# Hilbert's ramification theory in Galois extensions

*Written by GPT-6.1 Sol (OpenAI) in Codex, Ultra setting, October 2026. Self-checked by the writing AI. Independent full-lesson AI review is pending. Public domain (CC0).*

In a Galois extension, prime decomposition is controlled by subgroups of the Galois group. One subgroup preserves a chosen prime; a smaller subgroup acts trivially on its residue field. At an unramified prime, the residue field's power map lifts to a unique field automorphism. This is the Frobenius element, and its action often turns a calculation with polynomials into a calculation with permutations.

## What this lesson assumes

Let \(L/K\) be a finite Galois extension of number fields, let \(G=\operatorname{Gal}(L/K)\), and put \(A=\mathcal O_K\), \(B=\mathcal O_L\). Fix a nonzero prime \(\mathfrak p\) of \(A\). All primes discussed are finite primes.

We use [*Decomposition of primes in extensions*](decomposition-of-primes-in-extensions.md): finite integral closures, the fundamental identity \(\sum ef=[L:K]\), ideal norms, Kummer–Dedekind, and the discriminant criterion. We also use finite Galois theory: \(\#G=[L:K]\), the correspondence \(H\leftrightarrow L^H\), \([L:L^H]=\#H\), and the quotient group description when \(H\) is normal [Milne FGT, Theorem 3.17]. These are algebra prerequisites, not ramification results assumed in advance.

Write \(\kappa(\mathfrak p)=A/\mathfrak p\), \(\kappa(\mathfrak P)=B/\mathfrak P\), and \(q=\#\kappa(\mathfrak p)=N\mathfrak p\). The integer \(q\) is the size of the base residue field, not a residue degree. Finite residue fields are separable over their subfields, so unramifiedness here is equivalent to ramification index one.

## All primes above a prime are conjugate

Every \(\sigma\in G\) carries \(B\) to itself, since integrality is preserved, and sends a prime \(\mathfrak P\) above \(\mathfrak p\) to another such prime.

**Theorem 6.1.** The action of \(G\) on the primes above \(\mathfrak p\) is transitive. Their ramification indices have one common value \(e\), their residue degrees one common value \(f\), and, if there are \(g\) primes, then
\[
efg=[L:K]. \tag{1}
\]

**Proof.** Suppose that \(\mathfrak Q\) is not in the orbit of \(\mathfrak P\). Chinese remainders give \(x\in B\) with
\[
x\equiv0\pmod{\mathfrak Q},
\qquad x\equiv1\pmod{\sigma\mathfrak P}
\quad\text{for every }\sigma\in G.
\]
Its norm \(a=\prod_{\sigma\in G}\sigma(x)\) belongs to \(A\). Since one factor is \(x\), it belongs to \(\mathfrak Q\), and hence to \(\mathfrak Q\cap A=\mathfrak p\). But no \(\sigma(x)\) belongs to \(\mathfrak P\): membership would say \(x\in\sigma^{-1}\mathfrak P\). Primality implies \(a\notin\mathfrak P\), contradicting \(a\in\mathfrak p\).

An automorphism fixes the ideal \(\mathfrak pB\); uniqueness of its prime factorization makes the exponents equal on conjugate primes. It also induces an isomorphism \(B/\mathfrak P\to B/\sigma\mathfrak P\) fixing \(A/\mathfrak p\), so the residue degrees are equal. The fundamental identity now gives (1). \(\square\)

Define the **decomposition group**
\[
D_{\mathfrak P}=\{\sigma\in G\mid\sigma\mathfrak P=\mathfrak P\}.
\]
Orbit–stabilizer and (1) give
\[
[G:D_{\mathfrak P}]=g,\qquad
\#D_{\mathfrak P}=ef. \tag{2}
\]
Changing the chosen prime conjugates the subgroup:
\[
D_{\sigma\mathfrak P}=\sigma D_{\mathfrak P}\sigma^{-1}. \tag{3}
\]
These definitions agree with [Stacks, Tag 09EC].

## Residue fields and inertia

We first make the residue automorphism explicit. If \(E/k\) is an extension of finite fields, \(\#k=q\) and \([E:k]=f\), then \(E\) has \(q^f\) elements. The map
\[
\varphi:E\longrightarrow E,\qquad x\longmapsto x^q
\]
is a \(k\)-automorphism. It fixes \(k\), and it is bijective because it is an injective map of a finite field. Its \(f\)-th power is the identity. If its order were \(d<f\), every element of \(E\) would be a root of \(X^{q^d}-X\), contradicting the polynomial root bound. Thus its order is \(f\).

The field \(E\) is the splitting field of \(X^{q^f}-X\) over \(k\); this polynomial has derivative \(-1\). Therefore \(E/k\) is Galois, and its group, of order \(f\), is generated by \(\varphi\). We call \(\varphi\) the **arithmetic Frobenius**.

