# Unramified and totally ramified 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 not yet recorded. Public domain (CC0).*

*NT-ADL bundled edition, source-reconciled on 5 October 2026 by GPT-6.1 Sol (OpenAI), Ultra. This adaptation retains the provider lesson's mathematical scope and adds the source comparisons identified below. The upstream provider draft is unchanged. Original AI-written exposition is CC0; human reference works and genuinely reused human expression retain their own terms.*

A finite extension of a complete discretely valued field changes two things: its value group and its residue field. The unramified part changes only the residue field; the totally ramified part changes only the value group. Hensel lifting separates these contributions, and a carefully chosen integral element puts them back together.

Let \(K\) be complete for a nontrivial discrete valuation \(v_K\), normalized by \(v_K(K^\times)=\mathbf Z\). Write \(\mathcal O_K\) for its valuation ring, \(\mathfrak m_K\) for the maximal ideal, and \(\kappa\) for its residue field. Finite extensions carry their unique extended valuation, integer-normalized on the upper field. Thus
\[
v_L|_K=e(L/K)v_K,\qquad
f(L/K)=[\kappa_L : \kappa],\qquad
[L:K]=e(L/K)f(L/K).
\]
The last equality and the integral basis described below are imported from **Extensions of complete valued fields**, Theorem 3.1. They apply even to inseparable finite extensions.

We call \(L/K\) **unramified** if \(e=1\) and \(\kappa_L/\kappa\) is separable, and **totally ramified** if \(f=1\). These definitions agree with [Stacks, Tag 09E9](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/more-algebra.html#more-algebra-definition-types-of-extensions) in its finite separable setting. Residue separability belongs in the unramified definition even when it is automatic over a perfect field. In the statements about finite fields and norms, \(K\) will be a **nonarchimedean local field**, so that \(\kappa=\mathbf F_q\).

## 1. The integral basis we will use

Suppose \(L/K\) has ramification index \(e\) and residue degree \(f\). Choose a uniformizer \(\varpi\) of \(L\) and elements \(b_0,\ldots,b_{f-1}\in\mathcal O_L\) whose residues are a \(\kappa\)-basis of \(\kappa_L\). The integral basis theorem says that
\[
\{b_i\varpi^j\mid 0\le i<f,\ 0\le j<e\}
\tag{1.1}
\]
is an \(\mathcal O_K\)-basis of \(\mathcal O_L\), as well as a \(K\)-basis of \(L\).

Two features explain why it is useful. In a linear combination of the \(b_i\), dividing coefficients by one of least valuation leaves a nonzero residue, so that the valuation of the combination is that least coefficient valuation. Multiplication by \(\varpi^j\) then produces distinct classes modulo \(e\). No cancellation between those classes is possible. Completeness upgrades successive residue approximations to actual expansions in the finite module. We use the established theorem rather than assuming that an arbitrary algebraic field generator is an integral-ring generator.

Ramification indices and residue degrees multiply in towers. Indeed, indices of nested value groups multiply, and degrees of nested residue fields multiply. This also follows directly from the integer-normalized valuation restriction.

## 2. Lifting residue extensions and their maps

**Theorem 2.1 (unramified equivalence).** Reduction is an equivalence between the category of finite unramified extensions of \(K\), with \(K\)-embeddings, and the category of finite separable extensions of \(\kappa\), with \(\kappa\)-embeddings. No perfectness hypothesis on \(\kappa\) is necessary. Every finite unramified extension is separable over \(K\), has an unramified Galois closure, and its valuation ring is generated by a lift of a residue primitive element. Subextensions and composita of unramified extensions are unramified.

*Proof.* First construct an extension for a prescribed finite separable residue extension \(\lambda/\kappa\). Choose \(\bar a\) with \(\lambda=\kappa(\bar a)\), and let \(\bar F\in\kappa[X]\) be its monic irreducible polynomial of degree \(f\). Lift its coefficients to obtain a monic \(F\in\mathcal O_K[X]\).

