Semi-topological Galois theory and the inverse Galois problem
Hsuan-Yi Liaoa, Jyh-Haur Tehb,∗
aDepartment of Mathematics, National Tsing Hua University of Taiwan, Hsinchu, 30043, Taiwan. bDepartment of Mathematics, National Tsing Hua University of Taiwan, Hsinchu, 30043, Taiwan.
Abstract
We enhance the analogy between field extensions and covering spaces by introducing the concept of splitting covering which correspondences to the splitting field in Galois theory. We define semi-topological Galois groups for Weierstrass polynomials and prove the existence of a Galois correspondence. We use it to give a sufficient condition to realize a group as a Galois group over Q(i).
Keywords: Semi-topological Galois theory, inverse Galois problem, splitting covering.
1. Introduction
It is well known that there is a Galois correspondence between subgroups of the funda-mental group of a topological space X and covering spaces over it which is analogous to the Galois correspondence between field extensions and Galois groups. To make the anal-ogy stronger, we study some covering spaces defined by some polynomials in C(X)[z]. We introduce the splitting coverings for Weierstrass polynomials and define the semi-topological Galois groups for them. We have the following correspondences:
f ields ←→ topological spaces
splitting f ields ←→ splitting coverings
extension f ields ←→ covering spaces
Galois extensions ←→ Galois covering spaces
separable polynomials ←→ W eierstrass polynomials Galois groups ←→ semi − topological Galois groups
algebraic closures ←→ universal coverings
A Weierstrass polynomial f in C(X)[z] is a polynomial such that each root is of degree one. We construct its splitting covering p : Ef → X and show that it is smallest among
covering spaces that f splits. We define the semi-topological Galois group of f and show that it is isomorphic to the group of covering transformations of Ef → X. The proof uses a result
∗Corresponding author
Email addresses: [email protected] (Hsuan-Yi Liao), [email protected] (Jyh-Haur Teh)
of Chase-Harrison-Rosenberg about the existence of Galois correspondence in commutative rings. In the final section, we use semi-topological Galois groups to study the inverse Galois problem. We gives some criterion such that a finite group can be realized as the Galois group of some field extension over Q(i).
2. Algebraic closures and universal coverings
Throughout this article, unless otherwise stated, X, Y, Z will denote topological spaces which are Hausdorff, path-connected, locally path-connected and semi-locally simply con-nected. Let C(X) be the ring of all continuous functions from X to C and f = fx(z) =
an(x)zn + an−1(x)zn−1 + · · · + a0(x) be an element in C(X)[z], the polynomial ring with
coefficients in C(X). In general, there may no exist a continuous function α : X → C such that fx(α(x)) = 0 for all x ∈ X. For example, on the unit sphere S1, there is no continuous
function in C(S1) which satisfies the equation z2− x = 0.
Definition 2.1. Let n ∈ N and f = fx(z) = zn+ an−1(x)zn−1 + · · · + a0(x) ∈ C(X)[z]. f
is called a Weierstrass polynomial of degree n on X if for each x ∈ X, fx has distinct
n roots. For such f , E = {(x, z) ∈ X × C : fx(z) = 0} is called the solution space of f .
A root of f is a continuous function α : X → C such that fx(α(x)) = 0 for all x ∈ X. We
say that f splits in X if f has n distinct roots in X. A Weierstrass polynomial is called irreducible if it is irreducible as an element in the ring C(X)[z].
It is well known that the solution space of a Weierstrass polynomial under the first projection is a covering space over X, and the solution space of a Weierstrass polynomial is connected if and only if the Weierstrass polynomial is irreducible([5, Theorem 4.2, pg 141]). Since a Weierstrass polynomial f ∈ C(X)[z] may no have solutions in X, it is natural to ask if we can find solutions of f in some covering spaces over X. This is analogous to finding roots of a polynomial in some field extensions in Galois theory. We will see soon that the universal cover of X plays the role of algebraic closure.
