Algebraic Patching (lecture notes)

52 Pages·2013·0.385 MB·English

by  Moshe Jarden

Travaux math´ematiques, Volume 23 (2013), 87–137, (cid:13)c by the author
Algebraic Patching1
by Moshe Jarden
Abstract
The most effective challenge to the inverse problem of Galois theory has
been Hilbert Irreducibility Theorem. Indeed, one may use both arithmetic
andgeometryinordertorealizefinitegroupsoverQ(t). Oncethishasbeen
successfully done for a finite group G, HIT yields many specializations of
t to elements of Q that lead to a realization of G over Q. However, the
realization of G over Q(t) usually requires the existence of rational point
on a certain algebraic variety defined over Q. Unfortunately, one can not
alwaysguaranteetheexistenceofsuchapoint,sotheinverseGaloisproblem
over Q is still wide open.
The only known class of fields for which points that lead to realization
of all finite groups exist is that of “ample fields”. A field K is said to
be ample if every absolutely irreducible curve C defined over K with a
K-rational simple point has infinitely many K-rational points. Among
others,PACfields,Henselianfields,andrealclosedfieldsareample. Usinga
methodcalled“algebraicpatching”wewillprovethatifK isanamplefield,
then every finite split embedding problem over K(t) is properly solvable.
In particular, if K is countable and algebraically closed, this implies that
Gal(K) is isomorphic to the free profinite group Fˆ on countably many
ω
generators. Also, if K is PAC and countable, then K is Hilbertian if and
only if Gal(K) ∼= Fˆ .
ω
MSC (2010): 12E30.
1For more details, including exact references, see “Algebraic Patching”, Springer 2011, by
Moshe Jarden.
88 Moshe Jarden
Contents
1 Ample Fields 89
2 Examples of Ample Fields 91
3 Finite Split Embedding Problems 92
4 Axioms for Algebraic Patching 94
5 Galois Action on Patching Data 98
6 Normed Rings 100
7 Rings of Convergent Power Series 104
8 Convergent Power Series 108
9 Several Variables 112
10 A Normed Subring of K(x) 112
11 Mittag-Leffler Series 114
12 Fields of Mittag-Leffler Series 119
13 Factorization of Matrices over Complete Rings 120
14 Cyclic Extensions 123
15 Embedding Problems over Complete Fields 127
16 Embedding Problems over Ample Fields 129
17 PAC Hilbertian Fields are ω-Free 134
Algebraic Patching 89
Introduction
The ultimate goal of Galois theory is to describe the structure of the absolute
Galois group Gal(Q) of Q. This structure will be specified as soon as we know
which finite embedding problems can be solved over Q. If every finite Frattini
embedding problem and every finite split embedding problem is solvable, then
every embedding problem is solvable [9, Prop. 22.5.8]. However, not every finite
Frattini problem over Q can be solved. For example,
√ √
(Gal(Q) → Gal(Q( −1)/Q), Z/4Z → Gal(Q( −1)/Q))
is an unsolvable Frattini embedding problem. So, one may ask:
Problem A. Is every finite split embedding problem over Q solvable?
More generally, one would like to know:
Problem B. Let K be a Hilbertian field. Is every finite split embedding problem
over K solvable?
An affirmative answer to Problem B will follow from an affirmative answer to
theproblemforthesubfamilyofHilbertianfieldsconsistingofallrationalfunction
fields:
Problem C (D´ebes–Deschamps). Let K be a field and x a variable. Is every
finite split embedding problem over K(x) solvable?
1 Ample Fields
The most significant development around Problem C is its affirmative solution for
ample fields K. This family includes two subfamilies that seemed to have nothing
in common: PAC fields and Henselian fields. Indeed, if K is a PAC field and v is
a valuation of K, then the Henselization K of K at v is the separable closure K
v s
of K. In particular, if a PAC field is not separably closed, then it is not Henselian.
Florian Pop made a surprising yet simple and useful observation that both
PAC fields and Henselian fields are existentially closed in the fields of formal
power series over them. This property is one of several equivalent definitions of
an ample field.
Proposition 1.1. The following conditions on a field K are equivalent:
(a) For each absolutely irreducible polynomial f ∈ K[X,Y], the existence of a
point (a,b) ∈ K2 such that f(a,b) = 0 and ∂f(a,b) (cid:54)= 0 implies the existence
∂Y
of infinitely many such points.
