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Math Help - Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

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    Super Member Bernhard's Avatar
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    Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

    Can someone help me get started on the following problem.

    Determine the degree over  \mathbb{Q}   of  \ 2 + \sqrt{3} and of  1 + \sqrt[3]{2} + \sqrt[3]{2}

    Peter
    Last edited by Bernhard; September 28th 2013 at 01:06 AM.
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    Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

    Q \subset Q\left(2+\sqrt{3}\right)\subset Q\left(\sqrt{3}\right)

    So the degree of 2+\sqrt{3} is 2 since it cannot be 1

    Besides, 2+\sqrt{3} satisfies the irreducible polynomial x^2-4x+1
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    Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

    1+\sqrt[3]{2}+\sqrt[3]{2} = 1+2\sqrt[3]{2}. Are you sure that is what you want? Or did you mean 1+\sqrt[3]{2}+\sqrt[3]{3}? Assuming it is, 1+\sqrt[3]{2}+\sqrt[3]{3} \in \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}). Then, x^3-2 and x^3-3 generate that field extension. So, is x^3-3 reducible over \mathbb{Q}(\sqrt[3]{2})? If not, then \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}) has degree 3 over \mathbb{Q}(\sqrt[3]{2}). Try to figure out why the degree of \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}) over \mathbb{Q} equals the degree of \mathbb{Q}(\sqrt[3]{2}) over \mathbb{Q} times the degree of \mathbb{Q}(\sqrt[3]{2},\sqrt[3]{3}) over \mathbb{Q}(\sqrt[3]{2}).
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    Re: Field Theory - Algebrais Extensions - D&F - Section 13.2 - Exercise 4, page 530

    To see that x^3-3 is irreducible over \mathbb{Q}(\sqrt[3]{2}), assume that it is reducible. Then x^3-3 = (x-a)(x^2+bx+c) with a,b,c\in \mathbb{Q}(\sqrt[3]{2}). Multiply that out and try to solve for a,b,c. You find that b-a=0, c-ab=0, abc=3 (by equating coefficients). You find that any solution must have a=\sqrt[3]{3}, and since \sqrt[3]{3}\notin \mathbb{Q}(\sqrt[3]{2}), x^3-3 is indeed irreducible over \mathbb{Q}(\sqrt[3]{2}).
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