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Vector spaces as quotients of free modules

  1. Jul 18, 2011 #1
    1. The problem statement, all variables and given/known data
    Let R be a commutative ring, and let [itex] F = R^{\oplus B} [/itex] be a free R-module over R. Let m be a maximal ideal of R and take [itex] k = R/m [/itex] to be the quotient field. Show that [itex] F/mF \cong k^{\oplus B} [/itex] as k-vector spaces.

    3. The attempt at a solution

    If we remove the F and k notations, we essentially just want to show that
    [tex] R^{\oplus B}/ m R^{\oplus B} \cong (R/m)^{\oplus B} [/tex]
    and so it seems like we should use the first isomorphism theorem.

    Now we define [itex] R^{\oplus B} = \{ \alpha: B \to R, \alpha(x) = 0 \text{ cofinitely in } B \} [/itex]. If [itex] \pi : R \to R/m [/itex] is the natural projection map, define [itex] \phi: R^{\oplus B} \to (R/m)^{\oplus B} [/itex] by sending [itex] \alpha \mapsto \pi \circ \alpha [/itex]. This is an R-mod homomorphism, and is surjective by the surjectivity of the projection. Hence we need only show that the kernel of this map is given by [itex] mR^{\oplus B} [/itex]

    Now
    [tex]\begin{align*}
    \ker\phi &= \{ \alpha:B \to R, \forall x \in B \quad \pi(\alpha(x)) = 0_k \} \\
    &= \{ \alpha:B \to R, \forall x \in B, \alpha(x) \in m \} \end{align*}
    [/tex]
    where I may have skipped a few steps in this derivation, but I think this is right. Now it's easy to show that [itex] mF \subseteq \ker \phi [/itex] since m is an ideal. However, the other inclusion is where I'm having trouble.

    I guess maybe the whole question could be rephrased to avoid the baggage that comes with the question. Namely, if [itex] \alpha: B \to R [/itex] is such that [itex] \forall x \in B, \alpha(x) [/itex] lies in a maximal ideal of R, should that [itex] \alpha = r \beta [/itex] for some [itex] \beta: B \to R [/itex].

    Edit: I guess I'm hoping to show they're isomorphic as R-modules, and then push that over to k-vector spaces. Maybe this is where my mistake is coming from?
    Edit 2: Fixed mistake made in "without baggage" statement.
     
    Last edited: Jul 18, 2011
  2. jcsd
  3. Jul 18, 2011 #2
    I just had a thought, though I'm not sure if it's true and will have to think about it more. If m is a maximal ideal, does this pass a level of "maximality" to the submodule it generates? If this is true, then the statement that
    [tex] mF \subseteq \ker \phi \quad \Rightarrow \quad \ker\phi = mF [/tex]
    becomes trivial.
     
  4. Jul 18, 2011 #3

    micromass

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    Hi Kreizhn! :smile:

    So, you must show that [itex]Ker(\phi)\subseteq mR^{\oplus B}[/itex]. Now, working with maps [itex]\alpha:B\rightarrow R[/itex] is a bit confusing, so I'm going to work with tuples [itex](r_i)_{i\in B}[/itex].
    If such a tuple is in [itex]Ker(\phi)[/itex], the you have shown that each ri is in m.
    Now, the trick is to decompose your tuple as a sum, for example:

    [tex](1,2,1)=(1,0,0)+(0,2,0)+(0,0,1)[/tex]

    and show that each of the terms is in [itex]mR^{\oplus B}[/itex].
     
  5. Jul 18, 2011 #4
    Well what I had originally is let [itex] v \in k [/itex] so that we can write [itex] v= \sum r_i x_i [/itex], where in tuple notation this would be [itex] x_i \in m, \forall i [/itex] or in function notation,
    [tex] v = \sum r_i \alpha(i) [/tex]
    so that [itex] \alpha(i) \in m [/itex].

    From here each term is definitely in mF. But now I'm a little confused, but maybe I"m not confused, which is itself confusing.

    So in my head, mF consists of elements of the form [itex] \{ rf | r \in m, f \in F \} [/itex] and so the way I've written it above wouldn't be sufficient since the "coefficients" are not unique.

    But maybe this isn't necessary? Since each element individually is of this form, and since mF is a submodule, then the sum is also in the module and we're done. Does that make sense, or am I missing something here?
     
  6. Jul 18, 2011 #5

    micromass

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    The thing is that each [itex]r_ix_i[/itex] is in mF (and here the ri is unique), thus the sum is also in mF.
     
  7. Jul 18, 2011 #6
    Yeah, that fits with what I was thinking by the end. It just looks kinda weird from a "function" point of view, since it seems like we're saying that since each [itex] r_i\alpha(i) \in m [/itex] then there is an [itex] r \in R, \beta: B \to R [/itex] such that [itex] r \beta = \sum r_i \alpha [/itex]. I'll just have to think about it a little more, but thanks for your help micromass.

    Edit: No wait, is [itex] r = \sum r_i [/itex]? Then that's obvious!
    Edit 2: No wait, we need r such that [itex] r \beta(i) = r_i \alpha(i) [/itex] for each i. It's still hurting my brain, but I'll think about it more.
     
  8. Jul 18, 2011 #7

    micromass

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    I think you're kind of confused by the definition of mF. In general, if I is an ideal of R and if M is an R-module, then IM is the module generated by elements of the form im (with i in I and m in M).
     
  9. Jul 18, 2011 #8
    So it's more like the product of ideals than cosets of the group.
     
  10. Jul 18, 2011 #9
    Ah yes, just looked at the errata of the text I'm using. The definition of the text says it looks like cosets, but the errata corrects this to be the generating set.

    Yay for a great deal of confusion happening because of textbook errors!
     
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