# Ramification groups and the different of a local extension *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).* The ramification index separates tame extensions from wild ones, but it does not measure how far a wild automorphism moves an integral element. The lower ramification groups make that measurement. Their successive quotients constrain the Galois group, and the sum of their sizes gives the exponent of the different. Throughout the local discussion, \(L/K\) is a finite Galois extension of nonarchimedean local fields, in either characteristic. Write \(G=\operatorname{Gal}(L/K)\), \(\mathcal O_L\) for the valuation ring, \(\mathfrak P\) for its maximal ideal, and \(\kappa_L\) for its residue field. Normalize \(v_L(L^\times)=\mathbf Z\), put \(v_L(0)=+\infty\), and let \(p=\operatorname{char}\kappa_L\). Thus \(v_L|_K=e\,v_K\), where \(e=e(L/K)\). Every automorphism preserves \(v_L\). We use **Unramified and totally ramified extensions** for the maximal unramified subfield, integral monogenicity and Eisenstein extensions. **Tame ramification and the tame Galois group** supplies the meaning of tame ramification. Exact additional imports are listed at the end. ## 1. Measuring an automorphism intrinsically Choose \(\alpha\) with \(\mathcal O_L=\mathcal O_K[\alpha]\). For \(\sigma\in G\), define \[ i_G(\sigma)=v_L(\sigma\alpha-\alpha). \] In particular, \(i_G(1)=+\infty\). For integral \(i\geq-1\), set \[ G_i=\{\sigma\in G:i_G(\sigma)\geq i+1\}. \tag{1.1} \] **Proposition 1.1 (intrinsic lower groups).** The number \(i_G(\sigma)\) is independent of the chosen integral generator. More precisely, \[ i_G(\sigma)=\min_{x\in\mathcal O_L}v_L(\sigma x-x). \tag{1.2} \] The groups \(G_i\) are normal in \(G\), decrease with \(i\), and are trivial for sufficiently large \(i\). We have \(G_{-1}=G\), and \(G_0\) is the inertia group. If \(H\leq G\), the lower groups of \(L/L^H\), using the same normalized valuation on \(L\), satisfy \[ H_i=H\cap G_i. \tag{1.3} \] *Proof.* For \(x=P(\alpha)\), with \(P\in\mathcal O_K[X]\), the polynomial identity \(P(Y)-P(X)=(Y-X)Q(X,Y)\) gives \[ \sigma x-x=(\sigma\alpha-\alpha)Q(\alpha,\sigma\alpha). \] The second factor is integral. Hence every value in (1.2) is at least \(i_G(\sigma)\), and \(x=\alpha\) attains the bound. This proves (1.2), including the identity case, and removes dependence on \(\alpha\). For an integer \(i\geq0\), membership in \(G_i\) says exactly that \(\sigma\) acts trivially on every element of \(\mathcal O_L/\mathfrak P^{i+1}\). Consequently it is a subgroup. Explicitly, \[ \sigma\tau x-x=\sigma(\tau x-x)+(\sigma x-x), \] and preservation of valuation proves closure; replacing \(x\) by \(\sigma^{-1}x\) proves closure under inverses. Conjugation by \(\gamma\in G\) replaces the test element by \(\gamma^{-1}x\), which runs over all of \(\mathcal O_L\), and preserves its value. Thus \(G_i\) is normal. Decrease is immediate. Every difference of integral elements is integral, so \(G_{-1}=G\). Trivial action modulo \(\mathfrak P\) is precisely the definition of inertia, so \(G_0\) is inertia. If \(\sigma\ne1\), it cannot fix the generator \(\alpha\); its value \(i_G(\sigma)\) is finite. There are finitely many such automorphisms, and taking \(i\) above all their values makes \(G_i\) trivial. The extension \(L/L^H\) is again Galois and local. Its intrinsic test (1.2) uses the same ring \(\mathcal O_L\) and the same \(v_L\), so its condition on an element \(\sigma\in H\) is exactly the condition defining \(G_i\). This proves (1.3), without assuming \(H\) normal in \(G\). \(\square\) For real \(u\geq-1\), our convention is \[ G_u=G_{\lceil u\rceil}. \tag{1.4} \] For instance, \(G_u=G_0\) on \((-1,0]\) and \(G_u=G_1\) on \((0,1]\). At an integer jump the value is the group on its left. This convention will determine the endpoints in the upper numbering. ## 2. The graded pieces Let \(L_0\) be the maximal unramified subfield of \(L/K\). Its residue field is \(\kappa_L\), and \(L/L_0\) is totally ramified. The correspondence between unramified extensions and residue extensions identifies \(\operatorname{Gal}(L_0/K)\) with \(\operatorname{Gal}(\kappa_L/\kappa_K)\). The field \(L_0\) is stable under \(G\) by its defining maximality. Restriction from \(G\) onto \(\operatorname{Gal}(L_0/K)\) is surjective: an embedding of this intermediate field extends to an embedding of the finite normal extension \(L/K\). Its kernel is exactly \(G_0\), since the unramified correspondence identifies the action on \(L_0\) with the action on its residue field. In particular, \[ |G_0|=[L:L_0]=e. \tag{2.1} \] Fix a uniformizer \(\pi\) of \(L\). Total ramification gives \(\mathcal O_L=\mathcal O_{L_0}[\pi]\). Every \(\sigma\in G_0\) fixes \(L_0\), so the same polynomial-difference argument as before now gives \[ i_G(\sigma)=v_L(\sigma\pi-\pi) \qquad(\sigma\in G_0). \tag{2.2} \] This uniformizer formula is intended for inertia; a noninertial automorphism has \(i_G(\sigma)=0\). Write \(U_L^0=\mathcal O_L^\times\) and \(U_L^r=1+\mathfrak P^r\) for \(r\geq1\). **Proposition 2.1 (canonical graded injections).** There are injective homomorphisms \[ G_0/G_1\longrightarrow\kappa_L^\times, \qquad \sigma G_1\longmapsto\overline{\sigma\pi/\pi}, \tag{2.3} \] and, for \(r\geq1\), \[ G_r/G_{r+1}\longrightarrow U_L^r/U_L^{r+1}, \qquad \sigma G_{r+1}\longmapsto(\sigma\pi/\pi)U_L^{r+1}. \tag{2.4} \] Both maps are independent of \(\pi\). The group \(U_L^r/U_L^{r+1}\) is isomorphic to the additive group of \(\kappa_L\); that last identification uses \(\pi\). *Proof.* Put \(c_\sigma=\sigma\pi/\pi\). For inertia, \[ c_{\sigma\tau}=\sigma(c_\tau)c_\sigma. \tag{2.5} \] Inertia acts trivially on residue units, so reduction makes (2.5) multiplicative. Equation (2.2) says that the kernel is \(G_1\), proving (2.3). For \(r\geq1\), the map \[ U_L^r/U_L^{r+1}\longrightarrow(\kappa_L,+), \qquad (1+a\pi^r)U_L^{r+1}\longmapsto\bar a \tag{2.6} \] is well defined and bijective. Multiplication becomes addition because \(\mathfrak P^{2r}\subseteq\mathfrak P^{r+1}\). Every \(\sigma\in G_1\) acts trivially on this quotient. Indeed, \(\sigma a\equiv a\pmod{\mathfrak P}\), and \(\sigma\pi/\pi\equiv1\pmod{\mathfrak P}\), so \(\sigma(a\pi^r)\equiv a\pi^r\pmod{\mathfrak P^{r+1}}\). For \(\sigma,\tau\in G_r\), (2.2) puts \(c_\sigma,c_\tau\) in \(U_L^r\); (2.5) is therefore a homomorphism in this quotient. Its kernel is \(G_{r+1}\), again by (2.2). If \(\pi'=u\pi\), with \(u\in\mathcal O_L^\times\), then \[ \frac{\sigma\pi'}{\pi'}= \frac{\sigma u}{u}\frac{\sigma\pi}{\pi}. \] For \(\sigma\in G_r\), the intrinsic definition gives \(v_L(\sigma u-u)\geq r+1\). Thus \(\sigma u/u\in U_L^{r+1}\), also for \(r=0\) with the evident interpretation. Both (2.3) and (2.4) are unchanged. Changing \(\pi\) does change the coordinate (2.6): if \(\pi'=u\pi\), the residue coefficient changes by \(\bar u^{-r}\). The injection into the unit quotient is canonical; its residue-field coordinate need not be. \(\square\) **Corollary 2.2 (wild inertia and solvability).