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).

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.

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 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.

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 stronger equality is stated here with its precise classical locator: \[ N_{L_n/\mathbf Q_p}(L_n^\times) =p^{\mathbf Z}(1+p^n\mathbf Z_p). \tag{2.2} \] It is proved in Class field theory. For the multiplicative Lubin–Tate module over \(\mathbf Q_p\), with uniformizer \(p\), the division field of level \(n\) is \(L_n\) (Lubin–Tate division fields, Section 4, equation (16)). Corollary 9.2 of Explicit local reciprocity and the existence theorem says that its norm group is \(\langle p\rangle U^{(n)}\), which is the right-hand side of (2.2). Proposition 2.1 alone does not prove (2.2): it exhibits the norm \(p\) and computes the candidate subgroup's index, without identifying the norms of all units. 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

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

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 the unramified roots-of-unity description. Multiplication of ramification indices in 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. Upper cyclotomic groups are Proposition 5.1 of Herbrand's function and the 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. Simple-root lifting is its Corollary 1.2, and multiplicative residue lifts are Proposition 4.1.
• Unramified extensions correspond to finite residue extensions, and prime-to-\(p\) roots of unity generate unramified fields: Theorem 2.1 and Corollary 3.1 of Unramified and totally ramified extensions. Its Theorem 5.1 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.
• 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 and Theorem 14.3, 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 full cyclotomic norm-group equality (2.2) is proved in Class field theory: Corollary 9.2 of Explicit local reciprocity and the existence theorem, applied to the multiplicative Lubin–Tate module of Lubin–Tate division fields. 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. 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. Multiplication of ramification indices and the finite-Galois intersection facts of Graduate Algebra are used in the compositum and Kummer arguments.

References

J. S. Milne, Algebraic Number Theory, Chapter 6, “The basic results,” and Chapter 7.

Stacks, Tag 09I6, “Kummer extensions,” and Tag 09I7, “Artin–Schreier extensions.”