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Math Help - Subrings of the complex numbers

  1. #1
    Senior Member Pinkk's Avatar
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    Subrings of the complex numbers

    Let Q\[\sqrt{2},\sqrt{3}] denote the smallest subring of the complex numbers containing the rational numbers, \sqrt{2}, and \sqrt{3}. Is Q[\sqrt{2}+\sqrt{3}] =Q[\sqrt{2},\sqrt{3}]?

    How do I go about proving or disproving it? I'm totally lost...
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Pinkk View Post
    Let Q\[\sqrt{2},\sqrt{3}] denote the smallest subring of the complex numbers containing the rational numbers, \sqrt{2}, and \sqrt{3}. Is Q[\sqrt{2}+\sqrt{3}] =Q[\sqrt{2},\sqrt{3}]?

    How do I go about proving or disproving it? I'm totally lost...
    Can you give us any indication where you're having trouble? Do you think it's true?
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  3. #3
    Senior Member Pinkk's Avatar
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    I think it's true. I mean, unless I'm mistaken, every element of Q[\sqrt{2}, \sqrt{3}] is of the form a + b\sqrt{2} + c\sqrt{3} for a,b,c rational, and every element of Q[\sqrt{2} + \sqrt{3}] is of the form x + y(\sqrt{2} + \sqrt{3}) for x,y rational, but after this point I'm stuck.
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  4. #4
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    there are two ways you might proceed:

    one way is to show the sets contain the same elements, equivalently, that each set contains the other.

    a slightly more sophisticated way is to show that each ring is generated by the same same set (you should have 3 generators, the choice of which should be fairly obvious).

    you are mistaken that every element of Q[√2, √3] is of the form a + b√2 + c √3. for example, √6 is not of this form, yet is obviously in the ring being (√2)(√3).

    the key in all cases, is to see that √2+√3 satisfies some polynomial of minimal degree in Q[x]. what will the degree of this polynomial have to be? (hint: it is not of degree 2).
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  5. #5
    Senior Member Pinkk's Avatar
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    It is degree 4, if I'm not mistaken. I am not seeing how that helps though. Would the generators be 1, \sqrt{2}, and \sqrt{3}? Again though, not sure what connections I should be making.
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  6. #6
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    you're trying to prove Q(√2,√3) = Q(√2+√3).

    it is obvious that √2+√3 is in Q(√2,√3). what is NOT obvious is that √2 and √3 are in Q(√2+√3).

    now (√2+√3)^2 = 5+2√6, that's not terribly helpful.

    however, (√2+√3)^3 = (√2)^3 + 3(√12) + 3(√18) + (√3)^3 = 11√2 + 9√3. this IS helpful:

    (1/2)((√2+√3)^3 - 9(√2+√3)) = √2 and (1/2)(11(√2+√3) - (√2+√3)^3) = √3.

    alternatively, if you know that √2+√3 is a root of x^4 - 10x^2 + 1, then

    we know that (√2+√3)(√2+√3)^3 = 10(√2+√3)^2 - 1. working within the complex field, we know we can divide by √2+√3 to get:

    (√2+√3)^3 = 10(√2+√3) - (√3 - √2) = 11√2 + 9√3.
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