The polynomial \(F\) is irreducible over \(K\). Indeed, every root of a monic integral polynomial is integral: a root of absolute value greater than \(1\) would make its leading term strictly dominate all others. In a monic factorization over \(K\), each coefficient of each factor is an elementary symmetric expression in such roots, and is therefore integral. Reduction of those factors would factor \(\bar F\), a contradiction. Let \(a\) be a root and put \(E=K(a)\). It is integral. The residue of \(a\) is a root of \(\bar F\), so
\[
f\le[\kappa_E : \kappa]\le[E:K]=f.
\]
The degree identity forces \(e(E/K)=1\) and \(\kappa_E=\kappa(\bar a)\cong\lambda\). Moreover, \(F'(a)\) has nonzero residue because \(\bar F\) is separable, so \(E/K\) is separable. Applying (1.1), with residue basis \(1,\bar a,\ldots,\bar a^{f-1}\), gives
\[
\mathcal O_E=\mathcal O_K[a].
\tag{2.1}
\]

Conversely, let \(L/K\) be finite and unramified. Choose a primitive element \(\bar a\) of \(\kappa_L/\kappa\), lift its minimal polynomial to a monic \(F\), and apply simple-root Hensel lifting in the complete field \(L\). There is a unique root \(a\in\mathcal O_L\) of \(F\) with the chosen residue. The constructed subfield \(K(a)\) has degree \(f\). Since \(e=1\), the degree of \(L/K\) is also \(f\), so \(L=K(a)\), and (2.1) applies.

Now let \(L,L'\) be unramified and let
\(\bar\sigma : \kappa_L\to\kappa_{L'}\) be a \(\kappa\)-embedding. Present \(L=K(a)\) by the preceding polynomial \(F\). The image \(\bar\sigma(\bar a)\) is a simple root of \(\bar F\). Hensel lifting gives a unique root \(a'\) of \(F\) in \(\mathcal O_{L'}\) with that residue. Sending \(a\) to \(a'\) defines a \(K\)-embedding \(L\to L'\).

Every \(K\)-embedding preserves the uniquely extended absolute value, hence maps valuation rings and maximal ideals into each other and induces a residue embedding. Its image of \(a\) must be the Hensel lift just specified. Thus reduction gives a bijection on every set of embeddings. It respects identities and compositions because it is reduction of the actual field maps. We have proved essential surjectivity and full faithfulness, which are precisely the equivalence assertion.

We include the closure argument to account for composita inside a common separable closure. Let \(\lambda/\kappa\) be finite and separable, and let \(\widetilde\lambda/\kappa\) be its finite Galois closure. Lift \(\widetilde\lambda\) to an unramified \(M/K\). Full faithfulness gives
\[
\operatorname{Aut}_K(M)\cong
\operatorname{Gal}(\widetilde\lambda/\kappa).
\]
The latter has \([M:K]\) elements, so the finite separable extension \(M/K\) is Galois. Any unramified extension corresponding to \(\lambda\) embeds into \(M\). In a fixed separable closure, extend that embedding to an automorphism of the separable closure over \(K\). Normality makes the automorphism preserve \(M\), so the original unramified subfield itself lies in \(M\).

More generally, for two unramified fields choose a finite Galois residue extension containing normal closures of both residue extensions. The same argument puts both fields inside its unramified Galois lift. If \(K\subset E\subset M\) and \(M/K\) is unramified, multiplicativity gives \(e(E/K)=1\), and \(\kappa_E/\kappa\), as a subextension of a separable extension, is separable. Thus \(E/K\) is unramified. Applying this to the two fields and their compositum proves the last assertion. \(\square\)

For an unramified finite extension the reduction map on automorphisms is always an isomorphism. In particular, it is Galois if and only if its residue extension is Galois: the two automorphism groups have the same order, and the two field degrees agree.

Define \(K^{\mathrm{ur}}\) to be the union of the finite unramified fields in a fixed separable closure of \(K\). The closure result makes this a field. It is Galois, being the union of its finite unramified Galois subextensions. Its residue field is a separable closure of \(\kappa\), and reduction induces
\[
\operatorname{Gal}(K^{\mathrm{ur}}/K)
\cong\operatorname{Gal}(\kappa^{\mathrm{sep}}/\kappa).
\tag{2.2}
\]
Here the residue separable closure can be chosen as the residue field of \(K^{\mathrm{ur}}\). It contains every finite separable residue extension by construction. Passing the finite Galois isomorphisms to inverse limits proves (2.2), including its topology: fixing a finite subextension corresponds to fixing its residue field.

