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Math Help - Fields

  1. #1
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    Fields

    1) Find all
    c Z3 such that Z3 [x] /(x3 + x2 + c)is a field.
    2) Find the degree of and a basis for E = Q

    (√2, 3 2, 4 2)

    over Q.
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  2. #2
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    1) Find all
    c ∈ Z3 such that Z3 [x] /(x3 + x2 + c)is a field.
    \mathbb{Z}_3[x]/(x^3+x^2+c) is a field if and only if x^3+x^2+c is irreducible if and only if x^3+x^2+c has no zeros in \mathbb{Z}_3.

    2) Find the degree of and a basis for E = Q
    (\sqrt{2},\sqrt[3]{2},\sqrt[4]{2})
    Let \alpha = \sqrt[12]{2} then since x^{12}-2 is an irreducible polynomial which has \alpha as a root it means \{1,\alpha,\alpha^2,...,\alpha^{11}\} is a basis. This means \alpha^6,\alpha^4,\alpha^3 are linearly independent. Thus, \{ \sqrt{2}. \sqrt[3]{2}, \sqrt[4]{2} \} are linearly independent and so they spam a vector space of dimension three.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    [tex]
    Let \alpha = \sqrt[12]{2} then since x^{12}-2 is an irreducible polynomial which has \alpha as a root it means \{1,\alpha,\alpha^2,...,\alpha^{11}\} is a basis. This means \alpha^6,\alpha^4,\alpha^3 are linearly independent.
    Thus, \{ \sqrt{2}. \sqrt[3]{2}, \sqrt[4]{2} \} are linearly independent and so they spam a vector space of dimension three.

    that vaecor space is not \mathbb{Q}(\sqrt{2},\sqrt[4]{2},\sqrt[3]{2}).
    you see my comment in red! the degree is 12. let a=\sqrt{2}, \ b=\sqrt[4]{2}, \ c=\sqrt[3]{2}. then:

    [\mathbb{Q}(a,b,c):\mathbb{Q}]=[\mathbb{Q}(a,b)(c):\mathbb{Q}(a,b)][\mathbb{Q}(a)(b):\mathbb{Q}(a)][\mathbb{Q}(a):\mathbb{Q}]=3 \times 2 \times 2 = 12.

    a more constructive (and easier) way is to prove that \mathbb{Q}(\sqrt[12]{2})=\mathbb{Q}(a,b,c):

    clearly \mathbb{Q}(\sqrt[12]{2}) \supseteq \mathbb{Q}(a,b,c). the reverse inclusion is also clear because \sqrt[12]{2}=\frac{ab}{c^2}.
    Last edited by NonCommAlg; August 9th 2008 at 02:35 PM.
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