** The group \(G_0/G_1\) is cyclic of order prime to \(p\). The group \(G_1\) is the unique Sylow \(p\)-subgroup of \(G_0\). Every finite Galois group of nonarchimedean local fields is solvable. Moreover, \[ L/K\text{ is tame}\quad\Longleftrightarrow\quad G_1=1. \] *Proof.* The finite cyclic group \(\kappa_L^\times\) has order prime to \(p\). Equation (2.3) proves the first assertion. Each quotient \(G_r/G_{r+1}\), for \(r\geq1\), is a subgroup of the additive residue field, so it is an elementary abelian \(p\)-group. The filtration terminates; multiplying its quotient orders shows that \(G_1\) is a \(p\)-group. Its index in \(G_0\) is prime to \(p\), and its normality makes it the unique Sylow subgroup. The same terminating filtration has abelian quotients, so \(G_1\), and then \(G_0\), are solvable. Finally \(G/G_0\) is the cyclic finite-field Galois group described above. A group with a solvable normal subgroup and solvable quotient is solvable: lift a derived series for the quotient and then append one for the normal subgroup. This proves the assertion for \(G\). By (2.1), \(p\nmid e\) exactly when \(G_1=1\); residue extensions of local fields are separable, so this is precisely tameness. \(\square\) ## 3. Hilbert's formula For any finite separable extension of local fields, define its trace-dual lattice and different by \[ \mathcal O_L^*= \{x\in L : \operatorname{Tr}_{L/K}(x\mathcal O_L)\subseteq\mathcal O_K\}, \qquad \mathfrak D_{L/K}=(\mathcal O_L^*)^{-1}. \tag{3.1} \] The trace form is nondegenerate: [**Unramified and totally ramified extensions**, Lemma 4.0](NT-LOC-07.md) proves that every finite separable field extension has an element \(t\) of trace one. For any nonzero \(x\in L\), set \(y=t/x\); then \(\operatorname{Tr}_{L/K}(xy)=1\). This proves nondegeneracy in every characteristic. Its dual lattice is a fractional \(\mathcal O_L\)-ideal, and inversion is fractional-ideal inversion. If \(\mathcal O_L=\mathcal O_K[\alpha]\), with monic minimal polynomial \(f\), the monogenic different theorem gives \[ \mathfrak D_{L/K}=f'(\alpha)\mathcal O_L. \tag{3.2} \] This algebraic trace-dual theorem is proved in *Number fields*, lesson 14, **The different and the discriminant**, “The trace dual,” Euler's dual-basis lemma and Proposition 14.2. Its hypotheses are a Dedekind base ring and a finite separable extension with finite integral closure; the monogenic formula additionally requires the full integral closure to be \(\mathcal O_K[\alpha]\). Complete local integer rings satisfy these conditions by **Extensions of complete valued fields**, Theorem 3.1. The separable trace pairing is proved nondegenerate above, using the exact earlier trace-one lemma; [Stacks, Tag 0BIL](https://kokunoyumeto.github.io/stacks-zh-hans-cn/en/fields.html#fields-lemma-separable-trace-pairing) is a comparison. The ramification formula that follows is proved here. **Theorem 3.1 (Hilbert).** For the finite Galois extension \(L/K\), \[ d(L/K):=v_L(\mathfrak D_{L/K}) =\sum_{\sigma\ne1}i_G(\sigma) =\sum_{r\geq0}(|G_r|-1). \tag{3.3} \] Consequently a tame extension has different exponent \(e-1\). In the wild case, \[ d(L/K)\geq e+|G_1|-2\geq e+p-2. \tag{3.4} \] More precisely, \(d(L/K)=e+|G_1|-2\) if and only if \(G_2=1\), and \(d(L/K)=e+p-2\) if and only if \(G_2=1\) and \(|G_1|=p\). *Proof.