## 3. Finite residues and arithmetic Frobenius

Assume now \(\kappa=\mathbf F_q\), where \(q\) is a power of the residue characteristic \(p\).

**Corollary 3.1 (the unramified tower of a local field).** For every \(n\ge1\) there is a unique unramified extension \(K_n\) of degree \(n\) inside the fixed separable closure. It satisfies
\[
K_n=K(\mu_{q^n-1}),
\]
and is cyclic Galois, generated by its arithmetic Frobenius \(\phi_n\), characterized by
\[
\overline{\phi_n(x)}=\bar x^q\quad(x\in\mathcal O_{K_n}).
\]
The generators are compatible, and
\[
\operatorname{Gal}(K^{\mathrm{ur}}/K)\cong
\varprojlim_n\mathbf Z/n\mathbf Z=\widehat{\mathbf Z},
\qquad \phi\longmapsto1.
\]
The inverse limit runs over divisibility of positive integers.

*Proof.* We first justify the finite-field facts, independently of valuation theory. Inside an algebraic closure of \(\mathbf F_q\), let \(E_n\) be the roots of \(X^{q^n}-X\). Frobenius identities show that these roots are closed under sums, products and negatives; for a nonzero root, \(x^{q^n-1}=1\) also shows that its inverse is a root. Thus they form a field containing \(\mathbf F_q\). The derivative is \(-1\), so there are exactly \(q^n\) roots and \([E_n:\mathbf F_q]=n\).

Any degree-\(n\) extension of \(\mathbf F_q\) has \(q^n\) elements. Every nonzero element satisfies \(x^{q^n-1}=1\) by Lagrange's theorem, so the extension is exactly this root field in the fixed algebraic closure. This proves uniqueness. The map \(x\mapsto x^q\) is an automorphism of \(E_n\), and its \(n\)-th power is identity. Its order cannot be \(d<n\), since then all \(q^n\) elements would be roots of \(X^{q^d}-X\), contrary to the polynomial root bound. The field is a separable splitting field, hence Galois, and its group of order \(n\) is generated by this Frobenius. Cyclicity of its multiplicative group follows from [Stacks, Tag 09HX](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/fields.html#fields-lemma-cyclic): every \(d\)-torsion set has size at most \(d\) by that same root bound.

Theorem 2.1 lifts the field and its automorphisms. Since the lifted extension is Galois, uniqueness up to \(K\)-isomorphism is uniqueness as a subfield of the fixed separable closure.

Put \(m=q^n-1\). Every nonzero residue element in \(\mathbf F_{q^n}\) is a root of \(X^m-1\), and these roots are simple because \(p\nmid m\). Hensel lifting gives all \(m\) roots of unity in \(K_n\), with distinct residues. Thus
\[
K(\mu_m)\subset K_n.
\]
The residue field of the left side contains all the nonzero elements of \(\mathbf F_{q^n}\), so its residue degree is at least \(n\). Its field degree is at most \(n\), forcing equality and the asserted field identity.

If \(n\mid r\), the residue inclusion lifts to \(K_n\subset K_r\). Restriction sends the \(q\)-power Frobenius in degree \(r\) to that in degree \(n\). Every finite residue extension appears at one of these levels, so their union is \(K^{\mathrm{ur}}\). The inverse limit of the finite cyclic groups proves the assertion. The element \(1\in\widehat{\mathbf Z}\) is a topological generator; it does not mean that the profinite group is the discrete group \(\mathbf Z\). \(\square\)

The field \(K^{\mathrm{ur}}\) is an algebraic union, not its completion. For example, in equal characteristic it is
\(\bigcup_n\mathbf F_{q^n}((T))\); it is smaller than
\(\overline{\mathbf F}_q((T))\), whose series can have coefficients lying in no one finite subfield.

