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Thread: 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 $\displaystyle [K:F]=[K:E][E:F]$. In particular it shows that $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12$. 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 $\displaystyle \mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) = \mathbb{Q}(\sqrt[3]{2})(\sqrt[4]{3})$, can we just say that since the minimal polynomial of $\displaystyle \sqrt[4]{3}$ is $\displaystyle x^4-3$, $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})]=4$, then similar to obtain $\displaystyle [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$, which gives us $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3\cdot 4 = 12$?

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

    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 $\displaystyle [K:F]=[K:E][E:F]$. In particular it shows that $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=12$. 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 $\displaystyle \mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}) = \mathbb{Q}(\sqrt[3]{2})(\sqrt[4]{3})$, can we just say that since the minimal polynomial of $\displaystyle \sqrt[4]{3}$ is $\displaystyle x^4-3$, $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})]=4$, then similar to obtain $\displaystyle [\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$, which gives us $\displaystyle [\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[3]{2}, \sqrt[4]{3}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3\cdot 4 = 12$?

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

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