Here we follow the notation of the entry on the decomposition group. See also this entry.

Example 1

Let $K=\Rats(\sqrt{-7})$ ; then $\operatorname{Gal}(K/\Rats)=\{\operatorname{Id},\sigma\} \cong \Ints/2\Ints$ , where $\sigma$ is the complex conjugation map. Let $\mathcal{O}_K$ be the ring of integers of $K$ . In this case:$$ \mathcal{O}_K=\Ints\left[\frac{1+\sqrt{-7}}{2}\right]$$ The discriminant of this field is $D_{K/\Rats}=-7$ . We look at the decomposition in prime ideals of some prime ideals in $\Ints$ :

1. The only prime ideal in $\Ints$ that ramifies is $(7)$ : $$(7)\mathcal{O}_K=(\sqrt{-7})^2$$ and we have $e=2,f=g=1$ . Next we compute the decomposition and inertia groups from the definitions. Notice that both $\operatorname{Id}, \sigma$ fix the ideal $(\sqrt{-7})$ . Thus: $$D((\sqrt{-7})/(7))=\operatorname{Gal}(K/\Rats)$$ For the inertia group, notice that $\sigma\equiv \operatorname{Id}\ \operatorname{mod}\ (\sqrt{-7})$ . Hence: $$T((\sqrt{-7})/(7))=\operatorname{Gal}(K/\Rats)$$ Also note that this is trivial if we use the properties of the fixed field of $D((\sqrt{-7})/(7))$ and $T((\sqrt{-7})/(7))$ (see the section on ``decomposition of extensions'' in the entry on decomposition group), and the fact that $e\cdot f\cdot g=n$ , where $n$ is the degree of the extension ($n=2$ in our case).
2. The primes $(5),(13)$ are inert, i.e. they are prime ideals in $\mathcal{O}_K$ . Thus $e=1=g,f=2$ . Obviously the conjugation map $\sigma$ fixes the ideals $(5),(13)$ , so $$D(5\mathcal{O}_K/(5))=\operatorname{Gal}(K/\Rats)=D(13\mathcal{O}_K/(13))$$ On the other hand $\sigma(\sqrt{-7})\equiv-\sqrt{-7}\ \operatorname{mod} (5),(13)$ , so $\sigma\neq \operatorname{Id}\ \operatorname{mod} (5),(13)$ and $$T(5\mathcal{O}_K/(5))=\{\operatorname{Id}\}=T(13\mathcal{O}_K/(13))$$
3. The primes $(2),(29)$ are split: $$2\mathcal{O}_K={\left( 2,\frac{1+\sqrt{-7}}{2}\right)}{\left( 2,\frac{1-\sqrt{-7}}{2}\right)}=\mathcal{P}\cdot\mathcal{P'}$$ $$29\mathcal{O}_K=\left(29,14+\sqrt{-7}\right)\left(29,14-\sqrt{-7}\right)=\mathcal{R}\cdot\mathcal{R'}$$ so $e=f=1,g=2$ and $$D(\mathcal{P}/(2))=T(\mathcal{P}/(2))=\{\operatorname{Id}\}=D(\mathcal{R}/(29))=T(\mathcal{R}/(29))$$

Example 2

Let $\zeta_7=e^{\frac{2\pi i}{7}}$ , i.e. a $7^{th}$ -root of unity, and let $L=\Rats(\zeta_7)$ . This is a cyclotomic extension of $\Rats$ with Galois group $$\operatorname{Gal}(L/\Rats)\cong \left(\Ints/7\Ints\right)^{\times}\cong \Ints/6\Ints$$ Moreover $$\operatorname{Gal}(L/\Rats)=\{\sigma_a\colon L\to L \mid \sigma_a(\zeta_7)=\zeta_7^a,\quad a\in \left(\Ints/7\Ints\right)^{\times} \}$$

Galois theory gives us the subfields of $L$ :