**Example 3.2 (unramified quadratics).** For odd \(p\), let \(u\in\mathbf Z_p^\times\) have nonsquare residue. Then \(X^2-u\) has irreducible separable reduction, and
\(\mathbf Q_p(\sqrt u)\) is the unique unramified quadratic extension.

For \(p=2\), \(X^2+X+1\) has irreducible reduction and discriminant \(-3\); it gives \(\mathbf Q_2(\sqrt{-3})\). The polynomial \(X^2+3X+1\) has the same reduction and discriminant \(5\), so it gives \(\mathbf Q_2(\sqrt5)\). Uniqueness identifies these two quadratic subfields:
\[
\mathbf Q_2(\sqrt5)=\mathbf Q_2(\sqrt{-3}).
\]
The polynomials, rather than the reduction of \(X^2-5\), expose why the extension is unramified.

## 4. Norms from unramified local extensions

Write \(U_K=\mathcal O_K^\times\) and
\(U_K^{(r)}=1+\mathfrak m_K^r\) for \(r\ge1\).

**Proposition 4.1 (norm image).** If \(L/K\) is an unramified extension of nonarchimedean local fields of degree \(n\), then
\[
N_{L/K}(U_L)=U_K,\qquad
N_{L/K}(L^\times)=
\{\pi^{nk}u\mid k\in\mathbf Z,\ u\in U_K\},
\tag{4.1}
\]
where \(\pi\) is any uniformizer of \(K\). In fact
\(N_{L/K}(U_L^{(r)})=U_K^{(r)}\) for every \(r\ge1\).

*Proof.* Since \(e=1\), \(\pi\) is also a uniformizer of \(L\). By Corollary 3.1 the extension is Galois, and reduction identifies its \(n\) automorphisms with those of \(\mathbf F_{q^n}/\mathbf F_q\). Thus reduction of a unit norm is its finite-field norm.

We first verify both residue surjectivities used here. The multiplicative group of \(\mathbf F_{q^n}\) is cyclic, and its norm is the power
\[
x\longmapsto x^{1+q+\cdots+q^{n-1}}
=x^{(q^n-1)/(q-1)}.
\]
The image of a generator has order \(q-1\), so the norm is onto \(\mathbf F_q^\times\). The trace is the \(\mathbf F_q\)-linear map
\[
x\longmapsto x+x^q+\cdots+x^{q^{n-1}}.
\]
Its defining polynomial is nonzero and has degree \(q^{n-1}<q^n\), so it cannot vanish on all of \(\mathbf F_{q^n}\). Its image is a nonzero subspace of the one-dimensional space \(\mathbf F_q\), hence the trace is onto. This argument still works when \(p\mid n\).

For \(r\ge1\) and \(z\in\mathcal O_L\), expanding the product over automorphisms gives
\[
N(1+\pi^rz)\equiv
1+\pi^r\operatorname{Tr}_{L/K}(z)
\pmod{\pi^{r+1}\mathcal O_K}.
\tag{4.2}
\]
All terms involving two or more factors have valuation at least \(2r\ge r+1\). Reduction of the trace is the residue trace just proved onto.

Given \(a\in U_K\), residue norm surjectivity chooses \(b_1\in U_L\) with \(N(b_1)\equiv a\pmod\pi\). Inductively, suppose
\[
a/N(b_r)\equiv1\pmod{\pi^r}.
\]
Choose \(z_r\in\mathcal O_L\) whose residue trace is the coefficient of \(\pi^r\) in that quotient. Equation (4.2) shows that
\[
b_{r+1}=b_r(1+\pi^rz_r)
\]
has \(N(b_{r+1})\equiv a\pmod{\pi^{r+1}}\). The successive corrections tend to \(1\), making \(b_r\) a Cauchy sequence. Completeness gives a unit limit \(b\), and continuity of the norm, a polynomial in coordinates, gives \(N(b)=a\).

If \(a\in U_K^{(r)}\), start at \(b_r=1\) and make corrections from depth \(r\). The limit lies in the closed subgroup \(U_L^{(r)}\). Formula (4.2), or the product formula for the norm, also shows \(N(U_L^{(r)})\subset U_K^{(r)}\), proving equality.