Every member of \(D_{\mathfrak P}\) induces an automorphism of \(\kappa(\mathfrak P)\) over \(\kappa(\mathfrak p)\). Define the **inertia group** by
\[
I_{\mathfrak P}=
\ker\!\left(D_{\mathfrak P}\longrightarrow
\operatorname{Gal}(\kappa(\mathfrak P)/\kappa(\mathfrak p))\right).
\]
Equivalently, \(\sigma\in I_{\mathfrak P}\) exactly when
\(\sigma(x)\equiv x\pmod{\mathfrak P}\) for every \(x\in B\).

**Theorem 6.2.** The residue field extension is Galois, and there is an exact sequence
\[
1\longrightarrow I_{\mathfrak P}\longrightarrow D_{\mathfrak P}
\longrightarrow
\operatorname{Gal}(\kappa(\mathfrak P)/\kappa(\mathfrak p))
\longrightarrow1.
\]
In particular,
\[
\#D_{\mathfrak P}=ef,\qquad
\#I_{\mathfrak P}=e,\qquad
D_{\mathfrak P}/I_{\mathfrak P}\cong C_f. \tag{4}
\]

**Proof.** The finite-field argument proves the first assertion. To prove surjectivity, choose a nonzero element \(u\) generating \(\kappa(\mathfrak P)\) over \(\kappa(\mathfrak p)\). Such an element exists: the multiplicative group of the finite field is cyclic, by the elementary argument in the preceding lesson, and its generator generates the field. Chinese remainders give \(a\in B\) whose residue at \(\mathfrak P\) is \(u\), and whose residue at every other prime above \(\mathfrak p\) is zero.

The polynomial
\[
F(X)=\prod_{\sigma\in G}(X-\sigma(a))
\]
has coefficients in \(A\): they are integral and fixed by \(G\), hence belong to the integrally closed ring \(A\subset K\). Let \(\tau\) be an automorphism of \(\kappa(\mathfrak P)\) over \(\kappa(\mathfrak p)\). Since \(u\) is a root of \(\overline F\), so is \(\tau(u)\). The displayed product, reduced in \(\kappa(\mathfrak P)\), gives
\[
\tau(u)=\overline{\sigma(a)}
\quad\text{for some }\sigma\in G.
\]
This residue is nonzero. If \(\sigma\notin D_{\mathfrak P}\), then \(\sigma^{-1}\mathfrak P\ne\mathfrak P\), so \(a\in\sigma^{-1}\mathfrak P\) by construction and \(\sigma(a)\in\mathfrak P\), a contradiction. Thus \(\sigma\in D_{\mathfrak P}\), and its residue action agrees with \(\tau\) on the field generator \(u\). It agrees everywhere.

Surjectivity and (2) give \(\#I=ef/f=e\). The remaining assertions follow from the kernel definition and the finite-field group calculation. \(\square\)

For general discrete valuation rings the residue extension is normal but can be inseparable; [Stacks, Tag 09ED] retains this distinction. The formula \(\#I=e\) used here relies on separable residue fields, which number fields supply.

## The Frobenius element and its functoriality

**Proposition 6.3.** If \(\mathfrak p\) is unramified in \(L\), there is a unique
\(\operatorname{Frob}_{\mathfrak P}\in D_{\mathfrak P}\) satisfying
\[
\operatorname{Frob}_{\mathfrak P}(x)
\equiv x^q\pmod{\mathfrak P}
\quad(x\in B).
\]
It generates \(D_{\mathfrak P}\) and has order \(f\). For \(\sigma\in G\),
\[
\operatorname{Frob}_{\sigma\mathfrak P}
=\sigma\operatorname{Frob}_{\mathfrak P}\sigma^{-1}. \tag{5}
\]
If \(K\subseteq M\subseteq L\) and \(M/K\) is Galois, restriction sends
\(\operatorname{Frob}_{\mathfrak P}\) to
\(\operatorname{Frob}_{\mathfrak P\cap\mathcal O_M}\) for \(M/K\).

**Proof.** Unramifiedness gives \(e=1\), so Theorem 6.2 makes the residue map an isomorphism. The generator \(x\mapsto x^q\) has exactly one preimage, which generates \(D_{\mathfrak P}\) and has order \(f\).

For \(x\in B\), put \(y=\sigma^{-1}(x)\). The congruence at \(\mathfrak P\), transported by \(\sigma\), reads
\[
\sigma\operatorname{Frob}_{\mathfrak P}\sigma^{-1}(x)
\equiv\sigma(y^q)=x^q\pmod{\sigma\mathfrak P}.
\]
The conjugate preserves \(\sigma\mathfrak P\), so uniqueness proves (5).

We justify the restriction assertion, including the group maps it uses. Put \(H=\operatorname{Gal}(L/M)\), \(\mathfrak Q=\mathfrak P\cap\mathcal O_M\). For \(L/M\), the decomposition and inertia groups are respectively
\[
D_{\mathfrak P}\cap H,\qquad I_{\mathfrak P}\cap H. \tag{6}
\]
The first equality is the stabilizer definition; the second follows because inertia always means trivial action on the whole residue field \(\kappa(\mathfrak P)\).

