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Thread: 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 $\displaystyle \iff$ normal and separable.

    Then as a counterexample, in $\displaystyle \frac{F_3}{<x^2+x-1>}$, 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 $\displaystyle \iff$ normal and separable.

    Then as a counterexample, in $\displaystyle \frac{F_3[x]?}{<x^2+x-1>}$, 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, $\displaystyle K=\frac{\mathbb{F}_3[x]}{<x^2+x-1>}$ is a field with 9 elements. Let $\displaystyle \alpha=x+<x^2+x-1>$. Then $\displaystyle K=\{0, 1, 2, \alpha, \alpha+1, \alpha+2, 2\alpha, 2\alpha+1, 2\alpha+2\}$ and [K:F] is the Galois extension, where $\displaystyle F=\mathbb{F}_3$.

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