HENSELIAN IMPLIES LARGE
FLORIAN POP
Abstract. In this note we show that the quotient field of a domain which is Henselian
with respect to a non-trivial ideal is a large field, and give some applications of this fact,
using a specialization theorem for ramified covers of the line over (generalized) Krull fields.
1. Introduction
For a field K, let K(t) be the rational function field in t over K, and prt : GK(t) → GK
the corresponding canonical surjective projection between the corresponding absolute Galois
groups. Every finite split embedding problem EP = (γ : GK → A, α : B → A) for GK gives
rise to EPt := (γ ◦ prt : GK(t) → A, α : B → A), which is a finite split embedding problem
for GK(t) . The following are two main open (and equivalent) problems in Galois Theory:
Problem∞ . Let K is an arbitrary Hilbertian field. Then every finite split embedding
problem EP for GK has proper solutions.
Problem0 . Let K be an arbitrary field. Then for every finite split embedding problem EP
for GK , the corresponding EPt for GK(t) has proper solutions.
Positive answers to the above Problems would imply —among other things, the Inverse
Galois Problem and the Shafarevich Conjecture on the freeness of the kernel of the cyclotomic character. The above Problems have positive solutions over fields K which are large
fields, see e.g. [P], Main Theorems A and B. The large fields were introduced in loc.cit., and
proved to be the “right class” of fields over which one can do a lot of interesting mathematics, like (inverse) Galois theory, see e.g. Colliot-Thélène [CT], Pop [P], and the survey
article Harbater [Ha1], study torsors of finite groups Moret-Bailly [MB], study rationally
connected varieties Kollár [Ko], study the elementary theory of function fields Poonen–
Pop [P–P], etc.1 Recall that a field K is called a large field, if every smooth K-curve C
satisfies: If C(K) is non-empty, then C(K) is infinite. Examples of large fields are the PAC
fields, the complete fields like k((x)), the real/p-adically closed fields, and more general, the
Henselian valued fields, the p-fields, etc. See Pop [P] for more about large fields.
In recent years, Harbater–Stevenson solved Problem∞ over K = k((x, y)) in a stronger
form, see [H–S], Theorem 1.1, by showing that every non-trivial finite split embedding problem for GK has |K| distinct proper solutions. And very recently, Paran [Pa] solved Problem0
over K = Quot(R), where R is a complete Noetherian local ring (satisfying some further
1991 Mathematics Subject Classification. Primary 12E, 12F, 12G, 12J; Secondary 12E30, 12F10, 12G99.
Key words and phrases. Embedding problems, Henselian pairs, Hilbertian fields, large fields.
Supported by NSF grants DMS-0401056 and DMS 0801144.
1Maybe this is the reason why the “large fields” acquired several other names —google it: ample, AMPLE,
épais, fertile, weite Körper, anti-Mordellic, etc.
1
technical conditions). The methods of proof in both cases are ingenious and quite technical.
These results also seemed to give further new evidence for the fact that the Problems above
can be solved in general, as it was generally believed that the above fields K = k((x, y)),
and more general K = Quot(R) with R complete Noetherian local and Krull.dim(R) > 1,
were not large fields. Note that these fields are definitely not Henselian valued fields!
The first point of this short note is to prove that actually K = k((x, y)), and more
generally, K = Quot(R) with R a complete Noetherian ring, are large fields, and that the
class of large fields is much richer than previously believed. In particular, one can deduce
Paran [Pa] from the already known fact that Problem0 has a positive answer over large base
fields K. Second, I give a lower bound for the number of distinct solutions of a non-trivial
finite split embedding problem over a Hilbertian large field, a result which represents a wide
extension of Harbater–Stevenson [H–S]. Finally, using these results, one can generalize
Harbater’s result [Ha2], Theorem 4.6, see Theorem 1.3 below, thus giving new evidence for
(a stronger form of) Bogomolov’s Freeness Conjecture as presented in Positselski [Ps].
In order to announce the results of this note, let first recall that a commutative ring R
with identity is said to be Henselian with respect to an ideal a, or that R, a is a Henselian
pair, if denoting R := R/a and R[X] → R[X], f (X) 7→ f (X) the reduction map (mod a),
for every polynomial f (X) ∈ R[X] the following holds: If a ∈ R is a root of f (X) such
0
×
that f (a) ∈ R , then there exists a lifting a ∈ R of a such that f (a) = 0 and f 0 (a) ∈ R× .
Intuitively, this means that “simple roots” of f (X) lift to “simple roots” of f (X). See [Lf]
for basic facts about Henselian rings. The following remarks are in place here:
1) a-adically complete rings with a 6= (0), like the power series rings R = R0 [[x1 , . . . , xn ]]
where R0 is a domain and a = (x1 , . . . , xn ), are Henselian with respect to a.
2) If K is a Henselian field with respect to a valuation v, and Rv , mv are the corresponding
valuation ring and valuation ideal, then Rv , mv is a Henselian pair.
3) Nevertheless, if R, a is a Henselian pair, then K = Quot(R) is in general not a Henselian
valued field. This happens for instance if R is Noetherian and Krull.dim(R) > 1.
