# Cyclotomic, quadratic and Kummer extensions of local fields

*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 not yet recorded. Public domain (CC0).*

*Cyclotomic reciprocity specialization, exact prerequisite links and source-convention check by GPT-6 Astra (OpenAI), Ultra, 11 October 2026. CC0. This is a scoped self-check, not independent full-course review.*

We finish with extensions whose defining equations reveal their ramification. Roots of unity split into an unramified prime-to-\(p\) part and a totally ramified \(p\)-power part. Quadratic extensions can be read from square classes, but the integral generator changes in the unramified \(2\)-adic case. For tame Kummer extensions, the valuation of the radicand determines the ramification index exactly.

We use **Ramification groups and the different of a local extension** for cyclotomic \(p\)-power degrees and differents, and **The multiplicative group of a local field** for the unit filtration. The square criteria and unramified constructions come from the earlier lessons named in the precise imports below.

*Valuation-tower and cyclotomic-compositum proof connections by GPT-6 Astra (OpenAI), Ultra, 11 October 2026. Independent CC0 exposition; scoped self-check, not independent full-course review.*

## 1. Two kinds of cyclotomic extension

**Proposition 1.1.** Let \(m\geq2\) be prime to \(p\), and let \(\zeta_m\) be primitive of order \(m\). Then
\[
\mathbf Q_p(\zeta_m)/\mathbf Q_p
\]
is unramified of degree \(r=\operatorname{ord}_m(p)\), the least positive integer with \(p^r\equiv1\pmod m\). Its Galois group is cyclic of order \(r\), with arithmetic Frobenius \(\zeta_m\mapsto\zeta_m^p\). For \(m=1\), the extension is trivial of degree one.

For \(n\geq1\), the extension
\[
L_n=\mathbf Q_p(\zeta_{p^n})
\]
is totally ramified of degree \(N=(p-1)p^{n-1}\), has uniformizer \(1-\zeta_{p^n}\), and has Galois group
\[
(\mathbf Z/p^n\mathbf Z)^\times,\qquad
a: \zeta_{p^n}\longmapsto\zeta_{p^n}^a.
\tag{1.1}
\]
Its different exponent is \(np^n-(n+1)p^{n-1}\).

*Proof.* For the prime-to-\(p\) statement, reduction is injective on \(m\)-th roots of unity, by uniqueness of their simple-root Hensel lifts. The reduction of a primitive root therefore still has order \(m\). A finite field \(\mathbf F_{p^s}\) contains such a root exactly when \(m\mid p^s-1\); its multiplicative group is cyclic. Thus the residue field generated by that root has degree \(r\).

The unramified extension of degree \(r\) contains all the Hensel lifts of the \(m\)-th residue roots. It contains \(\zeta_m\), so the field generated by \(\zeta_m\) is unramified and has degree at most \(r\). Its residue field already has degree \(r\), forcing equality. The unramified correspondence makes its group cyclic, and uniqueness of lifts changes the residue Frobenius action into \(\zeta_m\mapsto\zeta_m^p\).

For the \(p\)-power statement, [Proposition 5.1 of **Ramification groups and the different of a local extension**](../NT-LOC-09.html#5-the-cyclotomic-filtration) proves that \(\Phi_{p^n}(1+Y)\) is Eisenstein, gives the degree and full exponent group, and computes the different. Its uniformizer \(\zeta_{p^n}-1\) differs from the one here by a unit \(-1\), so all assertions follow with the stated sign convention. \(\square\)

More generally, write an arbitrary positive order as \(p^n m\), \(p\nmid m\). Its root field is the compositum of these two root fields. The unramified and totally ramified fields intersect only in \(\mathbf Q_p\), since an intermediate field that is both has degree one. Since both are Galois, their compositum has the product Galois group, ramification index \((p-1)p^{n-1}\) for \(n\geq1\), and residue degree \(\operatorname{ord}_m(p)\) for \(m\geq2\); a missing factor has degree one. This follows also by adjoining the Eisenstein \(p\)-power generator over the unramified field, where \(p\) remains a uniformizer.

### Valuation towers

Here is the tower calculation needed both for the cyclotomic compositum and for the Kummer proof below. Let \(K\subseteq L\subseteq M\) be finite extensions of a complete discretely valued field. Use the unique extended valuation on each field, with each nonzero value group normalized to \(\mathbf Z\). The uniqueness and completeness assertions are [Extensions of complete valued fields, Theorem 1.2](../NT-LOC-04.html#1-a-distance-forced-by-a-polynomial) and [Proposition 2.1](../NT-LOC-04.html#2-coordinates-control-convergence); the normalization and degree identity are its [Theorem 3.1](../NT-LOC-04.html#3-an-integral-basis-from-residue-digits).