* Normality and separability give \[ f(X)=\prod_{\sigma\in G}(X-\sigma\alpha), \qquad f'(\alpha)=\prod_{\sigma\ne1}(\alpha-\sigma\alpha). \] Taking valuation in (3.2) proves the first sum. For a nonidentity automorphism, \(i_G(\sigma)\) is a nonnegative integer. It belongs to exactly \(i_G(\sigma)\) of the groups indexed \(r\geq0\), since their condition is \(r+1\leq i_G(\sigma)\). Summing these finitely many indicator functions proves the second sum. The \(r=0\) summand is \(e-1\). In the tame case all subsequent groups are trivial. In the wild case, Corollary 2.2 makes \(G_1\) a nontrivial \(p\)-group, so \(|G_1|\geq p\). Retaining its contribution gives \[ d(L/K)=(e-1)+(|G_1|-1)+\sum_{r\geq2}(|G_r|-1). \] The last sum is finite and nonnegative. It vanishes exactly when \(G_2=1\), because the filtration decreases. This proves the first bound and its equality criterion. The further inequality \(|G_1|\geq p\) gives the coarser bound; equality there also requires \(|G_1|=p\). In particular, a wild extension has \(d(L/K)=e\) exactly when \(p=2\), \(|G_1|=2\) and \(G_2=1\). \(\square\) **Example 3.2 (the wild bound is sharp for every prime).** Let \(K=\mathbf F_p((t))\), and choose a root \(x\) of \[ F(X)=X^p+tX^{p-1}-t. \] Every nonleading coefficient is divisible by \(t\), and the constant coefficient is not divisible by \(t^2\). Thus \(F\) is Eisenstein. By [**Unramified and totally ramified extensions**, Theorem 5.1](NT-LOC-07.md), the field \(L=K(x)\) is totally ramified of degree \(p\), \(x\) is a uniformizer, and \(\mathcal O_L=\mathbf F_p[[t]][x]\). In particular, \(v_L(x)=1\) and \(v_L(t)=p\). Put \(y=x^{-1}\). Dividing \(F(x)=0\) by \(tx^p\) gives \[ y^p-y=t^{-1}. \] The roots of this degree-\(p\) polynomial are exactly \(y+c\), with \(c\in\mathbf F_p\). They all lie in \(L=K(y)\), are distinct, and the derivative of the polynomial is \(-1\). Since \([K(y):K]=p\), it is the minimal polynomial of \(y\). Therefore \(L/K\) is cyclic Galois, with the translations \(\sigma_c(y)=y+c\). For \(c\ne0\), the unequal valuations of \(y\) and \(c\) give \(v_L(y+c)=v_L(y)=-1\). Hence \[ \sigma_c(x)-x=\frac{-c}{y(y+c)},\qquad v_L(\sigma_c(x)-x)=2. \] The uniformizer formula (2.2) now gives \[ G_0=G_1=G,\qquad G_2=1. \] Hilbert's formula yields \[ d(L/K)=(p-1)+(p-1)=2p-2=e+p-2, \] so the coarser wild bound is attained for every prime \(p\). The monogenic derivative formula (3.2) checks the exponent independently: \[ F'(x)=-tx^{p-2},\qquad v_L(F'(x))=p+(p-2)=2p-2. \] For \(p=2\), the factor \(x^{p-2}\) is \(1\), so the same computation gives \(d=2=e\). ## 4. Passing between number fields and completions Let \(E/F\) be a finite extension of number fields. Write \(\mathfrak D_{E/F}\) for the different defined by the global trace dual. For a finite prime \(\mathfrak p\) of \(F\) and each \(\mathfrak P\mid\mathfrak p\), let \(F_{\mathfrak p}\) and \(E_{\mathfrak P}\) be the completions. **Proposition 4.1 (local exponents determine the global different).** Put \[ d_{\mathfrak P}=v_{\mathfrak P} (\mathfrak D_{E_{\mathfrak P}/F_{\mathfrak p}}). \] Then \[ \mathfrak D_{E/F}=\prod_{\mathfrak P}\mathfrak P^{d_{\mathfrak P}}. \tag{4.1} \] Only finitely many factors are nontrivial. The relative discriminant ideal has exponent \[ v_{\mathfrak p}(\mathfrak d_{E/F}) =\sum_{\mathfrak P\mid\mathfrak p} f(\mathfrak P/\mathfrak p)d_{\mathfrak P}. \tag{4.2} \] *Proof.