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Math Help - ring -> field

  1. #1
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    ring -> field

    Let be a commutative ring with elements , . Suppose that if is not inversable, then . Prove that is a field.

    I know the following result:
    A finite ring with unit and with no non-zero nilpotent elements is a finite direct product of fields.How can this result help me ?
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  2. #2
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    Quote Originally Posted by petter View Post
    Let be a commutative ring with elements , . Suppose that if is not inversable, then . Prove that is a field.

    I know the following result:
    A finite ring with unit and with no non-zero nilpotent elements is a finite direct product of fields.How can this result help me ?
    suppose 0 \neq a \in A and a^k=0, \ k > 1. then obviously a is not invertible. thus a^2 = \pm a, which gives us: 0=a^k=\pm a. so a=0, i.e. A has no non-zero nilpotent element.

    hence A=\bigoplus_{i=1}^m Q_i, where Q_1, \cdots , Q_m are finite fields. if m=1, we're done. so suppose that m \geq 2. since |A| is odd, |Q_i| must be odd for all 1 \leq i \leq m. consider two cases:

    case 1: |Q_i| > 3, for some i. without loss of generality we assume that i=1. choose x \in Q_1 - \{0,1,-1 \} and let a=(x, 0, \cdots , 0). then a is clearly a non-invertible element

    of A, but a^2 = (x^2 , 0, \cdots , 0) \neq (\pm x, 0 , \cdots , 0)= \pm a, because x \notin \{0,1,-1 \}. contradiction!


    case 2: |Q_i|=3, \ \forall i. if m=2, then |A|=9, which is impossible because the problem assumes that |A| \geq 11. therefore m \geq 3. now let a=(1,-1, 0, \cdots , 0) \in A. then a is a

    non-invertible element of A and a^2=(1,1, 0, \cdots , 0) \neq \pm a, which is again a contradiction! Q.E.D.
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