Definition 2.2. Let λ : Y → X be a continuous map. The pullback λ∗ : C(X) → C(Y ) is defined by
λ∗(γ) := γ ◦ λ
which induces a ring homomorphism λ∗ : C(X)[z] → C(Y )[z] by
λ∗(anzn+ an−1zn−1+ · · · + a0) := (an◦ λ)zn+ (an−1◦ λ)zn−1+ · · · + (a0◦ λ).
The following basic facts about covering spaces can be found in [3] or [9]. We quote them here since we need to use them often and for the convenience of the reader.
Theorem 2.3. Let X, Y, E be connected and locally path-connected topological spaces and x0 ∈ X, y0 ∈ Y, e0 ∈ E. Suppose (E, e0)
p
→ (X, x0) is a covering space and f : (Y, y0) →
(X, x0) is a continuous map. Then there exists a lifting f
0
of f ,that is, pf0 = f if and only if
In particular, if Y is simply connected, then f always has a lifting. Theorem 2.4. (Unique lifting property) Let (E, e0)
p
→ (X, x0) be a covering space with based
points and f : (Y, y0) → (X, x0) be a continuous map. Assume Y is connected. If there exists
a continuous map f0 : (Y, y0) → (E, e0) such that pf
0
= f , then it is unique.
Proposition 2.5. If f is a Weierstrass polynomial of degree n and Y → X is a connectedp covering space, then any two roots of p∗f are either equal everywhere or equal nowhere; in particular, p∗f has at most n roots.
Proof. Suppose α, β are roots of p∗f . Let A = {y ∈ Y | α(y) = β(y)}. Then A is closed in Y as C is Hausdorff. Assume E is the solution space of f , and pr1, pr2 are the first and
second projections respectively. For y ∈ A, there is a neighborhood U of p(y) such that pr−11 (U ) = `n
i=1Ui is a trivial covering on U . Since C is Hausdorff, we may take a smaller
neighborhood if necessary such that pr2(U1), · · · , pr2(Un) lie in some disjoint open subsets
V1, · · · , Vnof C respectively. α(y) = β(y) lies in one of V1, · · · , Vn. Assume α(y) = β(y) ∈ V1.
Then W = α−1(V1)∩β−1(V1)∩p−1(U ) is an open neighborhood of y in Y and W ⊂ A. Hence
A is open in Y . Consequently, A is empty or whole Y . In other words, two roots of p∗f are either equal everywhere or equal nowhere; thus, p∗f has at most n roots.
Theorem 2.6. (Algebraic closure) Let f be a Weierstrass polynomial on X of degree n. Then f splits in eX where p : ( eX,ex0) → (X, x0) is the universal covering of X.
Proof. Let E1, ..., Ek be all path-connected components of the solution space π : E → X of
f . Then for each i, πi := π|Ei : Ei → X is a covering space of X. Let (πi)−1(x0) = {ei,1, ..., ei,ri}.
From Theorem 2.3 and the unique lifting theorem, for each ei,j, there exists a unique lifting,
˜
pi,j : ( ˜X, ˜x0) → (Ei, ei,j) of p. Define αi,j := qi◦ ˜pi,j which are roots of p∗f where qi : Ei → C
is the projection to the second factor. We have the following commutative diagram ˜ X p ˜ pi,j @@ @ @ @ @ @ αi,j ))S S S S S S S S S S S S S S S S S S S S S Ei πi ~~}}}} }}} qi // C X
Note that if (i, j) 6= (i0, j0), then qi(ei,j) 6= qi0(ei0,j0). Hence αi,j(˜x0) 6= αi0,j0(˜x0). Since
r1+ r2+ · · · + rk= n, the maps
αi,j, j = 1, ..., ri, i = 1, ..., k
3. Splitting coverings
Let Y → X be a covering space of X. We denote the group of covering transformationsp by A(Y /X), that is,
A(Y /X) = {Φ : Y → Y | Φ is a homeomorphism such that pΦ = p}.
Definition 3.1. Let Y → X be a covering space of X. Yp → X is called a Galois coveringp if A(Y /X) acts on a fibre of X transitively (hence all fibres).