When \(H\) is normal, restriction maps \(D_{\mathfrak P}\) onto the decomposition group at \(\mathfrak Q\) for \(M/K\). Indeed, lift a member of that group to \(\sigma\in G\). The primes \(\sigma\mathfrak P\) and \(\mathfrak P\) lie over the same \(\mathfrak Q\). Transitivity for the Galois extension \(L/M\) supplies \(h\in H\) with \(h\sigma\mathfrak P=\mathfrak P\), giving a lift in \(D_{\mathfrak P}\).

In any tower of these rings, ramification indices and residue degrees multiply:
\[
e(\mathfrak P\mid\mathfrak p)
=e(\mathfrak P\mid\mathfrak Q)e(\mathfrak Q\mid\mathfrak p),
\quad
f(\mathfrak P\mid\mathfrak p)
=f(\mathfrak P\mid\mathfrak Q)f(\mathfrak Q\mid\mathfrak p). \tag{7}
\]
The first equality follows by extending and factoring \(\mathfrak p\), then each intermediate prime; the second is the degree formula for the residue field tower. The image of \(I_{\mathfrak P}\) is contained in the inertia group at \(\mathfrak Q\). Its order, by (4), (6), and (7), is
\[
\frac{e(\mathfrak P\mid\mathfrak p)}
{e(\mathfrak P\mid\mathfrak Q)}
=e(\mathfrak Q\mid\mathfrak p).
\]
Thus restriction is also surjective on inertia groups. In particular, unramifiedness descends. The restricted Frobenius preserves \(\mathfrak Q\) and induces the \(q\)-power map on its residue field, so uniqueness identifies it. \(\square\)

For a ramified prime, Theorem 6.2 still gives one distinguished **Frobenius coset** in \(D_{\mathfrak P}/I_{\mathfrak P}\), the preimage of the arithmetic residue Frobenius. It gives no distinguished element of \(D_{\mathfrak P}\). When \(G\) is abelian and the prime is unramified, (5) shows that the element depends only on \(\mathfrak p\). We write the **Artin symbol**
\[
\left(\frac{L/K}{\mathfrak p}\right)
=\operatorname{Frob}_{\mathfrak P}.
\]
For nonabelian \(G\), varying \(\mathfrak P\) gives a conjugacy class. All conventions here use arithmetic Frobenius; geometric Frobenius is its inverse.

## Decomposition and inertia fields

Fix \(\mathfrak P\), abbreviate \(D=D_{\mathfrak P}\), \(I=I_{\mathfrak P}\), and put
\[
Z=L^D,\qquad T=L^I.
\]
Then \(K\subseteq Z\subseteq T\subseteq L\). Set
\(\mathfrak P_Z=\mathfrak P\cap\mathcal O_Z\) and
\(\mathfrak P_T=\mathfrak P\cap\mathcal O_T\).

**Proposition 6.4.** The chosen primes have the following invariants:
\[
\begin{array}{c|c|c}
\text{prime extension}&e&f\\ \hline
\mathfrak P_Z/\mathfrak p&1&1\\
\mathfrak P_T/\mathfrak P_Z&1&f\\
\mathfrak P/\mathfrak P_T&e&1.
\end{array}
\]
The extensions \(T/Z\) and \(L/T\) have degrees \(f\) and \(e\), respectively. There is a unique prime of \(L\) above \(\mathfrak P_Z\), and a unique prime at each step above it.

**Proof.** The extension \(L/Z\) is Galois with group \(D\). Its primes above \(\mathfrak P_Z\) form an orbit under \(D\), and every member of \(D\) fixes \(\mathfrak P\); hence there is only one such prime. Its \(e',f'\) satisfy
\[
e'f'=[L:Z]=\#D=ef.
\]
By (7), \(e'\leq e\) and \(f'\leq f\). Their product is already \(ef\), so \(e'=e,f'=f\), and the remaining indices for \(Z/K\) are both one.

For \(L/T\), the group is \(I\), and again \(\mathfrak P\) is the only prime above \(\mathfrak P_T\). Its residue action is trivial. Surjectivity in Theorem 6.2, applied to \(L/T\), makes its residue Galois group trivial. Thus \(f(\mathfrak P\mid\mathfrak P_T)=1\), and the fundamental identity gives \(e(\mathfrak P\mid\mathfrak P_T)=\#I=e\). Multiplicativity now gives the middle row. Since \(I\) is normal in \(D\), \(T/Z\) is Galois with group \(D/I\) and degree \(f\). The degree of \(L/T\) is \(\#I=e\). Every prime over \(\mathfrak P_Z\) extends to one in \(L\), so uniqueness in \(L\) also proves uniqueness in the intermediate fields. \(\square\)

Thus \(T/Z\) is unramified **at \(\mathfrak P_Z\)** and \(L/T\) is totally ramified **at \(\mathfrak P_T\)**. These are assertions about the selected primes. They do not assert unramifiedness at every other prime. Nor must \(\mathfrak p\) split completely in \(Z\) when \(D\) is not normal in \(G\): the proposition specifies the single prime \(\mathfrak P_Z\). If \(D\) is normal, \(Z/K\) is Galois, so transitivity makes every prime over \(\mathfrak p\) have \(e=f=1\); in that case it does split completely.

