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Thread: 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.
    $\displaystyle \mathbb{Z}_3[x]/(x^3+x^2+c)$ is a field if and only if $\displaystyle x^3+x^2+c$ is irreducible if and only if $\displaystyle x^3+x^2+c$ has no zeros in $\displaystyle \mathbb{Z}_3$.

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

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

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

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

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