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Thread: 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 $\displaystyle \mathbb{Q} $ of $\displaystyle \ 2 + \sqrt{3} $ and of $\displaystyle 1 + \sqrt[3]{2} + \sqrt[3]{2}$

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

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

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

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

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

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