## Subfields and double cosets

Let \(H\subseteq G\), \(M=L^H\), with no normality assumption on \(H\). The primes of \(M\) above \(\mathfrak p\) correspond to
\[
H\backslash G/D,\qquad
H\sigma D\longmapsto\sigma\mathfrak P\cap\mathcal O_M. \tag{8}
\]
For the prime attached to this double coset, put
\(D_\sigma=\sigma D\sigma^{-1}\) and
\(I_\sigma=\sigma I\sigma^{-1}\). Its invariants over \(K\) are
\[
e_M=\frac{\#I_\sigma}{\#(I_\sigma\cap H)},
\qquad
f_M=[D_\sigma:(D_\sigma\cap H)I_\sigma]. \tag{9}
\]

**Proof.** Right multiplication by \(D\) leaves \(\sigma\mathfrak P\) unchanged, and left multiplication by \(H\) leaves its contraction to \(M\) unchanged. Conversely, two top primes with the same contraction are conjugate under \(H\), by transitivity in \(L/M\). This proves the bijection.

At \(\sigma\mathfrak P\), the groups for \(L/M\) are \(D_\sigma\cap H\) and \(I_\sigma\cap H\). Divide the corresponding ramification indices using (7) and (4), obtaining the first formula. Dividing the residue degrees gives
\[
f_M=\frac{\#D_\sigma/\#I_\sigma}
{\#(D_\sigma\cap H)/\#(I_\sigma\cap H)}
=\frac{\#D_\sigma}{\#((D_\sigma\cap H)I_\sigma)}.
\]
The product in the denominator is a subgroup because \(I_\sigma\) is normal in \(D_\sigma\); its order is the usual product-of-subgroups formula. This proves (9). \(\square\)

In particular \(e_Mf_M=[D_\sigma:D_\sigma\cap H]\). Summing this over the double cosets gives \([G:H]=[M:K]\), since
\(\#(H\sigma D)=\#H\,\#D_\sigma/\#(H\cap D_\sigma)\).
This recovers the fundamental identity in \(M\).

There is also a useful tower formula for Frobenius. If \(\mathfrak p\) is unramified in \(L\), \(\mathfrak Q=\mathfrak P\cap\mathcal O_M\), then
\[
\operatorname{Frob}_{\mathfrak P,L/M}
=\operatorname{Frob}_{\mathfrak P,L/K}^{\,f(\mathfrak Q\mid\mathfrak p)}. \tag{10}
\]
Indeed, \(D\) is cyclic, generated by the Frobenius, and
\([D:D\cap H]=f(\mathfrak Q\mid\mathfrak p)\). Its indicated power generates \(D\cap H\), and acts on the top residue field by the power
\(q^{f(\mathfrak Q\mid\mathfrak p)}=\#\kappa(\mathfrak Q)\).
It is therefore exactly the relative Frobenius. Normality of \(M/K\) is unnecessary in (10).

### The largest fields with the prime properties

For the chosen contracted prime \(\mathfrak Q=\mathfrak P\cap\mathcal O_M\), formula (9) at \(\sigma=1\) gives
\[
e(\mathfrak Q\mid\mathfrak p)=1
\Longleftrightarrow I\subseteq H
\Longleftrightarrow M\subseteq T,
\]
and
\[
e(\mathfrak Q\mid\mathfrak p)=f(\mathfrak Q\mid\mathfrak p)=1
\Longleftrightarrow D\subseteq H
\Longleftrightarrow M\subseteq Z.
\]
Indeed the two indices are \([I:I\cap H]\) and \([D:D\cap H]\), the latter being the product \(ef\). Positive integer factors have product one exactly when both are one. Galois correspondence proves the field containments. Thus \(T\) and \(Z\) are the largest intermediate fields with these respective properties for this chosen prime.

To control every prime above \(\mathfrak p\), define the normal closures of the subgroups:
\[
N_I=\langle\sigma I\sigma^{-1}\mid\sigma\in G\rangle,\qquad
N_D=\langle\sigma D\sigma^{-1}\mid\sigma\in G\rangle.
\]
The double-coset parametrization and formula (9) show that all primes of \(M\) are unramified exactly when every conjugate of \(I\) lies in \(H\), or equivalently \(N_I\subseteq H\). Likewise all have \(e=f=1\) exactly when \(N_D\subseteq H\). In the latter case their number is \([M:K]\) by the fundamental identity, which is complete splitting. Consequently \(L^{N_I}\) is the largest intermediate field unramified at every prime above \(\mathfrak p\), and \(L^{N_D}\) is the largest where \(\mathfrak p\) splits completely. Both are Galois over \(K\), and
\[
K\subseteq L^{N_D}\subseteq L^{N_I}\subseteq L,\qquad
L^{N_D}\subseteq Z,\quad L^{N_I}\subseteq T.
\]
In particular, complete splitting in \(Z\) is equivalent to \(N_D\subseteq D\), hence to \(D\) being normal in \(G\). Unramifiedness at every prime of \(T\) is equivalent to \(N_I\subseteq I\), hence to \(I\) being normal in \(G\). Inertia is always normal in \(D\); it need not be normal in \(G\).