The generalization of Paran [Pa] is the following:
Theorem 1.1. Let R be a domain which is Henselian with respect to some ideal a 6= (0).
Then K = Quot(R) is a large field. In particular, Problem0 has a positive answer over K.
The generalization of Harbater–Stevenson [H–S], Theorem 1.1, is as follows: Let us
denote R = k[[x, y]] and K = k((x, y)) := Quot(R). First, K is a large field by Theorem 1.1
above, and K is Hilbertian by Weissauer’s [W] Theorem 7.2, applied to the Krull domain R.
Second, the set of discrete valuations V := {vp | p ∈ Spec(R), p minimal non-zero} satisfies:
i) The set Va := {v ∈ V | v(a) 6= 0} is finite for every a ∈ K × .
ii) If L|K is finite Galois, then VL|K := {v ∈ V | v totally split in L|K} has cardinality
|VL|K | = |K|, see e.g., Theorem 3.4.
A field K endowed with a set V of discrete valuations satisfying i), ii), is called a Krull field.
The point is that the property of K = k((x, y)) being a Hilbertian large Krull field implies
an even stronger/more precise result than [H–S], Theorem 1.1, as follows (see Section 4 for
definitions and Theorem 4.3, which strengthens and proves Theorem 1.2 below):
Theorem 1.2. Let K be a Hilbertian large Krull field. Then every non-trivial finite split
embedding problem for GK has |K| independent and totally ramified proper solutions.
2
Finally, the generalization of Harbater [Ha2], Theorem 4.6, is the following:
Theorem 1.3. Let R, m be an excellent two dimensional Henselian local ring with separably
closed residue field k such that K := Quot(R) has char(K) = char(k). If |k| < |R|, suppose
that there exists x ∈ m such that k[[x]] ⊂ R. Then GK ab is profinite free on |K ab | generators.
2. Proof of Theorem 1.1.
Let C → K be an integral curve over K with x ∈ C(K) a smooth K-rational point.
We show that actually |C(K)| = |K|, thus in particular, C(K) is infinite. Since any two
birationally equivalent curves have isomorphic Zariski open subsets, it is sufficient to prove
the above assertion for any particular K-curve which is K-birationally equivalent to C and
has a smooth K-rational point. Thus by general algebraic geometry non-sense, without loss
of generality, we can suppose the following: C ⊂ A2K , and x ∈ C(K) is the origin of A2K ,
and C = V (f ) is defined by an irreducible polynomial f (X1 , X2 ) ∈ K[X1 , X2 ] of the form
f (X1 , X2 ) = δX2 + fe, where δ 6= 0, and fe is a polynomial in X1 , X2 with vanishing terms
in degrees < 2. Moreover, since K = Quot(R) is the field of fractions of R, after “clearing
denominators”, we can suppose that f ∈ R[X1 , X2 ], hence δ ∈ R, δ 6= 0.
Let us consider the “change of variables” X1 = δY1 , X2 = δY2 . Then in the new variables
Y1 , Y2 the curve C is defined by g(Y1 , Y2 ) = 0, where
g(Y1 , Y2 ) = f (δY1 , δY2 ) = δ 2 Y2 + fe(δY1 , δY2 ) = δ 2 [ Y2 + ge(Y1 , Y2 ) ]
with ge ∈ R[Y1 , Y2 ] having vanishing terms in degrees < 2. Equivalently, the K-curve C is
defined in the (Y1 , Y2 )-affine plane by h(Y1 , Y2 ) := Y2 + ge(Y1 , Y2 ) = 0. And remark that
h(Y1 , Y2 ) = 0 defines a model, say Ch , of C over R, and the projection on the Y1 affine line
Ch = Spec R[Y1 , Y2 ]/(h) → A1R
is smooth in a neighborhood of the origin of A2R viewed as an R-rational point of Ch .
Coming back to the proof of Theorem 1.1, suppose that in the above context, R is Henselian
with respect to a. For every a ∈ a, let us set ha (X) := h(a, X). Then by the definition of h
and ha we get: ha (0) ∈ a, and h0a (0) ∈ 1 + a. Thus viewing this mod a, we get: 0 is a simple
root of ha ∈ R[X]. Since R is Henselian with respect to a 6= (0), there exists a (unique) b ∈ a
such that ha (b) = 0. Equivalently, h(a, b) = 0, i.e., (a, b) defines a K-rational point of C.
Moreover, the set of rational points of this form is in bijection with a. Thus since |a| = |R|,
and |R| = |K|, it follows that C(K) has cardinality |K|; in particular C(K) is in infinite,
and K is large. This concludes the proof to Theorem 1.1.
Note that in the first part of the argument above we did not use the fact that R is Henselian
with respect to a, and the above argument can be generalized to arbitrary dimensions:
Proposition 2.1. Let K = Quot(R) be the quotient field of some infinite domain, and
X → K be an integral d-dimensional K-variety with a smooth K-rational point x ∈ X(K).
which contains the origin,
Then X is birationally equivalent to a K-hypersurface H ⊂ Ad+1
K
and such that H is defined over R, and the projection on the first d-coordinates H → AdR is
smooth in a Zariski neighborhood of the origin viewed as an R-rational point of H.