90 Moshe Jarden
(b) Every absolutely irreducible K-curve C with a simple K-rational point has
infinitely many K-rational points.
(c) If an absolutely irreducible K-variety V has a simple K-rational point, then
V(K) is Zariski-dense in V.
(d) Every function field of one variable over K that has a K-rational place has
infinitely many K-rational places.
(e) K is existentially closed in each Henselian closure K(t)h of K(t) with respect
to the t-adic valuation.
(f) K is existentially closed in K((t)).
Proof. We only prove the implication “(f) =⇒ (a)” and refer the reader for the
other implications to [12, Lemma 5.3.1].
Inductively suppose there exist (a ,b ) ∈ K2, i = 1,...,n, such that f(a ,b ) =
i i i i
0 and a ,...,a are distinct. We choose a(cid:48) ∈ K[[t]], t-adically close to a such that
1 n
a(cid:48) (cid:54)= a , i = 1,...,n. Then f(a(cid:48),b) is t-adically close to 0 and ∂f(a(cid:48),b) (cid:54)= 0. Since
i ∂Y
K((t)) is Henselian, there exists b(cid:48) ∈ K[[t]] such that f(a(cid:48),b(cid:48)) = 0 and ∂f(a(cid:48),b(cid:48)) (cid:54)=
∂Y
0. Since K is existentially closed in K((t)), there exists a ,b ∈ K such that
n+1 n+1
f(a ,b ) = 0 and a (cid:54)= a ,...,a . This concludes the induction.
n+1 n+1 n+1 1 n
Corollary 1.2. Every ample field is infinite.
It is possible to strengthen Condition (b) of Proposition 1.1 considerably.
Lemma 1.3 (Fehm [7, Lemma 4]). Let K be an ample field, C an absolutely
irreducible curve defined over K with a simple K-rational point, and φ: C → C(cid:48)
a separable dominant K-rational map to an affine curve C(cid:48) ⊆ An defined over K.
Then, for every proper subfield K of K, card(φ(C(K))(cid:114)An(K )) = card(K).
0 0
The proof uses among others a trick of Jochen Koenigsmann that Pop applied
to prove Proposition 1.4(b) below.
Proposition 1.4. Let K be an ample field, V an absolutely irreducible variety
defined over K with a K-rational simple point, and K a subfield of K. Then:
0
(a) K = K (V(K)).
0
(b) card(V(K)) = card(K).
Proposition 1.5 (Pop [16, Prop. 1.2]). Every algebraic extension of an ample
field is ample.
Problem 1.6. Let L/K be a finite separable extension such that L is ample. Is
K ample?
Algebraic Patching 91
2 Examples of Ample Fields
The properties causing a field K to be ample vary from diophantine, arithmetic,
to Galois theoretic.
(a) PAC fields, in particular, algebraically closed fields.
(b) Henselian fields.
More generally, we say that a pair (A,a) consisting of a domain A and a
nonzero ideal a of A is Henselian if for each f ∈ A[X] satisfying
f(0) ≡ 0 mod a and f(cid:48)(0) is a unit mod a
there exists x ∈ a such that f(x) = 0.
Pop [17, Thm. 1.1] has observed that the proof that Henselian fields are ample
can be adjusted to a proof that if (A,a) is a Henselian pair, then Quot(A) is
ample.
(c) If A is complete with respect to a nonzero ideal a, then (A,a) is a Henselian
pair, hence Quot(A) is ample.
For example, K((X ,...,X )) is ample for every n ≥ 1 and any field K.
1 n
So is, for example, the field Quot(Z[[X ,...,X ]]). Note that if n ≥ 2, then
1 n
F = K((X ,...,X )) is Hilbertian (by Weissauer [9, Thm. 13.9.1]), hence F is
1 n
not Henselian (by Geyer [9, Lemma 15.5.4]), even though the ring K[[X ,...,X ]]
1 n
is complete and therefore Henselian.
(d) Real closed fields.
(e) Fields satisfying a local global principle.
Let K be a field and K be a family of field extensions of K. We say that K is
PKC (or also that K satisfies a local global principle with respect to K) if every
¯
nonempty absolutely irreducible variety defined over K with a simple K-rational
¯ ¯
point for each K ∈ K has a K-rational point. In this case, if each K ∈ K is ample,
then K is also ample.
For example, let K be a countable Hilbertian field and S a finite set of local
primes of K. Thus, each p ∈ S is an equivalence class of absolute values whose
ˆ ˆ
completion K is a local field. Let K = K ∩ K . Consider also an e-tuple
p p s p
σ = (σ ,...,σ ) taken in random in Gal(K)e (with respect to the Haar measure).
