# Decomposition and Inertia groups and Frobenius automorphism computations

Let $K = \mathbb{Q}[\sqrt{5}, \sqrt{-1}]$. Calculate the Frobenius automorphisms $\left(\frac{K/\mathbb{Q}}{p}\right)$ for $p$ prime distinct from $2$ and $5$ (which are the only primes that ramify in $K$). Calculate the decomposition and inertia groups for $2$ and $5$.

This is related to an exercise in Milne; it would help me a lot to understand better the concepts if someone is willing to do this particular case. Thanks in advance!

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I'll tackle the Frobenius elements first.

Let $K=\mathbb{Q}(i,\sqrt{5})$. We have $\operatorname{Gal}(K/\mathbb{Q})=\{\operatorname{id}_K,\rho,\tau,\rho\tau\}\cong\mathbb{Z}/2\mathbb{Z}\times\mathbb{Z}/2\mathbb{Z}$, where $\rho$ is complex conjugation and $\tau$ is the automorphism of $K$ switching $\sqrt{5}$ and $-\sqrt{5}$.

Fact 1: Given

• an extension of number fields $E/F$
• a prime $P\subset\mathcal{O}_F$ that doesn't ramify in $E$
• any prime $\mathcal{P}\subset\mathcal{O}_E$ above $P$

then for any intermediate field $L$ such that $L/F$ is Galois, letting $\mathbf{P}=\mathcal{P}\cap\mathcal{O}_L$ the prime of $L$ lying beneath $\mathcal{P}$, we have $$(\mathcal{P},E/F)|_L=(\mathbf{P},L/F).$$

Fact 2: For a squarefree integer $D$, the field $M=\mathbb{Q}(\sqrt{D})$ has discriminant $d_M=4D$ if $D\equiv 2,3\bmod 4$ and $d_M=D$ if $D\equiv 1\bmod 4$, and if $p\nmid d_M$ is an odd prime, $$p\text{ splits if }\left(\frac{D}{p}\right)=1\qquad\text{and}\qquad p\text{ is inert if }\left(\frac{D}{p}\right)=-1,$$ (these being Legendre symbols).

I can include proofs of these facts if you would like.

Let $p$ be a prime of $\mathbb{Q}$ that doesn't ramify in $K$, i.e. $p\neq 2,5$. By Fact 1, $$\textstyle\left.\left(\frac{K/\mathbb{Q}}{p}\right)\right|_{\mathbb{Q}(\sqrt{5})}=\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)\in\operatorname{Gal}(\mathbb{Q}(\sqrt{5})/\mathbb{Q})\quad\text{and}\quad\left.\left(\frac{K/\mathbb{Q}}{p}\right)\right|_{\mathbb{Q}(i)}=\left(\frac{\mathbb{Q}(i)/\mathbb{Q}}{p}\right)\in\operatorname{Gal}(\mathbb{Q}(i)/\mathbb{Q})$$ Thus, computing $\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)$ and $\left(\frac{\mathbb{Q}(i)/\mathbb{Q}}{p}\right)$ will allow us to find $\left(\frac{K/\mathbb{Q}}{p}\right)$, because an element of $\operatorname{Gal}(K/\mathbb{Q})$ is determined by what it does to $\sqrt{5}$ and $i$. Let $\mathfrak{P}\subset\cal{O}_{\mathbb{Q}(\sqrt{5})}$ and $\mathfrak{p}\subset\cal{O}_{\mathbb{Q}(i)}$ be any primes lying over $p$.