Moreover, if R is Henselian with respect to an ideal a 6= (0), then the image of the canonical
projection H(R) → Ad (R) = Rd contains ad .
3
3. Two basic Facts
A) Totally split primes/valuations
Notations 3.1. Let R be a domain, K := Quot(R), and L|K a finite Galois extension. Let
S ⊂ L a finite Gal(L|K)-invariant R-subalgebra such that L = Quot(S) and R = S ∩ K.
1) We denote by θ ∈ S a generator of L|K with minimal polynomial pθ (X) ∈ R[X], say
pθ (X) = X n + an−1 X n−1 + · · · + a0 , and discriminant δθ ∈ R.
2) By general Hilbert decomposition theory, the following are equivalent:
a) p ∈ Spec(R) is totally split in S|R.
b) There exists θ as above such that δθ 6∈ p and pθ (X) has a root in Rp .
c) There exists θ as above and α1 , . . . , αn ∈ Rp with αi 6≡ αj (mod pp ) such that
Q
p̃θ (X) := µ (X − αµ ) ≡ pθ (X) (mod pp ).
3) A way to generate the above context is as follows: Let θ ∈ S and pθ ∈ R[X] be as at
point 1) above. Set h(T, U ) = T n + an−1 T n−1 U + · · · + a1 T U n−1 + a0 U n . Then for every
r, s ∈ R, s 6= 0, one has sn pθ (r/s) = h(r, s). And if p ∈ Spec(R) satisfies: s, δθ 6∈ p and
h(r, s) ∈ p, then p is totally split in L|K.
We will apply the remarks above to get a lower bound for the cardinality of the set of
totally split prime ideals in L|K as follows:
4) Let m ∈ Spec R, κ ⊂ R a system of representatives for R/m, and κ•P
:= κ\m. For
a fixed non-zero x ∈ m, we say that a formal series of the form E(X) := n n X n with
n ∈ κ isP
x-adically convergent in R iff there exists rE(X) ∈ R such that for all n > 0 one has:
ν
n
rE(X)
P − nν<n ν xP∈ x R;n and if so, we say that rE(X) is an x-adic limit of E(X). Note that
if n n X and n ηn X are x-adically convergent series as above having a common limit
r ∈ R, then n = ηn for all n (proof by induction on n). In particular, if Eκ (X) is the set of
all the x-adically convergent series E(X) in R, and Eκ (x) ⊆ R contains exactly one x-adic
limit rE(X) for each E(X) ∈ Eκ (X), then Eκ (X) → Eκ (x), E(X) 7→ rE(X) , is one-to-one.
Therefore we have |Eκ (X)| ≤ |R|.
5) Let P be a set of prime ideals p ⊂ m, p 6= m, of R such that P(x) := {p ∈ P | x ∈ p } is
non-empty for every x ∈ m. Finally let us denote PL|K := {p ∈ P | p totally split in L|K }.
/ , where
Lemma 3.2. In the above Notations 3.1, let r, x ∈ m satisfy P(r) ∩ P(ax) = O
a := a0 δθ . Let Σ = Σm
⊂
R
be
an
infinite
subset
satisfying
the
following
conditions:
a,r,x
/ for all u ∈ Σ.
i) P(u) ∩ P(arx) = O
ii) P(u − v) ⊆ P(x) for all distinct u, v ∈ Σ.
Then PL|K has cardinality |PL|K | ≥ |Σ|.
Proof. Since h(T, U ) = T n + an−1 T n−1 U + · · · + a0 U n ∈ R[T, U ], and r, x ∈ m, we must
have h(ru, ax) ∈ m for all u ∈ Σ. Hence by the hypotheses on P, there exists pu ∈ P such
that h(ru, ax) ∈ pu . We first claim that r, u, a, x 6∈ pu . Indeed, since h(T, U ) ∈ R[T, U ],
and h(ru, ax) ∈ pu , we have: If ax ∈ pu , then (ru)n ∈ pu , hence ru ∈ pu ; and if ru ∈ pu ,
then a0 (ax)n ∈ pu , hence ax ∈ pu , because a0 |a in R; thus finally ru ∈ pu iff ax ∈ pu . Since
/ and P(u) ∩ P(arx) = O
/ by hypothesis, we finally must have ru, ax 6∈ pu .
P(r) ∩ P(ax) = O
We conclude that h(ru, ax) ∈ pu implies r, u, a, x 6∈ pu , and in particular, δθ 6∈ pu . Therefore, by point 3) above we get: If h(ru, ax) ∈ pu , then pu is totally split in L|K.
4
/.
