Galois groups of finite fields

Each $ \sigma\in D=D_\mathfrak{p}$ acts in a well-defined way on the finite field $ k_{\mathfrak{p}} = \O_K/\mathfrak{p}$, so we obtain a homomorphism

$\displaystyle \varphi :D_\mathfrak{p}\to \Gal (k_{\mathfrak{p}}/\mathbf{F}_p).

We pause for a moment and derive a few basic properties of $ \Gal (k_{\mathfrak{p}}/\mathbf{F}_p)$, which are general properties of Galois groups for finite fields. Let $ f=[ k_{\mathfrak{p}}:\mathbf{F}_p]$.

The group $ \Gal (k_{\mathfrak{p}}/\mathbf{F}_p)$ contains the element $ \Frob _p$ defined by

$\displaystyle \Frob _p(x) = x^p,$

because $ (xy)^p = x^p y^p$ and

$\displaystyle (x+y)^p = x^p + px^{p-1}y + \cdots + y^p \equiv

The group $ k_{\mathfrak{p}}^*$ is cyclic (see proof of Lemma 8.1.7), so there is an element $ a\in k_{\mathfrak{p}}^*$ of order $ p^f-1$, and $ k_{\mathfrak{p}}=\mathbf{F}_p(a)$. Then $ \Frob _p^n(a) = a^{p^n} = a$ if and only if $ (p^f-1)\mid p^n-1$ which is the case precisely when $ f\mid n$, so the order of $ \Frob _p$ is $ f$. Since the order of the automorphism group of a field extension is at most the degree of the extension, we conclude that $ \Aut (k_{\mathfrak{p}}/\mathbf{F}_p)$ is generated by $ \Frob _p$. Also, since $ \Aut (k_{\mathfrak{p}}/\mathbf{F}_p)$ has order equal to the degree, we conclude that $ k_{\mathfrak{p}}/\mathbf{F}_p$ is Galois, with group $ \Gal (k_{\mathfrak{p}}/\mathbf{F}_p)$ cyclic of order $ f$ generated by $ \Frob _p$. (Another general fact: Up to isomorphism there is exactly one finite field of each degree. Indeed, if there were two of degree $ f$, then both could be characterized as the set of roots in the compositum of $ x^{p^f}-1$, hence they would be equal.)

William Stein 2012-09-24