Remark 3.2. From theory of covering spaces ([3, pg 25]), the above definition is equivalent to
p∗π1(Y, y0) π1(X, x0),
where x0 ∈ X, y0 ∈ Y , and p(y0) = x0. In particular, the universal covering is a Galois
covering.
3.1. The existence and uniqueness of splitting coverings
Definition 3.3. Let f be a Weierstrass polynomial of degree n on X and Y → X be ap covering space where Y is path-connected. Y is said to be a splitting covering of f if
1. f splits in Y ,
2. Y is the smallest among such coverings, that is, if Y0 p 0
→ X is another covering where f splits, then there exists a covering map π : Y0 → Y such that the diagram
Y0 π // p0 A A A A A A A Y p ~~~~~~ ~~ X commutes.
Construction of a splitting covering of f : Let h0 be an irreducible component of f in
C(X)[z]. Let E1 p1
→ X be the solution space of h0 and π1 : E1 → C be the projection to the
second component. Hence
(p∗1f )(z) = (z − π1)g1
in (p∗1C(X))[z]. Inductively, assume for i < n, we have
(p∗if )(z) = (z − qi∗· · · q∗2π1) · · · (z − q∗iπi−1)(z − πi)gi in (p∗iC(X))[z], where Ei qi // pi **V V V V V V V V V V V V V V V V V V V V V V V V V V Ei−1 qi−1 // · · · q2 // E1 p1 X
and πj : Ej → C is the projection of the last component, j = 1, ..., i. Note that gi is
a Weierstrass polynomial in Ei. For i + 1, let hi be an irreducible component of gi in
(p∗iC(X))[z], Ei+1 qi+1
→ Ei be the solution space of hi, pi+1= p1q2· · · qi+1 and πi+1: Ei+1→ C
be the projection of the last component. Hence
(p∗i+1f )(z) = (z − q∗i+1· · · q∗2π1) · · · (z − q∗i+1πi)(z − πi+1)gi+1
in (p∗i+1C(X))[z]. By induction, we have Ef := En q:=pn −→ X and (q∗f )(z) = (z − α1) · · · (z − αn−1)(z − αn) in (q∗C(X))[z] where αj := qn∗· · · q ∗ j+1πj,
j = 1, ..., n. Hence Ef is connected, and f splits in Ef. Note that an element in Ef is
of the form (· · · (((x, z1), z2) · · · , zn). We identify it as (x, z1, ..., zn). Then αj becomes the
projection of the (j + 1)th component, q becomes the projection of the first component and Ef ⊂ Sf
where
Sf := {(x, z1, ..., zn) ∈ X × Cn: fx(zi) = 0, i = 1, ..., n, and zi 6= zj if i 6= j}.
It is clear that (pr1, pr2) : Ef → E is a covering map where pr1, pr2 are the projection to the
first and second factor respectively.
Theorem 3.4. Let f be a Weierstrass polynomial of degree n in X. 1. Ef
q
→ X is a splitting covering.
2. Splitting covering is unique up to covering isomorphisms. 3. Ef
q
→ X is a Galois covering.
Proof. 1. Let Y → X be a covering space such that f splits in Y with roots βp 1, · · · , βn.
Let x0 ∈ X, (x0, z0,1, · · · , z0,n) ∈ Ef and y0 be any element in p−1(x0). After reordering
β1, · · · , βn if necessary, we can assume β1(y0) = z0,1, · · · , βn(y0) = z0,n. Define π : Y →
Ef by
π(y) = (p(y), β1(y), · · · , βn(y)).
Then
p = q ◦ π. (1)
For any e = (x, z1, · · · , zn) ∈ Ef there exists a path-connected open neighborhood U
of x in X such that q−1(U ) =`
iUi, p
−1(U ) =`
jVj, all Ui and Vj are open in Ef and
by (1), π(Vj) ⊂ Ui for some i. Hence
π|Vj = (q|Ui)
−1◦ (p|
Vj) (2)
is a homeomorphism. Therefore, if π(Y ) ∩ Ui 6= φ, then π(Vj) ∩ Ui 6= φ for some j, and
hence Ui = π(Vj) ⊂ π(Y ). In other words, either Ui ⊂ π(Y ) or Ui ⊂ π(Y )c. Therefore,
π(Y ) is open and closed in Ef. Since (x0, z0,1, · · · , z0,n) ∈ π(Y ) and Ef is connected,
π is surjective. Hence by (2), Y → Eπ f is a covering space.