The discriminant of the extension ${L/\Rats}$ is $D_{L/\Rats}=-7^5$ . Let $\mathcal{O}_L$ denote the ring of integers of $L$ , thus $\mathcal{O}_L=\Ints[\zeta_7]$ . We use the results of this entry to find the decomposition of the primes $2,5,7,13,29$ :

1. The prime ideal $7\Ints$ is totally ramified in $L$ , and the only prime ideal that ramifies: $$7\mathcal{O}_L=(1-\zeta_7)^6=\mathfrak{T}^6$$ Thus $$e(\mathfrak{T}/(7))=6,\quad f(\mathfrak{T}/(7))=g(\mathfrak{T}/(7))=1$$ Note that, by the properties of the fixed fields of decomposition and inertia groups, we must have $L^{T(\mathfrak{T}/(7))}=\Rats=L^{D(\mathfrak{T}/(7))}$ , thus, by Galois theory, $$D(\mathfrak{T}/(7))=T(\mathfrak{T}/(7))=\operatorname{Gal}(L/\Rats)$$
2. The ideal $2\Ints$ factors in $K$ as above, $2\mathcal{O}_K=\mathcal{P}\cdot\mathcal{P'}$ , and each of the prime ideals $\mathcal{P},\mathcal{P'}$ remains inert from $K$ to $L$ , i.e. $\mathcal{P}\mathcal{O}_L=\mathfrak{P}$ , a prime ideal of $L$ . Note also that the order of $2\ \operatorname{mod}\ 7$ is $3$ , and since $g$ is at least $2$ , $2\cdot3=6$ , so $e$ must equal $1$ (recall that $efg=n$ ): $$e(\mathfrak{P}/(2))=1,\quad f(\mathfrak{P}/(2))=3,\quad g(\mathfrak{P}/(2))=2$$ Since $e=1$ , $L^{T(\mathfrak{P}/(2))}=L$ , and $[L\colon L^{D(\mathfrak{P}/(2))}]=3$ , so $$D(\mathfrak{P}/(2))=<\sigma_2>\cong \Ints/3\Ints,\quad T(\mathfrak{P}/(2))=\{\operatorname{Id}\}$$
3. The ideal $(5)$ is inert, $5\mathcal{O}_L=\mathfrak{S}$ is prime and the order of $5$ modulo $7$ is $6$ . Thus: $$e(\mathfrak{S}/(5))=1,\quad f(\mathfrak{S}/(5))=6,\quad g(\mathfrak{S}/(5))=1$$ $$D(\mathfrak{S}/(5))=\operatorname{Gal}(L/\Rats),\quad T(\mathfrak{S}/(5))=\{\operatorname{Id}\}$$
4. The prime ideal $13\Ints$ is inert in $K$ but it splits in $L$ , $13\mathcal{O}_L=\mathfrak{Q}_1\cdot\mathfrak{Q}_2\cdot\mathfrak{Q}_3$ , and $13\equiv 6\equiv -1\ \operatorname{mod}\ 7$ , so the order of $13$ is $2$ : $$e(\mathfrak{Q}_i/(13))=1,\quad f(\mathfrak{Q}_i/(13))=2,\quad g(\mathfrak{Q}_i/(13))=3$$ $$D(\mathfrak{Q}_i/(13))=<\sigma_6>\cong\Ints/2\Ints,\quad T(\mathfrak{Q}_i/(13))=\{\operatorname{Id}\}$$
5. The prime ideal $29\Ints$ is splits completely in $L$ , $$29\mathcal{O}_L=\mathfrak{R}_1\cdot\mathfrak{R}_2\cdot\mathfrak{R}_3\cdot\mathfrak{R'}_1\cdot\mathfrak{R'}_2\cdot\mathfrak{R'}_3$$ Also $29\equiv 1\ \operatorname{mod}\ 7$ , so $f=1$ , $$e(\mathfrak{R}_i/(29))=1,\quad f(\mathfrak{R}_i/(29)=1,\quad g(\mathfrak{R}_i/(29))=6$$ $$D(\mathfrak{R}_i/(29))=T(\mathfrak{R}_i/(29))=\{\operatorname{Id}\}$$