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    A finite separable splitting field $L/K$ has $[L:K]=|\gal(L/K)|$

    P-X7AK4

    Area
    Algebra
    Topics
    Galois Theory, Separability, Splitting Fields
    Status
    draft

    Let \(L/K\) be a finite separable extension which is the splitting field of a polynomial in \(K[x]\). Prove \[ [L:K]=|\operatorname{Gal}(L/K)|. \]

    Solution

    Because \(L/K\) is finite and separable, the primitive element theorem gives \[ L=K(\alpha) \] for some \(\alpha\in L\). Let \(m_\alpha(x)\) be its minimal polynomial over \(K\). Then \[ \deg m_\alpha=[L:K]. \]

    A \(K\)-embedding \[ \sigma:L\hookrightarrow\overline K \] is determined by the image of \(\alpha\), and \(\sigma(\alpha)\) can be any root of \(m_\alpha\). Since the extension is separable, \(m_\alpha\) has exactly \([L:K]\) distinct roots. Hence there are exactly \([L:K]\) \(K\)-embeddings of \(L\) into \(\overline K\).

    Because \(L/K\) is a splitting field, it is normal. Therefore every \(K\)-embedding of \(L\) into \(\overline K\) has image equal to \(L\), so every such embedding is a \(K\)-automorphism of \(L\).

    Thus \[ |\operatorname{Gal}(L/K)|=[L:K]. \]

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