Every \(x\in L^\times\) is \(\pi^ku\), with \(u\in U_L\). Since \(\pi\in K\), \(N(\pi)=\pi^n\), and the first assertion gives precisely (4.1). \(\square\)

The finite residue hypothesis in this proposition matters. For example,
\(\mathbf C((T))/\mathbf R((T))\) is an unramified degree-two extension of complete discretely valued fields, but the residue of a unit norm is a positive real number. The unit \(-1\) is not a norm. The unramified equivalence of Theorem 2.1 has wider scope than this norm-surjectivity statement.

## 5. Uniformizers and Eisenstein polynomials

**Theorem 5.1 (totally ramified extensions).** A finite extension \(L/K\) is totally ramified of degree \(e\) if and only if it is generated by a root of an Eisenstein polynomial of degree \(e\). In a totally ramified extension every uniformizer \(\varpi\) of \(L\) satisfies
\[
L=K(\varpi),\qquad
\mathcal O_L=\mathcal O_K[\varpi].
\tag{5.1}
\]
The statement allows inseparable extensions.

*Proof.* If \(F\) is Eisenstein of degree \(e\), the Eisenstein theorem of **Extensions of complete valued fields**, Corollary 5.2, makes it irreducible and gives its root valuation \(1/e\) when the valuation extends \(v_K\) literally. The ramification index is therefore at least \(e\). The field degree is \(e\), so the degree identity forces ramification index \(e\) and residue degree \(1\). The root has integer-normalized upper valuation \(1\), making it a uniformizer. Applying the integral basis (1.1), with the sole residue basis element \(1\), gives the ring equality.

Conversely, suppose \(f=1\), so \([L:K]=e\), and choose any uniformizer \(\varpi\). The values of
\(1,\varpi,\ldots,\varpi^{e-1}\) lie in distinct classes modulo the value group of \(K\). They are linearly independent over \(K\), and hence form a basis. In particular, \(K(\varpi)=L\).

Write the resulting minimal-polynomial relation as
\[
\varpi^e+a_{e-1}\varpi^{e-1}+\cdots+a_0=0.
\]
The valuations of the nonzero lower terms are \(e\,v_K(a_i)+i\). For \(0<i<e\), these are distinct from one another and from the valuations of both the constant and leading terms modulo \(e\). A vanishing sum cannot have a unique term of least valuation. The only possible repeated minimum therefore comes from the leading term, of value \(e\), and the constant term. It follows that
\[
v_K(a_0)=1,\qquad
e\,v_K(a_i)+i>e\quad(0<i<e).
\]
Thus \(v_K(a_i)\ge1\) for all \(i<e\), and the polynomial is Eisenstein. The same residue-basis application gives (5.1). Nothing in the argument assumed distinct conjugates. \(\square\)

**Example 5.2.** For every positive integer \(n\), \(X^n-p\) is Eisenstein over \(\mathbf Q_p\). Therefore \(\mathbf Q_p(p^{1/n})/\mathbf Q_p\) is totally ramified of degree \(n\), and its integral ring is \(\mathbf Z_p[p^{1/n}]\). When \(p\mid n\) this is wild ramification, rather than a failure of the Eisenstein argument.

## 6. Separating a general extension and finding one integral generator

**Proposition 6.1 (maximal unramified subfield).** Let \(L/K\) be finite and suppose \(\kappa_L/\kappa\) is separable. There is a unique largest unramified intermediate field \(L_0\), with
\[
[L_0:K]=f(L/K),\qquad
\kappa_{L_0}=\kappa_L.
\]
The extension \(L/L_0\) is totally ramified.

*Proof.* Choose a residue primitive element \(\bar a\), lift its monic minimal polynomial to \(F\in\mathcal O_K[X]\), and Hensel-lift \(\bar a\) to a root \(a\in L\). The construction in Theorem 2.1 gives an unramified field \(L_0=K(a)\) of degree \(f\) with the desired residue field. Its residue equality makes \(L/L_0\) totally ramified.