In the \(S_3\) example below at \(5\), \(D=\langle c\rangle\) and \(I=1\). Its decomposition field is \(\mathbf Q(\alpha)\), \(\alpha=\sqrt[3]2\), with the two primes
\[
(5,\alpha-3),\qquad(5,\alpha^2+3\alpha+4),
\]
of respective invariants \((e,f)=(1,1),(1,2)\), as the polynomial factorization and integral basis show. The first chosen prime has both indices one, while \(5\) does not split completely in that field. The conjugate transpositions generate all of \(S_3\), so \(N_D=G\); the largest completely split intermediate field is \(\mathbf Q\). Since \(N_I=1\), the largest unramified one is the whole splitting field. At \(2\), \(D=G\) and \(I=A_3\) are normal; the largest fields are respectively \(\mathbf Q\) and \(\mathbf Q(\omega)\). At \(3\), both largest fields are \(\mathbf Q\).

## When a prime splits completely

**Corollary 6.5.** For an unramified prime \(\mathfrak p\) in a Galois extension \(L/K\),
\[
\mathfrak p\text{ splits completely}
\quad\Longleftrightarrow\quad
\operatorname{Frob}_{\mathfrak P}=1.
\]
For finite extensions \(L_1,L_2\) of \(K\) inside one algebraic closure, a prime splits completely in \(L_1L_2\) exactly when it splits completely in each factor.

**Proof.** At an unramified prime \(e=1\) and \(D\) is cyclic of order \(f\), generated by Frobenius. This element is one exactly when \(f=1\); then (1) gives \(g=[L:K]\), which is complete splitting.

First suppose both factors are Galois. Their compositum is Galois. Restriction embeds its Galois group into the product of the two groups: an automorphism fixing both fields fixes their compositum. If the prime splits in both factors, the decomposition group of any top prime maps to the trivial decomposition group in each factor, by Proposition 6.3's stabilizer restriction argument. It is consequently trivial, which by (2) means complete splitting in the compositum. Complete splitting descends to every subfield, since each prime in that subfield extends to a top prime and (7) forces its indices to be one.

To handle arbitrary factors, we establish the Galois closure criterion. If \(M/K\) is finite and \(N/K\) its Galois closure, put \(G=\operatorname{Gal}(N/K)\), \(H=\operatorname{Gal}(N/M)\). If \(\mathfrak p\) splits completely in \(M\), (9) says \(D_\sigma\subseteq H\) for every \(\sigma\). Thus
\[
D\subseteq\bigcap_{\sigma\in G}\sigma^{-1}H\sigma=\{1\}.
\]
The last equality holds because a member of this intersection fixes every conjugate of \(M\), and these generate \(N\). Hence \(\mathfrak p\) splits completely in \(N\). The reverse implication is descent. Apply this criterion to both factors, the proved Galois compositum assertion to their closures, and descent to \(L_1L_2\). \(\square\)

### Unramifiedness in closures and composita

**Corollary 6.6.** A finite prime \(\mathfrak p\) of \(K\) is unramified in a finite extension \(M/K\) if and only if it is unramified in its Galois closure. For finite extensions \(L_1,L_2\) in one algebraic closure, it is unramified in \(L_1L_2\) if and only if it is unramified in both factors. Unramified means that every prime above \(\mathfrak p\) has ramification index one.

**Proof.** Let \(N/K\) be the Galois closure of \(M/K\), and write \(G=\operatorname{Gal}(N/K)\), \(H=\operatorname{Gal}(N/M)\). Fix a prime of \(N\) over \(\mathfrak p\), with inertia group \(I\). If every prime of \(M\) is unramified, formula (9) says
\[
\sigma I\sigma^{-1}\subseteq H\qquad(\sigma\in G).
\]
The formula covers every double coset, and its index is unchanged by changing the representative; hence it holds for every \(\sigma\). Therefore
\[
I\subseteq\bigcap_{\sigma\in G}\sigma^{-1}H\sigma=\{1\}.
\]
As in the preceding proof, this intersection fixes all conjugates of \(M\), which generate \(N\). Theorem 6.2 now gives unramifiedness in \(N\). Conversely unramifiedness descends by the multiplicative ramification formula (7), since all primes of a subfield extend to \(N\).

