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    Degree of $\QQ(7^{1/5},5^{1/4})$ over $\QQ$

    P-CASNS

    Area
    Algebra
    Topics
    Fields, Galois Theory
    Seen at
    UW
    Years
    2018
    Status
    draft

    Let \(\alpha,\beta\) denote the unique positive real \(5^{\text{th}}\) root of 7 and \(4^{\text{th}}\) root of 5, respectively. Determine the degree of \(\mathbb Q(\alpha,\beta)\) over \(\mathbb Q\).

    Solution

    We have \(\alpha = 7^{1/5}\) and \(\beta = 5^{1/4}\).

    1. Consider the polynomial \(f(x) = x^5 - 7 \in \QQ[x]\). By Eisenstein’s criterion at the prime \(p = 7\), \(f(x)\) is irreducible over \(\QQ\). Since \(\alpha\) is a root of \(f(x)\), the minimal polynomial of \(\alpha\) over \(\QQ\) is \(x^5 - 7\), so \([\QQ(\alpha) : \QQ] = 5\).

    2. Similarly, consider \(g(x) = x^4 - 5 \in \QQ[x]\). By Eisenstein’s criterion at the prime \(p = 5\), \(g(x)\) is irreducible over \(\QQ\), so \([\QQ(\beta) : \QQ] = 4\).

    3. Since \(\gcd([\QQ(\alpha) : \QQ], [\QQ(\beta) : \QQ]) = \gcd(5, 4) = 1\), and both extensions are subfields of \(\QQ(\alpha, \beta)\), the degree of the compositum satisfies: \[ [\QQ(\alpha, \beta) : \QQ] = [\QQ(\alpha) : \QQ] \cdot [\QQ(\beta) : \QQ] = 5 \cdot 4 = 20. \]

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