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Math Help - Method for finding a Minimal Polynomial?

  1. #1
    Junior Member Dark Sun's Avatar
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    Method for finding a Minimal Polynomial?

    Hello, I cannot seem to find in my notes nor in any text a method of finding minimal polynomials. Most minimal polynomials I've encountered are somewhat trivial, but I came across one that I may be tested on that I am unable to solve using a brute-force method:

    m_{\sqrt 2+i} over \mathbb Q

    The reason I ask about a method of finding the min. poly. is because I am taking an oral test, and I don't want to be floundering about if I am asked something different than this one in particular.

    Anyway, I would appreciate any help. Thanks!

    P.s.: That's m_{\sqrt 2 +i}; sorry that the "i" is so tiny.
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  2. #2
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    Quote Originally Posted by Dark Sun View Post
    Hello, I cannot seem to find in my notes nor in any text a method of finding minimal polynomials. Most minimal polynomials I've encountered are somewhat trivial, but I came across one that I may be tested on that I am unable to solve using a brute-force method:

    m_{\sqrt 2+i} over \mathbb Q

    The reason I ask about a method of finding the min. poly. is because I am taking an oral test, and I don't want to be floundering about if I am asked something different than this one in particular.

    Anyway, I would appreciate any help. Thanks!

    P.s.: That's m_{\sqrt 2 +i}; sorry that the "i" is so tiny.
    x=\sqrt{2}+i

    x^2=(\sqrt{2}+i)^2 \iff x^2=2+2i\sqrt{2}-1 \iff x^2-1=2i\sqrt{2}

    (x^2-1)^2=(2i\sqrt{2})^2

    x^4-2x^2+1=-8

    x^4-2x^2+9=0
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  3. #3
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    Quote Originally Posted by TheEmptySet View Post
    x=\sqrt{2}+i

    x^2=(\sqrt{2}+i)^2 \iff x^2=2+2i\sqrt{2}-1 \iff x^2-1=2i\sqrt{2}

    (x^2-1)^2=(2i\sqrt{2})^2

    x^4-2x^2+1=-8

    x^4-2x^2+9=0
    you also need to prove that p(x)=x^4 - 2x^2 + 9 is irreducible over \mathbb{Q}: we only need to show that it cannot be factored into two quadratic polynomials over \mathbb{Z}. why?

    suppose p(x)=(x^2+ax+b)(x^2+cx+d), for some a,b,c,d \in \mathbb{Z}. then we'll have bd=9 and b+d+2=a^2, which is easily seen to have no solutions in \mathbb{Z}.
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  4. #4
    Junior Member Dark Sun's Avatar
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    Because irreducible over \mathbb Z implies irreducible over \mathbb Q.

    x^4-2x^2+9 clearly does not have any degree 1 factors. Also, a degree 3 factor would imply a degree 1 factor, so if a factor exists, then it will be degree 2, which you have just shown is a contradiction.

    Thank you EmptySet and NonCommAlg for presenting me with this truly powerful method!
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