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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).

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 in its finite separable setting. Residue separability belongs in the unramified definition even when it is automatic over a perfect field. Section 3 and the finite-residue consequence of Section 4 take \(K\) to be a nonarchimedean local field, so that \(\kappa=\mathbf F_q\). The general norm-image theorem and integral-generator construction allow arbitrary residue fields with the separability explicitly stated.

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 is proved in Hensel's lemma, squares and roots of unity in p-adic fields, Lemma 4.0.

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 extensions

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

We will need a trace fact for arbitrary finite separable residue extensions. Here is its complete algebraic proof.

Lemma 4.0 (a residue element of trace one). Let \(\lambda/\kappa\) be finite and separable of degree \(n\). Its trace is onto \(\kappa\), even when the characteristic of \(\kappa\) divides \(n\).

Proof. Choose a primitive element \(a\) and its monic separable minimal polynomial \(G\) of degree \(n\). Write its distinct roots in a splitting field as \(a_1,\ldots,a_n\). The polynomials on the two sides of \[ X^{n-1}=\sum_{i=1}^n \frac{a_i^{n-1}}{G'(a_i)}\frac{G(X)}{X-a_i} \] have degree at most \(n-1\) and agree at every \(a_i\), hence agree identically. Their leading coefficients give \[ \sum_{i=1}^n\frac{a_i^{n-1}}{G'(a_i)}=1. \] Evaluation at these roots has invertible Vandermonde matrix in the basis \(1,a,\ldots,a^{n-1}\). It diagonalizes multiplication by any element of \(\lambda\) after extending scalars to the splitting field. Thus its trace is the sum of the conjugate values, and \[ \operatorname{Tr}_{\lambda/\kappa} \left(\frac{a^{n-1}}{G'(a)}\right)=1. \] Multiplying this element by any scalar in \(\kappa\) proves surjectivity. The argument also covers \(n=1\); it never divides by \(n\). \(\square\)

Proposition 4.1 (exact norm image). Let \(L/K\) be any finite unramified extension of complete discretely valued fields, of degree \(n\), with residue extension \(\lambda/\kappa\). Fix a uniformizer \(\pi\) of \(K\), and put \[ R=N_{\lambda/\kappa}(\lambda^\times)\subset\kappa^\times. \] Then \[ N_{L/K}(U_L^{(r)})=U_K^{(r)}\quad(r\ge1), \] \[ N_{L/K}(U_L)=\{u\in U_K\mid\bar u\in R\}, \qquad N_{L/K}(L^\times)= \{\pi^{nk}u\mid k\in\mathbf Z,\ u\in U_K,\ \bar u\in R\}. \tag{4.1} \] No Galois or finite-residue hypothesis is needed.

Proof. Since \(e=1\), \(\pi\) is also a uniformizer of \(L\). Choose lifts \(b_1,\ldots,b_n\) of a \(\kappa\)-basis of \(\lambda\). By (1.1) these are an \(\mathcal O_K\)-basis of \(\mathcal O_L\). For \(z\in\mathcal O_L\), multiplication by \(z\) has an integral matrix \(M_z\) in that basis. Its reduction is the multiplication matrix of \(\bar z\) on \(\lambda\). Consequently trace and determinant reduce to the residue trace and norm. In particular, the residue of every unit norm lies in \(R\).

The multiplication matrix of \(1+\pi^r z\) is \(I+\pi^rM_z\). In its determinant, the linear term is \(\pi^r\operatorname{Tr}(M_z)\), and each higher term is divisible by \(\pi^{2r}\). Since \(2r\ge r+1\) for \(r\ge1\), this gives \[ N(1+\pi^rz)\equiv 1+\pi^r\operatorname{Tr}_{L/K}(z) \pmod{\pi^{r+1}\mathcal O_K}. \tag{4.2} \] The same determinant expansion shows \(N(U_L^{(r)})\subset U_K^{(r)}\). By Lemma 4.0 the residue trace is onto.