Claim. Let I ⊂ Σ be a finite set of cardinality |I| > n. Then ∩u∈I P h(ru, ax) = O
By contradiction, let p ∈ P h(ru, ax) for all u ∈ I. By Notations 3.1, 2) and 3), applied
Q
to p, there exist α1 , . . . , αn ∈ Rp such that h̃(T, U ) := µ (T − αµ U ) ∈ Rp [T, U ] satisfies:
h(T, U ) − h̃(T, U ) ∈ pp [T, U ]. Since h(ru, ax) ∈ p for all u ∈ I, it follows that h̃(ru, ax) ∈ pp
Q
for all u ∈ I. On the other hand, h̃(ru, ax) = µ (ru − axαµ ), hence for every u ∈ I there
exists µu such that ru − axαµu ∈ pp . Since |I| > n, there exists u 6= v in I such that µu = µv ,
and ru − axαµu, rv − axαµv ∈ pp . Hence ru − rv = r(u − v) ∈ pp , i.e., r(u − v) ∈ p. Since
r, u, a, x 6∈ p by the discussion above, we get u − v ∈ p. But P(u − v) ⊆ P(x) by hypothesis,
hence x ∈ p, contradiction! The Claim is proved.
To conclude, let PΣ := ∪u∈Σ P h(ru, ax) , and Σp := {u ∈ Σ | h(ru, ax) ∈ p} for p ∈ PΣ .
Then the map ϕ : PΣ → 2Σ , p 7→ Σp , has the properties: ∪p∈PΣ Σp = Σ; and |Σp | ≤ n for all
p ∈ PΣ . By cardinal arithmetic, and taking into account that Σ is infinite, it follows that
the set {Σp | p ∈ PΣ } has cardinality |Σ|, thus concluding the proof.
Lemma 3.3. In the above Notations 3.1, suppose that for every non-zero r0 ∈ m, there
/ . Then for every non-zero x ∈ m, the following
exists r1 ∈ m such that P(r0 ) ∩ P(r1 ) = O
holds: |PL|K | ≥ |Eκ (X)| ≥ ℵ0 |R/m|.
Proof. For L|K and the corresponding a := a0 δθ as in Lemma 3.2, choose r ∈ m such that
/ . We claim that the set Σ := κ• + xEκ (x) satisfies the conditions i), ii),
P(r) ∩ P(ax) = O
from loc.cit.: First, if u := 0 + xE(x) ∈ Σ with 0 6∈ m, xE(x) ∈ m, then u 6∈ m. Thus
/ , and Σ satisfies i). Second, for u 6= v in Σ, one has u − v = ±xµ [(µ − ηµ ) + xr]
P(u) = O
for some µ ≥ 0 with µ 6= ηµ in κ, and r ∈ R. Since µ 6= ηµ , we get µ 6= η µ in R/m, hence
µ − η µ 6= 0 in R/m. But then µ − ηµ 6∈ m, and therefore, (µ − ηµ ) + xr 6∈ m. Therefore,
P(u − v) = P(xµ ) ⊆ P(x), hence Σ satisfies condition ii). Conclude by taking into account
that |Σ| ≥ |xEκ (x)| = |Eκ (X)|, and by applying Lemma 3.2.
Theorem 3.4. Let R be a domain whose interal closure R̃ in K := Quot(R) is a Krull
domain, e.g. R is Noetherian, or itself a Krull domain. Let V be the set of valuations v on
K defined by the localizations of R̃ at its minimal non-zero prime ideals. Then K endowed
with V is a Krull field, provided there exists a prime ideal m ⊂ R of height > 1, a set of
representatives κ ⊂ R for R/m, and a non-zero x ∈ m, such that |Eκ (x)| = |R|. This holds,
if one of the following is true:
P
i) |R| = ℵ0 ; or |R| = |R/m|; or R ≤ 2ℵ0 and all
X n , N ⊆ N, belong to Eκ (X).
n∈N
P
n
ii) m is countably generated, and ∩n mn = (0), and all
n n X , n ∈ κ, belong to Eκ (X).
The hypothesis ii) holds if R is complete with respect to a finitely generated non-zero ideal
a ⊆ m, and R/a is either Noetherian or a Krull domain, e.g., R = R0 [[X1 , . . . , Xn ]], where
R0 is a Noetherian or a Krull domain such that n + Krull.dim(R0 ) > 1.
Proof. First we prove that any of the conditions i), ii), implies |Eκ (x)| = |R|: Let κ ⊂ R
be a system of representatives for R/m, which in case ii) equals the given one, respectively
such that 0, 1 ∈ κ in case i). Then in the case i), it follows directly from the hypothesis and
(elementary) cardinal arithmetic that |Eκ (x)| = |R|. In case ii), let (xi )i∈I be a system of
generators of m with |I| ≤ ℵ0 , and let M be the set of all the (formal) monomials in the
b is injective, because
xi ’s. Then |M | = ℵ0 . Further, the m-adic completion morphism R → R
n
b is represented by a series of the form P au u with u ∈ M
∩n m = (0). Since every α
b∈R
u
5
b ≤ |κM | ≤ |κ| ℵ0 . On the other hand, |Eκ (x)| = |κ| ℵ0 , and
and au ∈ κ, we get: |R| ≤ |R|
|Eκ (x)| ≤ |R|. Finally, |R| = |κ| ℵ0 = |Eκ (x)|, as claimed.
Next we prove that K endowed with V is a Krull field. By hypothesis we have: Every
v ∈ V is a discrete valuation with valuation ring of the form Ov := R̃q with q ⊂ R̃ a minimal
non-zero prime ideal; and every non-zero r ∈ R̃ is contained in only finitely many q as above.
