ALGEBRAIC EXTENSIONS OF GRADED AND VALUED FIELDS

Y.-S. Hwang^{1}

Department of Mathematics College of Science Korea University 5–1, Anam-dong, Sungbuk-ku Seoul, 136–701 Korea

                    e-mail: yhwang@semi.korea.ac.kr

and

A. R. Wadsworth^{2}

Department of Mathematics, 0112 University of California, San Diego 9500 Gilman Drive La Jolla, California 92093-0112 USA

                     e-mail: arwadsworth@ucsd.edu

If F is a field with a (Krull) valuation, then the filtration of F induced by the valuation yields an associated graded ring, which is a graded field. Conversely, if R is a graded field with totally ordered grade group, then R is an integral domain and there is a canonically associated valuation on the quotient field of R. The processes of passing from valued field to graded field and vice versa are not quite inverses of each other, but many properties in one setting are well-reflected in the other.

The goal of this paper is to describe an algebraic extension theory for graded fields analogous to what is known for valued fields, and then to spell out the correspondence between tame extensions of graded fields and Henselian valued fields. This has the benefit that graded fields are easier to work with for many purposes than valued fields. But beyond this, there is a similar correspondence between graded division rings and valued division rings, where the graded objects seem to be significantly easier to work with than the valued objects. We first learned of this correspondence from a paper by M. Boulagouaz [B_{2}]. The correspondence for division rings is actually far more extensive than what was described by Boulagouaz, and we pursue that subject in a sequel to this paper [HW]. The choice of topics to treat here was influenced by the needs of the study of division rings. But, we feel that the commutative theory presented here is of interest in its own right.

1Supported in part by the Non-directed Research Fund, Korea.

2Supported in part by the NSF.

§1 Graded fields (with totally ordered grade group)

Let (Γ, +) be an abelian group and let R = ⨁_{γ∈Γ} R_{γ} be a graded ring with

respect to Γ (i.e., R is a ring with 1, such that R_{γ} · R_{δ} ⊆ R_{γ+δ} for all γ, δ ∈ Γ). We set

Γ_{R} = supp(R) = {γ ∈ Γ | R_{γ} ≠ (0)}.

Also, let R^{h} = ⋃_{γ∈Γ} R_{γ}, the set of homogeneous elements of R. If r ∈ R_{γ}, r ≠ 0, we

write deg(r) = γ.

The graded ring R is said to be a graded field if R is commutative with 1 ≠ 0 and every nonzero homogeneous element of R is a unit. When this occurs, Γ_{R} is a subgroup of Γ, and we call it the grade group of R. We will be interested exclusively in the case where Γ_{R} is totally ordered. We adopt as a standing hypothesis throughout the paper that all the graded fields R we consider are equipped with a total ordering on Γ_{R}. (But, note that any torsion-free abelian group admits a total ordering. Therefore, all the results in §§1-3 below on extensions of graded fields hold if we merely assume that Γ_{R} is torsion-free abelian. Only when we wish to build a valuation ring on the quotient field of R will we need to specify an ordering on Γ_{R}.) A graded isomorphism R → R′ of graded fields consists of an order-preserving group isomorphism α: Γ_{R} → Γ_{R′} and a ring isomorphism β: R → R′ such that β(R_{γ}) = R′_{α(γ)} for all γ ∈ Γ_{R}. When such a graded isomorphism exists, we write R ≅_{g} R′.

It follows from the total ordering of Γ_{R} that a graded field R is an integral domain and that its group of units R^{∗} = R^{h} − {0}. Let

QR = quotient field of R.

Since Γ_{R} is totally ordered, it is a torsion-free abelian group, and we set

Δ_{R} = ℚ ⊗_{ℤ} Γ_{R}, the divisible hull of Γ_{R}.

We identify Γ_{R} with its isomorphic image in Δ_{R} (γ ↔ 1 ⊗ γ). The total ordering on Γ_{R} extends uniquely to a total ordering on Δ_{R}. Moreover, if Λ is any torsionfree abelian group containing Γ_{R} as a subgroup with Λ/Γ_{R} torsion, then there is a unique monomorphism Λ → Δ_{R} extending the embedding Γ_{R} ↪ Δ_{R}. We will thus routinely view any such Λ as a subgroup of Δ_{R}. The ordering on Γ_{R} extends uniquely to an ordering on Λ.

Since R is a graded field, it is clear that R_{0} is a field and each R_{γ} is a 1dimensional vector space over R_{0}. For each γ ∈ Γ_{R}, fix some nonzero t_{γ} ∈ R_{γ}. Then for each γ, δ ∈ Γ_{R} there is c_{γ,δ} ∈ R_{0}^{∗}, such that t_{γ}t_{δ} = c_{γ,δ}t_{γ+δ}. We call {c_{γ,δ} | γ, δ ∈ Γ_{R}} a family of structure constants of R. The commutativity and associativity of R imply

c_{γ,δ} = c_{δ,γ} and c_{γ,δ}c_{γ+δ,ρ} = c_{δ,ρ}c_{γ,δ+ρ} for all γ, δ, ρ ∈ Γ_{R}.  (1.1)

Conversely, it is clear that given any field F, any totally ordered abelian group Γ, and any function Γ×Γ → F^{∗} ((γ, δ) ↦ c_{γ,δ}) satisfying the conditions in (1.1), there is a graded field R with R_{0} = F, Γ_{R} = Γ, and structure constants given by the specified function.

We say that a graded field R is of group-ring type if there is a family of structure constants of R with c_{γ,δ} = 1 for all γ, δ ∈ Γ_{R}. Clearly, R is of group-ring type iff R is isomorphic (as a graded ring) to the group ring R_{0}[Γ].

Proposition 1.1. A graded field R is of group-ring type iff the canonical short exact sequence of abelian groups

0 → R_{0}^{∗} → R^{∗} → Γ_{R} → 0

is split exact.

Proof. Clearly, the exact sequence splits iff there is a family {t_{γ} | γ ∈ Γ_{R}} with each t_{γ} ∈ R_{γ} − {0} and t_{γ}t_{δ} = t_{γ+δ}. This is exactly what is needed so that the structure constants relative to {t_{γ}} will all be 1. □

Prop. 1.1 shows that if Γ_{R} ≅ ℤ, or Γ_{R} is any free abelian group, then the graded field R is of group-ring type (and hence determined up to isomorphism by R_{0} and Γ_{R}). But, not every graded field is of group-ring type, as the following example shows.