For a target \(a\in U_K^{(r)}\), start with \(b_r=1\). Inductively suppose \[ a/N(b_j)\equiv1\pmod{\pi^j}\qquad(j\ge r). \] Choose \(z_j\in\mathcal O_L\) whose residue trace is the residue of \((a/N(b_j)-1)/\pi^j\). Equation (4.2) shows that \[ b_{j+1}=b_j(1+\pi^jz_j) \] has \(a/N(b_{j+1})\equiv1\pmod{\pi^{j+1}}\). For \(m>j\), the ratio \(b_m/b_j\) belongs to \(1+\pi^j\mathcal O_L\), because each successive factor does. Thus \((b_j)\) is Cauchy. Completeness supplies its limit in the closed subgroup \(U_L^{(r)}\). The norm is a continuous polynomial in coordinates, so the limit has norm \(a\). This proves equality on every principal-unit subgroup.

Now take \(u\in U_K\) with \(\bar u=N_{\lambda/\kappa}(\bar c)\), and lift \(\bar c\ne0\) to \(c\in U_L\). Reduction of the norm gives \(u/N(c)\in U_K^{(1)}\). The result just proved supplies \(d\in U_L^{(1)}\) with \(N(d)=u/N(c)\). Then \(N(cd)=u\). Together with the necessary residue condition, this proves the exact unit image.

Every \(x\in L^\times\) is \(\pi^kc\), with \(c\in U_L\). Since \(\pi\in K\), its multiplication matrix is \(\pi I\) and \(N(\pi)=\pi^n\). The unit image therefore gives the last assertion of (4.1), including negative \(k\). \(\square\)

Corollary 4.2 (finite residues). For an unramified degree-\(n\) extension of nonarchimedean local fields, \[ N(U_L)=U_K,\qquad N(L^\times)=\{\pi^{nk}u\mid k\in\mathbf Z,\ u\in U_K\}. \]

Proof. If \(\kappa=\mathbf F_q\), Corollary 3.1 identifies \(\lambda=\mathbf F_{q^n}\) and its arithmetic Frobenius. The residue norm is \[ x\longmapsto x^{1+q+\cdots+q^{n-1}} =x^{(q^n-1)/(q-1)}. \] The image of a multiplicative generator has order \(q-1\), so \(R=\mathbf F_q^\times\). Proposition 4.1 proves the assertions. One can also see residue trace surjectivity directly: \(X+X^q+\cdots+X^{q^{n-1}}\) is nonzero of degree less than \(q^n\), so its image is a nonzero \(\mathbf F_q\)-subspace of \(\mathbf F_q\). This remains valid when \(p\mid n\). \(\square\)

The finite-residue condition matters for full unit surjectivity, rather than for principal-unit surjectivity. For example, \(\mathbf C((T))/\mathbf R((T))\) is unramified of degree two, and its residue norm is \(z\mapsto z\bar z\), with image the positive real numbers. Its exact norm image consists of elements of even \(T\)-valuation with positive leading unit coefficient. Thus \(-1\) is not a norm, although every principal unit is a norm.

Reduction also identifies \(U_K/N(U_L)\) with \(\kappa^\times/R\): reduction is onto and Proposition 4.1 identifies its kernel. Choosing \(\pi\) identifies the full cokernel with \[ K^\times/N(L^\times)\cong (\mathbf Z/n\mathbf Z)\times(\kappa^\times/R), \qquad \pi^m u\longmapsto(m\bmod n,[\bar u]). \] The displayed splitting depends on \(\pi\). In the real/complex Laurent-series example the two factors are \(\mathbf Z/2\mathbf Z\) and the two signs.

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 any finite extension and \(\kappa_L/\kappa\) is separable, then \[ \mathcal O_L=\mathcal O_K[\beta] \] for some \(\beta\in\mathcal O_L\). The field extension \(L/K\) may be inseparable; 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.