* Set \(A=(\mathcal O_F)_{\mathfrak p}\), \(B=(\mathcal O_E)_{\mathcal O_F\setminus\mathfrak p}\), and let \(\widehat A=\mathcal O_{F_{\mathfrak p}}\). The ring \(B\) is a finite torsion-free \(A\)-module. Since \(A\) is a DVR, \(B\) is free. Ideal factorization and the Chinese remainder theorem give \[ B/\mathfrak p^nB \simeq \prod_{\mathfrak P\mid\mathfrak p} \mathcal O_E/\mathfrak P^{n e(\mathfrak P/\mathfrak p)}. \] Localizing the factors at \(\mathfrak P\) changes none of these quotients. Their inverse limits are the completed local integer rings: the subsequence of powers \(n e(\mathfrak P/\mathfrak p)\) is cofinal among all powers. Taking inverse limits, or completing a finite free \(A\)-basis, proves \[ B\otimes_A\widehat A \simeq\prod_{\mathfrak P\mid\mathfrak p}\mathcal O_{E_{\mathfrak P}}. \tag{4.3} \] Here is why the trace dual commutes with this operation. Choose an \(A\)-basis \(b_1,\ldots,b_m\) of \(B\), and let \(T=(\operatorname{Tr}_{E/F}(b_jb_k))\). It is invertible over \(F\), by separability. In coordinates, \[ B^*=T^{-1}A^m, \qquad B^*\otimes_A\widehat A=T^{-1}\widehat A^m. \] The trace of multiplication is preserved by scalar extension. The completion decomposition of **Places of number fields in extensions and the product formula** identifies it with the sum of the local traces. The dual of the product lattice (4.3) for this sum is therefore \[ B^*\otimes_A\widehat A \simeq \prod_{\mathfrak P\mid\mathfrak p} \mathcal O_{E_{\mathfrak P}}^*. \tag{4.4} \] Indeed, testing elements supported in a single factor tests each local trace condition separately. The global trace dual localizes to \(B^*\): in a finite basis its trace integrality conditions are a finite system of linear equations, and localization clears their finitely many denominators. Fractional ideals in the Dedekind ring \(B\) are invertible and become principal at each maximal ideal. Completion preserves the exponent of that principal ideal, and inversion negates its exponent. Inverting (4.4) thus shows that the exponent of the global different at \(\mathfrak P\) is exactly \(d_{\mathfrak P}\). Unique factorization of fractional ideals proves (4.1); a fractional ideal has finite support. Finally the algebraic discriminant theorem identifies \(\mathfrak d_{E/F}\) with the ideal norm of \(\mathfrak D_{E/F}\). Since the norm of \(\mathfrak P\) is \(\mathfrak p^{f(\mathfrak P/\mathfrak p)}\), taking norms in (4.1) gives (4.2). \(\square\) The proposition applies to extensions that are not Galois. Hilbert's formula computes each local exponent directly when that local extension is Galois; the monogenic derivative formula also applies in the separable nongalois case. **Corollary 4.2 (a weighted wild discriminant bound).