Results 1 to 4 of 4
Like Tree3Thanks
  • 1 Post By Vlasev
  • 2 Post By johnsomeone

Math Help - decomposition of a module

  1. #1
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    decomposition of a module

    Here is the problem I'm working on:

    Let R be an associative ring with 1, M a right R-module, F:M\rightarrow{R} a homomorphism of R-modules with f(M)=R. Prove that there is a decomposition M=K\oplus{L} with f(K)=0 and f|_L:L\rightarrow{R} is an isomorphism.

    I think that K should be the kernel of F. Then L should be something like M/K, but it seems like I need something that's actually in M instead. What should I use for L, and how do I show that M=K\oplus{L} and f|_L is an isomorphism?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member
    Joined
    Jul 2010
    From
    Vancouver
    Posts
    432
    Thanks
    17

    Re: decomposition of a module

    First off, if x \in K, then f(x) = 0, so x \in ker(f). So K, if it exists, is a subset of ker(f). So, it definitely feels like K = ker(f) is the right choice. As for the rest, there is an isomorphism theorem for modules that is much like the ring and group isomorphism theorems.

    It basically says that im(f) \cong M/ker(f). Since im(f) = R you have that im(f) = R \cong M/ker(f). Now, you are right that M/ker(f) does not reside within M so maybe you can try to find a submodule L of M that is isomorphic to M/ker(f) and then show that M = K \oplus L. From there it'll follow that f|_L : L \to R is indeed an isomorphism.
    Last edited by Vlasev; September 10th 2012 at 10:20 PM.
    Thanks from hollywood
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Mar 2010
    Posts
    993
    Thanks
    244

    Re: decomposition of a module

    So I need g:M\rightarrow{M} with kernel K=ker{f}. Then the image will be a submodule of M isomorphic to M/K.

    I think I can fix m_0 in f^{-1}(1) and make g(m)=f(m)m_0 for all m\in{M}. The kernel will be the same as the kernel of f, so that works.

    So my answer is:
    Fix m_0\in{f^{-1}(1)}, which is not empty since f(M)=R. Let g:M\rightarrow{M} be given by g(m)=f(m)m_0 for all m\in{M}. This map is a module homomorphism since:
    g(rm)=f(rm)m_0=rf(m)m_0=rg(m) and
    g(m+n)=f(m+n)m_0=(f(m)+f(n))m_0 =f(m)m_0+f(n)m_0=g(m)+g(n).
    Let L=g(M) be the image of this map and K=ker(g) be the kernel. Then I claim that M=K\oplus{L}.

    Since both K and L are submodules of M, K\oplus{L}\subseteq{M}. If m\in{M}, then it can be represented as m=(m-g(m))+g(m), where m-g(m)\in{K} since f(m-g(m))=f(m)-f(g(m))=f(m)-f(f(m)m_0) =f(m)-f(m)f(m_0)=f(m)-f(m)=0, and of course g(m)\in{L}=g(M). So M\subseteq{K}\oplus{L} and by double containment M={K}\oplus{L}.

    Given M={K}\oplus{L} as described above, f(K)=0 since K is the kernel of f. If f(m)=0 and m\in{L}, m is of the form rm_0 for some r\in{R}, so 0=f(m)=f(rm_0)=rf(m_0)=r, giving m=rm_0=0, so f|_L is injective. And every r\in{R} has a preimage (namely rm_0) in L, so f|_L is surjective. Therefore f|_L is an isomorphism.

    Thank you for giving me the push I needed.

    - Hollywood
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member
    Joined
    Sep 2012
    From
    Washington DC USA
    Posts
    525
    Thanks
    147

    Re: decomposition of a module

    What follows is basically identical to your work, except that it avoids your function g. Notice that your L = g(M) is the same as L = m_0R, since f(M) = R by assumption.
    (One other minor point - the problem said that M is a right R module, but your work treated it as if ti were a left R module. It doesn't make a difference here - you just flip the order in which you need to write those things down, and nothing is really impacted.)

    Let m_0 \in f^{-1}(1) (possible since f is surjective). Let L = m_0R. Let K = ker(f). Both are submodules of M. Let M' = L + K, also a submodule.

    Let m \in K \cap L. Then m \in L so \exists r \in R \ni m = m_0r. Also, m \in K, so f(m) = 0, so f(m_0r) = 0, so f(m_0)r = 0, so 1r = 0, so r = 0.

    Thus m \in K \cap L \implies m = m_0(0) = 0. Thus M' = K \oplus L.

    To show M' = M, let m \in M. Note that m = (m - m_0f(m)) + m_0f(m).

    Since f(m - m_0f(m)) = f(m) - f(m_0f(m)) = f(m) - f(m_0)f(m),

    have f(m - m_0f(m)) = f(m) - (1)f(m) = 0, so it follows that (m - m_0f(m)) \in ker(f) = K.

    Obviously m_0f(m) \in L. Thus m = (m - m_0f(m)) + m_0f(m) \in K + L = M'. Thus M \subset M'. Since M' \subset M, have M' = M.

    Since m \in L means \exists r \in R \ni m = m_0r, have that f|L: L \rightarrow R by f(m) = f(m_0r) = f(m_0)r = 1r = r.

    Thus ker(f|L) = \{m_0r \in L | f(m_0r) = 0 \} = \{m_0r \in L | r = 0 \} = \{ 0 \}. Also, Im( f|L ) = R, since if r \in R, then (f|L)(m_0r) = r.

    Thus f|L is an isomorphism.

    Have shown that \exists K, L submodules of M \ni M = K \oplus L, and f|K \equiv 0, and f|L is an isomorphism onto R.
    Last edited by johnsomeone; September 11th 2012 at 03:40 PM.
    Thanks from Deveno and hollywood
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: November 30th 2011, 03:50 AM
  2. Module Help
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: November 5th 2011, 04:13 PM
  3. k[x]-module
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: May 20th 2010, 04:03 PM
  4. Jordan Decomposition to Schur Decomposition
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 30th 2009, 02:52 PM
  5. Module 77
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: March 26th 2008, 02:30 PM

Search Tags


/mathhelpforum @mathhelpforum