For an inseparable instance, take \(K=\mathbf F_p((t))\), \(L=\mathbf F_p((s))\), with the embedding \(t=s^p\). The polynomial \(X^p-t\) is Eisenstein and has derivative zero, so \(L/K\) is purely inseparable of degree \(p\). Its residue extension is the identity. Grouping series exponents modulo \(p\) gives \[ \sum_{j\ge0}c_js^j =\sum_{i=0}^{p-1}s^i \left(\sum_{k\ge0}c_{pk+i}t^k\right). \] Thus \(\mathcal O_L=\mathcal O_K[s]\). This also illustrates directly why the field-separability hypothesis is unnecessary.

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 for arbitrary residue fields, including a proof of residue trace surjectivity and convergence of the successive corrections. Deduce Corollary 4.2, accounting for \(p\mid[L:K]\).

  4. Prove Corollary 6.2 for any finite \(L/K\) with separable residue extension, including inseparable \(L/K\), 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. Let \(\lambda/\kappa\) be the separable residue extension. For a primitive element \(a\) with monic minimal polynomial \(G\), interpolation of \(X^{n-1}\) at the distinct roots \(a_i\) gives \[ \sum_i\frac{a_i^{n-1}}{G'(a_i)}=1. \] The sum is the trace of \(a^{n-1}/G'(a)\), since evaluation at the roots diagonalizes multiplication. Thus residue trace is onto. This argument does not require Galois symmetry or division by \(n\).

Lift a residue basis to an integral basis of \(\mathcal O_L\). The integral multiplication matrix \(M_z\) reduces to the residue multiplication matrix; its trace and determinant therefore reduce correctly. The identity \[ N(1+\pi^jz)=\det(I+\pi^jM_z) \equiv1+\pi^j\operatorname{Tr}(z)\pmod{\pi^{j+1}} \] holds for \(j\ge1\), since terms of degree at least two have valuation at least \(2j\ge j+1\). It also shows \(N(U_L^{(r)})\subset U_K^{(r)}\).

For \(a\in U_K^{(r)}\), start at \(b_r=1\). If \(a/N(b_j)\) is \(1+\pi^jc_j\) modulo \(\pi^{j+1}\), choose integral \(z_j\) with residue trace \(\bar c_j\), and put \(b_{j+1}=b_j(1+\pi^jz_j)\). The displayed congruence improves agreement by one depth. For \(m>j\), \(b_m/b_j\in1+\pi^j\mathcal O_L\); ultrametricity therefore makes the sequence Cauchy. Completeness gives a limit in \(U_L^{(r)}\), and continuity of the norm polynomial gives norm exactly \(a\).

A unit \(u\) is a norm only if \(\bar u\in N(\lambda^\times)\). Conversely choose a unit \(c\) lifting a residue preimage of \(\bar u\). Then \(u/N(c)\in U_K^{(1)}\), and principal-unit surjectivity supplies \(d\) with that norm, so \(u=N(cd)\). Writing \(x=\pi^kc\) and using \(N(\pi)=\pi^n\) proves the full norm-image formula.

For finite residues the norm exponent \((q^n-1)/(q-1)\) maps a multiplicative generator to an element of order \(q-1\), so every unit residue is allowed. The trace polynomial \(X+X^q+\cdots+X^{q^{n-1}}\) is nonzero of degree less than \(q^n\), giving a second proof of trace surjectivity, even when \(p\mid n\). Tracing \(1\) alone would fail in that case.

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 \(K(\beta)=L\). Hensel lifting used completeness of \(L\) and separability of the residue polynomial; neither it nor the integral basis requires \(L/K\) to be separable.

9. What this lesson does not prove

The complete finite separable primitive-element proof is Extensions of complete valued fields, Proposition 0.1. The complete finite multiplicative-subgroup cyclicity proof is Hensel's lemma, squares and roots of unity in p-adic fields, Lemma 4.0. Stacks, Tags 030N and 09HX remain comparison sources. 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, 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, Chapters 3–5.

Stacks, Tag 09E9, for the definitions of unramified, tame and totally ramified DVR extensions.

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