In particular, since R̃ is infinite, |V| ≤ |R̃|. Let n ⊂ R̃ be a prime ideal above m having height
> 1. Since R/m ⊆ R̃/n canonically, we can choose a set of representatives ω ⊂ R̃ for R̃/n
containing the above set of representatives κ for R/m. Let P be the set of all the minimal
non-zero prime ideals q ⊂ n of R̃, and W ⊆ V be the set of all the valuation in V defined
by the q ∈ P. Since R̃ is a Krull domain, the hypothesis of Lemma 3.3 is satisfied for n
and P. Hence by loc.cit. we have: If L|K is a finite Galois extension, then |PL|K | ≥ |Eω (x)|,
or equivalently, |WL|K | ≥ |Eω (x)|. On the other hand, since κ ⊆ ω, one obviously has
|Eω (x)| ≥ |Eκ (x)|. Further, since W ⊆ V, one has WL|K ⊆ VL|K . Thus taking into account all
the above (in)equalities we finally get |R̃| ≥ |V| ≥ |VL|K | ≥ |WL|K | ≥ |Eω (x)| ≥ |Eκ (x)| = |R|.
Since |R| = |R̃| = |K|, conclude that |VL|K | = |K|, as claimed.
B) Specializations of Galois covers
Notations 3.5. Let K be a base field, and B a finite group. Let ϕ : X → P1K be a finite
ramified B-cover, with branch locus S ⊂ P1K . Suppose that X is smooth, and P1K is the
projective t-line, i.e., P1K = SpecK[t] ∪ SpecK[ 1t ] such that S ⊂ SpecK[t]. Let κ(X)
be the
function field of X, hence κ(X)|K(t) is a Galois extension with Gal κ(X)|K(t) = B.
1) Let KA be the relative algebraic closure of K in κ(X).
Then A := Gal(KA |K) shall be
called the arithmetical quotient of B, and C := Aut P1K X the geometric part of B. One has:
A
a) The A-cover P1KA → P1K is étale.
b) The C-cover X → P1KA is such that X is geometrically integral over KA .
Hence, first, the inertia groups of ϕ are contained in C, and second, they generate C.
2) Let Xram ⊂ X be the ramification locus of ϕ, and Xs ⊂ Xram be the fiber of ϕ at s ∈ S;
and let es := |Ix | > 1 be the order of the inertia group Ix at x 7→ s.
• From now on suppose that K is Hilbertian and κ(x)|K is separable for all x ∈ Xram .
3) Let Kϕ |K be a minimal Galois extension such that Xram ⊂ X(Kϕ ). For s ∈ S, consider
the set of valuations Vs = {v | v totally split in Kϕ |K, and vK 6= ` · vK for `|es , ` > 1}.
4) Let Hϕ ⊂ K be a Hilbertian set such that for all b ∈ Hϕ , the fiber of ϕ at t = b is
irreducible, and the resulting Galois extension
Kb |K is linearly disjoint from Kϕ over KA .
Hence Gal(Kb |K) = B = Gal κ(X)|K(t) in a canonical way.
5) Finally, for b ∈ Hϕ , and v ∈ Vs , let Vv := {w | w prolongs v to Kb }, and for every
w ∈ Vv , let Iw be the inertia group at w|v.
Theorem 3.6. There exists a finite subset Σϕ ⊂ K × such that for every system of independent rank one valuations (vs )s∈S with vs ∈ Vs and vs (Σϕ ) = 0, there exists b ∈ Hϕ satisfying:
1) For every s ∈ S, one has { Iw | w ∈ Vvs } = { Ix | x ∈ Xs } canonically inside C.
2) In particular, Gal(Kb |KA ) is generated by the Iw with w ∈ Vvs and s ∈ S.
Proof. We begin by a preparation along the following three main steps:
6
Step 1. Let A0 := Gal(Kϕ |K), and B 0 := B ×A A0 . Then setting X 0 := X ×KA Kϕ , the
resulting ϕ0 : X 0 → P1K is a ramified B 0 -cover dominating both the étale A0 -cover P1Kϕ → P1K ,
and the ramified B-cover ϕ : X → P1K . The geometric part of ϕ0 is C 0 = C ×A {1} = C, and
under this identification, the inertia groups of ϕ0 are identified with those of ϕ; precisely, if
X 0 3 x0 7→ x ∈ X are above s ∈ S, then Ix0 = Ix ×A {1} = Ix .
In the same way, on the valuation theoretical side one has: Let Kb0 := Kϕ Kb be the
compositum of Kb and Kϕ . Since Kϕ |K and Kb |K are linearly disjoint over KA , we have
Gal(K 0 |K) = B ×A A0 . Let vs ∈ Vs , and w0 |vs a prolongation to Kb0 , and w := w0 |Kb . Then
by general decomposition theory, Iw0 projects onto Iw under B 0 → B. On the other hand,
since vs ∈ Vs is totally split in Kϕ |K (by the definition of Vs ), hence in KA = Kb ∩ Kϕ too,
we have: Iw0 ⊂ C 0 , and Iw ⊂ C. Since C 0 = C ×A {1} = C canonically, we have Iw0 = Iw .
