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Math Help - finite Extension of Fp Galois

  1. #1
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    finite Extension of Fp Galois

    Are all finite Extensions of Fp (the field w/ p elements) Galois?

    My understanding is no, since finite Galois \iff normal and separable.

    Then as a counterexample, in \frac{F_3}{<x^2+x-1>}, x^2+x-1 has one root (x or equivalently x+<x^2+x-1>) but no others (right?) so it is not normal and thus not Galois.
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  2. #2
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    Quote Originally Posted by alighieri View Post
    Are all finite Extensions of Fp (the field w/ p elements) Galois?

    My understanding is no, since finite Galois \iff normal and separable.

    Then as a counterexample, in \frac{F_3[x]?}{<x^2+x-1>}, x^2+x-1 has one root (x or equivalently x+<x^2+x-1>) but no others (right?) so it is not normal and thus not Galois.
    I don't fully understand what your argument is, but the answer is yes.
    In your example, K=\frac{\mathbb{F}_3[x]}{<x^2+x-1>} is a field with 9 elements. Let \alpha=x+<x^2+x-1>. Then K=\{0, 1, 2, \alpha, \alpha+1, \alpha+2, 2\alpha, 2\alpha+1, 2\alpha+2\} and [K:F] is the Galois extension, where F=\mathbb{F}_3.

    Generally speaking, if F=\mathbb{F}_p (p is a prime number here) and [K:F]=n, then K is the splitting field of x^{p^n}-x over F (link). Thus K is the Galois extension of F. In fact, |\text{Gal}(K/F)|=n, where |\text{Gal}(K/F)| is a cyclic group generated by \sigma_{p} defined by \sigma_{p}(\alpha)=\alpha^{p} for \alpha \in K. Verify that \sigma_p has an order n such that (\sigma_{p})^{i}(\alpha)=\alpha^{p^i}.
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