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Corollary to Correspondence Theorem for Modules

  1. Nov 5, 2015 #1
    I am reading Joseph J. Rotman's book: Advanced Modern Algebra and I am currently focused on Section 6.1 Modules ...

    I need some help with the proof of Corollary 6.25 ... Corollary to Theorem 6.22 (Correspondence Theorem) ... ...

    Corollary 6.25 and its proof read as follows:


    ?temp_hash=3e7e3658af5514dc0972081c2217eed4.png


    Can someone explain to me exactly how Corollary 6.25 follows from the Correspondence Theorem for Modules ...?

    Hope that someone can help ...

    Peter


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    *** EDIT ***

    The above post refers to the Correspondence Theorem for Modules (Theorem 6.22 in Rotman's Advanced Modern Algebra) ... so I am proving the text of the Theorem from Rotman's Advanced Modern Algebra as follows:


    ?temp_hash=3e7e3658af5514dc0972081c2217eed4.png
     

    Attached Files:

  2. jcsd
  3. Nov 6, 2015 #2

    fresh_42

    Staff: Mentor

    If I is a maximal ideal of R and M := R / I then a submodule of M is of the form S / I with I ⊆ S ⊆ R (6.22). The module property makes S an ideal of R. So either S = I, i.e. S / I = 0, or S = R, i.e. S / I = M, which means M is simple. On the other hand, if M is a simple R left module then we can pick an element m ∈ M, m ≠ 0. Then φ : R → M with φ(r) := r.m defines a ring homomorphism. φ cannot be 0, because otherwise {m} would be a non-zero submodule of M. (We suppose that either R has a 1 and 1.m = m or more generally require that R doesn't operate trivially on M, i.e. R.M may not be {0}.) Since im φ is a non-zero submodule of M and M is simple, φ has to be surjective (im φ = M) with kernel I:= ker φ.
    Therefore R / I = R / ker φ ≅ im φ = M.
    According to Theorem 6.22 the absence of submodules of M ≅ R / I implies the absence of ideals in R containing I, i.e. I is a maximal left ideal.
     
    Last edited: Nov 6, 2015
  4. Nov 10, 2015 #3
    Thanks so much for your analysis Fresh 42 ... most helpful ...

    Peter
     
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