Therefore, mutatis mutandis, we can and will suppose that Kϕ = KA , i.e., all ramified
points of X → P1K are KA -rational. Set S 0 := S ×K KA .
Step 2. Let R be the integral closure of KA [t] in κ(X). For s0 ∈ S 0 above s ∈ S, let x ∈ Xs
be a fixed point above s0 . Since R is a Dedekind ring, we can choose u ∈ R satisfying:
• vpx (u) = 1, and vpxσ (u − 1) = 1 for σ ∈ C\Ix ; and κ(X) = KA (t)[u].
Hence for s ∈ S, s0 ∈ S 0 and x ∈ Xs as above, one has: vpx (σu) = 1 for all σ ∈ Ix , and
vpx σ(u − 1) = 1 for σ ∈ C\Ix . Let f (U, t) ∈ KA (t)[U ] be the minimal polynomial of u
over KA (t). Since u ∈ R, one has f (U, t) := U d + ad−1 (t)U d−1 + · · · + a0 (t) ∈ KA [U, t]; and
recalling that es = |Ix |, the following hold:
(∗) vps0 a0 (t) = 1; and vps0 am (t) ≥ 1 for m < es ; and vps0 aes(t) = 0.
Let ps0 (t) := t − θs0 ∈ KA [t] define s0 . Then the θs0 ∈ KA , s0 ∈ S 0 , are distinct simple roots of
a0 (t) by condition (∗) above. In particular, the following hold:
(∗∗) aes(θs0 ) 6= 0; and a0 (t) = ps0 (t) bs0 (t) in KA [t] with bs0 (θs0 ) 6= 0 in KA .
In particular, setting Rs0 := R[1/bs0 (t)], we have: x is the only zero of u in SpecRs0 , and
u is a uniformizing parameter at x in Rs0 . Therefore, u is a prime element of Rs0 , and a
uniformizing parameter at x ∈ SpecRs0 . Hence there exist integers µ, ν ≥ 0, and uσs0 ∈ R for
σ ∈ C, such that the following hold:
a) If σ ∈ Ix , then σ(u) = u uσs0/bs0 (t)µ in Rs0 , with uσs0/bs0 (t)µ ∈ Rs×0 . In particular, there
exits ũσs0 ∈ R such that uσs0 ũσs0 = bs0 (t)ν (for ν large enough).
b) If σ ∈ C\Ix , then σ(u) = 1 + u uσs0/bs0 (t)µ in Rs0 , with uσs0/bs0 (t)µ ∈ Rs0 .
And since uσs0, ũσs0 ∈ R, their minimal polynomials satisfy fσs0(U, t), f˜σs0(U, t) ∈ KA [U, t].
Let o = Z[α1 , . . . , αr ] ⊂ K with αi 6= 0 be a Z-algebra of finite type such that denoting by
oKA its integral closure in KA , the following hold: First, the ramified B-cover ϕ : X → P1K
is defined over o. Second, f (U, t), ps0 (t), fσs0(U, t), f˜σs0(U, t) ∈ oKA [U, t] for all s0 ∈ S 0 and
σ ∈ C. Third, θs0 , aes(θs0 ), bs0 (θs0 ), 1/aes(θs0 ), 1/bs0 (θs0 ) ∈ oKA for all s0 ∈ S 0 and s ∈ S.
Definition of the set Σϕ : We define Σϕ ⊂ K × to be the set of generators Σϕ := {α1 , . . . , αr }.
0
0
Notice that aes(θs0 ), bs0 (θs0 ) ∈ o×
KA for all s ∈ S . Hence if v ∈ Vs satisfies v(Σϕ ) = 0,
a
then o ⊂ Ov ; and if v prolongs v to an algebraic closure K a of K, then oKA ⊂ Ova , and in
particular, aes(θs0 ), bs0 (θs0 ) are v a -units.
7
Step 3. Recall that for v ∈ Vs , one has by definition: First, vK 6= ` · vK for `|es ; hence
there exists πs ∈ K × such that v(πs ) > 0 and v(πs ) has order es in vK/(` · vK). Second, v
is totally split in KA |K, hence since v has rank one, K is dense in KA endowed with v a . By
Geyer [Ge], we have: Since (vs )s∈S are independent by hypothesis, there exists b ∈ Hϕ such
that vsa (b − θs0 ) = vs (πs ) > 0, s ∈ S. Further, since aes(t), bs0 (t) have vsa -integral coefficients,
and vsa (b − θs0 ) > 0, and aes(θs0 ), bs0 (θs0 ) are vsa -units, it follows that aes(b), bs0 (b) are vsa -units
too. Recalling that Kb ←- K with Gal(Kb |K) = B is the fiber of X → P1K at t = b, we have:
Claim. Let w be the restriction of vsa to Kb . Then Ix = Iw inside B.
Indeed, since a0 (b) = (b − θs0 ) bs0 (b), and w(b − θs0 ) = vsa (b − θs0 ) = vs (πs ) > 0, one has
w a0 (b) = w(πs ) > 0. Hence f (U, b) = U n + · · · + a0 (b) has a root ξs0 such that w(ξs0 ) > 0.
