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Math Help - Field Theory - Algebraic Extensions - D&F - Section 13.2 - Exercise 7, page 530

  1. #1
    Super Member Bernhard's Avatar
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    Field Theory - Algebraic Extensions - D&F - Section 13.2 - Exercise 7, page 530

    Exercise 7 in Section 13.2 Algebraic Extensions, page 530 of Dummit and Foote states the following:

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    7. Prove that  \mathbb{Q} ( \sqrt{2} +  \sqrt{3} ) = \mathbb{Q} ( \sqrt{2} , \sqrt{3} ).

    Conclude that  [\mathbb{Q} ( \sqrt{2} +  \sqrt{3} ) \ : \  \mathbb{Q} ]  = 4 .

    Find an irreducible polynomial satisfied by  \sqrt{2} +  \sqrt{3}

    --------------------------------------------------------------------------------------------------

    I am somewhat overwhelmed by this problem ... can someone advise me on an approach ... and, indeed, get me started?

    Peter
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    Re: Field Theory - Algebraic Extensions - D&F - Section 13.2 - Exercise 7, page 530

    Let u=\sqrt{2}+\sqrt{3}

    \frac{1}{2}u^3-\frac{9}{2}u=\sqrt{2}

    \frac{11}{2}u-\frac{1}{2}u^3=\sqrt{3}

    These equations show that \sqrt{2},\sqrt{3}\in Q(u)
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    Re: Field Theory - Algebraic Extensions - D&F - Section 13.2 - Exercise 7, page 530

    For the irreducible polynomial, take powers.

    \begin{align*}(\sqrt{2}+\sqrt{3})^2 & = 5+2\sqrt{6} \\ (\sqrt{2}+\sqrt{3})^3 & = ? \\ (\sqrt{2}+\sqrt{3})^4 & = ? \\ & \vdots \end{align*}

    From those, you should be able to find a relationship such that if x=\sqrt{2}+\sqrt{3}, then x^n+a_{n-1}x^{n-1}+\ldots + a_1x+a_0=0. Hint: the problem tells you you don't need to go past x^4.
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