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I Simple Modules and Right Annihilator Ideals ...

  1. Feb 6, 2017 #1
    I am reading Paul E. Bland's book, "Rings and Their Modules".

    I am focused on Section 6.1 The Jacobson Radical ... ...

    I need help with the proof of Proposition 6.1.7 ...


    Proposition 6.1.7 and its proof read as follows:


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    ?temp_hash=7f07df23b94ee18b2335eb75acdc280b.png



    In the above text from Bland ... in the proof of (1) we read the following:


    " ... ... we see that ##\text{ann}_r( R / \mathfrak{m} ) = \text{ann}_r(S)##. But ##a \in \text{ann}_r( R / \mathfrak{m} )## implies that ##a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0## , so ##a \in \mathfrak{m}##. "


    Can someone please explain exactly why ##a \in \text{ann}_r( R / \mathfrak{m} )## implies that ##a + \mathfrak{m} = ( 1 + \mathfrak{m} ) a = 0## ... ... ?

    Can someone also explain how this then implies that ##a \in \mathfrak{m}## ... ?


    Hope someone can help ...

    Peter
     

    Attached Files:

  2. jcsd
  3. Feb 6, 2017 #2

    andrewkirk

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    Assume ##a \in \text{ann}_r( R / \mathfrak{m} )##. That means that, for any element of ##R/\mathfrak m##, say ##s+\mathfrak m##, we have
    $$(s+\mathfrak m)a=0_{R/\mathfrak m}\ \ \ \ \ (1)$$
    Hence, in particular, we have
    $$(1+\mathfrak m)a=0_{R/\mathfrak m}\ \ \ \ (2)$$
    (I assume, since the proof refers to ##1##, that ##R## is assumed to contain a multiplicative identity, ie it is a 'unitary ring')

    Now by definition of multiplication in a quotient:
    $$(1+\mathfrak m)a=1\times a+\mathfrak m=a+\mathfrak m\ \ \ \ (3)$$
    and by (2) this is equal to ##0_{R/\mathfrak m}##, which is ##\mathfrak m##.

    Thus
    $$a+\mathfrak m=\mathfrak m\ \ \ \ (4)$$
    which means that ##a\in\mathfrak m##.
     
  4. Feb 7, 2017 #3
    Thanks Andrew ... most clear and very helpful ...

    Appreciate your help...

    Peter
     
  5. Feb 7, 2017 #4
     
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