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Math Help - Degree of an extension - easier way?

  1. #1
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    Degree of an extension - easier way?

    My book gives an example in which they calculate the degree of a field extension using [K:F]=[K:E][E:F]. In particular it shows that [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12. I feel like I might have an easier way to do it than the book does, and I was hoping someone could confirm that this works, or if it does not, why it does not.

    Since \mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) = \mathbb{Q}(\sqrt[3]{2})(\sqrt[4]{3}), can we just say that since the minimal polynomial of \sqrt[4]{3} is x^4-3, [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})]=4, then similar to obtain [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3, which gives us [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3\cdot 4 = 12?

    The book first points out that we know that both 3 and 4 divide [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}] so it's at least 12, but then the minimum polynomial of \sqrt[4]{3} is x^4-3 so [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})] \le 4 (when is it less?) and so [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}] \le 12, yielding of course [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12.

    Thanks.
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  2. #2
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    Re: Degree of an extension - easier way?

    Quote Originally Posted by AlexP View Post
    My book gives an example in which they calculate the degree of a field extension using [K:F]=[K:E][E:F]. In particular it shows that [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12. I feel like I might have an easier way to do it than the book does, and I was hoping someone could confirm that this works, or if it does not, why it does not.

    Since \mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) = \mathbb{Q}(\sqrt[3]{2})(\sqrt[4]{3}), can we just say that since the minimal polynomial of \sqrt[4]{3} is x^4-3, [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})]=4, then similar to obtain [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3, which gives us [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3\cdot 4 = 12?

    The book first points out that we know that both 3 and 4 divide [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}] so it's at least 12, but then the minimum polynomial of \sqrt[4]{3} is x^4-3 so [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})] \le 4 (when is it less?) and so [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}] \le 12, yielding of course [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12.

    Thanks.
    by Eisenstein's criterion, the minimal polynomial of \sqrt[4]{3} over \mathbb{Q} is x^4 - 3 but [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})] is the degree of the minimal polynomial of \sqrt[4]{3} over \mathbb{Q}(\sqrt[3]{2}) not \mathbb{Q}. so, the problem is that you didn't prove that x^4-3 is irreducible over \mathbb{Q}(\sqrt[3]{2}).
    your textbook's approach is avoiding the above problem and i think it's a good approach.
    the reason that [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})] \le 4 is that \sqrt[4]{3} is a zero of the polynomial x^4-3 \in \mathbb{Q}(\sqrt[3]{2})[x] and thus the degree of the minimal polynomial of \sqrt[4]{3} over \mathbb{Q}(\sqrt[3]{2}) is at most \deg (x^4-3)=4.
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  3. #3
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    Re: Degree of an extension - easier way?

    To clarify... Is it the fact that the minimal polynomial of \sqrt[4]{3} has degree 4 in \mathbb{Q} that gives us an upper bound of degree 4 on the minimal polynomial of \sqrt[4]{3} in \mathbb{Q}(\sqrt[3]{2})?
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