**Tower lemma.** In this setting,
\[
e(M/K)=e(M/L)e(L/K),\qquad
f(M/K)=f(M/L)f(L/K).
\]
No separability hypothesis, either on the field extensions or on the residue extensions, is needed.

*Proof.* Uniqueness of the extended valuation makes the restrictions compatible. For every \(x\in K^\times\), integer normalization gives
\[
v_M(x)=e(M/L)v_L(x)=e(M/L)e(L/K)v_K(x).
\]
Evaluate at a uniformizer of \(K\), whose \(v_K\)-value is one. By the definition of \(e(M/K)\), its \(v_M\)-value is \(e(M/K)\), proving the first identity.

The inclusions of valuation rings induce injections
\[
\kappa_K\longrightarrow\kappa_L\longrightarrow\kappa_M.
\]
Indeed, an element of a lower valuation ring has positive value in the upper field exactly when it has positive value in the lower field, so the kernel on residues is zero. Let \((b_i)_{i=1}^s\) be a basis of \(\kappa_L\) over \(\kappa_K\), and \((c_j)_{j=1}^t\) a basis of \(\kappa_M\) over \(\kappa_L\). Their products \(b_i c_j\) span \(\kappa_M\) over \(\kappa_K\): first expand in the \(c_j\), then expand each coefficient in the \(b_i\). In a relation \(\sum_{i,j}a_{ij}b_i c_j=0\), with \(a_{ij}\in\kappa_K\), independence of the \(c_j\) gives \(\sum_i a_{ij}b_i=0\) for every \(j\), and independence of the \(b_i\) gives every \(a_{ij}=0\). Thus the residue degree is \(st\), as asserted. This is the tower argument stated in [Unramified and totally ramified extensions, section 1](../NT-LOC-07.html#1-the-integral-basis-we-will-use). \(\square\)

### The compositum and its automorphisms

We now prove the preceding compositum assertion without leaving a general intersection theorem implicit. Put \(K=\mathbf Q_p\), \(E=K(\zeta_m)\), \(F=K(\zeta_{p^n})\) and \(M=EF\), where \(n\geq1\) and \(p\nmid m\); allow \(m=1\), so \(E=K\). Set \(d=[E:K]\) and \(N=(p-1)p^{n-1}\). Choose compatible roots: for a primitive \(\zeta\) of order \(p^n m\), take \(\zeta_{p^n}=\zeta^m\) and \(\zeta_m=\zeta^{p^n}\). Integers \(u,v\) with \(um+vp^n=1\) give \(\zeta=\zeta_{p^n}^{u}\zeta_m^{v}\). This proves \(M=K(\zeta)\), including both containments.

Let \(\theta=\zeta_{p^n}-1\). [Ramification groups and the different of a local extension, Proposition 5.1](../NT-LOC-09.html#5-the-cyclotomic-filtration) gives the monic degree-\(N\) polynomial
\[
P(Y)=\Phi_{p^n}(1+Y)\in\mathbf Z_p[Y]
\]
and proves it is Eisenstein at \(p\). Since \(E/K\) is unramified, \(p\) still has integer-normalized value one in \(E\); the same coefficient inequalities make \(P\) Eisenstein over \(E\). By [Extensions of complete valued fields, Corollary 5.2](../NT-LOC-04.html#5-root-valuations-in-the-coefficients), and [Unramified and totally ramified extensions, Theorem 5.1](../NT-LOC-07.html#5-uniformizers-and-eisenstein-polynomials), it remains irreducible and
\[
M=E(\theta),\qquad [M:E]=e(M/E)=N,\qquad
f(M/E)=1,\qquad
\mathcal O_M=\mathcal O_E[\theta].
\]
In particular, \(1,\theta,\ldots,\theta^{N-1}\) is an \(E\)-basis. The [tower lemma](#valuation-towers) now gives
\[
[M:K]=Nd,\qquad e(M/K)=N,\qquad f(M/K)=d.
\]
For completeness it also proves the earlier intersection assertion. If \(D=E\cap F\), multiplicativity and \(e(E/K)=1\) force \(e(D/K)=1\), while \(f(F/K)=1\) forces \(f(D/K)=1\). The complete-valued degree identity \([D:K]=e(D/K)f(D/K)\) then gives \(D=K\). We will construct the product action directly, rather than use this intersection alone as an unexplained degree argument.