Example 1.2. A graded field R with R_{0} = ℚ and Γ_{R} = ℤ[\frac{1}{2}], and R not of group-ring type. Let t be an indeterminate, and let A = ℚ[t, t^{−1}], the Laurent polynomial ring in t over the rational numbers ℚ. This A is the group ring ℚ[ℤ], so a graded field, with Γ_{A} = ℤ and A_{i} = ℚt^{i}, i ∈ ℤ. Let p be any prime number. In an algebraic closure of the quotient field of A, there are y_{1}, y_{2}, … satisfying y_{1}^{2} = pt, y_{2}^{2} = p^{2}y_{1}, y_{3}^{2} = p^{4}y_{2}, …, y_{i}^{2} = p^{2^{i−1}}y_{i−1}, . . .. Let B_{1} = A[y_{1}], B_{2} = B_{1}[y_{2}], …, B_{i} = B_{i−1}[y_{i}], …, and let R = ⋃_{i=1}^{∞} B_{i}. Note that, as an A-module, B_{1} = A ⊕ y_{1}A, and B_{1} is a graded field with (B_{1})_{0} = ℚ, Γ_{B_{1}} = \frac{1}{2}ℤ, and for j ∈ ℤ, (B_{1})_{j/2} = ℚy_{1}^{j}. Also, the gradings on A and B_{1} are compatible, so A is a graded subfield of B_{1}. Proceeding inductively, we have for each i, B_{i} = B_{i−1} ⊕B_{i−1}y_{i}, B_{i} is a graded field with (B_{i})_{0} = ℚ, Γ_{B_{i}} = 2^{−i}ℤ, (B_{i})_{j/2^{i}} = ℚy_{i}^{j} for all j ∈ ℤ, and B_{i−1} is a graded subfield of B_{i}. Hence, R is a graded field with R_{0} = ℚ and Γ_{R} = ℤ[\frac{1}{2}] (the additive group of the ring ℤ[\frac{1}{2}], with the usual ordering). If R were of group-ring type, then there would exist nonzero s_{1} ∈ R_{1} = ℚt and s_{2^{−i}} ∈ R_{2^{−i}} = ℚy_{i} such that (s_{2^{−i}})^{2^{i}} = s_{1}, for all integers i > 0. However, observe that y_{i}^{2^{i}} = p^{(4^{i}−1)/3}t. Hence, s_{1} ∈ ⋂_{i=1}^{∞}(R_{2^{−i}})^{2^{i}} = ⋂_{i=1}^{∞} ℚ^{2^{i}}p^{(4^{i}−1)/3}t = (0), a contradiction. Hence, R is not of group-ring type.

Corollary 1.3. Let R be a graded field. Then R is integrally closed, and R_{0} is algebraically closed in QR.

Proof. If R were not integrally closed, this would be detected by an equation involving only finitely many elements of R, hence only finitely many homogeneous elements. Thus, there is a finitely generated subgroup Λ of Γ_{R} such that for the graded subfield A = R|_{Λ} = ⨁_{λ∈Λ} R_{λ} of R, A is not integrally closed. Note that Γ_{A} = Λ and A_{0} = R_{0}. As Λ is finitely generated and torsion-free, Λ ≅ ℤ^{n} for some n. So, Prop. 1.1 shows A ≅ A_{0}[Λ] ≅ A_{0}[x_{1}, x_{1}^{−1}, …, x_{n}, x_{n}^{−1}], where x_{1}, …, x_{n} are independent indeterminates. Since A is a localization of a polynomial ring over a field, A is integrally closed, a contradiction. Hence, R must be integrally closed. That R_{0} is algebraically closed in QR is proved in the same way, as R_{0} = A_{0}, which is algebraically closed in the rational function field QA ≅ R_{0}(x_{1}, …, x_{n}). □

Let R be a graded field. By a graded R-module, we mean an R-module M such that M has a direct sum decomposition as abelian groups M = ⨁_{μ∈Γ_{M}} M_{μ}, where Γ_{R} acts freely on the set Γ_{M}, and for γ ∈ Γ_{R}, μ ∈ Γ_{M}, we have R_{γ} · M_{μ} ⊆ M_{γ+μ}. (Here γ + μ denotes the image of μ under the action of γ.) There is no group structure assumed on Γ_{M}. That Γ_{R} acts freely on Γ_{M} means that for all γ, δ ∈ Γ_{R}, μ ∈ Γ_{M}, we have γ + μ ≠ δ + μ whenever γ ≠ δ. Observe that every graded module M over any graded field R is a free R-module, with a homogeneous base.

Indeed, any maximal R-linearly independent homogeneous subset of M is a base (cf. [B_{1}, Th. 3]). We write dim_{R}(M) for the rank of M as a free R-module. Note that if N is any graded R-submodule of M then, as M/N is a graded, hence free, R-module, we have M ≅ N ⊕ (M/N); so dim_{R}(M) = dim_{R}(N) + dim_{R}(M/N). In particular, if N is a proper graded R-submodule of M and dim_{R}(M) < ∞ then dim_{R}(N) < dim_{R}(M).

§2 Graded algebraic extensions of graded fields

Let R ⊆ S be graded fields (i.e., R is a graded subfield of S). We set [S : R] = dim_{R}(S).

Clearly, R_{0} is a subfield of S_{0}, Γ_{R} is a subgroup of Γ_{S}, and QR is a subfield of QS. The following proposition is easy but important.

Proposition 2.1. (cf. [B_{2}, p. 4278]) Let R ⊆ S be graded fields with [S : R] < ∞. Then,

[S : R] = [S_{0} : R_{0}]|Γ_{S} : Γ_{R}| = [QS : QR],

and QS ≅ QR ⊗_{R} S.

Proof. If {s_{i}}_{i∈I} is an R_{0}-base of S_{0}, and {t_{j}}_{j∈J} ⊆ S^{∗} with {deg(t_{j})} a set of representatives for the cosets of Γ_{R} in Γ_{S}. It is easy to check that {s_{i}t_{j}} is an R-base of S. This gives the first formula for [S : R], and the second follows from QS ≅ QR ⊗_{R} S. This isomorphism holds as S is an integral domain finitely generated as an R-module; so S (a torsion-free R-module) embeds in its localization QR ⊗_{R} S which is a field, since it is an integral domain finite-dimensional over the field QR. □

We will want to consider infinite degree extensions of graded fields which are algebraic in an appropriate sense. For this, we first look at gradings on the polynomial ring R[x] over a graded field R. For any δ ∈ Δ_{R}, there is a unique grading on R[x] extending that on R, such that x ∈ R[x]_{δ}. We call this the δ-grading of R[x]. An f ∈ R[x] is said to be homogenizable if there is δ ∈ Δ_{R} such that f is homogeneous with respect to the δ-grading of R[x] (cf. [B_{1}, §3], [vGvO, p. 274]). The following proposition is essentially in [vGvO, pp. 274–276] for Γ_{R} ≅ ℤ.

Proposition 2.2. Let R be a graded field which is a subring of the (ungraded) field F. For any α ∈ F, the following are equivalent:

(i) α is algebraic over QR, the minimal polynomial of α over QR, denoted m_{QR,α}, lies in R[x] and m_{QR,α} is homogenizable.

(ii) R[α] is a graded field extension of R, with α ∈ R[α]^{h} and [R[α] : R] < ∞. (iii) There is a graded field extension S of R, with R ⊆ S ⊆ F, [S : R] < ∞,

and α ∈ S^{h}.

Proof. (i) ⇒ (ii) Suppose α is algebraic over QR and that m_{QR,α} lies in R[x] and is homogeneous with respect to the δ-grading on R[x], for some δ ∈ Δ_{R}. Then (m_{QR,α}) is a homogeneous ideal R[x], so the ring R[x]/(m_{QR,α}) inherits a grading from R[x] which extends the grading on R. The canonical evaluation homomorphism R[x] → F given by f ↦ f(α) has image R[α] and kernel {f ∈ R[x] | f(α) = 0} = (m_{QR,α}QR[x]) ∩ R[x] = m_{QR,α}R[x]. (The last equality holds as m_{QR,α} is monic in R[x].) Hence, the isomorphism R[α] ≅ R[x]/(m_{QR,α}) allows us to define a grading on R[α] extending that on R, with respect to which α ∈ R[α]^{h} since α is the image of x ∈ R[x]^{h}. Also, [R(α) : R] = [R[x]/(m_{QR,α}) : R] = deg(m_{QR,α}) < ∞. Because R[α] is a graded integral domain (since it lies in F), with [R[α] : R] < ∞, R[α] is actually a graded field. (For, if β ∈ R[α]^{h} , β ≠ 0, then the map λ_{β} : R[α] → R[α] given by c ↦ βc is an injective graded R-module homomorphism. Then, as im(λ_{β}) is a graded R-submodule of R[α] with dim_{R}(im(λ_{β})) = [R(α) : R] < ∞, so λ_{β} is surjective by dimension count. (See the remarks at the end of §1.) Hence β ∈ R[α]^{∗}.

