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    Field homomorphisms from $\mathbb{Q}(\alpha)$ into $\mathbb{C}$, $\mathbb{R}$, and $\overline{\mathbb{Q}}$ for a root $\alpha$ of $x^{17}-2$

    P-XIB32

    Area
    Algebra
    Topics
    Field Extensions, Roots of Unity, Splitting Fields
    Status
    draft

    Suppose that \(\alpha\) is a root in \(\mathbb C\) of \(P(x) = x^{17} - 2\). How many field homomorphisms are there from \(\mathbb Q (\alpha)\) to:

    1. \(\mathbb C\),

    2. \(\mathbb R\),

    3. \(\overline{\mathbb Q}\), the algebraic closure of \(\mathbb Q\)?

    Let \(\alpha\in\CC\) satisfy \(\alpha^{17}=2\). How many field homomorphisms are there from \(\QQ(\alpha)\) to

    1. \(\CC\),
    2. \(\RR\),
    3. \(\overline{\QQ}\)?
    Solution

    The polynomial \[ f(x)=x^{17}-2 \] is irreducible over \(\QQ\) by Eisenstein’s criterion at \(2\). Hence it is the minimal polynomial of \(\alpha\) and \[ [\QQ(\alpha):\QQ]=17. \] Since the characteristic is \(0\), \(f\) is separable and has \(17\) distinct roots in \(\CC\).

    Every field homomorphism from \(\QQ(\alpha)\) fixing \(\QQ\) is determined by the image of \(\alpha\), and that image may be any root of \(f\) lying in the target field.

    1. Into \(\CC\):All \(17\) roots lie in \(\CC\), so there are\[ 17 \]homomorphisms.

    2. Into \(\RR\):The function \(x\mapsto x^{17}\) is strictly increasing on \(\RR\), so \(x^{17}=2\) has exactly one real root. Hence there is exactly\[ 1 \]homomorphism into \(\RR\).

    3. Into \(\overline{\QQ}\):All roots of \(f\) are algebraic over \(\QQ\), hence lie in \(\overline{\QQ}\). Therefore there are again\[ 17 \]homomorphisms.

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