Both \(E\) and \(F\) are splitting fields over \(K\) of their respective separable root-of-unity polynomials. Their compositum is the splitting field of \(X^{p^n m}-1\), so it too is Galois. Every \(K\)-automorphism of \(M\) preserves each of \(E,F\), giving the restriction homomorphism
\[
\operatorname{res}:\operatorname{Gal}(M/K)
\longrightarrow\operatorname{Gal}(E/K)\times\operatorname{Gal}(F/K),
\qquad g\longmapsto(g|_E,g|_F).
\]
It is injective because \(E\) and \(F\) generate \(M\). Here is its explicit inverse. For \(\sigma\in\operatorname{Gal}(E/K)\) and \(\tau\in\operatorname{Gal}(F/K)\), define
\[
T_{\sigma,\tau}:M\longrightarrow M,\qquad
\sum_{j=0}^{N-1}a_j\theta^j
\longmapsto\sum_{j=0}^{N-1}\sigma(a_j)\tau(\theta)^j,
\qquad a_j\in E.
\]
Uniqueness of the basis expansion makes the formula unambiguous. To check multiplication, map \(E[Y]\) to \(M\) by applying \(\sigma\) to coefficients and evaluating at \(\tau(\theta)\). All coefficients of \(P\) lie in \(K\), and \(P(\tau(\theta))=\tau(P(\theta))=0\). Thus the ideal \((P)\) lies in the kernel, so the map descends through the field presentation \(E[Y]/(P)=M\). It is a unital \(K\)-algebra homomorphism, restricts to \(\sigma\) on \(E\), and restricts to \(\tau\) on \(F=K(\theta)\).

Constructing \(T_{\sigma^{-1},\tau^{-1}}\) by the same formula gives an inverse: both composites fix \(E\) and \(F\) pointwise, hence fix their generated field \(M\). The composition of two such maps restricts to the coordinatewise composites on \(E,F\), so these formulas respect the group law. Consequently
\[
\operatorname{Gal}(M/K)\simeq
\operatorname{Gal}(E/K)\times\operatorname{Gal}(F/K)
\]
with the specified restriction map and inverse, not merely an abstract equality of group orders. For \(m=1\) the first factor is trivial. For \(p=2,n=1\), the polynomial \(P(Y)=Y+2\) has degree one, \(F=K\), and the second factor is trivial. If \(n=0\), omit \(F\) and the Eisenstein step entirely: \(M=E\), with the identity restriction to the sole factor. \(\square\)

**Example 1.2.** The field \(\mathbf Q_3(\zeta_9)\) has degree \(6\) and is totally ramified. Its Galois group is \((\mathbf Z/9\mathbf Z)^\times\), cyclic of order \(6\), and its different exponent is \(2\cdot9-3\cdot3=9\). In contrast, \(\mathbf Q_3(\zeta_8)\) is unramified of degree \(2\), since \(3^2\equiv1\pmod8\) and \(3\not\equiv1\pmod8\).

## 2. An elementary norm and a class-field-theoretic statement

**Proposition 2.1.** With \(L_n\) as above,
\[
N_{L_n/\mathbf Q_p}(1-\zeta_{p^n})=p,
\qquad
[\mathbf Q_p^\times:p^{\mathbf Z}(1+p^n\mathbf Z_p)]
=(p-1)p^{n-1}.
\tag{2.1}
\]

*Proof.* The conjugates of \(\zeta_{p^n}\) are all the primitive roots. Their product of differences from \(1\) is
\[
\prod_{a\in(\mathbf Z/p^n\mathbf Z)^\times}(1-\zeta_{p^n}^a)
=\Phi_{p^n}(1)
=\sum_{j=0}^{p-1}1=p.
\]
This is the norm, with its exact sign.

The decomposition \(\mathbf Q_p^\times=p^{\mathbf Z}\times\mathbf Z_p^\times\) reduces the index to that of \(1+p^n\mathbf Z_p\) in \(\mathbf Z_p^\times\). The first residue quotient has order \(p-1\), and each subsequent quotient has order \(p\). There are \(n-1\) subsequent quotients, proving the index. For \(p=2,n=1\), the index is one, as it should be for the trivial extension. \(\square\)

The full norm group is
\[
N_{L_n/\mathbf Q_p}(L_n^\times)
=p^{\mathbf Z}(1+p^n\mathbf Z_p).
\tag{2.2}
\]
*Proof of (2.2), using the explicit reciprocity theorem.* For the local field \(\mathbf Q_p\), the [multiplicative Lubin–Tate module](../../NT-CFT/formal-groups-and-lubin-tate-modules.html#6-the-multiplicative-example-over-mathbf-z_p) has formal law \(X+Y+XY\) and multiplication by \(p\) given by \((1+X)^p-1\). Its level-\(n\) division points are \(\zeta-1\), where \(\zeta^{p^n}=1\); hence its division field is precisely \(L_n\), as proved in [Lubin–Tate division fields, section 4](../../NT-CFT/lubin-tate-division-fields.html#4-two-concrete-descriptions).