There are two elements of $\operatorname{Gal}(\mathbb{Q}(\sqrt{5})/\mathbb{Q})$, $\operatorname{id}_{\mathbb{Q}(\sqrt{5})}$ and $\tau$, where $\tau(\sqrt{5})=-\sqrt{5}$. Note that $\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)=\operatorname{id}_{\mathbb{Q}(\sqrt{5})}$ if and only if $\operatorname{id}_{\mathbb{Q}(\sqrt{5})}$ induces the Frobenius automorphism on $(\cal{O}_{\mathbb{Q}(\sqrt{5})}/\mathfrak{P})/(\mathbb{Z}/p\mathbb{Z})$, i.e. every element of $\cal{O}_{\mathbb{Q}(\sqrt{5})}/\mathfrak{P}$ is its own $p$th power, which is the case if and only if $\cal{O}_{\mathbb{Q}(\sqrt{5})}/\mathfrak{P}\cong\mathbb{F}_p$, i.e. $f(\mathfrak{P}/p)=1$. Recall that $p$ does not ramify in $\mathbb{Q}(\sqrt{5})$ by assumption, so that $e(\mathfrak{P}/p)=1$. Because $\mathbb{Q}(\sqrt{5})/\mathbb{Q}$ is Galois of prime degree, $e(\mathfrak{P}/p)=1$ and $f(\mathfrak{P}/p)=1$ if and only if $p$ splits in $\mathbb{Q}(\sqrt{5})$. Thus, we have shown that $\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)=\operatorname{id}_{\mathbb{Q}(\sqrt{5})}$ if and only if $p$ splits in $\mathbb{Q}(\sqrt{5})$, which Fact 2 tells us happens if and only if $\left(\frac{5}{p}\right)=1$. By quadratic reciprocity, this is the case if and only if $\left(\frac{p}{5}\right)=1$, i.e. $p\equiv1,4\bmod 5$. Thus $$\left(\frac{\mathbb{Q}(\sqrt{5})/\mathbb{Q}}{p}\right)=\begin{cases}\operatorname{id}_{\mathbb{Q}(\sqrt{5})}\quad\text{ if }p\equiv1,4\bmod 5\\ \,\,\,\tau\quad\quad\quad\quad\;\text{ if }p\equiv 2,3\bmod 5\end{cases}$$

A similar analysis with $\frak{p}$ shows that $$\left(\frac{\mathbb{Q}(i)/\mathbb{Q}}{p}\right)=\begin{cases}\operatorname{id}_{\mathbb{Q}(i)}\quad\text{ if }p\equiv1\bmod 4\\ \,\,\,\rho\quad\quad\quad\;\text{ if }p\equiv 3\bmod 4\end{cases}$$

Because

$$\begin{array}{c|c|c} & p\equiv1\bmod4 & p\equiv 3\bmod 4 \\\hline p\equiv 1,4\bmod 5 & p\equiv 1,9\bmod 20 & p\equiv11,19\bmod 20\\\hline p\equiv 2,3\bmod 5 & p\equiv 17,13\bmod 20 & p\equiv 7,3\bmod 20\\\hline \end{array}$$ we have that $$\left(\frac{K/\mathbb{Q}}{p}\right)=\begin{cases}\operatorname{id}_K\quad\text{ if } p\equiv 1,9\bmod 20\\ \tau\quad\quad\;\;\text{ if } p\equiv 13,17\bmod20 \\ \rho\quad\quad\;\;\text{ if } p\equiv 11,19\bmod 20\\ \rho\tau\quad\quad\text{ if }p \equiv 3,7\bmod 20\end{cases}$$

Now to find the decomposition and inertia groups for $2$ and $5$. We will need

Fact 3: Given

• an abelian Galois extension of number fields $L/F$
• a prime $P\subset\mathcal{O}_F$
• a prime $\mathcal{P}\subset\mathcal{O}_L$ lying over $P$

then

• the inertia field of $\mathcal{P}$ over $P$, i.e. the fixed field of $I(\mathcal{P}/P)$, is the largest intermediate field in which $P$ does not ramify
• the decomposition field of $\mathcal{P}$ over $P$, i.e. the fixed field of $D(\mathcal{P}/P)$, is the largest intermediate field in which $P$ splits completely.