(ii) ⇒ (iii) Take S = R[α].

(iii) ⇒ (i) Assume α ∈ S^{h} for a graded field S ⊇ R, with [S : R] < ∞. Then α is integral over R, since [S : R] < ∞. So α is algebraic over QR, and m_{QR,α} ∈ R[x], as R is integrally closed by Cor. 1.3. Write m_{QR,α} = x^{n}+c_{n−1}x^{n−1}+…+c_{0}. If we take the n deg(α) homogeneous component of the equation α^{n} + c_{n−1}α^{n−1} + … + c_{0}, we get another monic polynomial in α of degree n with coefficients in R, which equals 0. The uniqueness of m_{QR,α} assures that this polynomial must coincide with m_{QR,α}. Hence, each c_{i} ∈ R_{(n−i)deg(α)}. So m_{QR,α} is homogeneous for the deg(α)-grading of R[x], and deg(α) = \frac{1}{n}deg(c_{0}) ∈ Δ_{R}. □

Definition 2.3. An α satisfying the equivalent conditions of Prop. 2.2 is said to be gr-algebraic over the graded field R. For a graded field extension S of R, we say S is gr-algebraic over R if each α ∈ S^{h} is gr-algebraic over R. (In particular, by Prop. 2.2, if [S : R] < ∞, then S is gr-algebraic over R.)

Corollary 2.4. Suppose α is gr-algebraic over a graded field R. Then, (a) m_{QR,α} determines deg(α).

(b) Γ_{R[α]} = ⟨deg(α)⟩ + Γ_{R}.

(c) deg(α) = 0 iff m_{QR,α} ∈ R_{0}[x], iff α is algebraic over R_{0}.

(d) R[α]_{0} = R_{0} iff α^{m} ∈ R, where m is the order of the image of deg(α) in Δ_{R}/Γ_{R}. This occurs iff m_{QR,α} = x^{n} − d_{0}, where d_{0} ∈ R^{h} and deg(d_{0}) has

order n in Γ_{R}/nΓ_{R}. Then n = m = [R[α] : R].

Proof. (a) The proof of Prop. 2.2 shows deg(α) = \frac{1}{n}deg(c_{0}), where n = deg(m_{QR,α}) and c_{0} is the constant term of m_{QR,α}. Part (b) holds since α ∈ R[α]^{h}. For (c), if α is algebraic over R_{0}, then α ∈ R[α]_{0} by Cor. 1.3; the rest of (c) is clear from the proof of Prop. 2.2.

(d) By (b), m = |Γ_{R[α]}/Γ_{R}|. If R[α]_{0} = R_{0}, then R = ⨁_{γ∈Γ_{R}} R[α]_{γ}, by Prop. 2.1, since these graded fields have the same grade group and the same degree 0 component. Since deg(α^{m}) ∈ Γ_{R}, this shows α^{m} ∈ R. Now, suppose we just know α^{m} ∈ R. Let c_{0} = α^{m}, so m_{QR,α} | (x^{m} − c_{0}). But, using Prop. 2.1, deg(x^{m} − c_{0}) = m = |Γ_{R[α]}/Γ_{R}| ≤ [QR(α) : QR] = deg(m_{QR,α}). Hence, m_{QR,α} = x^{m}−c_{0}. Since deg(α) has order m in Δ_{R}/Γ_{R}, we have deg(c_{0}) = m deg(α) has order m in Γ_{R}/mΓ_{R}. Now, suppose instead that m_{QR,α} = x^{n} − d_{0}, where d_{0} ∈ R^{h} and deg(d_{0}) has order n in Γ_{R}/nΓ_{R}. Then \frac{1}{n}deg(d_{0}) has order n in Δ_{R}/Γ_{R}. But deg(α) = \frac{1}{n}deg(d_{0}). Hence, m = n. Since m = |Γ_{R[α]}/Γ_{R}| and n = [R[α] : R], the equality m = n forces R[α]_{0} = R_{0}, by Prop. 2.1. □

Corollary 2.5. Let R ⊆ S be graded fields with S gr-algebraic over R. Then, (a) S is the integral closure of R in QS.

(b) QR ∩ S = R.

(c) If α ∈ S and a is gr-algebraic over R, then α ∈ S^{h}.

(d) If τ is any QR-automorphism of QS, then τ(S) = S and τ|_{S}: S → S is a graded field isomorphism.

Proof. (a) S is integral over R by Prop. 2.2, and is integrally closed by Cor. 1.3.

(b) is immediate from (a), as R is integrally closed.

(c) By Prop. 2.2 there is a grading on R[x] with Γ_{R[x]} ⊆ Δ_{R} and x ∈ R[x]^{h}, such that m_{QR,α} ∈ R[x]^{h}. This grading extends (uniquely) to a grading on S[x] which extends the grading on S. Since Γ_{S[x]} ⊆ Δ_{S}, Γ_{S[x]} is totally ordered. Hence, the factor x − α of m_{QR,α} in S[x] must be homogeneous, as m_{QR,α} ∈ S[x]^{h}. So, α ∈ S^{h}.

(d) For any QR-automorphism τ of QS, we have τ(S) = S as S is the integral closure of R in QS. Take any α ∈ S^{h}. Then m_{QR,τ(α)} = m_{QR,α}. Since α is gralgebraic over R, Prop. 2.2 shows τ(α) is also gr-algebraic over R. Then τ(α) ∈ S^{h} by (c) above, and deg(τ(α)) = deg(α) by Cor. 2.4(a). So the ring isomorphism τ|_{S} maps each homogeneous component of S onto itself. □

Corollary 2.6. Let R ⊆ S be graded fields, and let K be a field containing S. If α ∈ K is gr-algebraic over R, then α is gr-algebraic over S.

Proof. There is a grading on R[x] extending that on R with x ∈ R[x]^{h}, such that m_{QR,α} ∈ R[x]^{h}. This grading extends (uniquely) to a grading on S[x] extending that on S, with Γ_{S[x]} ⊆ Δ_{S}. Since α is integral over S, we have m_{QS,α} ∈ S[x]; then m_{QS,α} | m_{QR,α} in S[x], since this divisibility holds in QS[x], and m_{QS,α} is monic. Since m_{QR,α} ∈ S[x]^{h} and Γ_{S[x]} is totally ordered, we must have m_{QS,α} ∈ S[x]^{h}. Hence, α is gr-algebraic over S. □

The following corollary can now be proved just as in the ungraded case:

Corollary 2.7. Let R be a graded field which is a subring of the ungraded field F. Then,

(a) If α_{1}, …, α_{n} ∈ F and each α_{i} is gr-algebraic over F, then R[α_{1}, …, α_{n}] is a graded field graded algebraic over R, with

[R[α_{1}, …, α_{n}] : R] ≤ ∏_{i=1}^{n}[R[α_{i}] : R] < ∞ .

(b) If S is a graded field, R ⊆ S ⊆ F, with S gr-algebraic over R, and if α ∈ F is gr-algebraic over S, then α is gr-algebraic over R.

(c) Let A = R[{α ∈ F | α is gr-algebraic over R}]. Then A is a graded field which is gr-algebraic over R, and for every graded field S with R ⊆ S ⊆ F and S gr-algebraic over R, S is a graded subfield of A.