Write \(a=u p^r\), uniquely with \(u\in\mathbf Z_p^\times\) and \(r\in\mathbf Z\). Normalize the Artin map by arithmetic Frobenius on unramified extensions. [Explicit local reciprocity and the existence theorem, Theorem 9.1](../../NT-CFT/explicit-local-reciprocity-and-existence.html#theorem-9-1-explicit-local-reciprocity), then gives the typed action
\[
\operatorname{Art}_{L_n/\mathbf Q_p}:\mathbf Q_p^\times
   \longrightarrow\operatorname{Gal}(L_n/\mathbf Q_p),
\qquad
\operatorname{Art}_{L_n/\mathbf Q_p}(u p^r)(\zeta_{p^n})
   =\zeta_{p^n}^{\,u^{-1}\bmod p^n}.
\]
The exponent means the inverse of the reduction of \(u\) in \((\mathbf Z/p^n\mathbf Z)^\times\). Since the primitive root generates \(L_n\), this automorphism is the identity exactly when \(u\equiv1\pmod{p^n}\); the integer \(r\) is unrestricted. Thus its kernel is \(p^{\mathbf Z}(1+p^n\mathbf Z_p)\). [Local reciprocity and norm groups, Theorem 6.3](../../NT-CFT/local-reciprocity-and-norm-groups.html#theorem-6-3-the-local-reciprocity-law), identifies that kernel with \(N_{L_n/\mathbf Q_p}(L_n^\times)\), proving (2.2). This is exactly the multiplicative specialization of [Corollary 9.2](../../NT-CFT/explicit-local-reciprocity-and-existence.html#corollary-9-2-norm-groups-of-the-explicit-fields): its subgroup \(\langle p\rangle U^{(n)}\) is the right-hand side of (2.2), since \(U^{(n)}=1+p^n\mathbf Z_p\). For \(p=2,n=1\), every unit is one modulo two, the action is trivial, and the kernel is all of \(\mathbf Q_2^\times\), as required. \(\square\)

The kernel theorem is the additional input: Proposition 2.1 exhibits the norm \(p\) and calculates the index of a proposed subgroup, but does not identify all unit norms. The proof uses finite-level reciprocity, not the later local existence theorem or the description of the entire abelian closure. None of the ramification or quadratic proofs below assumes (2.2).

## 3. Quadratic extensions from square classes

First recall an elementary field fact. Over a field of characteristic different from \(2\), every quadratic extension has the form \(K(\sqrt a)\) for a nonsquare \(a\). Completing the square in a quadratic minimal polynomial proves this. Moreover,
\[
K(\sqrt a)=K(\sqrt b)
\quad\Longleftrightarrow\quad a/b\in K^{\times2}
\tag{3.1}
\]
for nonsquares \(a,b\). Indeed write \(\sqrt b=x+y\sqrt a\). The coefficient of \(\sqrt a\) in its square is \(2xy\), so \(xy=0\). Since \(\sqrt b\notin K\), \(y\ne0\) and \(x=0\); hence \(b=ay^2\). The converse is immediate.

**Proposition 3.1 (all quadratic extensions and their discriminants).** For odd \(p\), choose a unit \(u\) with nonsquare residue. The three quadratic extensions of \(\mathbf Q_p\) and their discriminant exponents are

| Field | Ramification | Discriminant exponent |
| --- | --- | --- |
| \(\mathbf Q_p(\sqrt u)\) | Unramified | \(0\) |
| \(\mathbf Q_p(\sqrt p)\) | Totally ramified, tame | \(1\) |
| \(\mathbf Q_p(\sqrt{up})\) | Totally ramified, tame | \(1\) |

For \(p=2\), the seven extensions are

| Field | Integral generator | Discriminant exponent |
| --- | --- | --- |
| \(\mathbf Q_2(\sqrt5)\) | \((1+\sqrt5)/2\) | \(0\) |
| \(\mathbf Q_2(\sqrt{-1})\) | \(\sqrt{-1}\) | \(2\) |
| \(\mathbf Q_2(\sqrt3)\) | \(\sqrt3\) | \(2\) |
| \(\mathbf Q_2(\sqrt{2u})\), \(u\in\{1,-1,5,-5\}\) | \(\sqrt{2u}\) | \(3\) |

The first \(2\)-adic extension is unramified; the other six are totally ramified and wild. Also \(\mathbf Q_2(\sqrt5)=\mathbf Q_2(\sqrt{-3})\).