2. Let x0 ∈ X and e0 ∈ q−1(x0) Suppose Y p
→ X is also a splitting covering. Then by the proof of part one, there exists a covering map π : Ef → Y and a covering map
π0 : Y → Ef such that the diagram
Ef π // q A A A A A A A A Y π0 ll p X
commutes, and π0(π(e0)) = e0. By the unique lifting theorem,
π ◦ π0 = idY, π
0
◦ π = idEf. Hence the coverings Y → X and Ep f
q
→ X are isomorphic.
3. Let x0 ∈ X and e0, e1 ∈ q−1(x0). By the argument in the first part, we have π0 :
(Ef, e0) → (Ef, e1) and π1 : (Ef, e1) → (Ef, e0) such that
q = q ◦ πi, i = 0, 1.
By the unique lifting theorem,
π1◦ π0 = π0◦ π1 = idEf,
and hence π0 ∈ A(Ef/X). Therefore, A(Ef/X) acts transitively on q−1(x0).
Recall that in Galois theory, the splitting field of a separable polynomial is Galois over the base field. The above result makes a parallel correspondence between splitting fields and splitting coverings.
Proposition 3.5. Let f be an irreducible Weierstrass polynomial on X of degree n with solution space E → X. Suppose p : Y → E is a covering space and q : Y → X is a Galoisπ covering of X where q = π ◦ p. Then f splits in Y .
Proof. Let x0 ∈ X, π−1(x0) = {e1, · · · , en} and y0 ∈ p−1(e1). Assume γi is a path from e1
to ei in E. Since q∗π1(Y, y0) = π∗p∗π1(Y, y0) ⊂ π∗π1(E, e1) and q∗π1(Y, y0) / π1(X, x0),
Consequently, there are pi : (Y, y0) → (E, ei) such that (Y, y0) q piJJJ%%J J J J J J **V V V V V V V V V V V V V V V V V V V V V V V V V (E, ei) π yytttttt ttt pr2 //C (X, x0)
commutes. Therefore, q∗f = (z − pr2 ◦ p1) · · · (z − pr2 ◦ pn) where pr2 : E → C is the
projection to the second factor.
Corollary 3.6. Let f be an irreducible Weierstrass polynomial of degree n on X and E → Xπ be its solution space. Suppose E → X is a Galois covering. Then Eπ → X is a splittingπ covering.
3.2. Another construction of splitting coverings
Recall that any symmetric polynomial in n variables can be written as a unique polyno-mial in the elementary symmetric polynopolyno-mials, s0, · · · , sn−1 where
n Y i=1 (z − zi) = zn+ n−1 X i=0 (−1)n−isi(z1, · · · , zn)zi.
Hence there is a unique polynomial in n variables δ(a0, · · · , an−1) such that
δ(−sn−1(z1, · · · , zn), · · · , (−1)n−isi(z1, · · · , zn), · · · , (−1)ns0(z1, · · · , zn)) =
Y
16i<j6n
(zi− zj).
The polynomial δ(a0, · · · , an−1) is called the discriminant polynomial. Define
Bn := Cn− Z(δ) where Z(δ) is the set of zeros of δ.
Lemma 3.7. 1. Let S := {(a0, · · · , an−1, z1, · · · , zn) ∈ Bn× Cn : zn+ n−1 X i=0 aizi = n Y i=1 (z − zi)}
and π be the projection to Bn. Then π : S → Bn is an n!-fold covering space.
2. Let Sf := {(x, z1, · · · , zn) ∈ X × Cn : fx(zi) = 0, i = 1, · · · , n, and zi 6= zj if i 6= j}.
Then Sf ˜ q
→ X is an n!-fold covering space where ˜q is the projection to X. Proof. 1. Similar to [5, pg 88, Lemma 2.2].
2. Let f = zn+Pn−1
i=0 aiz
i and a : X → Bn:
a(x) := (a0(x), · · · , an−1(x)).
