# examples of prime ideal decomposition in number fields

Here we follow the notation of the entry on the decomposition group  . See also http://planetmath.org/encyclopedia/PrimeIdealDecompositionInQuadraticExtensionsOfMathbbQ.htmlthis entry.

Example 1

Let $K=\mathbb{Q}(\sqrt{-7})$; then $\operatorname{Gal}(K/\mathbb{Q})=\{\operatorname{Id},\sigma\}\cong\mathbb{Z}/2% \mathbb{Z}$, where $\sigma$ is the complex conjugation map. Let $\mathcal{O}_{K}$ be the ring of integers  of $K$. In this case:

 $\mathcal{O}_{K}=\mathbb{Z}\left[\frac{1+\sqrt{-7}}{2}\right]$

The discriminant   of this field is $D_{K/\mathbb{Q}}=-7$. We look at the decomposition in prime ideals   of some prime ideals in $\mathbb{Z}$:

1. 1.

The only prime ideal in $\mathbb{Z}$ 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:

 $D((\sqrt{-7})/(7))=\operatorname{Gal}(K/\mathbb{Q})$

For the inertia group, notice that $\sigma\equiv\operatorname{Id}\ \operatorname{mod}\ (\sqrt{-7})$. Hence:

 $T((\sqrt{-7})/(7))=\operatorname{Gal}(K/\mathbb{Q})$

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. 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

 $D(5\mathcal{O}_{K}/(5))=\operatorname{Gal}(K/\mathbb{Q})=D(13\mathcal{O}_{K}/(% 13))$

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. 3.

The primes $(2),(29)$ are split:

 $2\mathcal{O}_{K}={\left(2,\frac{1+\sqrt{-7}}{2}\right)}{\left(2,\frac{1-\sqrt{% -7}}{2}\right)}=\mathcal{P}\cdot\mathcal{P^{\prime}}$
 $29\mathcal{O}_{K}=\left(29,14+\sqrt{-7}\right)\left(29,14-\sqrt{-7}\right)=% \mathcal{R}\cdot\mathcal{R^{\prime}}$

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 $L=\mathbb{Q}(\zeta_{7})$. This is a cyclotomic extension of $\mathbb{Q}$ with Galois group  $\operatorname{Gal}(L/\mathbb{Q})\cong\left(\mathbb{Z}/7\mathbb{Z}\right)^{% \times}\cong\mathbb{Z}/6\mathbb{Z}$

Moreover

 $\operatorname{Gal}(L/\mathbb{Q})=\{\sigma_{a}\colon L\to L\mid\sigma_{a}(\zeta% _{7})=\zeta_{7}^{a},\quad a\in\left(\mathbb{Z}/7\mathbb{Z}\right)^{\times}\}$

Galois theory  gives us the subfields  of $L$: $\xymatrix@dr@C=1pc{L=\mathbb{Q}(\zeta_{7})\ar@{-}[r]\ar@{-}[d]&\mathbb{Q}(% \zeta_{7}+\zeta_{7}^{6})\ar@{-}[d]\\ \mathbb{Q}(\sqrt{-7})\ar@{-}[r]&\mathbb{Q}}$

The discriminant of the extension ${L/\mathbb{Q}}$ is $D_{L/\mathbb{Q}}=-7^{5}$. Let $\mathcal{O}_{L}$ denote the ring of integers of $L$, thus $\mathcal{O}_{L}=\mathbb{Z}[\zeta_{7}]$. We use the results of http://planetmath.org/encyclopedia/PrimeIdealDecompositionInCyclotomicExtensionsOfMathbbQ.htmlthis entry to find the decomposition of the primes $2,5,7,13,29$:

$\xymatrix{{L=\mathbb{Q}(\zeta_{7})}\ar@{-}[d]^{3}&{(1-\zeta_{7})^{6}}\ar@{-}[d% ]&{\mathfrak{P}\cdot\mathfrak{P}^{\prime}}\ar@{-}[d]&{(5)}\ar@{-}[d]&{% \mathfrak{Q}_{1}\cdot\mathfrak{Q}_{2}\cdot\mathfrak{Q}_{3}}\ar@{-}[d]\\ {K=\mathbb{Q}(\sqrt{-7})}\ar@{-}[d]^{2}&{(\sqrt{-7})^{2}}\ar@{-}[d]&{\left(2,% \frac{1+\sqrt{-7}}{2}\right)}{\left(2,\frac{1-\sqrt{-7}}{2}\right)}\ar@{-}[d]&% (5)\ar@{-}[d]&(13)\ar@{-}[d]\\ \mathbb{Q}&(7)&(2)&(5)&(13)}$

1. 1.