1 e
Let K (σ) be the fixed field in K of σ ,...,σ and let K [σ] be the maximal
s s 1 e s
Galois extension of K in K (σ). Then the field
s
(cid:92) (cid:92)
K [σ] = K [σ]∩ Kρ
tot,S s p
p∈Sρ∈Gal(K)
92 Moshe Jarden
is PKC with K = {Kρ| p ∈ S, ρ ∈ Gal(K)}. Since each of the fields Kσ is
p p
Henselian or real closed, it is ample (by (b) and (d) above). Hence, K [σ] is
tot,S
ample (Geyer-Jarden [12, Example 5.6.6]).
(f) Fields with a pro-p absolute Galois group (Colliot-Th´el`ene [2], Jarden [11]).
Problem 2.1. Let K be a field such that the order of Gal(K) is divisible by only
finitely many prime numbers. Is K ample?
3 Finite Split Embedding Problems
The raison d’etre of ample fields is that they are the only known fields for which
Problem C has an affirmative answer.
Theorem 3.1 (Pop [Main Thm. A], Haran-Jarden [3, Thm. A]). Let K be an
ample field, L a finite Galois extension of K, and x a variable. Suppose Gal(L/K)
acts on a finite group H. Then K(x) has a Galois extension F that contains L
and there is a commutative diagram
Gal(F/K(x))
(cid:118)(cid:118)(cid:108)(cid:108)(cid:108)(cid:108)(cid:108)(cid:108)γ(cid:108)(cid:108)(cid:108)(cid:108)(cid:108)(cid:108)(cid:108) (cid:15)(cid:15)res
Gal(L/K)(cid:110)H α (cid:47)(cid:47)Gal(L/K)
in which α is the projection on the first component and γ is an isomorphism.
The proof of Theorem 3.1 is carried out in two steps. First one solves the
ˆ
corresponding embedding problem over the field K = K((t)) using patching.
ˆ
Then one reduces the solution obtained over K(x) to a solution over K(x), using
ˆ
that K is existentially closed in K.
The most striking application of Theorem 3.1 is a solution of a problem of
Field Arithmetic that stayed open for a long time:
Theorem 3.2 ([12, Thm. 5.10.2]). Every PAC Hilbertian field K is ω-free (that
is, every finite embedding problem over K is solvable). In particular, if K is
ˆ
countable, then Gal(K) is isomorphic to the free profinite group F on countably
ω
many generators.
For the next application we need an improvement of Theorem 3.1.
Theorem 3.3 (Harbater-Stevenson [10, Thm. 4.3], Pop [17, Thm. 1.2], Jarden
[12, Prop. 8.6.3]). Let K be an ample field and x a variable. Then every finite
split embedding problem over K(x) has as many solutions as the cardinality of K.
Algebraic Patching 93
In particular, this theorem applies when K is algebraically closed. Since
Gal(K(x)) is projective, we get the following generalization of a theorem that
was proved in characteristic 0 using Riemann existence theorem.
Corollary 3.4 (Harbater [6, Cor. 4.2], Pop [15, Geometric (SC)], Haran-Jarden
[4, Main Theorem]). Let K be an algebraically closed field of cardinality m. Then
∼ ˆ
Gal(K(x)) = F .
m
Actually, Theorem3.3wasprovedinastrongerform, inwhichK(x)isreplaced
by an arbitrary function field E of one variable over K and the solution field is
regular over the field of constants of E.
Harbater-Stevenson proved that every finite split embedding problem over
K((t ,t )) has as many solutions as the cardinality of K. Moreover, this property
1 2
is inherited by K((t ,t )) . Finally, the absolute Galois group of the latter field
1 2 ab
is projective. Together, this proves the following result:
Theorem 3.5. Let K be a separably closed field and E = K((t ,t )). Then
1 2
ˆ
Gal(E ) is isomorphic to the free profinite group F of cardinality m = card(E).
ab m
Theorem 3.3 can be improved further.
Theorem 3.6 (Bary-Soroker, Haran, Harbater [1, Thm. 7.2]; Jarden
[12, Thm. 11.7.1]). Let E be a function field of one variable over an ample field
K. Then Gal(E) is semi-free. That is, every finite split embedding problem over
E
(res: Gal(E) → Gal(F/E), α: G → Gal(F/E))
has card(E)-linearly disjoint solution fields F (i.e. the fields F are linearly
i i
disjoint extensions of F.)
