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Math Help - Minimum polynomial

  1. #1
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    Minimum polynomial

    Let K = \mathbb {Q} [ \sqrt {m} ] , where m is a square-free integer. Let a=r+s \sqrt {m} be an algebraic integer in K.

    a) Show that the minimum polynomial of a is x^2-2rx+r^2-ms^2

    Now, I have verified that (a)^2 - 2r(a)+r^2-ms^2 = 0

    And I have to prove it is irreducible over K, should I use the norm, but the norm is not defined here yet, can I use N(ax+b) = a^2 + b^2?

    I factored the polynomial into [x-(r- \sqrt {m}s)][x+(r- \sqrt {m}s)] - 2rx, will this help?

    b) Show that 2r, r^2-ms^2 are integers.
    Last edited by tttcomrader; February 9th 2008 at 03:42 PM.
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  2. #2
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    Quote Originally Posted by tttcomrader View Post
    Let K = \mathbb {Q} [ \sqrt {m} ] , where m is a square-free integer. Let a+r+s \sqrt {m} be an algebraic integer in K.

    a) Show that the minimum polynomial of a is x^2-2rx+rs^2
    I assume you mean  a = r+s\sqrt{m}.
    But then, (r+s\sqrt{m})^2 - 2r (r+s\sqrt{m}) + rs^2 \not =  0 if r\not = 0. Thus, it cannot be a minimal polynomial.
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  3. #3
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    When I plug in a into the poly, I have:

    (a)^2 - 2r(a) + r^2 - ms^2 = (r+s \sqrt {m})^2-2r(r+s \sqrt {m}) + r^2 - ms^2
    =r^2+2sr \sqrt {m} + s^2m - 2r^2 - 2s \sqrt {m}r + r^2 - ms^2 = 0

    So it does equal to zero.
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  4. #4
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    Quote Originally Posted by tttcomrader View Post
    When I plug in a into the poly, I have:

    (a)^2 - 2r(a) + r^2 - ms^2 = (r+s \sqrt {m})^2-2r(r+s \sqrt {m}) + r^2 - ms^2
    =r^2+2sr \sqrt {m} + s^2m - 2r^2 - 2s \sqrt {m}r + r^2 - ms^2 = 0

    So it does equal to zero.
    Okay, so then you should change x^2 - 2rx + rs^2 to x^2 - 2rx + (r^2 - ms^2) in your first post. Now, you are saying, "minimal polynomial" remember you need to specify over what field it is a minimal polynomial over, I assume it should be \mathbb{Q}. In that case you have to prove that this polynomial has no rational zeros. Can you show that?
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    Since m is a square-free integer,  \sqrt {m} is not a real number, so it is not a rational number. However, the only way to eliminate the m in the polynomial is to let x=a+b \sqrt {m} , otherwise the m would remain. So there cannot be any rational zeros.

    Is that sufficient to prove that ain't any rational zeros? One problem is I know it is true, I can explain it in words, but I don't know how to prove it mathematically.
    Last edited by tttcomrader; February 10th 2008 at 10:07 AM.
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