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Math Help - Fields and linear algebra

  1. #1
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    Fields and linear algebra

    What type of argument could be used to show there exists a field with a certain amount of elements, "x".

    If your given the integers modulo 4 how can you show it is a field. Ive done the cayley table but how do you interpret that and show the properties of a field how true?
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  2. #2
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    Quote Originally Posted by GreenandGold View Post
    What type of argument could be used to show there exists a field with a certain amount of elements, "x".

    If your given the integers modulo 4 how can you show it is a field. Ive done the cayley table but how do you interpret that and show the properties of a field how true?
    Not always can you construct a field with x elements where x is an assinged number.
    Also, \mathbb{Z}_4 is not a field.
    There is no inverse for 2.
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  3. #3
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    how could you show that for say Z_6 and could you elborate more
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    Quote Originally Posted by GreenandGold View Post
    how could you show that for say Z_6 and could you elborate more
    To be a field you need all non-zero elements to have inverses. This, \mathbb{Z}_4 is not a field because what is the inverse of 2 mod 4? Just try it out, 2\cdot 1\equiv 2 (\bmod 4), 2\cdot 2\equiv 0 (\bmod 4), 2\cdot 3\equiv 2(\bmod 4). We see that it has no inverse.

    And the same with \mathbb{Z}_6. See that 2,3,4 fail to have inverse.
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  5. #5
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    so Z_9 fails because 3 and 6 dont have a 1 on the cayley table under multiplication?
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  6. #6
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    Quote Originally Posted by GreenandGold View Post
    so Z_9 fails because 3 and 6 dont have a 1 on the cayley table under multiplication?
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  7. #7
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    Is that enough to say that Z_5 is a field?
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  8. #8
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    Quote Originally Posted by GreenandGold View Post
    Is that enough to say that Z_5 is a field?
    Yes. In general if p is a prime then \mathbb{Z}_p is a field.

    To see you need to know if p\not|a \text{ then }ax\equiv 1(\bmod p) is solvable.
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