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Math Help - irreductible polinomials

  1. #1
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    Red face irreductible polinomials

    hi! i have to find the polinomial ireductible over the rationals asosiated with -1+2^[1/2] can any body heplme?
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  2. #2
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    Re: irreductible polinomials

    You're looking for a polynomial with rational coefficients that has -1 + \sqrt{2} as a root.
    That means, in some extension field, you can factor out x - (-1 + \sqrt{2}) from the polynomial.
    Can you think of another root, say u (which may be in the extension), so that when you expand (x - (-1 + \sqrt{2}))(x-u), the polynomial will have rational coefficients?
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  3. #3
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    Re: irreductible polinomials

    note that if x = √2 - 1

    x2 = (√2 - 1)2 = 2 - 2√2 + 1 = 3 - 2√2

    so x2 + 2x = 3 - 2√2 + 2√2 - 2 = 3 - 2 = 1

    thus: x2 + 2x - 1 = 0

    that is, √2 - 1 is a root of x2 + 2x - 1 in Q[x].

    note that if x2 + 2x - 1 is reducible over Q, it must factor as x2 + 2x - 1 = (x - a)(x - b) for rational a, b.

    by a theorem of Gauss, a and b must actually be integers, so either:

    a = 1, b= -1, or a = -1, b = 1, both of which lead to a contradiction.
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