The intermediate fields of $K/\mathbb{Q}$ are, of course, $$K$$ $$\text{ / }\qquad |\qquad \text{ \ }$$ $$\quad\quad\mathbb{Q}(\sqrt{5})\quad\mathbb{Q}(i\sqrt{5})\,\,\,\,\quad \mathbb{Q}(i)\quad\quad$$ $$\text{ \ } \qquad |\qquad \text{ / }$$ $$\mathbb{Q}$$ which have discriminants $$\begin{array}{ccccc} & & 400 & & \\ & & & & \\ 5 & & -20 & & -4\\ & & & & \\ & & 1 & & \end{array}$$ (sorry about the ugly diagram, the LaTeX on the site has some limitations)

Because a prime ramifies if and only if it divides the discriminant, we see that the inertia field of $2$ is $\mathbb{Q}(\sqrt{5})$, and the inertia field of $5$ is $\mathbb{Q}(i)$. Thus, $I(\mathcal{P}/2)=\{\operatorname{id}_K,\rho\}$ for any prime $\mathcal{P}\subset\mathcal{O}_K$ lying over $2$, and $I(\mathfrak{P}/5)=\{\operatorname{id}_K,\tau\}$ for any prime $\mathfrak{P}\subset\mathcal{O}_K$ lying over $5$.

Because the decomposition group contains the inertia group, the inertia field contains the decomposition field. Thus, the decomposition field for $2$ is either $\mathbb{Q}(\sqrt{5})$ or $\mathbb{Q}$, and the decomposition field for $5$ is either $\mathbb{Q}(i)$ or $\mathbb{Q}$. By Fact 2, $2$ is inert in $\mathbb{Q}(\sqrt{5})$ and $5$ splits in $\mathbb{Q}(i)$ (indeed $5=(2+i)(2-i)$), so the decomposition field for $2$ is $\mathbb{Q}$ and the decomposition field for $5$ is $\mathbb{Q}(i)$. Thus, $D(\mathcal{P}/2)=\{\operatorname{id}_K,\rho,\tau,\rho\tau\}=\operatorname{Gal}(K/\mathbb{Q})$ for any prime $\mathcal{P}\subset\mathcal{O}_K$ lying over $2$, and $D(\mathfrak{P}/5)=\{\operatorname{id}_K,\tau\}$ for any prime $\mathfrak{P}\subset\mathcal{O}_K$ lying over $5$.

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This is a really nice answer, Zev. –  Keenan Kidwell Nov 28 '11 at 21:39
Thank you for the kind words :) –  Zev Chonoles Nov 29 '11 at 3:29
It really is. Thanks a lot for your help! –  Anna Nov 30 '11 at 18:11
@ZevChonoles, this is really a wonderful answer –  Sushma Dec 6 '13 at 5:49

Instead of thinking in terms of decomposition group and inertia group, sometimes it's easier to think in terms of the corresponding fields (via the usual Galois correspondence). For abelian Galois groups (as is the case for $K$ here), it's particularly simple:

The decomposition group corresponds to the largest subextension of $K/\mathbb{Q}$ over which $p$ splits completely. (This subextension is usually called the decomposition field.)

The inertia group corresponds to the largest subextension of $K/\mathbb{Q}$ over which $p$ is unramified. (This subextension is usually called the inertia field.)

(Note: In the non-abelian case, it's a little more complicated. The decomposition and inertia groups are only determined up to conjugacy by the lower prime $p$; to get a specific subgroup, we need to choose a prime $P$ in $K$ lying above $p$. In the above statements, we would need to change "largest subextension" to "largest subextension such that $P$ lies over the subextension".)

Now we can tackle $K = \mathbb{Q}(\sqrt{5}, \sqrt{-1})$. Consider first the case $p \ne 2,5$. Note that $K/\mathbb{Q}$ has 3 proper subextensions: $\mathbb{Q}(\sqrt{-1})$, $\mathbb{Q}(\sqrt{-5})$, $\mathbb{Q}(\sqrt{5})$. Check that $p$ either splits in all 3 of these subextensions, in which case the decomposition field is all of $K$ (so the decomposition group is trivial), or else $p$ splits in exactly one of the subextensions, in which case the decomposition field is that subextension (the decomposition group is the corresponding subgroup of order 2). The Frobenius automorphism is a generator of the decomposition group, which is uniquely determined in these cases since the decomposition groups all have order 1 or 2.

I'll leave the ramified cases $p=2,5$ to you for now.

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