Definition 2.8. We call the A of Cor. 2.7(c) the graded algebraic closure of R in F.

Note that even if F is algebraic over QR, the field QA may be a proper subfield of F. The following example is given in [vGvO, Ex. 3.10.2]: Let R = K[t, t^{−1}], where K is any field of characteristic not 2, t is transcendental over K, and R is graded with R_{0} = K, Γ_{R} = ℤ, R_{i} = Kt^{i}; so QR = K(t). Let F = K(t)(\sqrt{t + 1}). For α ∈ F −QR write α = r+s\sqrt{t + 1}, with r, s ∈ QR, s ≠ 0; so, m_{QR,α} = x^{2} −2rx+(r^{2} −s^{2}(t+1)). If α is integral over R, then r, s ∈ R, but then m_{QR,α} is not homogenizable, since its constant term is not homogeneous in R. The gr-algebraic closure of R in F is in this case R itself.

Now, for any graded field R, let QR_{alg} denote the algebraic closure of QR. Let A be the graded algebraic closure of R in QR_{alg}. It is clear that for any gr-algebraic extension field S of R, S is graded R-isomorphic to a graded subfield of A. Also, from Cor. 2.7(b) it follows easily that A_{0} ≅ R_{0alg} and Γ_{A} = Δ_{R}. Also, QA is normal over QR, since for each α ∈ A^{h}, Prop. 2.2 and Cor. 2.4 show that each root of m_{QR,α} in QR_{alg} actually lies in A^{h}. However, QA need not be Galois over QR. We call A the gr-algebraic closure of R, denoted R_{gr-alg}.

§3. Unramified, totally ramified, and tame graded field extensions

We now look at some specific types of gr-algebraic extensions of graded fields. Let R ⊆ S be graded fields with S gr-algebraic over R. The torsion group Γ_{S}/Γ_{R} is called the ramification group of S over R, and |Γ_{S}/Γ_{R}| the ramification index of S over R. We say S is totally ramified over R if S_{0} = R_{0}. At the other extreme, we say S is unramified over R if Γ_{S} = Γ_{R} and S_{0} is separable over R_{0}. (This is analogous to the terminology for extensions of valued fields.) Note that every graded extension of graded fields R ⊆ S has a unique subextension A with Γ_{A} = Γ_{R} and S totally ramified over A, namely A = S_{0} ⊗_{R_{0}} R.

Remark 3.1. If Γ_{S} = Γ_{R}, then clearly S ≅_{g} S_{0}⊗_{R_{0}} R, and QS ≅ S_{0}⊗_{R_{0}} QR. For any intermediate graded field A, R ⊆ A ⊆ S, necessarily Γ_{A} = Γ_{R}, so A ≅_{g} A_{0} ⊗_{R_{0}} R. Thus, there is a one-to-one correspondence between intermediate graded fields A, R ⊆ A ⊆ S, and intermediate fields A_{0}, R_{0} ⊆ A_{0} ⊆ S_{0}. Further QS is separable (resp. purely inseparable, resp. Galois) over QR iff S_{0} is separable (resp. purely inseparable, resp. Galois) over R_{0}. For the Galois case, use Cor. 2.5; note that then G(QS/QR) ≅ G(S_{0}/R_{0}).

For any graded field R and any subgroup Λ of Γ_{R}, we write R|_{Λ} for ⨁_{γ∈Λ} R_{γ}, a graded subfield of R.

Proposition 3.2. Let R ⊆ S be graded fields with S gr-algebraic over R. Suppose S is totally ramified over R. Then,

(a) R = S|_{Γ_{R}}.

(b) Every intermediate graded field A, R ⊆ A ⊆ S has the form A = S|_{Λ}, where Λ is a group, Γ_{R} ⊆ Λ ⊆ Γ_{S}. Thus, there is a one-to-one correspondence between subgroups of Γ_{S}/Γ_{R} and intermediate graded fields.

(c) Suppose [S : R] = n < ∞, so Γ_{S}/Γ_{R} ≅ ℤ/t_{1}ℤ×…×ℤ/t_{k}ℤ with t_{1} … t_{k} = n. Pick any s_{i} ∈ S^{h} − {0} with deg(s_{i}) mapping to a generator of the i-th component in the given cyclic decomposition of Γ_{S}/Γ_{R}, 1 ≤ i ≤ k. Then, s_{i}^{t_{i}} ∈ R^{h}, [R[s_{i}] : R] = t_{i}, and S ≅_{g} R[s_{1}] ⊗_{R} … ⊗_{R} R[s_{k}]. Likewise, QS ≅ QR(s_{1}) ⊗_{QR} … ⊗_{QR} QR(s_{k}), with [QR(s_{i}) : QR] = t_{i} and s_{i}^{t_{i}} ∈ QR.

Proof. (a) is clear since (S|_{Γ_{R}})_{0} = S_{0} = R_{0} and also S|_{Γ_{R}} and R have the same grade group. (b) is immediate from (a), since for R ⊆ A ⊆ S, S is totally ramified over A, since it is totally ramified over R. For (c), take any s_{i} as described in (c). Since t_{i}deg(s_{i}) ∈ Γ_{R}, we have s_{i}^{t_{i}} ∈ S_{t_{i}deg(s_{i})} = R_{t_{i}deg(s_{i})} (see (a)). Then [R[s_{i}] : R] = t_{i} by Cor. 2.4(d). The graded ring homomorphism ρ: R[s_{i}] ⊗_{R} … ⊗_{R} R[s_{k}] → S has image a graded subfield of S with Γ_{im(ρ)} = Γ_{S}. Hence im(ρ) = S by (b). Because [S : R] = t_{i} … t_{n} = dim_{R}(R[s_{1}] ⊗_{R} … ⊗_{R} R[s_{k}]), ρ must be an isomorphism. The isomorphism for QS follows from this by Prop. 2.1. □

The Galois case deserves special attention:

Proposition 3.3. Let R ⊆ S be graded fields with [S : R] < ∞ and S totally ramified over R. Let ℓ be the exponent of Γ_{S}/Γ_{R}, and let μ_{ℓ} denote the group of all ℓ-th roots of unity in QR_{alg}. Then,

(a) QS is Galois over QR iff char(R_{0}) ∤ [S : R] and μ_{ℓ} ⊆ R_{0}.

(b) When QS is Galois over QR, there is a (well-defined) perfect pairing

Γ_{S}/Γ_{R} × G(QS/QR) → μ_{ℓ}

given by (γ + Γ_{R}, τ) ↦ a/τ(a), for any a ∈ S^{h}, a ≠ 0, with deg(a) = γ. Furthermore, QS is an ℓ-Kummer extension of QR, and if B = {b ∈ QS^{∗} | b^{ℓ} ∈ QR^{∗}}, then B/QR^{∗} ≅ Γ_{S}/Γ_{R} and there is a commutative diagram

Γ_{S}/Γ_{R} × G(QS/QR) ⟶ μ_{ℓ}
≅↓  ∥
B/QR^{∗} × G(QS/QR) ⟶ μ_{ℓ}

with the lower row the perfect Kummer pairing, given by (b QR^{∗}, τ) ↦ b/τ(b).

Proof. (a) From the description of QS in Prop. 3.2(c) as a radical extension of QR, it is clear that QS is separable over QR iff char(R_{0}) ∤ [S : R]; further, as ℓ = lcm(t, …, t_{k}) for the t_{i} of Prop. 3.2(c), QS is Galois over QR iff additionally QR contains the required roots of unity, i.e., μ_{ℓ} ⊆ QR. When this occurs, μ_{ℓ} ⊆ R_{0} by Cor. 1.3, since μ_{ℓ} is algebraic over R_{0}. Also, QS is clearly an ℓ-Kummer extension of QR.