For Galois \(L_1,L_2\), their compositum \(L\) is Galois and restriction injects \(\operatorname{Gal}(L/K)\) into the product of their groups. An inertia element at a prime of \(L\) restricts to an inertia element in each factor: it fixes the contracted prime and acts trivially on its residue subfield. If both factors are unramified, both images are trivial, so the inertia element itself is trivial. This proves unramifiedness in \(L\); the reverse implication is descent. For arbitrary factors, use the just-proved closure criterion, apply the Galois assertion to their closures, and descend to \(L_1L_2\). \(\square\)

This corollary supplies the unramified-compositum step used in the coprime-discriminant argument of *Cyclotomic fields*. Its hypothesis concerns all primes over the selected base prime, not only one chosen prime.

## Abelian examples, including one ramified odd prime

In \(\mathbf Q(i)\), an odd prime has Frobenius the identity if \(p\equiv1\pmod4\), and complex conjugation \(c\) if \(p\equiv3\pmod4\), by the quadratic law and Corollary 6.5.

For \(V=\mathbf Q(\sqrt2,\sqrt3)\), the Galois group is \(\{1,u,v,uv\}\), where \(u\) negates \(\sqrt2\) and \(v\) negates \(\sqrt3\). The degree is four, since \(\sqrt3\notin\mathbf Q(\sqrt2)\): squaring a hypothetical \(a+b\sqrt2\), \(a,b\in\mathbf Q\), would force \(ab=0\), then either \(a^2=3\) or \(b^2=3/2\), neither a rational square.

The integral basis calculation gives discriminant \(2304=2^8 3^2\), so every prime other than \(2,3\) is unramified. Restriction to the quadratic subfields gives
\[
\operatorname{Frob}_p(\sqrt2)=(2/p)\sqrt2,\qquad
\operatorname{Frob}_p(\sqrt3)=(3/p)\sqrt3.
\]
The following table covers the first ten odd primes. The signs can be checked directly by listing squares modulo \(p\).

| \(p\) | Frobenius in \(\mathbf Q(i)\) | \((2/p)\) | \((3/p)\) | Frobenius in \(V\) |
|---|---|---|---|---|
| \(3\) | \(c\) | \(-1\) | \(0\) | ramified; no Frobenius element |
| \(5\) | \(1\) | \(-1\) | \(-1\) | \(uv\) |
| \(7\) | \(c\) | \(1\) | \(-1\) | \(v\) |
| \(11\) | \(c\) | \(-1\) | \(1\) | \(u\) |
| \(13\) | \(1\) | \(-1\) | \(1\) | \(u\) |
| \(17\) | \(1\) | \(1\) | \(-1\) | \(v\) |
| \(19\) | \(c\) | \(-1\) | \(-1\) | \(uv\) |
| \(23\) | \(c\) | \(1\) | \(1\) | \(1\) |
| \(29\) | \(1\) | \(-1\) | \(-1\) | \(uv\) |
| \(31\) | \(c\) | \(1\) | \(-1\) | \(v\) |

Every nonidentity element of this group has order two. Thus a nonidentity Frobenius gives \(e=1,f=2,g=2\); identity gives \(e=f=1,g=4\).

At \(3\), the subfield \(\mathbf Q(\sqrt2)\) has \(e=1,f=2\), and \(\mathbf Q(\sqrt3)\) has \(e=2,f=1\). Tower multiplicativity forces \(e\geq2,f\geq2\) in \(V\); the degree four forces \(e=2,f=2,g=1\). Hence \(D=G\), \(I=\langle v\rangle\): inertia must fix the unramified subfield \(\mathbf Q(\sqrt2)\), and has order two. The distinguished residue Frobenius is the coset \(uI=\{u,uv\}\).

For \(\omega=\zeta_3\), the quadratic field \(\mathbf Q(\omega)\) has discriminant \(-3\). At \(p\ne3\), its Frobenius sends \(\omega\) to \(\omega^p\). Both sides are conjugate cube roots and have the same residue by the defining congruence; their residues are distinct at such a prime, so they are equal. This includes \(p=2\), where Frobenius is conjugation. For \(\zeta_4=i\), the same argument at odd \(p\) gives \(i\mapsto i^p\). The general cyclotomic decomposition law is proved in *Cyclotomic fields*.

## A nonabelian example: the splitting field of \(X^3-2\)

Let \(\alpha=\sqrt[3]{2}\) be real and \(L=\mathbf Q(\alpha,\omega)\), \(\omega=\zeta_3\). Since \(\mathbf Q(\alpha)\) is real and \(\omega\) is not, \([L:\mathbf Q]=6\). This is the splitting field of \(X^3-2\). Its group is \(S_3\), generated by
\[
r(\alpha)=\omega\alpha,\quad r(\omega)=\omega,
\qquad
c(\alpha)=\alpha,\quad c(\omega)=\omega^2,
\qquad crc=r^{-1}.
\]
The element \(r\) cycles the three roots; \(c\) fixes one and exchanges two.