Combining this proposition with results of Bary-Soroker-Haran-Harbater,
Efrat, and Pop, we were able to prove the following result.
Theorem 3.7 (Jarden [12, Thm. 11.7.6]). Let K be a PAC field of cardinality
c and x a variable. For each irreducible polynomial p ∈ K[x] and every positive
√ √ √
integer n satisfying char(K) (cid:45) n let n p be an nth root of p such that ( mn p)m = n p
√
for all m,n. Let F be the compositum of all fields K( n p). Then, F is Hilbertian
∼ ˆ
and Gal(F) = F .
c
Here is a special case:
Corollary 3.8 (Jarden[12,Example11.7.8]). Let K be an PAC field of cardinality
∼
m and x a variable. Suppose K contains all root of unity. Then Gal(K(x) ) =
ab
ˆ
F .
m
And here is another example of a semi-free absolute Galois group:
94 Moshe Jarden
Theorem 3.9 (Pop[12, Thm.12.4.4]). Each of the following fields K is Hilbertian
and ample. Moreover, Gal(K) is semi-free of rank card(K).
(a) K = K ((X ,...,X )), where K is an arbitrary field and n ≥ 2.
0 1 n 0
(b) K = Quot(R [[X ,...,X ]]), where R is a Noetherian domain which is not
0 1 n 0
a field and n ≥ 1.
More about Theorem 3.9 can be found in Chapter 12 of [12].
Problem 3.10. Give an example of a non-ample field K such that every finite
split embedding over K(x) is solvable.
Note that the existence of an example as in Problem 3.10 will give a negative
answer to Problem C. Conversely, a positive answer to Problem C is a negative
answer to Problem 3.10.
4 Axioms for Algebraic Patching
Let E be a field, G a finite group, and (G ) a finite family of subgroups of G
i i∈I
that generates G. Suppose that for each i ∈ I we have a finite Galois extension
F of E with Galois group G . We use these extensions to construct a Galois
i i
extension F of E (not necessarily containing F ) with Galois group G. First we
i
‘lift’ each F /E to a Galois field extension Q /P , where P is an appropriate field
i i i i
extension of E (that we refer to as “analytic”) such that P ∩F = E and all of
i i
the Q ’s are contained in a common field Q. Then we define F to be the maximal
i
(cid:84)
subfield contained in Q on which the Galois actions of Gal(Q /P ) combine
i∈I i i i
to an action of G.
P Gi Q Q
i i
E Gi F
i
The construction works if certain patching conditions on the initial data are
satisfied.
Definition 4.1 (Patching data). Let I be a finite set with |I| ≥ 2. Patching
data
E = (E,F ,P ,Q;G ,G)
i i i i∈I
consists of fields E ⊆ F ,P ⊆ Q and finite groups G ≤ G, i ∈ I, such that the
i i i
following conditions hold.
(4.1a) F /E is a Galois extension with Galois group G , i ∈ I.
i i
(4.1b) F ⊆ P(cid:48), where P(cid:48) = (cid:84) P , i ∈ I.
i i i j(cid:54)=i j
Algebraic Patching 95
(cid:84)
(4.1c) P = E.
i∈I i
(4.1d) G = (cid:104)G | i ∈ I(cid:105).
i
(4.1e) (Cartan’s decomposition) Let n = |G|. Then for every B ∈ GL (Q) and
n
each i ∈ I there exist B ∈ GL (P ) and B ∈ GL (P(cid:48)) such that B = B B .
1 n i 2 n i 1 2
We extend E by more fields. For each i ∈ I let Q = P F be the composi-
i i i
tum of P and F in Q. Conditions (4.1b) and (4.1c) imply that P ∩ F = E.
i i i i
Hence, Q /P is a Galois extension with Galois group isomorphic (via restriction
i i
of automorphisms) to G = Gal(F /E). We identify Gal(Q /P ) with G via this
i i i i i
isomorphism.
Definition 4.2 (Compound). The compound of the patching data E is the set
(cid:84) (cid:84)
F of all a ∈ Q for which there exists a function f: G → Q such that
i∈I i i∈I i
(4.2a) a = f(1) and
(4.2b) f(ζτ) = f(ζ)τ for every ζ ∈ G and τ ∈ (cid:83) G .
i∈I i
Note that f is already determined by f(1). Indeed, by (4.1d), each τ ∈ G
(cid:83)
can be written as τ = τ τ ···τ with τ ,...,τ ∈ G . Hence, by (4.2b),
1 2 r 1 r i∈I i
f(τ) = f(1)τ1···τr.