For (b) assume now that QS is Galois over QR. It is easy to check that the map Γ_{S}/Γ_{R} × G(QS/QR) → μ_{ℓ} is a well-defined pairing of finite abelian groups.

Also, there is a well-defined group homomorphism β: Γ_{S}/Γ_{R} → B/QR^{∗} given by γ + Γ_{R} ↦ a QR^{∗} for any a ∈ S_{γ}, a ≠ 0. This β is clearly injective, and is an isomorphism since |Γ_{S}/Γ_{R}| = [S : R] = [QS : QR] = |B/QR^{∗}|, the last equality given by Kummer theory. The diagram of the proposition is clearly commutative, and its vertical lines are isomorphisms. Since the bottom row of the diagram is the perfect pairing of Kummer theory, the top row must also be a perfect pairing. □

The graded analogue to a tamely ramified extension of a valued field is of particular interest here.

Definition 3.4. Let R ⊆ S be graded fields with S gr-algebraic over R. Then S is said to be tame over R if char(R_{0}) = 0 or char(R_{0}) = p ≠ 0 and S_{0} is separable over R_{0} and Γ_{S}/Γ_{R} has no p-torsion.

Clearly, for graded fields R ⊆ A ⊆ S, S is tame over R iff S is tame over A and A is tame over R.

Proposition 3.5. (cf. [B_{1}, Th. 4]) Let R ⊆ S be a gr-algebraic extension of graded fields. Then S is tame over R iff QS is separable over QR.

Proof. Since this is clear if char(R_{0}) = 0, assume char(R_{0}) = p ≠ 0. Let I = S_{0} ⊗_{R_{0}} R. Since QI ≅ S_{0} ⊗_{R_{0}} QR, I_{0} = S_{0}, and Γ_{I} = Γ_{R}, we have I is tame over R iff I_{0} is separable over R_{0} iff QI is separable over QR. Since S is totally ramified over I, Prop. 3.2 shows that for every finite degree subextension S′ of S, QS′ is a radical extension of QI obtained by adjoining t_{i}-th roots of homogeneous elements of I, where lcm({t_{i}}) = exp(Γ_{S′}/Γ_{I}). Thus, QS is separable over QI iff each QS′ is separable over QI iff each Γ_{S′}/Γ_{I} has no p-torsion iff Γ_{S}/Γ_{I} has no p-torsion iff S is tame over I. The transitivity of tameness and of separable field extensions completes the proof. □

Lemma 3.6. Let R ⊆ S be a graded algebraic extension of graded fields with char(R_{0}) = p. There is no proper tame extension of R in S iff S_{0} is purely inseparable over R_{0} and Γ_{S}/Γ_{R} is a p-primary torsion group, iff QS is purely inseparable over QR.

Proof. In the first equivalence, ⇐ is clear. For ⇒, suppose R has no proper tame extension in S. Let A_{0} be the separable closure of R_{0} in S_{0}. Since A_{0} ⊗_{R_{0}} R is a tame extension of R in S, we must have A_{0} = R_{0}, so S_{0} is purely inseparable over R_{0}. Suppose Γ_{S}/Γ_{R} is not p-primary. Then, there is γ ∈ Γ_{S} − Γ_{R} with image of order n in Γ_{S}/Γ_{R}, where p ∤ n. Take any s ∈ S_{γ}, s ≠ 0. Then deg(s^{n}) ∈ Γ_{R}, so if we take any r ∈ R_{nγ}, r ≠ 0, we have s^{n}/r ∈ S_{0}. Because S_{0} is purely inseparable over R_{0}, (s^{n}/r)^{p^{k}} ∈ R_{0} for some k. So, (s^{p^{k}})^{n} = (s^{n}/r)^{p^{k}}r^{p^{k}} ∈ R_{p^{k}nγ}. Since deg(s^{p^{k}}) = p^{k}γ, which has order n in Γ_{S}/Γ_{R}, R[s^{p^{k}}] is a proper tame and totally ramified extension of R, by Cor. 2.4(b) and (d), a contradiction. Hence, Γ_{S}/Γ_{R} must be p-primary. This proves the first equivalence.

Let A = S_{0} ⊗_{R_{0}} R. Then, QA is purely inseparable over QR iff S_{0} is purely inseparable over R_{0} (see Remark 3.1). Also, as S is totally ramified over A and Γ_{A} = Γ_{R}, QS is purely inseparable over QA iff Γ_{S}/Γ_{R} is p-primary torsion, as QS is built from QA by radical extensions of degrees the orders of elements in Γ_{S}/Γ_{R}, by Prop. 3.2(a). □

Proposition 3.7. Let R ⊆ S be a gr-algebraic extension of graded fields. Let T = L ∩ S, where L is the separable closure of QR in QS. Then T is a graded field gr-algebraic and tame over R. Also, for any graded field A with R ⊆ A ⊆ S, A is tame over R iff A ⊆ T.

Proof. Let U be a maximal tame graded field extension of R in S. Such a U exists by Zorn’s Lemma, using the transitivity of the property of tameness. Then QU is separable over QR, by Prop. 3.5, and QS is purely inseparable over QU by Lemma 3.6. Hence, QU = L, so that by Cor. 2.5(b), U = QU ∩ S = L ∩ S = T. This shows T is a graded field tame over R. Also, QT = QU = L. If A is any graded field, R ⊆ A ⊆ S and A is tame over R, then QA is separable over QR by Prop. 3.5. Hence, by Cor. 2.5(b) again, A = QA ∩ S ⊆ L ∩ S = T. Conversely, if A ⊆ T, then A is tame over R, since T is tame over R. □

Definition 3.8. The T of Prop. 3.7 is called the tame closure of R in S. Note that by Lemma 3.6, T_{0} is the separable closure of R_{0} in S_{0}, and (if char(R_{0}) = p ≠ 0) Γ_{T} /Γ_{R} is the prime-to-p part of Γ_{S}/Γ_{R}. Also, QT is the separable closure of QR in QS, as we saw in the proof of Prop. 3.7.

Theorem 3.9. Let R be a graded field. Let T be the tame closure of R in R_{gr-alg}. Then,

(a) QT is the separable closure of QR in Q(R_{gr-alg}), so QT is Galois over QR.

(b) There is a one-to-one correspondence between the graded fields A such that R ⊆ A ⊆ R_{gr-alg} with A tame over R and the fields L with QR ⊆ L ⊆ QT. (The correspondence is given by A ↦ QA and L ↦ T ∩ L.)

Proof. (a) Since Q(R_{gr-alg}) is normal over QR, as noted after Definition 2.8, and QT is the separable closure of QR in Q(R_{gr-alg}), QT is Galois over QR (typically of infinite degree).