If \(E\subset L\) is any unramified intermediate field, the residue inclusion \(\kappa_E\subset\kappa_L\) lifts by Theorem 2.1 to an embedding \(E\to L_0\). Its composition into \(L\) is the actual inclusion of \(E\): both maps have the same residue embedding, and full faithfulness for maps from an unramified field into \(L\) follows by the identical simple-root uniqueness argument, even though \(L/K\) need not be unramified. Thus \(E\subset L_0\). This proves maximality and uniqueness. \(\square\)

**Corollary 6.2 (one generator of the integral ring).** If \(L/K\) is a finite separable extension and \(\kappa_L/\kappa\) is separable, then
\[
\mathcal O_L=\mathcal O_K[\beta]
\]
for some \(\beta\in\mathcal O_L\). Perfectness of the entire residue field is unnecessary.

*Proof.* Use \(a,F,L_0\) from Proposition 6.1 and let \(e=e(L/K)\), \(f=f(L/K)\). If \(e=1\), Theorem 2.1 gives the result with \(\beta=a\).

For \(e>1\), choose a uniformizer \(\varpi\) of \(L\), and put
\[
\beta=a+\varpi,\qquad \gamma=F(\beta).
\]
Since \(F(a)=0\) and \(F'(a)\) is a unit, Taylor expansion with integral coefficients gives
\[
F(a+\varpi)=F'(a)\varpi+\varpi^2c,\qquad
c\in\mathcal O_L.
\]
Hence \(\gamma\) has valuation \(1\) and is a uniformizer. The residues of
\(1,\beta,\ldots,\beta^{f-1}\) form a basis of \(\kappa_L/\kappa\). Formula (1.1), using this basis and the uniformizer \(\gamma\), says that
\[
\{\beta^i\gamma^j\mid 0\le i<f,\ 0\le j<e\}
\]
is an \(\mathcal O_K\)-basis of \(\mathcal O_L\). Every one of these elements belongs to \(\mathcal O_K[\beta]\), since \(\gamma=F(\beta)\). This ring consequently contains the entire basis and hence all of \(\mathcal O_L\); the reverse inclusion follows from integrality. \(\square\)

This proof explains the adjustment in the generator: a residue lift alone does not see ramification, and a uniformizer alone may not see the residue extension. Their sum makes its residue carry one part and the value of \(F(\beta)\) carry the other.

## 7. Exercises

1. Find the unramified quadratic extensions of \(\mathbf Q_2\) and \(\mathbf Q_3\), and justify their residue degrees.

2. For odd \(p\) and a unit \(u\) with nonsquare residue, show that \(\mathbf Q_p(\sqrt p,\sqrt u)\) has \(e=f=2\) over \(\mathbf Q_p\).

3. Prove Proposition 4.1, accounting for the case \(p\mid[L:K]\) and the convergence of the successive corrections.

4. Prove Corollary 6.2 for its stated separable residue extension, without assuming that an arbitrary primitive field element generates the integral ring.

## 8. Complete solutions

**Solution 1.** Over \(\mathbf F_2\), \(X^2+X+1\) has no root and derivative \(1\), so it is irreducible separable. Lifting it gives an unramified quadratic field whose discriminant presentation is \(\mathbf Q_2(\sqrt{-3})\). Lifting instead \(X^2+3X+1\) gives \(\mathbf Q_2(\sqrt5)\), since its discriminant is \(5\) and its reduction is again \(X^2+X+1\). Both have \(e=1,f=2\); uniqueness gives the equality in Example 3.2.

Over \(\mathbf F_3\), the only nonzero square is \(1\). The polynomial \(X^2-2\) has irreducible separable reduction, so \(\mathbf Q_3(\sqrt2)\) is the unramified quadratic extension, also with \(e=1,f=2\). One may equivalently use \(X^2+1\) and write \(\mathbf Q_3(i)\); uniqueness identifies the fields.

