    Hurkyl

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    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].

    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: Mar 27, 2009
  18. Mar 28, 2009 #17

    quasar987

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    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' ?


    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!)
     
  19. Mar 28, 2009 #18

    Hurkyl

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    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].
     
  20. Mar 30, 2009 #19

    quasar987

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    Ok, I understand now, thank you both.
     
  21. Apr 7, 2009 #20

    quasar987

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    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.
     
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