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Math Help - G is isomorphic to a field

  1. #1
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    G is isomorphic to a field

    Hello

    }}
    Consider G, a Domain which contains the one and a finite number of elements. Now Show that G is isomorphic to a field containing q^n elements where q is a prime and n is a natural number.
    }}

    A domain firstly means that G is a ring containing no left- or right- zerodivisors but the 0 itself. So intuitively if there is a number m that divides q there exists an element which can be multiplicated with q and since the characteristic of \mathbb{Z}_{q^n} is q that has to be zero?

    Yet I don't see a way to Show this in an understandable, formal correct way. Could please someone give me a hint what to do in this case here?
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  2. #2
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    Re: G is isomorphic to a field

    Hi,
    I think the attachment solves your problem:

    G is isomorphic to a field-mhfrings9.png
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