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    $[\QQ(2^{3/2}):\QQ]$

    P-JWYGQ

    Area
    Algebra
    Topics
    Field Extensions
    Status
    draft

    Compute \[ [\QQ(2^{3/2}):\QQ]. \]

    Solution

    Since \[ 2^{3/2}=2\sqrt2, \] and \(2\in\QQ^\times\), we have \[ \QQ(2^{3/2})=\QQ(2\sqrt2)=\QQ(\sqrt2). \] The polynomial \(x^2-2\) is irreducible over \(\QQ\) by the rational-root test (or Eisenstein at \(2\)), so \[ [\QQ(\sqrt2):\QQ]=2. \] Therefore \[ [\QQ(2^{3/2}):\QQ]=2. \]

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