The prime ideal $7\mathbb{Z}$ 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 $L^{T(\mathfrak{T}/(7))}=\mathbb{Q}=L^{D(\mathfrak{T}/(7))}$, thus, by Galois theory,

 $D(\mathfrak{T}/(7))=T(\mathfrak{T}/(7))=\operatorname{Gal}(L/\mathbb{Q})$
2. 2.

The ideal $2\mathbb{Z}$ factors in $K$ as above, $2\mathcal{O}_{K}=\mathcal{P}\cdot\mathcal{P^{\prime}}$, and each of the prime ideals $\mathcal{P},\mathcal{P^{\prime}}$ 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 $2$, $2\cdot 3=6$, 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

 $D(\mathfrak{P}/(2))=<\sigma_{2}>\cong\mathbb{Z}/3\mathbb{Z},\quad T(\mathfrak{% P}/(2))=\{\operatorname{Id}\}$
3. 3.

The ideal $(5)$ is inert, $5\mathcal{O}_{L}=\mathfrak{S}$ is prime and the order of $5$ modulo $7$ is $6$. Thus:

 $e(\mathfrak{S}/(5))=1,\quad f(\mathfrak{S}/(5))=6,\quad g(\mathfrak{S}/(5))=1$
 $D(\mathfrak{S}/(5))=\operatorname{Gal}(L/\mathbb{Q}),\quad T(\mathfrak{S}/(5))% =\{\operatorname{Id}\}$
4. 4.

The prime ideal $13\mathbb{Z}$ 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 $2$:

 $e(\mathfrak{Q}_{i}/(13))=1,\quad f(\mathfrak{Q}_{i}/(13))=2,\quad g(\mathfrak{% Q}_{i}/(13))=3$
 $D(\mathfrak{Q}_{i}/(13))=<\sigma_{6}>\cong\mathbb{Z}/2\mathbb{Z},\quad T(% \mathfrak{Q}_{i}/(13))=\{\operatorname{Id}\}$
5. 5.

The prime ideal $29\mathbb{Z}$ is splits completely in $L$,

 $29\mathcal{O}_{L}=\mathfrak{R}_{1}\cdot\mathfrak{R}_{2}\cdot\mathfrak{R}_{3}% \cdot\mathfrak{R^{\prime}}_{1}\cdot\mathfrak{R^{\prime}}_{2}\cdot\mathfrak{R^{% \prime}}_{3}$

Also $29\equiv 1\ \operatorname{mod}\ 7$, so $f=1$,

 $e(\mathfrak{R}_{i}/(29))=1,\quad f(\mathfrak{R}_{i}/(29)=1,\quad g(\mathfrak{R% }_{i}/(29))=6$
 $D(\mathfrak{R}_{i}/(29))=T(\mathfrak{R}_{i}/(29))=\{\operatorname{Id}\}$
 Title examples of prime ideal decomposition in number fields Canonical name ExamplesOfPrimeIdealDecompositionInNumberFields Date of creation 2013-03-22 13:53:05 Last modified on 2013-03-22 13:53:05 Owner alozano (2414) Last modified by alozano (2414) Numerical id 12 Author alozano (2414) Entry type Example Classification msc 11S15 Related topic DecompositionGroup Related topic Discriminant Related topic NumberField Related topic PrimeIdealDecompositionInQuadraticExtensionsOfMathbbQ Related topic PrimeIdealDecompositionInCyclotomicExtensionsOfMathbbQ Related topic ExamplesOfRamificationOfArchimedeanPlaces