(b) If A is a graded field with R ⊆ A ⊆ R_{gr-alg} and A tame over R, then A ⊆ T by Prop. 3.7, so QA ⊆ QT and A = T ∩ QA, by Cor. 2.5(b). On the other hand, let L be any field with QR ⊆ L ⊆ QT. So QT is Galois over L. Let H = G(QT/L). By Cor. 2.5(d), each τ ∈ H maps T to T by a graded field isomorphism. Hence, as T ∩ L = T^{H}, the H-fixed points of T, T ∩ L is a graded integral domain, which is a graded field, since it is gr-algebraic over R. For any ℓ ∈ L, since ℓ is algebraic over QR, there is r ∈ R, r ≠ 0 with rℓ integral over R. Then, as T is the integral closure of R in QT, rℓ ∈ T ∩ L. This shows L = Q(T ∩ L), which proves the one-to-one correspondence asserted in the Theorem. □

Note one somewhat surprising consequence of the theorem just proved: If t ∈ T, then QR(t) ∩ T is a graded field containing t, so all the homogeneous components of t lie in QR(t). This requires the tameness of R[t] over R, as the next example illustrates:

Example 3.10. Let R_{0} be a field, char(R_{0}) = p ≠ 0, and let R = R_{0}[s, t, s^{−1}, t^{−1}], where s and t are algebraically independent over R_{0}. Then R is a graded field with deg(s) = (1, 0) and deg(t) = (0, 1) in Γ_{R} = ℤ × ℤ, and QR is the rational function field R_{0}(s, t). Let S = R[\sqrt[p]{s}, \sqrt[p]{t}], which is a totally ramified but not a tame gr-algebraic graded field extension of R. Let u = \sqrt[p]{s} + \sqrt[p]{t} = \sqrt[p]{s + t} ∈ S. Then, \sqrt[p]{s}, \sqrt[p]{t} ∉ QR(u), since otherwise QR(u) = QS. This cannot occur, as [QR(u) : QR] = p while [QS : QR] = p^{2}. Here, QR(u) ∩ S is not a graded subring of S since it contains u but not its homogeneous components \sqrt[p]{s} and \sqrt[p]{t}. In fact, the kind of one-to-one correspondence described in the theorem for the tame case fails dramatically here, since there are infinitely many fields L with QR ⊆ L ⊆ QS, but by Prop. 3.2(b) there are only finitely many graded fields A with R ⊆ A ⊆ S.

Theorem 3.11. Let R ⊆ S be graded fields with [S : R] < ∞. Then, (a) S is separable over R iff S is tame over R iff QS is separable over QR.

(b) S is Galois over R iff QS is Galois over QR. When this occurs, G(S/R) ≅ G(QS/QR).

(c) If S is separable over R, there is a graded field A ⊇ S with [A : S] < ∞ and A Galois over R.

Proof. (b) If S is Galois over R, with group N, then by base extension (cf. [G, p. 5, Lemma 1.11]) QS is Galois over QR, as QS ≅ QR ⊗_{R} S by Prop. 2.1. Also, the general base extension result shows G(QS/QR) ≅ N. Conversely, suppose QS is Galois over QR. Let G = G(QS/QR). By Cor. 2.5, each τ ∈ G restricts to a graded automorphism of S. Furthermore, S^{G} = S ∩ QR = R by Cor. 2.5(b). For τ ∈ G, τ ≠ id_{QS} there is t ∈ S^{h}, t ≠ 0 with τ(t) ≠ t. Since τ preserves degrees τ(t) − t ∈ S^{h} − {0} = S^{∗}. So, there is no maximal ideal M of S containing {τ(s) − s | s ∈ S}. This together with S^{G} = R shows that S is Galois over R with respect to G, by [CHR, Th. 1.3] or [G, pp. 2–3, Th. 1.6].

(a) and (c) Now, assume QS is separable over QR. Then S is tame over R by Prop. 3.5, so we can assume S lies in the tame closure T of R in R_{gr-alg}. Since QT is Galois over QR by Th. 3.9, the normal closure L of QS over QR lies in QT. Furthermore, L is Galois over QR and [L : QR] < ∞. Let A = T ∩ L, which is a graded subfield of T with QA = L, by Th. 3.9, and [A : R] = [L : QR] < ∞ by Prop. 2.1. So, A is Galois over R with respect to (the restriction to A of) G(L/QR), as we just proved. Let H = G(L/QS) ⊆ G(L/QR). Since A^{H} = A ∩ QS = S by Cor. 2.5(b), S is separable over R by [CHR, Th. 2.2] or [G, p. 7, Th. 2.2]. On the other hand, if S is separable over R, then QS ≅ QR ⊗_{R} S is separable over QR as separability is preserved under base extension. By Prop. 3.5 this is equivalent to S being tame over R. This yields (a), and also (c) along the way. □

§4. Tame extensions of valued fields

We now recall some facts about tame extensions of valued fields, which will be needed for our comparison of valued fields and graded fields in the next section. Everything we mention in this section is known, but a concise summary seems worthwhile.

Let F be a field, Γ a totally ordered abelian group, and v: F^{∗} → Γ a (Krull) valuation on F. (That is, (i) v(αβ) = v(α)+v(β), and (ii) v(α+β) ≥ min(v(α), v(β)) for all α, β ∈ F^{∗} (with β ≠ −α in (ii)).) Let Γ_{F} = im(v), the value group of v; let V_{F} be the valuation ring of v; M_{F} the unique maximal ideal of V_{F}; and \overline{F} = V_{F} /M_{F}, the residue field of v. The indexing by F will cause no confusion, because we will never consider more than one valuation at a time on any given field. Let \overline{p} = char(\overline{F}).

Now, let F ⊆ K be fields with K algebraic over F, and let v be a valuation on F which has a unique extension (also called v) to K. If [K : F] < ∞, we say K is tame over F (with respect to v) if \overline{K} is separable over \overline{F}, \overline{p} ∤ |Γ_{K} : Γ_{F} |, and [K : F] = [\overline{K} : \overline{F}]|Γ_{K} : Γ_{F} |. In the terminology of Endler [E, pp. 178-180], K is tamely ramified and defectless over F. If [K : F] = ∞, we say that K is tame over F if for every field L with F ⊆ L ⊆ K and [L : F] < ∞ we have L is tame over F (with respect to the restriction of v from K to L). Recall that if K is tame over F, then K is separable over F. This follows at once from the fact that if L is purely inseparable over F, then \overline{L} is purely inseparable over \overline{F} and Γ_{L}/Γ_{F} is a \overline{p}-group (cf. [E, Ex. III-6, pp. 228-229]). Note also that if [K : F] < ∞ and N is any field with F ⊆ N ⊆ K, then K is tame over F iff N is tame over F and K is tame over N. This is immediate from the fundamental inequality [E, (13.10)].

Still assuming v has a unique extension from F to K, suppose [K : F] < ∞ and K is Galois over F. Let G be the Galois group G(K/F). Let G^{V} = {σ ∈ G | σ(c)/c ∈ M_{K} for all c ∈ K^{∗}} the ramification subgroup (= Verzweigungsgruppe) of v, and let F^{V} be the fixed field K^{G^{V}}, which is called the ramification field of K over F. Then it is known (cf. [E, (20.11), (19.12), (20.20), (19.14)]) that G^{V} is a \overline{p}-group (if \overline{p} = 0 this means |G^{V} | = 1), G^{V} is a normal subgroup of G, so F^{V} is Galois over F, and that F^{V} is tame over F (since F^{V} is tame over the inertia field F^{T} and F^{T} is tame over F), and that \overline{K} purely inseparable over \overline{F^{V}} and Γ_{K}/Γ_{F^{V}} is a \overline{p}-group (so Γ_{F^{V}} = Γ_{K} if \overline{p} = 0). Hence, F^{V} can have no proper tame extension in K, so F^{V} is a maximal tame extension of F in K. For any field L with F ⊆ L ⊆ K, let H be the corresponding group G(K/L). Since H^{V} = G^{V} ∩ H, we have L^{V} = L · F^{V}. If L is tame over F, then L^{V} is also tame over F, by the transitivity of tameness, so L^{V} = F^{V}, and hence L ⊆ F^{V}. Thus, F^{V} is the unique maximal tame extension of F, and an intermediate field L is tame over F iff L ⊆ F^{V}. It follows that if intermediate fields L_{1}, L_{2} are each tame over F, then L_{1} · L_{2} is tame over F. Thus, if N is any intermediate field, then there is a unique maximal tame extension N_{0} of F inside N, and we have N_{0} = F^{V} ∩ N. We call N_{0} the tame closure of F in N.