**Solution 2.** Put \(U=\mathbf Q_p(\sqrt u)\). Its residue field has degree \(2\), and its ramification index is \(1\), by Example 3.2. The element \(p\) remains a uniformizer of \(U\). Therefore \(X^2-p\) is Eisenstein over \(U\), and adjoining its root gives a totally ramified degree-two extension \(L/U\). Tower multiplication now yields
\[
[L : \mathbf Q_p]=4,\qquad
e(L/\mathbf Q_p)=2\cdot1=2,\qquad
f(L/\mathbf Q_p)=1\cdot2=2.
\]
By its generators \(L\) is the field in the exercise. This computation also proves that \(\sqrt p\) did not already belong to \(U\).

**Solution 3.** Reduction identifies the \(n\) automorphisms of \(L/K\) with those of its finite residue extension. The norm on residues is onto because it takes a multiplicative generator to an element of order \(q-1\). The trace on residues is onto because its polynomial
\(X+X^q+\cdots+X^{q^{n-1}}\) is nonzero of degree less than \(q^n\). In particular, one must not try to obtain surjectivity merely by tracing \(1\): that trace is zero when \(p\mid n\).

Choose a unit with the required residue norm. If its norm agrees with the target unit through depth \(r-1\), divide the target by that norm to obtain \(1+\pi^rc\) modulo \(\pi^{r+1}\). Choose an integral \(z\) whose residue trace is \(\bar c\). Expanding the \(n\) conjugate factors of \(1+\pi^rz\) shows that its norm has that correction modulo \(\pi^{r+1}\), since products of two correction terms have valuation at least \(2r\ge r+1\). Multiplying by \(1+\pi^rz\) therefore improves the agreement by one depth.

Repeating this constructs a Cauchy sequence of units: its successive ratios lie in \(1+\pi^r\mathcal O_L\), and ultrametricity bounds every later difference by the first omitted depth. The complete field \(L\) supplies a unit limit. The norm is a continuous polynomial, so its limit is exactly the target. Starting from \(1\) at depth \(r\) also proves surjectivity on \(U_L^{(r)}\). Finally, write any nonzero element as \(\pi^ku\); its norm is \(\pi^{nk}N(u)\). This proves both equalities of Proposition 4.1.

**Solution 4.** Choose \(\bar a\) generating the separable extension \(\kappa_L/\kappa\). Let \(F\in\mathcal O_K[X]\) be a monic lift of its minimal polynomial, of degree \(f\). Its root \(a\in\mathcal O_L\) in the prescribed residue class is supplied by Hensel, and \(F'(a)\) is a unit. If \(e=1\), the residues of \(1,a,\ldots,a^{f-1}\) form a basis, and the integral basis theorem gives \(\mathcal O_L=\mathcal O_K[a]\).

If \(e>1\), set \(\beta=a+\varpi\) for any upper uniformizer. The expansion
\[
F(\beta)=F'(a)\varpi+\varpi^2c
\]
shows that \(F(\beta)\) is another uniformizer. The residues of the \(f\) powers of \(\beta\) are the same residue basis as those of \(a\). Consequently the \(ef\) elements
\[
\beta^iF(\beta)^j\quad(0\le i<f,\ 0\le j<e)
\]
form an integral basis by (1.1). All are polynomials in \(\beta\) with integral base coefficients, so their full integral span lies in \(\mathcal O_K[\beta]\). That span is \(\mathcal O_L\), proving the claim. This checks the ring equality directly, rather than inferring it from \(K(\beta)=L\).

## Source comparison for this edition

Sutherland, Lecture 10, Theorem 10.13, and Milne 7.50–7.54 treat the unramified equivalence. Milne's initial perfect-residue presentation is accompanied by the explicit nonperfect-residue extension in footnote 10: unramified means \(e=1\) with separable residue extension. Sections 2 and 6 prove the assertions with those hypotheses, without silently imposing perfectness. The totally ramified uniformizer and Eisenstein route, compared with Milne 7.55–7.56, retains its stated applicability to inseparable finite extensions. The full integral-ring equality in Corollary 6.2 is justified by its \(ef\)-element integral basis; it is not inferred solely from a primitive field element.