*Proof.* The square criterion for odd \(p\) says that an element is a square exactly when its valuation is even and its unit residue is a square. Thus the four square classes have representatives \(1,u,p,up\). Equation (3.1) proves completeness and distinctness of the three nontrivial extensions.

The polynomial \(X^2-u\) has irreducible separable reduction, giving the unramified quadratic field and its integral basis \(1,\sqrt u\). The other two polynomials are Eisenstein; they give total quadratic fields with integral bases \(1,\sqrt p\) and \(1,\sqrt{up}\). For a quadratic integral basis \(1,\sqrt a\), the trace determinant is \(4a\). Its valuation is zero for the unit \(u\) and one for \(p,up\). Their degree \(2\) is prime to odd \(p\), so the ramification is tame.

At \(2\), the square criterion says that \(2^r w\), with \(w\) an odd unit, is a square exactly when \(r\) is even and \(w\equiv1\pmod8\). The eight square classes can therefore be represented by
\[
1,\ -1,\ 3,\ 5,\ 2,\ -2,\ 10,\ -10.
\tag{3.2}
\]
The four odd representatives have distinct residues \(1,7,3,5\) modulo \(8\), and multiplying each class by \(2\) gives the four odd-valuation classes. For the residue-three class, \(3\) and \(-5\) represent the same class. Equation (3.1) again proves there are exactly seven quadratic extensions.

Let \(\omega=(1+\sqrt5)/2\). It satisfies
\[
\omega^2-\omega-1=0.
\]
The reduction \(X^2+X+1\) is irreducible over \(\mathbf F_2\), so this is the unramified quadratic extension, with integer ring \(\mathbf Z_2[\omega]\). Its basis discriminant is \(5\), a unit. Furthermore \((-3)/5\equiv1\pmod8\), so the square criterion and (3.1) give the claimed equality with \(\mathbf Q_2(\sqrt{-3})\).

For \(a=-1\) or \(3\), put \(\pi=\sqrt a-1\). Its equation is
\[
Y^2+2Y+(1-a)=0.
\]
The constant coefficient is respectively \(2\) or \(-2\), so both polynomials are Eisenstein. Thus each field is totally ramified, with
\(\mathcal O_L=\mathbf Z_2[\pi]=\mathbf Z_2[\sqrt a]\).
The discriminant \(4a\) has exponent \(2\).

For \(a=2u\), \(u\in\{1,-1,5,-5\}\), the polynomial \(X^2-2u\) is Eisenstein. Its root is a uniformizer and gives the full integer ring. The basis discriminant \(8u\) has exponent \(3\). These six fields have ramification index \(2\), divisible by their residue characteristic, so are wild.

These discriminant computations also check the different. In the ramified odd-unit \(2\)-adic fields, the derivative \(2\sqrt a\) has normalized upper value \(2\); in the even-radicand fields it has value \(2+1=3\). Each residue degree is one, so these are also the base discriminant exponents. In the unramified field the derivative \(2\omega-1=\sqrt5\) of the integral generator's polynomial is a unit. \(\square\)

The integral generator in the first row matters: the power basis \(1,\sqrt5\) has discriminant \(20\), and is not an integral basis of that unramified extension.

## 4. Tame Kummer ramification

Let \(K\) now be any nonarchimedean local field, in either characteristic, with residue characteristic \(p\), and let \(n\geq1\) be prime to \(p\). Assume \(\mu_n\subseteq K\). For \(a\in K^\times\), choose \(\alpha\) with \(\alpha^n=a\), and put \(L=K(\alpha)\).

**Proposition 4.1.** The extension \(L/K\) is cyclic Galois of degree dividing \(n\), and is tame. If \(r=v_K(a)\), then
\[
e(L/K)=\frac n{\gcd(n,r)}.
\tag{4.1}
\]
Here \(\gcd(n,0)=n\). In particular,
\[
L/K\text{ is unramified}
\quad\Longleftrightarrow\quad n\mid v_K(a)
\quad\Longleftrightarrow\quad
a\in U_K K^{\times n}.
\tag{4.2}
\]

*Proof.* The polynomial \(X^n-a\) is separable, and all its roots \(\zeta\alpha\) lie in \(L\), because \(\mu_n\subseteq K\). Thus it splits there and \(L/K\) is Galois. The ratio map
\[
\operatorname{Gal}(L/K)\longrightarrow\mu_n,\qquad
\sigma\longmapsto\sigma(\alpha)/\alpha
\]
is an injective homomorphism, since its ratios lie in \(K\) and \(\alpha\) generates \(L\). It proves cyclicity and that the degree divides \(n\).

