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Math Help - Prime Degree Polynomial

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    Prime Degree Polynomial

    I discovered an elegant theorem. But it is easy to prove using some heavy altirary.

    Theorem: Let f(x) be a polynomial with rational coefficients having a prime degree p. If f(x) is irreducible (cannot be factored in terms of other non-constant polynomials having rational coefficients) then f(x) has exactly p complex zeros.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    I discovered an elegant theorem. But it is easy to prove using some heavy altirary.

    Theorem: Let f(x) be a polynomial with rational coefficients having a prime degree p. If f(x) is irreducible (cannot be factored in terms of other non-constant polynomials having rational coefficients) then f(x) has exactly p complex zeros.
    Correct me if I'm wrong, but
    x^3 + x^2 - x + 1 = 0
    does not have 3 complex zeros.

    -Dan
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    Quote Originally Posted by topsquark View Post
    Correct me if I'm wrong, but
    x^3 + x^2 - x + 1 = 0
    does not have 3 complex zeros.

    -Dan
    When I say "complex" I allow real also. Since \mathbb{R} < \mathbb{C}.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    When I say "complex" I allow real also. Since \mathbb{R} < \mathbb{C}.
    Ah, so if I'm not mistaken (again) do you mean to say that these p zeros are distinct?

    -Dan
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    Super Member Rebesques's Avatar
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    When I say "complex" I allow real also.

    Then, isn't asking for it to be irreducible over Q just too demanding?
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    Quote Originally Posted by topsquark View Post
    Ah, so if I'm not mistaken (again) do you mean to say that these p zeros are distinct?
    Yes, exactly.


    We can state this theorem in a different way (equivalent):
    Let f(x) be a polynomial with integer coefficients and prime degree p so that f(x) is irreducible (cannot be factord into non-constant polynomials with integer coefficients) then f(x) has exactly p zeros.

    Quote Originally Posted by Rebesques View Post
    Then, isn't asking for it to be irreducible over Q just too demanding?
    Why?
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    Super Member Rebesques's Avatar
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    Because the thing I found most intriguing in the original statement, was the promise that the roots are not real Can we have that?
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    Quote Originally Posted by Rebesques View Post
    Because the thing I found most intriguing in the original statement, was the promise that the roots are not real Can we have that?
    In topsquark's example, that polynomial satisfies my conditions. Now, as promised, it has exactly 3 zeros (1 real 2 not). But, yes, we need that condition that they can be complex too.


    Note, it should not be supprising because in field theory we usually deal in one big extension field, usually the algebraic closure. Which in this case is the closure of \mathbb{Q} which are the algebraic numbers.
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