We call f the expansion of a and denote it by f . Thus, f (1) = a and
a a
f (ζτ) = f (ζ)τ for all ζ ∈ G and τ ∈ (cid:83) G .
a a i∈I i
We list some elementary properties of expansions:
Lemma 4.3. Let F be the compound of E. Then:
(a) Every a ∈ E has an expansion, namely the constant function ζ (cid:55)→ a.
(b) Let a,b ∈ F. Then a+b,ab ∈ F; in fact, f = f +f and f = f f .
a+b a b ab a b
(c) If a ∈ F×, then a−1 ∈ F. More precisely: f (ζ) (cid:54)= 0 for all ζ ∈ G, and
a
ζ (cid:55)→ f (ζ)−1 is the expansion of a−1.
a
(d) Let a ∈ F and σ ∈ G. Then f (σ) ∈ F; in fact, f (ζ) = f (σζ).
a fa(σ) a
Proof. Statement (a) holds, because aτ = a for each τ ∈ (cid:83) G . Next observe
i∈I i
that the sum and the product of two expansions is again an expansion. Hence,
Statement(b)followsfromtheuniquenessofexpansionsandfromtheobservations
(f )(1) = a+b = f (1)+f (1) = (f +f )(1) and f (1) = (f f )(1).
a+b a b a b ab a b
Next we consider a nonzero a ∈ F and let τ ∈ G. Using the notation of
Definition 4.2, we have f (τ) = (cid:0)(aτ1)τ2···(cid:1)τr (cid:54)= 0. Since taking the inverse in
a
(cid:84) Q commutes with the action of G, the map ζ (cid:55)→ f (ζ)−1 is the expansion of
i∈I i a
a−1. This proves (c).
Finally, one can check that the map ζ → f (σζ) has the value f (σ) at ζ = 1
a a
and it satisfies (4.2b). Hence, that map is an expansion of f (σ), as claimed in
a
(d).
96 Moshe Jarden
Definition 4.4 (G-action on F). For a ∈ F and σ ∈ G put
(4.3) aσ = f (σ),
a
where f is the expansion of a.
a
Lemma 4.5. The compound F of the patching data E is a field on which G acts
by (4.3) such that FG = E. Moreover, for each i ∈ I, the restriction of this action
to G coincides with the action of G = Gal(Q /P ) on F as a subset of Q .
i i i i i
Proof. By Lemma 4.3(a),(b),(c), F is a field containing E. Furthermore, (4.3)
defines an action of G on F. Indeed, a1 = f (1) = a. Moreover, if ζ is another
a
element of G, then by (4.3) and Lemma 4.3(d), (aσ)ζ = f (σ)ζ = f (ζ) =
a fa(σ)
f (σζ) = a(σζ).
a
Claim: FG = E. Indeed, by Lemma 4.3(a), elements of E have constant
expansions, hence are fixed by G. Conversely, let a ∈ FG. Then for each i ∈ I we
have a ∈ QGi = P . Hence, by (4.1c), a ∈ E.
i i
Finally, let τ ∈ G and a ∈ F. Then, f (τ) = f (1)τ = aτ, where τ acts as an
i a a
element of G = Gal(Q /P ). Thus, that action coincides with the action given by
i i i
(4.3).
The next goal is to prove that F/E is a Galois extension with Galois group G.
ToachievethisgoalweintroducemoreobjectsandinvokeCartan’sdecomposition.
Let
(cid:110)(cid:88) (cid:111)
(4.4) N = a ζ| a ∈ Q
ζ ζ
ζ∈G
be the vector space over Q with basis (ζ| ζ ∈ G), where G is given some fixed
ordering. Thus, dim N = |G|. For each i ∈ I we consider the following subset of
Q
N:
(cid:110)(cid:88) (cid:111)
(4.5) N = a ζ ∈ N | a ∈ Q , aη = a for all ζ ∈ G, η ∈ G .
i i i
ζ ζ ζ ζη
ζ∈G
It is a vector space over P .
i
N N
i
P Gi Q Q
i i
F
(cid:124)
(cid:124)
(cid:124)
(cid:124)
(cid:124)
(cid:124)
(cid:124)
(cid:124)
E Gi F
i