Thread: finite Extension of Fp Galois

1. 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 has one root (x or equivalently x+<x^2+x-1>) but no others (right?) so it is not normal and thus not Galois.

2. Originally Posted by alighieri
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 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, $K=\frac{\mathbb{F}_3[x]}{}$ is a field with 9 elements. Let $\alpha=x+$. 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 $(\sigma_{p})^{i}(\alpha)=\alpha^{p^i}$.