Write \(a=\pi^r u\), with \(u\in U_K\). At a finite unramified level \(K'/K\), the residue unit \(\bar u\) has an \(n\)-th root. Simple-root Hensel lifting then gives \(\beta\in U_{K'}\) with \(\beta^n=u\). The field \(K'\) contains \(\mu_n\). Choosing \(\rho^n=\pi\), we obtain
\[
K'L=K'(\rho^r),
\tag{4.3}
\]
since \(\alpha/(\beta\rho^r)\) is an \(n\)-th root of unity.

Put \(g=\gcd(n,r)\), \(e_0=n/g\), \(r_0=r/g\). The element \(\gamma=\rho^g\) satisfies \(\gamma^{e_0}=\pi\), and \(\rho^r=\gamma^{r_0}\). As \(\gcd(e_0,r_0)=1\), integers \(A,B\) with \(A r_0+B e_0=1\) give
\(\gamma=(\gamma^{r_0})^A\pi^B\).
Thus the fields in (4.3) equal \(K'(\gamma)\). The Eisenstein polynomial \(X^{e_0}-\pi\) shows that this field is totally ramified of degree \(e_0\) over \(K'\); this includes \(e_0=1\).

The extension \(K'L/L\) is unramified. Indeed the finite unramified field \(K'\) is generated by prime-to-\(p\) roots of unity, and adjoining those roots to the local field \(L\) again gives an unramified extension, by [Unramified and totally ramified extensions, Corollary 3.1](../NT-LOC-07.html#3-finite-residues-and-arithmetic-frobenius), applied over the local field L. The [tower lemma proved above](#valuation-towers), applied to the two towers, therefore gives
\[
e(L/K)=e(K'L/K)=e(K'L/K')=e_0.
\]
This proves (4.1). Since \(e_0\mid n\), it is prime to \(p\); residue fields are finite, so the extension is tame.

It is unramified exactly when \(e_0=1\), or \(n\mid r\). Finally \(a\in U_K K^{\times n}\) exactly when its valuation is divisible by \(n\): divide it by \(\pi^{nk}\) when \(r=nk\), and use valuation for the converse. This proves (4.2). \(\square\)

The degree need not be \(n\): a radicand already an \(n\)-th power gives the trivial extension. The formula concerns the ramification index and retains that case.

For characteristic \(p\), the different kind of cyclic extension is stated from [Stacks, Tag 09I7]: adjoining a root of \(X^p-X-a\) gives a Galois extension of degree \(1\) or \(p\), with automorphisms acting by additions from \(\mathbf F_p\); conversely every cyclic degree-\(p\) extension has such a generator. This Artin–Schreier theorem is a cited comparison, not a prerequisite of Proposition 4.1.

## 5. A norm obstruction at two

In \(\mathbf Q_2(i)\),
\[
N(x+iy)=x^2+y^2.
\]
Thus \(5=1^2+2^2\) is a norm. To show that \(3\) is not a norm, the possibility of denominators must be ruled out before reducing modulo \(4\).

Suppose \(x^2+y^2=3\), and let \(m=\min(v_2(x),v_2(y))\). Neither coordinate need individually be nonzero, but the minimum is finite. After factoring out \(2^m\), at least one scaled coordinate is odd. If exactly one is odd, the scaled sum has valuation zero; if both are odd, their squares sum to \(2\) modulo \(8\), so it has valuation one. Hence the valuation of the original sum is \(2m\) or \(2m+1\). If \(m<0\), both are negative, contradicting \(v_2(3)=0\).

Therefore \(x,y\in\mathbf Z_2\). Their squares modulo \(4\) lie in \(\{0,1\}\), whose two-element sums cannot equal \(3\). This contradiction proves that \(3\) is not a norm, without using reciprocity or Hilbert symbols.

## 6. Exercises

1. Show that \(\mathbf Q_p(\zeta_p)=\mathbf Q_p((-p)^{1/(p-1)})\), including \(p=2\).

2. List the quadratic extensions of \(\mathbf Q_3\), giving integral bases and discriminant exponents.

3. Prove that \(5\) is a norm from \(\mathbf Q_2(i)\), and \(3\) is not.

4. Prove the full \(2\)-adic quadratic list in Proposition 3.1, with its discriminant exponents.

## 7. Complete solutions

**Solution 1.** Put \(\delta=\zeta_p-1\). Its translated cyclotomic equation gives
\[
\delta^{p-1}=-p\,u,\qquad
u=1+\sum_{j=1}^{p-2}\frac{\binom p{j+1}}p\delta^j.
\]
The displayed coefficients are integral and \(u\equiv1\) in the upper residue field. Since \(p\nmid p-1\), Hensel lifting at the residue root \(1\) gives \(b\equiv1\) with \(b^{p-1}=u\). Hence \(\gamma=\delta/b\) satisfies \(\gamma^{p-1}=-p\). It is an upper uniformizer, so it generates the totally ramified field; alternatively its Eisenstein equation has the same degree \(p-1\). Thus the two fields coincide. All choices of the radical differ by \(\mu_{p-1}\subseteq\mathbf Q_p\), so the field is independent of that choice. At \(p=2\), the sum is empty, \(\delta=-2\), and both fields are simply \(\mathbf Q_2\).