All of the preceding paragraph extends readily to the case of infinite degree algebraic extensions (see [E, (20.12), (20.17), (20.18) ]) and leads to the following in the Henselian case:

Suppose F is a field with a Henselian valuation v. We work in some fixed algebraic closure F_{alg} of F. That v is Henselian means that there is a unique extension of v to F_{alg}.

Proposition 4.1. Suppose a field F has a Henselian valuation v. Then,

(a) There is a unique maximal tame extension F_{t} of F in F_{alg}, and F_{t} is the compositum of all the finite degree tame extensions of F in F_{alg}.

(b) \overline{F_{t}} = (\overline{F})_{sep}, Γ_{F_{t}} = Γ_{F} ⊗_{ℤ} ℤ_{(\overline{p})} (if \overline{p} = 0, then Γ_{F_{t}} = Γ_{F} ⊗_{ℤ} ℚ).

(c) For any field L, F ⊆ L ⊆ F_{alg}, we have L is tame over F iff L ⊆ F_{t}.

(d) For any field N, F ⊆ N ⊆ F_{alg}, we have N_{t} = F_{t} · N, and N ∩ F_{t} is the unique maximal tame extension of F in N.

Proof. Since any tame extension of F is separable over F, this follows easily from the preceding discussion, when we take F_{t} = F^{V}, the ramification field with respect to the (usually infinite degree) Galois extension F_{sep} over F, where F_{sep} is the separable closure of F in F_{alg}. □

Definition 4.2. We call the F_{t} of Prop. 4.1 the tame closure of F and N ∩ F_{t} the tame closure of F in N. Note that \overline{N ∩ F_{t}} is the separable closure of \overline{F} in \overline{N}, and Γ_{N∩F_{t}}/Γ_{F} is the prime-to-\overline{p} part of Γ_{N}/Γ_{F}. Also, [N : N ∩ F_{t}] is a power of \overline{p} (N = N ∩ F_{t} if \overline{p} = 0).

§5 Correspondence between graded fields and valued fields

We now recall how valued fields can be obtained from graded fields, and vice versa. We will then prove correspondences for the tame extensions of each kind of field.

Let R = ⨁_{γ∈Γ_{R}} R_{γ} be a graded field (with Γ_{R} totally ordered, as we are always

assuming). Define a function v: R − {0} → Γ_{R} by

v( ∑_{γ} r_{γ}) = the least δ such that r_{δ} ≠ 0.

So, for a, a′ ∈ R − {0}, we clearly have (i) v(aa′) = v(a) + v(a′); (ii) v(a + a′) ≥ min(v(a), v(a′)), if a′ ≠ −a.

This function v extends canonically to a function v: QR − {0} → Γ_{R} by v(rs^{−1}) = v(r)−v(s), for r, s ∈ R−{0}. Property (i) shows that v is well-defined on QR−{0}. Properties (i) and (ii) hold for all a, a′ ∈ QR−{0}. Thus v is a valuation on QR, for which clearly Γ_{QR} = Γ_{R}. Also, for the residue field, \overline{QR} = R_{0}. For, the canonical injection R_{0} → \overline{QR} is onto, since if v(rs^{−1}) = 0, then v(r) = v(s), so r = r_{γ} + higher degree terms, s = s_{γ} + higher degree terms, for γ = v(r) = v(s)), yielding \overline{rs^{−1}} = \overline{r_{γ}s_{γ}^{−1}} ∈ im(R_{0}).

Let HR denote the Henselization of QR with respect to v (see, e.g. [E, p. 131] for the Henselization of a valued field). If S is a gr-algebraic extension graded field of R, then QS ⊇ QR canonically, and while HS is determined only up to isomorphism in QS_{alg}, we will assume HS has been chosen to be QS · HR.

Now, suppose instead we start with a field F with a valuation v on F. For each γ ∈ Γ_{F}, define

W^{γ} = {c ∈ F^{∗} | v(c) ≥ γ} ∪ {0}
W^{>γ} = {c ∈ F^{∗} | v(c) > γ} ∪ {0}.

Then set R_{γ} = W^{γ}/W^{>γ}. Define multiplication R_{γ} × R_{δ} → R_{γ+δ} by, for a ∈ W^{γ} and b ∈ W^{δ}: (a + W^{>γ}) · (b + W^{>δ}) = ab + W^{>γ+δ}. This is well-defined, and extends to a multiplication on all of ⨁_{γ∈Γ_{F}} R_{γ}, making it a graded field. We denote

this graded field by GF. Clearly, Γ_{GF} = Γ_{F} and GF_{0} = \overline{F}. Note that if K is a field with a valuation extending v on F, then GK is a graded field extension of GF.

If we start with a graded field R, and build the valuation on QR as described above, then form the associated graded field GQR, then GQR ≅_{g} R, canonically. For, each R_{γ} maps bijectively onto (GQR)_{γ}. Likewise, GHR ≅_{g} R canonically. On the other hand, if we start with a valued field F, we need not have QGF ≅ F, nor HGF ≅ F, even if the valuation on F is Henselian. (For example, char(QGF) = char(\overline{F}), which need not equal char(F). Also, if \overline{F} and Γ_{F} are countable, then QGF is countable, though F might be uncountable.)

Proposition 5.1. Let R be a graded field, and let S be the tame closure of R in R_{gr-alg}. Then HS is the maximal tame extension of HR, and G(HS/HR) ≅ G(QS/QR). Hence, there are one-to-one correspondences

tame gr-algebraic graded field extensions of R in R_{gr-alg} ↔ field extensions of QR in QS ↔ tame field extensions of HR in (HR)_{alg}.

Proof. Let T be a graded field with R ⊆ T ⊆ S and [T : R] < ∞. So, T is tame over R by Prop. 3.7. Then, \overline{QT} = T_{0} is separable over \overline{QR} = R_{0}, and

[QT : QR] = [T : R] = [T_{0} : R_{0}]|Γ_{T} : Γ_{R}| = [\overline{QT} : \overline{QR}]|Γ_{QT} : Γ_{QR}|;

hence QT is a tame valued field extension of QR. Since by convention HT = QT · HR, we have

[HT : HR] ≤ [QT : QR] = [\overline{QT} : \overline{QR}]|Γ_{QT} : Γ_{QR}|
= [\overline{HT} : \overline{HR}]|Γ_{HT} : Γ_{HR}| ≤ [HT : HR] ;

the last inequality is the fundamental inequality for valued field extensions [E, (13.10)]. So, equality holds throughout. Since \overline{HT} = T_{0} is separable over \overline{HR} = R_{0}, this shows HT is tame over HR. Also, the equality just proved shows QT and HR are linearly disjoint over QR. Since S is the union of such finite-degree extensions as T, HS is tame over HR, and QS and HR are linearly disjoint over QR. Furthermore, as \overline{HS} = S_{0} is separably closed and (Γ_{HS} ⊗_{ℤ} ℚ)/Γ_{HS} = Δ_{R}/Γ_{S} is p-primary where p = char(R_{0}) = char(\overline{HS}), HS can have no proper tame field extensions. Hence, HS is the maximal tame extension of the valued field HR. From the linear disjointness, we have G(HS/HR) = G(QS · HR/HR) ≅ G(QS/QR).

