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Math Help - Factor x^3 - 1 in Q[√3]

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    Factor x^3 - 1 in Q[√3]

    How can you factor x^3 - 1 completely into linear factors in Q[sqrt(3)]?
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    Quote Originally Posted by elemental View Post
    How can you factor x^3 - 1 completely into linear factors in Q[sqrt(3)]?
    Dear elemental,

    I dont understand what you mean by "Q[\sqrt 3]". Can you please elaborate? However if you factor this,

    x^3-1=(x-1)(x^2+x+1)=(x-1)((x+\frac{1}{2})^2+\frac{3}{4})=(x-1)((x+\frac{1}{2})^2-\frac{3i^2}{4})~where~i=\sqrt -1

    x^3-1=(x-1)(x+\frac{1}{2}-\frac{i\sqrt 3}{2})(x-\frac{1}{2}+\frac{i\sqrt 3}{2})

    Hope this will help you.
    Last edited by Sudharaka; January 2nd 2011 at 05:16 AM.
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  3. #3
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by Sudharaka View Post
    Dear elemental,
    I dont understand what you mean by "Q[\sqrt 3]". Can you please elaborate?
    It is the algebraic extension

    \mathbb{Q}[\sqrt{3}]=\mathbb{Q}[x]/(x^2-3)


    x^3-1=(x-1)(x^2+x+1)=(x-1)((x+\frac{1}{2})^2+1)
    It should be:

    x^3-1=(x-1)(x^2+x+1)=(x-1)\left(\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\r  ight)=\ldots

    Fernando Revilla
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by elemental View Post
    How can you factor x^3 - 1 completely into linear factors in Q[sqrt(3)]?
    Possibly you meant \mathbb{Q}[\;\sqrt{3}\;i\;] instead of \mathbb{Q}[\;\sqrt{3}\;]

    Fernando Revilla
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    " Q[\sqrt{3}]" can also be expressed as "the smallest field containing all rational numbers and \sqrt{3}". It can be shown that all such numbers are of the form p+ q\sqrt{3} for some rational numbers p and q.

    x^3- 1= (x- 1)(x^2+ x+ 1) and that is as far as it can be factored in Q or, for that matter, Q[\sqrt{3}].

    However, x^2+ x+ 1= (x^2+ x+ 1/4- 1/4)+ 1= x^2+ x+ 1/4+ 3/4= (x+ 1/2)^2+ 3/4
    = (x+ 1/2)^2-(-3/4)= (x+1/2- i\sqrt{3}{2})(x+ 1/2+ i\sqrt{3}{2}
    which are in Q[i\sqrt{3}]

    So: x^3- 1= (x-1)(x^2+ x+ 1) in Q[\sqrt{3}] but x^3- 1= (x- 1)(x+ 1/2- i\sqrt{3}{2})(x+ 1/2+ i\sqrt{3}/2) in Q[i\sqrt{3}].
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  6. #6
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    Sorry! I did mean -3, not 3. Thank you very much.
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