Fesenko–Vostokov, Chapter III, §1, Proposition (1.2), compares the norm on the valuation quotient, the residue units and the successive principal-unit quotients with multiplication by the degree, the residue norm and the residue trace. That section develops its proof in the prime-degree setting. Section 4 here expands all \(n\) conjugate factors directly and proves the needed finite-field norm and trace surjectivities for every finite degree \(n\). For each depth \(r\ge1\), terms with two corrections have valuation at least \(2r\ge r+1\), so the residue trace supplies the next correction even when \(p\mid n\). Completeness gives the convergent sequence of corrections. This proves all the stated principal-unit norm equalities. The finite residue hypothesis for unit-norm surjectivity and the \(\mathbf C((T))/\mathbf R((T))\) counterexample are retained; the more general unramified equivalence has a wider scope.

## 9. What this lesson does not prove

- Unique extension of the absolute value, completeness of finite extensions, the degree identity and the lifted residue-basis integral basis are Theorems 1.2 and 3.1 and Proposition 2.1 of **Extensions of complete valued fields**. Its Corollary 5.2 gives the Eisenstein criterion and root valuation.
- Simple-root Hensel lifting is Corollary 1.2 of **Hensel's lemma, squares and roots of unity in p-adic fields**. The present lesson uses it in each finite complete upper field.
- The primitive element theorem for finite separable extensions is Theorem 5.1 of Milne's *Fields and Galois Theory*. Existence of a finite Galois closure, the Galois criterion by the number of automorphisms, and finite Galois correspondence are the results of its Chapter 3, especially Corollary 3.12 and Theorem 3.17.
- The classification of finite fields and their Frobenius Galois groups are Propositions 4.20 and 4.23 of the same work. Their cyclic multiplicative groups are the finite-subgroup fact used in Proposition 4.19, with its algebra exercise locator 1-3. Their norm and trace surjectivities are proved in Section 4 here.
- The DVR structure and uniformizer expressions are recalled in Section 2 of **Absolute values, valuations and Ostrowski's theorem**, with the exact ring characterization [Stacks, Tag 00PD]. The monic integral factorization needed here is proved directly in Section 2.
- The topology of inverse limits of finite Galois groups is the usual definition of the Krull topology. The identifications in Sections 2 and 3 are proved at every finite level before passing to that topology.

The finite separable primitive-element theorem is supplied by the open proof at [Stacks, Tag 030N](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/fields.html#fields-lemma-primitive-element). Finite multiplicative-subgroup cyclicity follows from [Tag 09HX](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/fields.html#fields-lemma-cyclic), since a degree-\(d\) polynomial has at most \(d\) roots. The residue extension and Frobenius assertions needed here are justified directly in the proof of Corollary 3.1.

## References

J. S. Milne, [*Algebraic Number Theory*](https://www.jmilne.org/math/CourseNotes/ANT.pdf), version 3.08 (19 July 2020), Chapter 7, “Unramified extensions of a local field” and “Totally ramified extensions of \(K\),” Propositions 7.50 and 7.55, Corollaries 7.51–7.52, Remarks 7.53 and 7.56, and Example 7.54.

J. S. Milne, [*Fields and Galois Theory*](https://www.jmilne.org/math/CourseNotes/FT.pdf), version 5.10 (September 2022), Chapters 3–5.

A. V. Sutherland, MIT 18.785 Number Theory I lecture notes, Fall 2021: Lecture 10, [*Extensions of complete DVRs*](https://math.mit.edu/classes/18.785/2021fa/LectureNotes10.pdf), Theorems 10.12, 10.13 and 10.23 and Corollary 10.15; Lecture 11, [*Totally ramified extensions and Krasner's lemma*](https://math.mit.edu/classes/18.785/2021fa/LectureNotes11.pdf), Theorems 11.5 and 11.10.

[Stacks, Tag 09E9](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/more-algebra.html#more-algebra-definition-types-of-extensions), for the definitions of unramified, tame and totally ramified DVR extensions.

I. B. Fesenko and S. V. Vostokov, [*Local Fields and Their Extensions*](https://ivanfesenko.org/wp-content/uploads/2021/10/vol.pdf), second edition (2002), approved author-hosted copy, Chapter III, §1, (1.1)–(1.2), printed pp. 68–69, for the norm expansion and its graded trace/norm comparison; the scope of this comparison is explained above.