Every \(p\ne2,3\) is unramified in \(L\). To prove this without an integral power basis for \(L\), let \(\mathfrak P\mid p\). The polynomial discriminant \(-108\) is a unit modulo \(\mathfrak P\), so all differences of distinct roots remain nonzero there. An inertia element sends a root to a root with the same residue, and must therefore fix that root. It fixes all three roots, which generate \(L\), so inertia is trivial.

For a polynomial with distinct roots modulo \(p\), Frobenius cycles these roots in lengths equal to the degrees of its irreducible factors over \(\mathbf F_p\). Here is why. A residue root \(a\) has orbit \(a,a^p,\ldots\) under Frobenius. The product over its distinct orbit has coefficients fixed by Frobenius, hence in \(\mathbf F_p\); the fixed field is exactly \(\mathbf F_p\), by the root bound for \(X^p-X\). Its minimal polynomial divides this product, while every point of the orbit is a root of the minimal polynomial. The two degree inequalities force equality. Distinct reduction of the original roots transfers these cycles to the field automorphism.

For \(p\ne2,3\) with \(p\equiv2\pmod3\), restriction to \(\mathbf Q(\omega)\) is nontrivial, so Frobenius is a transposition. The cube map is bijective on \(\mathbf F_p^\times\), since \(3\) is relatively prime to \(p-1\); \(X^3-2\) has one root and an irreducible quadratic factor. If \(p\equiv1\pmod3\), Frobenius is even. When \(2\) is a cube modulo \(p\), one root and the two nontrivial cube roots of unity give three linear factors, so Frobenius is the identity. Otherwise the cubic has no root, is irreducible, and Frobenius is a three-cycle.

In particular,
\[
\begin{aligned}
X^3-2&=(X-3)(X^2+3X+4)&&\text{over }\mathbf F_5,\\
X^3-2&\text{ is irreducible}&&\text{over }\mathbf F_7,\\
X^3-2&=(X-4)(X-7)(X-20)&&\text{over }\mathbf F_{31}.
\end{aligned}
\]
At \(5\), choose a prime where \(\alpha\) reduces to \(3\); its Frobenius fixes \(\alpha\), sends \(\omega\) to \(\omega^2\), and is \(c\). At \(7\), choose a prime with \(\omega\equiv4\). Since \(\alpha^7=4\alpha\), its Frobenius is \(r\); the conjugate prime with \(\omega\equiv2\) has Frobenius \(r^2\). At \(31\) every Frobenius is the identity.

## Exercises

1. **Easy.** Show that in an abelian Galois extension all primes over \(\mathfrak p\) have the same decomposition and inertia groups.
2. **Medium.** In \(L=\mathbf Q(\sqrt[3]{2},\zeta_3)\), determine \(D_{\mathfrak P}\), \(I_{\mathfrak P}\), and Frobenius at \(p=2,3,5,7,31\). At ramified primes, specify the Frobenius coset instead of an element.
3. **Medium.** Prove the Galois closure criterion for complete splitting, including a non-Galois extension. Show that an extension with noncyclic Galois group has no unramified inert prime.
4. **Hard.** Prove the double coset correspondence and formulas (8)–(9), and apply them to \(M=\mathbf Q(\sqrt[3]{2})\) inside its splitting field.

## Complete solutions

**1.** Transitivity writes every top prime as \(\sigma\mathfrak P\). Equations (3) and (5)'s kernel version give
\[
D_{\sigma\mathfrak P}=\sigma D_{\mathfrak P}\sigma^{-1},
\qquad
I_{\sigma\mathfrak P}=\sigma I_{\mathfrak P}\sigma^{-1}.
\]
In an abelian group both conjugates equal the original subgroup. This holds at ramified primes too; the existence of a Frobenius element requires the additional unramified hypothesis.

**2.** Use the generators \(r,c\) above. At \(2\), the pure cubic subfield has \(e=3,f=1\), while \(\mathbf Q(\omega)\) is inert with \(e=1,f=2\). The tower identities force \(e\) to be divisible by three and \(f\) by two. Equation (1), with total degree six, forces \(e=3,f=2,g=1\). Hence \(D=S_3\), \(I=\langle r\rangle=A_3\), and the Frobenius coset is the set of all three transpositions.

At \(3\), the cubic subfield has \(e=3\), while the quadratic subfield has \(e=2\). Both divide the top ramification index, so \(e=6,f=g=1\). Here \(D=I=S_3\); the residue group is trivial, and its Frobenius coset is the whole group. There is no unique Frobenius element at either \(2\) or \(3\).