Let R ⊂ R be the integral closure of Ovs [t] in R. Since oKA is integral over o, and u, uσs0, ũσs0
are integral over oKA [t] (by the definition of o and oKA ), it follows that u, uσs0, ũσs0 are integral
over o[t] ⊂ Ovs [t], hence u, uσs0, ũσs0 ∈ R. Let Ob be the integral closure of Ovs in Kb . Then
B acts on Ob , and the B-equivariant projection Ψ : R → Kb defined by (u, t) 7→ (ξs0 , b), has
a B-equivariant restriction ψ : R → Ob . Recall that a0 (t) = (t − θs0 ) bs0 (t) in oKA [t], hence in
Ow [t], and bs0 (b) ∈ Ow× . Therefore, the canonical B-equivariant projections Ψ and ψ, have
canonical prolongations to C-equivariant projections Ψs0 : Rs0 → Kb , and ψs0 : Rs0 → Ob ,
where Rs0 := R[1/bs0 (t)]. Hence we have:
a)0 If σ ∈ Ix , then σ(ξs0 ) = ξs0 ψ(uσs0)/bs0 (b)µ in Ob and ψ(uσs0)/bs0 (b)µ ∈ Ob × .
b)0 If σ ∈ C\Ix , then σ(ξs0 ) = 1 + ξs0 ψ(uσs0)/bs0 (b)µ in Ob and ψ(uσs0)/bs0 (b)µ ∈ Ob .
And for σ ∈ Ix , we have ψ(uσs0) ψ(ũσs0) = bs0 (b)ν ∈ Ow× , hence ψ(uσs0) ∈ Ow× . Therefore:
a)00 If σ ∈ Ix , then w σ(ξs0 ) = w(ξs0 ) > 0.
b)00 If σ ∈ C\Ix , then w σ(ξs0 ) = 0.
Q
On the other hand, a0 (b) = NKb |KA (ξs0 ) = σ∈C σ(ξs0 ), and w a0 (b) = w(πs ). Hence the
following hold: First, w(πs ) = w a0 (b) = |Ix | w(ξs0 ); hence since w(πs ) = vs (πs ) has order
es = |Ix | in vs K/(es · vs K) weget e(w|vs ) ≥ |Ix |; thus |Iw | ≥ |Ix |. Second, if σ ∈ C\Ix , then
by b)00 we have: 0 = w σ(ξs0 ) = (w ◦ σ)(ξs0 ), hence σ 6∈ Dw because w(ξs ) > 0; and since vs
is totally split in KA |K, one has Dw ⊆ C, thus Dw ⊆ Ix . Hence since |Dw | ≥ |Iw | ≥ |Ix |, we
conclude that |Dw | = |Ix |. Therefore, Dw = Iw = Ix ⊆ Dx , thus proving the Claim.
To 1): Recall that by Hilbert decomposition theory, Vvs ∼
= B/Dw , and Xs ∼
= B/Dx as
B-sets. Since Dw = Iw = Ix , we see that Vvs projects B-equivariantly onto Xs , the fibers
being isomorphic to Dx /Ix . This concludes the proof of assertion 1).
To 2): Since Ix with x ∈ Xs , s ∈ S, generate C, the same is true for Iw with w ∈ Vvs and
s ∈ S, by assertion 1). Hence by Hilbert decomposition theory, Kb |KA has no non-trivial
subextension in which all the vs , s ∈ S, are unramified.
4. A generalization of Theorem 1.2
Definition/Remarks 4.1. Let ℵ be an infinite cardinal.
1) A field K endowed with a set of non-equivalent valuations V shall be called a generalized
ℵ Krull field, respectively a generalized Krull field provided ℵ = |K|, if the following hold:
i) If Σ ⊂ K × has cardinality |Σ| < ℵ, then VΣ := {v ∈ V | v(Σ) 6= 0} has |VΣ | < ℵ.
ii) For every finite Galois extension L|K, and every integer n > 1, the set
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VL|K,n := {v ∈ V | v totally split in L|K, vK 6= ` · vK for ` |n, ` > 1} has |VL|K,n | ≥ ℵ.
Note that in particular, the Krull fields are exactly the generalized ℵ0 Krull fields.
2) Prominent examples of Krull fields are the following:
a) The global fields (by the Chebotarev Density Theorem).
b) The function fields K|k with tr.deg(K|k) > 0. (Indeed, by point 3 below, it is sufficient
to consider the case K = k(t1 , . . . , td ) is a rational function field, etc.)
c) The quotient fields of domains R as in Theorem 3.4.
3) The class of generalized ℵ Krull fields is closed under finite field extensions.
Definitions/Notations 4.2. For an embedding problem EP = (γ : GK → A, α : B → A)
over K, let KA be the fixed field of ker(γ), hence Gal(KA |K) = A canonically. And for proper
solutions β of EP, let Kβ be the fixed field of ker(β), hence Gal(Kβ |K) = B canonically.
1) A family of proper solutions {βj }j∈J of EP is called independent, if for all j ∈ J one
has: Kβj and the compositum Lj := ∪j 0 6=j Kj 0 are linearly disjoint over KA .