**Solution 2.** The nonsquare unit class in \(\mathbf Q_3\) can be represented by \(-1\). The four square classes are \(1,-1,3,-3\), so the three quadratic fields are
\[
\mathbf Q_3(i),\quad
\mathbf Q_3(\sqrt3),\quad
\mathbf Q_3(\sqrt{-3}).
\]
The polynomial \(X^2+1\) has irreducible separable reduction over \(\mathbf F_3\), so the first is unramified with integral basis \(1,i\) and discriminant \(-4\), of exponent zero. The other polynomials \(X^2-3\) and \(X^2+3\) are Eisenstein, so their roots give integral bases and total quadratic extensions. Their basis discriminants are \(12\) and \(-12\), each with exponent one. Equation (3.1) and the square criterion prove distinctness and completeness.

**Solution 3.** Multiplying \(1+2i\) by its conjugate gives \(5\). Suppose instead that \(x+iy\) had norm \(3\). With \(m=\min(v_2(x),v_2(y))\), scale the coordinates by \(2^{-m}\); their squared sum has valuation zero if just one is odd and one if both are odd. Thus \(v_2(x^2+y^2)\) is \(2m\) or \(2m+1\). A negative \(m\) would make that valuation negative, impossible for \(3\). Both coordinates are therefore integral. Squares modulo \(4\) are zero or one, and the sum of two cannot be three. This contradiction excludes every possible norm representation, including those with denominators.

**Solution 4.** For a nonzero \(2\)-adic number, remove an even power of \(2\). Its remaining valuation is zero or one. Remove an odd-unit square as well; an odd unit is a square exactly when it is one modulo \(8\). The four distinct odd unit classes are represented by \(1,-1,3,5\); multiplying them by \(2\) gives four more distinct classes. The class of \(3\) equals that of \(-5\), since their ratio is one modulo \(8\). Thus the nonidentity classes are exactly those in the \(2\)-adic table.

Every quadratic field arises by adjoining a square root of one of them, by completing the square; equation (3.1) proves these seven fields are distinct. For \(5\), use \(\omega=(1+\sqrt5)/2\), whose polynomial \(X^2-X-1\) has irreducible reduction \(X^2+X+1\). It is unramified, has integer ring \(\mathbf Z_2[\omega]\), and discriminant \(5\).

For \(-1\) and \(3\), the shifted roots \(\sqrt a-1\) satisfy the Eisenstein polynomials \(Y^2+2Y+2\) and \(Y^2+2Y-2\). The rings generated by the shifted and original roots are equal, so \(1,\sqrt a\) are integral bases. Their discriminants \(-4\) and \(12\) have exponent two.

The four remaining radicands \(2,-2,10,-10\) are \(2u\) with \(u\in\{1,-1,5,-5\}\). Their polynomials \(X^2-2u\) are Eisenstein, giving integral bases \(1,\sqrt{2u}\). Each discriminant \(8u\) has exponent three. All six ramified fields have \(e=2\) and \(f=1\); the unramified field has \(e=1,f=2\). Finally \((-3)/5\equiv1\pmod8\) proves \(\mathbf Q_2(\sqrt5)=\mathbf Q_2(\sqrt{-3})\), as claimed.