The unramified calculations in the preceding section give the remaining rows:

| \(p\) | \(e\) | \(f\) | \(g\) | \(D_{\mathfrak P}\) | \(I_{\mathfrak P}\) | Frobenius or its coset |
|---|---|---|---|---|---|---|
| \(2\) | \(3\) | \(2\) | \(1\) | \(S_3\) | \(\langle r\rangle\) | transposition coset |
| \(3\) | \(6\) | \(1\) | \(1\) | \(S_3\) | \(S_3\) | whole group |
| \(5\) | \(1\) | \(2\) | \(3\) | \(\langle c\rangle\), for the chosen prime | \(1\) | \(c\) |
| \(7\) | \(1\) | \(3\) | \(2\) | \(\langle r\rangle\) | \(1\) | \(r\) or \(r^2\), as specified above |
| \(31\) | \(1\) | \(1\) | \(6\) | \(1\) | \(1\) | \(1\) |

Each row has \(efg=6\). Other primes at \(5\) conjugate \(\langle c\rangle\) and its generator. At \(2\), the inertia field is \(\mathbf Q(\omega)\) and the decomposition field is \(\mathbf Q\); their extension is unramified at \(2\), but ramified at \(3\). This illustrates the prime-specific qualification in Proposition 6.4.

**3.** Let \(N/K\) be the Galois closure of \(M/K\). Complete splitting in \(N\) descends by (7). Conversely, complete splitting in \(M\) makes \(e_Mf_M=1\) for every prime. Under the double coset correspondence this says \(D_\sigma\subseteq H=\operatorname{Gal}(N/M)\) for every \(\sigma\). Thus \(D\) fixes every conjugate of \(M\), and therefore their compositum \(N\). It is trivial, so the prime splits completely in \(N\). No prior unramifiedness assumption on \(N\) was needed.

If a prime were unramified and inert in a Galois extension \(L/K\), then \(e=g=1\), \(f=[L:K]\), and (2) would give \(D=G\). Proposition 6.3 makes \(D\) cyclic, so a noncyclic \(G\) excludes such a prime. Consequently, any prime with only one prime above it in such an extension must belong to its finite ramified set.

**4.** Equality of contractions means that the top primes are conjugate under \(H\), since \(L/M\) is Galois. Stabilizing the reference prime contributes the right factor \(D\). This gives exactly \(H\sigma D\), as proved in (8). At \(\sigma\mathfrak P\), the top groups for \(L/M\) are the intersections with \(H\); divide their orders using multiplicativity of \(e,f\). The resulting formulas are (9), including the subgroup \((D_\sigma\cap H)I_\sigma\) rather than an intersection in its denominator.

For \(M=\mathbf Q(\alpha)\), take \(H=\langle c\rangle\), the stabilizer of the root \(\alpha\). At \(2\) and \(3\), \(D=S_3\), so there is one double coset. At \(2\), \(I=A_3\) meets \(H\) trivially, giving \(e_M=3,f_M=1\); at \(3\), \(I=S_3\) gives \(e_M=[S_3:H]=3,f_M=1\).

At \(5\), \(I=1\) and \(D=\langle c\rangle\) for the chosen prime. There are two double cosets: \(H\) and \(HrH\). The first has \(D\cap H=D\), giving \(e_M=f_M=1\). For the second, the conjugate transposition subgroup meets \(H\) trivially, giving \(e_M=1,f_M=2\). These are the linear and quadratic factors modulo \(5\).

At \(7\), \(D=A_3\) and \(I=1\). Since \(HA_3=S_3\), there is one double coset and \(H\cap A_3=1\), giving \(e_M=1,f_M=3\). At \(31\), \(D=I=1\); the three left cosets of \(H\) give three primes, each with \(e_M=f_M=1\). These recover exactly the pure cubic decompositions already computed by Kummer–Dedekind.

## What this lesson does not prove

Finite Galois correspondence, subgroup degrees, and the quotient description are imported from [Milne FGT, Theorem 3.17, pp. 39–40]. Prime factorization, the fundamental identity, integral bases used in the examples, and the discriminant criterion are imported from the preceding lessons. The degree and maximality calculation for the biquadratic integral basis has its full proof in *Algebraic integers and rings of integers*.

The identification of a decomposition group with the Galois group of a completion belongs to the local-field course and is not used here. Higher ramification groups and their relation to the different are treated in *The different and the discriminant*. The general cyclotomic calculation is reserved for *Cyclotomic fields*. Distribution results about Frobenius classes, including density statements, are not asserted.

## References

- **[Milne ANT]** J. S. Milne, *Algebraic Number Theory*, version 3.08 (2020), Chapter 8, Propositions 8.10–8.13, properties 8.14–8.17, Examples 8.18–8.19 and Corollary 8.22, pp. 139–145. [Lecture notes](https://www.jmilne.org/math/CourseNotes/ANT.pdf).
- **[Milne FGT]** J. S. Milne, *Fields and Galois Theory*, version 5.10 (2022), Theorem 3.17 and Remark 3.18, pp. 39–40. [Lecture notes](https://www.jmilne.org/math/CourseNotes/FT.pdf).
- **[Stacks]** The Stacks project, [Tag 09EC](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/more-algebra.html#more-algebra-definition-decomposition-inertia), decomposition and inertia definitions, and [Tag 09ED](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/more-algebra.html#more-algebra-lemma-galois-galois), normality and the residue action.