** Let \(E/F\) be an extension of number fields and \(\mathfrak p\) a finite prime of residue characteristic \(p\). Suppose that every completion \(E_{\mathfrak P}/F_{\mathfrak p}\), for \(\mathfrak P\mid\mathfrak p\), is Galois. Write \(e_{\mathfrak P}=e(\mathfrak P/\mathfrak p)\), \(f_{\mathfrak P}=f(\mathfrak P/\mathfrak p)\), and \(G_{r,\mathfrak P}\) for its lower ramification groups. Let \(W\) be the set of wild primes \(\mathfrak P\) above \(\mathfrak p\). Then \[ \begin{aligned} v_{\mathfrak p}(\mathfrak d_{E/F}) &\geq \sum_{\mathfrak P\mid\mathfrak p} f_{\mathfrak P}(e_{\mathfrak P}-1) +\sum_{\mathfrak P\in W} f_{\mathfrak P}(|G_{1,\mathfrak P}|-1)\\ &\geq \sum_{\mathfrak P\mid\mathfrak p} f_{\mathfrak P}(e_{\mathfrak P}-1) +(p-1)\sum_{\mathfrak P\in W}f_{\mathfrak P}. \end{aligned} \tag{4.5} \] Equality in the first bound holds if and only if \(G_{2,\mathfrak P}=1\) at every wild prime above \(\mathfrak p\). Equality between the left side and the final, coarser bound holds if and only if, in addition, \(|G_{1,\mathfrak P}|=p\) at every such prime. *Proof.* Proposition 4.1 gives \(v_{\mathfrak p}(\mathfrak d_{E/F}) =\sum_{\mathfrak P\mid\mathfrak p}f_{\mathfrak P}d_{\mathfrak P}\). Theorem 3.1 gives \(d_{\mathfrak P}=e_{\mathfrak P}-1\) in the tame case, and the two bounds in (3.4) in the wild case. Multiply each by \(f_{\mathfrak P}\) and sum. Every weight is positive and every discarded summand is nonnegative, so equality is exactly the stated collection of local equality conditions. \(\square\) ## 5. The cyclotomic filtration Let \(n\geq1\), let \(\zeta\) be a primitive \(p^n\)-th root of unity, and put \(L=\mathbf Q_p(\zeta)\). **Proposition 5.1.** The extension is totally ramified of degree \[ N=(p-1)p^{n-1}, \] with uniformizer \(\pi=\zeta-1\) and Galois group \[ G=(\mathbf Z/p^n\mathbf Z)^\times, \qquad \sigma_a(\zeta)=\zeta^a. \] Its lower groups are \(G_0=G\), and \[ G_i=\{a:a\equiv1\pmod{p^k}\} \quad\text{if}\quad p^{k-1}\leq i\leq p^k-1,\quad 1\leq k\leq n. \tag{5.1} \] In particular \(G_i=1\) for \(i\geq p^{n-1}\), and \[ d(L/\mathbf Q_p)=np^n-(n+1)p^{n-1}. \tag{5.2} \] *Proof.* The cyclotomic polynomial here is \[ \Phi_{p^n}(X)=\sum_{j=0}^{p-1}X^{j p^{n-1}}. \] Modulo \(p\), \(\Phi_{p^n}(1+Y)=Y^N\); its constant coefficient over \(\mathbf Z_p\) is \(p\). Thus \(\Phi_{p^n}(1+Y)\) is Eisenstein. It proves the degree, total ramification, and uniformizer assertions, with \(\mathcal O_L=\mathbf Z_p[\zeta]\). All primitive roots \(\zeta^a\) already lie in \(L\); the polynomial is separable, so the extension is Galois and all \(N\) residue-unit exponents give distinct automorphisms. For \(a\ne1\) modulo \(p^n\), let \(k=v_p(a-1)\), so \(0\leq k1\), so \(d>e+|G_1|-2\). For \(p=2,n=2\), it gives \(|G_1|=2\) and \(G_2=1\), so \(d=e+p-2=2\). For \(p=2,n\geq3\), it gives \(|G_2|=2^{n-2}>1\), and again the first wild bound is strict. ## 6. Three quadratic examples **Example 6.1.** In \(L=\mathbf Q_2(i)\), the element \(\pi=i-1\) satisfies the Eisenstein polynomial \(Y^2+2Y+2\). Thus \(v_L(2)=2\), and conjugation gives \[ \sigma\pi-\pi=-2i,\qquad i_G(\sigma)=2. \] The only nonidentity element belongs to \(G_0,G_1\), so \[ G_0=G_1=G,\quad G_2=1,\quad d=2. \] The integral basis \(1,i\) has discriminant \(-4\), whose \(2\)-adic exponent is \(2\). **Example 6.2.** In \(L=\mathbf Q_2(\sqrt2)\), \(\pi=\sqrt2\) is an Eisenstein uniformizer. Conjugation changes it by \(-2\sqrt2\), of value \(2+1=3\). Therefore \[ G_0=G_1=G_2=G,\quad G_3=1,\quad d=3. \] The integral basis \(1,\sqrt2\) has discriminant \(8\), again with the predicted exponent. **Example 6.3.