## 8. What this lesson does not prove

- The cyclotomic \(p\)-power Eisenstein equation, its degree, Galois group, lower groups and different exponent are [Proposition 5.1 of **Ramification groups and the different of a local extension**](../NT-LOC-09.html#5-the-cyclotomic-filtration). Upper cyclotomic groups are [Proposition 5.1 of **Herbrand's function and the upper numbering**](../NT-LOC-10.html#5-cyclotomic-upper-numbering); they are not needed for the elementary norm calculation.
- The odd-prime and \(2\)-adic square criteria and complete square-class representatives are [Proposition 2.1 of **Hensel's lemma, squares and roots of unity in p-adic fields**](../NT-LOC-03.html#2-squares-and-their-obstructions). Simple-root lifting is its [Corollary 1.2](../NT-LOC-03.html#1-a-root-from-a-shrinking-error), and multiplicative residue lifts are [Proposition 4.1](../NT-LOC-03.html#4-residues-lift-multiplicatively).
- Unramified extensions correspond to finite residue extensions, and prime-to-\(p\) roots of unity generate unramified fields: [Theorem 2.1](../NT-LOC-07.html#2-lifting-residue-extensions-and-their-maps) and [Corollary 3.1](../NT-LOC-07.html#3-finite-residues-and-arithmetic-frobenius) of **Unramified and totally ramified extensions**. Its [Theorem 5.1](../NT-LOC-07.html#5-uniformizers-and-eisenstein-polynomials) gives the Eisenstein integral-basis theorem. The unit quotients used in (2.1) are [Proposition 1.1 of **The multiplicative group of a local field**](../NT-LOC-11.html#1-three-factors-and-the-unit-filtration).
- The monogenic different is generated by the derivative of the minimal polynomial, and its ideal norm is the discriminant. The written programme proofs are *Number fields*, lesson 14, **The different and the discriminant**, [Proposition 14.2](../../NT-ANT/the-different-and-the-discriminant.html#euler-s-lemma-and-a-power-basis) and [Theorem 14.3](../../NT-ANT/the-different-and-the-discriminant.html#norm-and-discriminant), for a finite separable field extension with finite integral closure over a Dedekind base; the derivative formula additionally requires the full integral closure to be monogenic. A basis \(1,\sqrt a\) has trace Gram matrix \(\operatorname{diag}(2,2a)\), whose determinant \(4a\) was used directly here.
- The proof of (2.2) above imports the written finite reciprocity and explicit Lubin–Tate action, with the exact links and specialization given there. [Corollary 9.5](../../NT-CFT/explicit-local-reciprocity-and-existence.html#corollary-9-5-abelian-extensions-of-mathbf-q_p) also records the cyclotomic action. Those class-field-theory proofs build on the earlier local-field constructions, not on the norm equality (2.2). Artin–Schreier theory has an accessible open proof at [Stacks, Tag 09I7]; it is cited as a comparison, not used by the Kummer ramification argument.
- The finite-field multiplicative group is cyclic by [**Hensel's lemma, squares and roots of unity in p-adic fields**, Lemma 4.0](../NT-LOC-03.html#4-residues-lift-multiplicatively). Stacks, Tag 09HX remains a comparison source. Finite-field Frobenius and extensions are supplied by the residue-field argument in **Unramified and totally ramified extensions**, Corollary 3.1. See also Milne, *Fields and Galois Theory*, Propositions 4.19–4.20 and 4.23. The [valuation-tower proof](#valuation-towers) above gives both multiplicativity identities at their full complete-discrete-valuation scope. The [compositum proof](#the-compositum-and-its-automorphisms) gives the intersection and the actual product automorphisms directly from the earlier Eisenstein theorem, with their domains, coefficient action and inverse. No unnamed Graduate Algebra intersection theorem is needed for those steps.

## Source comparison for this edition

J. S. Milne, *Class Field Theory*, version 4.03 (6 August 2020), Chapter I, Examples 2.18 and 3.8, gives the multiplicative formal group and its cyclotomic division fields. Example 3.13(a)–(b), printed page 44, uses arithmetic Frobenius and the inverse-unit action used here. The proof outline in Chapter I, section 1 distinguishes constructing the explicit action from proving that its kernel is the norm group; the latter is an essential input to (2.2).

There is a typographical error in Example 3.13(c) of that edition: for \(a=u p^t\), the prime-to-\(p\) root must map to \(\zeta_m^{p^t\bmod m}\), not \(\zeta_m^t\). This follows by iterating arithmetic Frobenius; for negative \(t\), use the inverse of \(p\) modulo \(m\). For instance, \(a=1\) has \(t=0\) and must fix the root, whereas the printed exponent zero would send it to one. The programme's [Corollary 9.5](../../NT-CFT/explicit-local-reciprocity-and-existence.html#corollary-9-5-abelian-extensions-of-mathbf-q_p) already has the correct formula. This correction concerns that exponent, not the theorem identifying the norm group.

## References
J. S. Milne, [*Algebraic Number Theory*](https://www.jmilne.org/math/CourseNotes/ANT.pdf), Chapter 6, “The basic results,” and Chapter 7.

[Stacks, Tag 09I6](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/fields.html#fields-section-Kummer), “Kummer extensions,” and [Tag 09I7](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/fields.html#fields-section-Artin-Schreier), “Artin–Schreier extensions.”

J. S. Milne, [*Class Field Theory*, version 4.03, 6 August 2020](https://www.jmilne.org/math/CourseNotes/CFT.pdf), Chapter I, section 1 and Examples 2.18, 3.8, 3.13. The comparison above refers to that edition.