2) If K endowed with V is a generalized ℵ Krull field, a proper solution β of EP is called
totally ramified, if Kβ |KA has no proper subextension in which all the v ∈ V are unramified.
Theorem 4.3. Let K endowed with a set of rank one valuations V be a generalized ℵ Krull
field. Suppose that K is large and Hilbertian. Then every non-trivial finite split embedding
problem for GK has at least ℵ independent and totally ramified proper solutions.
Proof. Let EP = (γ : GK → A, α : B → A) be a non-trivial finite split embedding problem
over K, and EPt = (γ ◦prt : GK(t) → A, α : B → A) be the non-trivial finite split embedding
problem for GK(t) . By Theorem A of Pop [P], EPt has proper regular solutions βt , which
means that if KA ⊆ K(t)βt are the fixed fields of ker(γ) in K s , respectively of ker(βt ) in
K(t)s , then K(t)βt ∩ K s = KA . Moreover, if ϕ : X → P1K is the B-ramified cover defining
βt : GK(t) → B, then sorting through the proof of loc.cit., one can see that the ramification
points of ϕ are actually Kϕ -rational, where Kϕ |KA is some cyclotomic extension. Hence
Theorem 3.6 is applicable here.
Using Zorn’s Lemma, let {βj }j∈J be a maximal set of independent proper solutions of EP,
given by specializing ϕ as in Theorem 3.6; and note that these solutions are totally ramified
by assertion 2) of loc.cit. We claim that |J| ≥ ℵ. Indeed, by contradiction, suppose that
|J| < ℵ. For every βj , set Kβj = K[ξj ], and let pj (T ) = T n + aj,n−1 T n−1 + · · · + aj,0 ∈ K[t] be
the minimal polynomial of ξj , and δj ∈ K × its discriminant. If v ∈ V satisfies: pj (T ) ∈ Ov [T ]
and v(δj ) = 0, then by Hilbert decomposition theory, v is unramified in Kβ |K. Hence if
Σj := {δj , aj,0 , . . . , aj,n−1 } ∩ K × , we have: If v is ramified in Kβj |K, then v(Σj ) 6= 0; or
equivalently, Vj := {v ∈ V | v is ramified in Kβj |K} ⊆ {v | v(Σj ) 6= 0} =: VΣj .
Hence setting ΣJ := ∪j∈J Σj , we have VΣJ := {v ∈ V | v(ΣJ ) 6= 0} = ∪j∈J VΣj . Therefore
we get VJ := ∪j∈J Vj ⊆ ∪j∈J VΣj =: VΣJ . Since each Σj is finite, and we supposed that
|J| < ℵ, it follows that ΣJ = ∪j∈J Σj has cardinality |ΣJ | < ℵ. But then by condition i)
in the definition of V, we have |VΣJ | < ℵ; hence |VJ | < ℵ because VJ ⊆ VΣJ . Hence
by condition ii) in the definition of V, and with Kϕ and es as in Notations 3.5, one has:
|VKϕ |K,es \VJ | ≥ ℵ for each s ∈ S. Since S is finite, we can choose a system of independent
valuations (vs )s∈S with vs ∈ VKϕ |K,es \VJ . For this system (vs )s∈S , consider a solution β of EP
as given by Theorem 3.6. Let LJ |K be the compositum of all the Kβj , j ∈ J. We claim that
LJ ∩ Kβ = KA . Indeed, since vs 6∈ VJ , the vs , s ∈ S, are unramified in Kβj |K, for all j ∈ J;
9
hence in LJ |K. Hence the vs are unramified in LJ ∩ Kβ too. But then by assertion 2) of
Theorem 3.6, we get LJ ∩ Kβ = KA . Thus the family of distinct totally ramified solutions
{β} ∪ {βj }j∈J is independent and contradicts the maximality of {βj }j∈J .
5. Proof of Theorem 1.3.
First, K is large, by Theorem 1.1; and Hilbertian by Weissauer [W], Theorem 7.2, because
the integral closure of R is a Krull domain with Krull.dim > 1. Hence every split non-trivial
embedding problem for GK has |K| proper solutions by Theorems 1.2 and 3.4. Second, the
same is true correspondingly for GK ab , by [Ha2], Theorem 2.4. Finally, cd(K ab ) ≤ 1, by
Colliot-Thélène–Ojanguren–Parimala [COP], Theorem 2.2, and [Ha2], Theorem 4.4, if
char(K) > 0. One concludes by applying [H–S], Theorem 2.1.
Thanks: I would like to thank D. Harbater, M. Jarden, and others for inspiration and discussions; and E. Paran for pointing out a gap in the first version of the manuscript, and the referee
for his useful comments.
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[MB]
[CT]
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[Ge]
[Ha1]
[Ha2]
[H–S]
[Ko]
[ Lf ]
[ Pa ]
[P-P]
[P]
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Department of Mathematics, University of Pennsylvania
DRL, 209 S 33rd Street, Philadelphia, PA 19104, USA
E-mail address: [email protected]
URL: http://math.penn.edu/~pop
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