** In \(L=\mathbf Q_3(\sqrt3)\), the same difference \(-2\sqrt3\) has value \(1\), since \(2\) is a unit. Thus \[ G_0=G,\quad G_1=1,\quad d=1=e-1. \] This is tame. The discriminant of \(1,\sqrt3\) is \(12\), whose \(3\)-adic exponent is \(1\). Each residue degree is \(1\), so the local different exponent equals the base discriminant exponent, as (4.2) predicts. The wild equality case distinguishes the first two examples. For \(\mathbf Q_2(i)/\mathbf Q_2\), one has \(|G_1|=2\) and \(G_2=1\), so \(d=e+p-2=2\). For \(\mathbf Q_2(\sqrt2)/\mathbf Q_2\), the group \(G_2\) is nontrivial, so \(d=3>e+p-2=2\). The third example is tame and retains \(d=e-1=1\). ## 7. Exercises 1. Compute the lower groups of \(\mathbf Q_2(i)/\mathbf Q_2\) and verify Hilbert's formula. 2. Compute them for \(\mathbf Q_2(\sqrt2)/\mathbf Q_2\) and \(\mathbf Q_3(\sqrt3)/\mathbf Q_3\), including their differents. 3. Prove the subgroup rule \(H_i=H\cap G_i\), explaining why no normality assumption on \(H\) is needed. 4. Prove the cyclotomic filtration and different formula (5.1)–(5.2), including \(p=2\) and \(n=1\). ## 8. Complete solutions **Solution 1.** Substitute \(i=1+\pi\) into \(i^2+1=0\), obtaining \(\pi^2+2\pi+2=0\). Eisenstein gives a totally ramified quadratic extension with uniformizer \(\pi\), so \(v_L(2)=2\). Its conjugate is \(-i-1\), and the difference is \(-2i\). Since \(i\) is a unit, its value is \(2\). By (2.2), the nonidentity conjugation belongs exactly to \(G_r\) with \(r+1\leq2\). Thus \(G_{-1}=G_0=G_1=G\) and all \(G_r\), \(r\geq2\), are trivial. Hilbert's sum is \((2-1)+(2-1)=2\). Directly, \(f'(i)=2i\) generates the different and has the same value; the integral power basis discriminant is \(-4\). **Solution 2.** The Eisenstein polynomials \(X^2-2\) and \(X^2-3\) give total quadratic extensions with uniformizers their respective square roots. For the first, \(v_L(2)=2\), so conjugation moves the uniformizer by \(-2\sqrt2\), of value \(3\). The groups through index \(2\) equal the order-two group, and those from index \(3\) onward are trivial. Hilbert gives \(d=3\), also the value of \(2\sqrt2\); the basis discriminant is \(8\). For the second, \(2\) is a \(3\)-adic unit, so the conjugation difference \(-2\sqrt3\) has value \(1\). Only \(G_0\) among the groups indexed at least zero is nontrivial. Hilbert gives \(d=1\), also the value of the derivative \(2\sqrt3\). The basis discriminant is \(12\), of \(3\)-adic exponent \(1\). The index \(2\) is prime to the residue characteristic \(3\), consistent with the tame formula. **Solution 3.** The fixed-field theorem makes \(L/L^H\) finite Galois with group \(H\); its upper field and normalized valuation are still \(L,v_L\). Integral monogenicity holds over the local base \(L^H\), so the polynomial-difference proof of (1.2) expresses its ramification number as the minimum of \(v_L(\sigma x-x)\) over all \(x\in\mathcal O_L\). This is exactly the same minimum used for \(L/K\). Membership in its \(i\)-th group therefore means \(\sigma\in H\) and \(i_G(\sigma)\geq i+1\), which is \(H\cap G_i\). Normality of \(H\) is relevant to whether \(L^H/K\) is Galois, and is unnecessary for \(L/L^H\). The argument also covers real indices by the ceiling convention. **Solution 4.** Reducing \(\sum_{j=0}^{p-1}(1+Y)^{j p^{n-1}}\) modulo \(p\) gives \(Y^{(p-1)p^{n-1}}\), and its constant term is exactly \(p\). Eisenstein proves the degree \(N=(p-1)p^{n-1}\), total ramification and uniformizer \(\zeta-1\). The roots \(\zeta^a\), for unit exponents \(a\), all lie in the field, giving the stated Galois group. If \(a\ne1\) and \(k=v_p(a-1)