We have seen in Th. 3.9(b) the one-to-one correspondence between tame gralgebraic graded field extensions of R and field extensions of QR in QS. The isomorphism of Galois groups gives the one-to-one correspondence between field extensions of QR in QS and the field extensions of HR in HS. The latter are precisely the tame field extensions of HR in (HR)_{alg}. □

Theorem 5.2. Let F be a field with Henselian valuation v, and let K be the maximal tame extension of F in F_{alg} re v. Then GK is the tame closure of GF in GF_{gr-alg}, and G(QGK/QGF) ≅ G(K/F). There are one-to-one correspondences:

tame field extensions of F in F_{alg} re v ↔ tame graded field extensions of GF in (GF)_{gr-alg} ↔ field extensions of QGF in QGK.

Proof. Let \overline{p} = char(\overline{F}) = char(GF). Since GK_{0} = \overline{K} = \overline{F}_{sep} = (GF_{0})_{sep} and Γ_{GK}/Γ_{GF} = Γ_{K}/Γ_{F}, which is the prime-to-\overline{p} part of Δ_{GF} /Γ_{GF}, GK is tame over GF and has no proper tame extensions. Hence, GK is the tame closure of GF.

We define a homomorphism α: G(K/F) → G(QGK/QGF) as follows: For σ ∈ G(K/F), σ(V_{K}) = V_{K} since V_{K} is the unique extension of the Henselian valuation ring V_{F} to K. Moreover, as Γ_{K}/Γ_{F} is a torsion group, σ must induce the identity automorphism on Γ_{K}. Hence, σ(W_{K}^{γ}) = W_{K}^{γ} and σ(W_{K}^{>γ}) = W_{K}^{>γ} for each γ ∈ Γ_{K}. Thus, σ induces a graded ring isomorphism σ′: GK → GK which is the identity on GF. This isomorphism extends to an isomorphism σ̃: QGK → QGK of the quotient field. Define α(σ) = σ̃ ∈ G(QGK/QGF).

To show that α is an isomorphism, we proceed in stages. First, let I be the maximal unramified extension of F in K, re v. Then \overline{I} = \overline{F}_{sep} and Γ_{I} = Γ_{F}, and I is Galois over F with G(I/F) ≅ G(\overline{I}/\overline{F}) canonically. So, GI is the maximal unramified graded field extension of GF, since GI_{0} = \overline{I} = (\overline{F})_{sep} = (GF_{0})_{sep} and Γ_{GI} = Γ_{I} = Γ_{F} = Γ_{GF}. Hence, QGI ≅ GI_{0} ⊗_{GF_{0}} QGF, by Remark 3.1, showing that QGI is Galois over QGF, with

G(QGI/QGF) ≅ G(GI_{0}/GF_{0}) ≅ G(\overline{F}_{sep}/\overline{F}) ≅ G(I/F) .

The inverse of this isomorphism corresponds to the mapping induced by α.

The field extension K/I is tame and totally ramified. Let Ω = {ω ∈ I_{sep} | ω^{n} = 1 for some n ∈ ℕ with \overline{p} ∤ n\}. Since V_{I} is Henselian and \overline{I} is separably closed, Ω ⊆ I and Ω maps injectively to \overline{I}, let Ω′ be the image of Ω in \overline{I} = GI_{0}. Let L be any finite-degree field extension of I in K. Then, L is tame and totally ramified over I. Because further V_{I} is Henselian and I contains all ℓ-th roots of unity for ℓ = [L : I], L is an ℓ-Kummer extension of I (cf. [S, p. 64, Th. 3] or [TW, Prop. 1.4(iii)]). Likewise, QGL is an ℓ-Kummer extension of QGI, by Prop. 3.3. Moreover, by [E, (20.11)] or [TW, Prop. 1.4(i)] there is a perfect pairing Γ_{L}/Γ_{I} × G(L/I) → Ω′ given by (γ +Γ_{I}, σ) ↦ \overline{a/σ(a)} for any a ∈ L with v(a) = γ. With respect to the canonical isomorphism Γ_{L}/Γ_{I} → Γ_{GL}/Γ_{GI} and the map G(L/I) → G(QGL/QGI) induced by α, the following diagram is evidently commutative,

Γ_{L}/Γ_{I} × G(L/I)  ⟶ Ω′
↓  ∥
Γ_{GL}/Γ_{GI} × G(QGL/QGI) ⟶ Ω′

where the bottom row is the perfect pairing of Prop. 3.3(b). Because both rows of the diagram are perfect pairings, the map G(L/I) → G(QGL/QGI) must be an isomorphism. Since K is the union of fields such as L and QGK is the union of the corresponding fields QGL, the restriction of α mapping G(K/I) to G(QGK/QGI) is an isomorphism. Thus, we have a commutative diagram

1 ⟶  G(K/I)  ⟶  G(K/F)  ⟶  G(I/F)  ⟶ 1
≅↓  ^{α}↓  ≅↓
1 ⟶ G(QGK/QGI) ⟶ G(QGK/QGF) ⟶ G(QGI/QGF) ⟶ 1

The 5-lemma shows that α is an isomorphism. The isomorphism α of Galois groups gives the one-to-one correspondence between tame field extensions of F in F_{alg} (i.e., subfields of K) and field extensions of QGF in QGK. The correspondence between these fields and the tame graded field extensions of GF in GF_{gr-alg} was given in Th. 3.9. □

In the setting of Th. 5.2, it was shown in [B_{1}, Th. 5] that if L is a field extension of F with [L : F] = n < ∞, then L is tame over F iff QGL is separable over QGF with [QGL : QGF] = n.

Corollary 5.3. Let K be the maximal tame extension of a Henselian valued field F. Then HGK is the maximal tame extension of HGF, and G(HGK/HGF) ≅ G(K/F).

Proof. This is immediate from Prop. 5.1 and Th. 5.2. □

References

[B_{1}] M. Boulagouaz, The graded and tame extensions, pp. 27–40 in Commutative Ring Theory (Fès, 1992) (P. J. Cahen, et. al., eds.), Lecture Notes in Pure and Applied Math., No. 153, Marcel Dekker, New York, 1994.

[B_{2}] M. Boulagouaz, Le graduè d’une algèbre à division valuée, Comm. Algebra, 23 (1995), 4275–4300.

[CHR] S. U. Chase, D. K. Harrison, and A. Rosenberg, Galois Theory and Galois Cohomology of Commutative Rings, Mem. Amer. Math. Soc., No. 52, Amer. Math. Soc., Providence, R.I., 1965.

[E] O. Endler, Valuation Theory, Springer, New York, 1972.

[G] C. Greither, Cyclic Galois Extensions of Commutative Rings, Lecture Notes in Math., No. 1534, Springer, Berlin, 1992.

[HW] Y.-S. Hwang and A. R. Wadsworth, Correspondences between valued division algebras and graded division algebras, preprint, UCSD, 1997.

[S] O. F. G. Schilling, The Theory of Valuations, Math. Surveys, No. 4, Amer. Math. Soc., Providence, R.I., 1950.

[TW] J.-P. Tignol and A. R. Wadsworth, Totally ramified valuations on finitedimensional division algebras, Trans. Amer. Math. Soc., 302 (1987), 223– 249.

[vGvO] J. van Geel and F. van Oystaeyen, About graded fields, Indag. Math., 43 (1981), 273–286.