Galois groups of finite fields

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

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

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

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

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

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

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

By Exercise 29 (see Chapter 22), the group $\mathbf{F}_\mathfrak{p}^*$ is cyclic, so there is an element $a\in\mathbf{F}_\mathfrak{p}^*$ of order $p^f-1$, and $\mathbf{F}_\mathfrak{p}=\mathbf{F}_p(a)$. Then $\Frob _p^n(a) = a^{p^n} = a$ if and only if $(p^f-1)\mid p^n-1$ which is the case preciselywhen $f\mid n$, 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 ${\mathrm{Aut}}(\mathbf{F}_\mathfrak{p}/\mathbf{F}_p)$ is generated by $\Frob _p$. Also, since ${\mathrm{Aut}}(\mathbf{F}_\mathfrak{p}/\mathbf{F}_p)$ has order equal to the degree, we conclude that $\mathbf{F}_\mathfrak{p}/\mathbf{F}_p$ is Galois, with group ${\mathrm{Gal}}(\mathbf{F}_\mathfrak{p}/\mathbf{F}_p)$ cyclic of order $f$ generated by $\Frob _p$. (Anther 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.)