Then we get the induced fibre bundle
a∗(S) a∗ // π∗ S π X a //Bn. Define a0 : Sf → a∗(S) by a0(x, z1, · · · , zn) := (x, a(x), z1, · · · , zn).
Then a0 is a homeomorphism such that the diagram
Sf a 0 // ˜ q@@@ @ @ @ @ @ a∗(S) π∗ ||yyyyyy yy X commutes. Hence Sf ˜ q
→ X is an n!-fold covering space.
Proposition 3.8. Let E0 be a connected component of Sf and q := ˜q|E0. Then E0 → X isq a splitting covering.
Proof. By the previous lemma, E0 → X is a covering space. Moreover, f splits in Eq 0 with
roots α1, · · · , αn where αi is the projection of the (i + 1)th component, i = 1, · · · , n. The
result follows as in the proof of Theorem 3.4.
Observe that S → Bπ nis a Galois covering since the map (a = (a
0, · · · , an−1), z1, · · · , zn) 7→
(a, zσ(1), · · · , zσ(n)) is a covering transformation for each σ ∈ Sn. Consequently, S π
→ Bn
be-comes a locally trivial principal A(S/Bn)-bundle (A(S/Bn) with discrete topology). Let’s recall a classical result about principal G-bundles ([7, pg 51, Theorem 9.9]).
Proposition 3.9. Let ξ be a numerable principal G-bundle over B (which is true whenever B is Hausdorff and paracompact), and let ft : B0 → B be a homotopy. Then the principal
G-bundles f0∗(ξ) and f1∗(ξ) are isomorphic over B0. Corollary 3.10. Let f = zn+Pn−1
i=0 aizi and g = zn+
Pn−1
i=0 bizi be two Weierstrass
poly-nomials on a Hausdorff and paracompact space X and a, b : X → Bn:
Let Ea qa
→ X and Eb qb
→ X be the splitting cover of f and g respectively. If a and b are homotopic as maps of X into Bn, then E
a qa
→ X and Eb qb
→ X are equivalent covering spaces. 4. Semi-topological Galois groups
All rings are assumed to be commutative rings with identity if without mentioned explic-itly.
Definition 4.1. Let ¯T be a ring and T be a subring of ¯T . Define AutT( ¯T ) = {φ : ¯T → ¯T | φ is a T − algebra automorphism}
= {φ : ¯T → ¯T | φ is a ring automorphism such that φ(x) = x, ∀x ∈ T }. Let f = fx(z) = zn+ an−1(x)zn−1+ · · · + a0(x) be a Weierstrass polynomial on X, and
let Ef q
→ X be the splitting covering. We define
R = q∗C(X) = {γ ◦ q : γ ∈ C(X)},
which is a subring of C(Ef). The semi-topological Galois group of f is defined to be
Gf := AutR(R[α1, · · · , αn])
where α1, · · · , αn: Ef → C are the solutions of f.
Although the domain of solutions is mentioned to be the splitting covering in the defini-tion of semi-topological group of f , it is, in fact, not important. To be precise, we have the following proposition.
Proposition 4.2. Let Ef q
→ X be the splitting covering of f with roots α1, · · · , αn : Ef → C.
Suppose f splits on Y → X with solutions αp 0
1, · · · , α0n : Y → C. Then there exists an
isomorphism Φ : q∗C(X)[α1, · · · , αn] → p∗C(X)[α01, · · · , α 0 n] such that Φ(q ∗C(X)) = p∗C(X), Φ({α1, · · · , αn}) = {α01, · · · , α 0 n}, and hence, Gf ∼= Autp∗C(X)p∗C(X)[α01, · · · , α0n].
Proof. By the definition of splitting coverings, there is a covering map π : Y → Ef such that
Y p ? ? ? ? ? ? ? ? π //Ef q ~~}}}} }}} X
commutes. Observe that π∗α1, · · · , π∗αn are all roots as for i = 1, · · · , n, (p∗f )(π∗αi) =
(π∗q∗f )(π∗αi) = π∗((q∗f )(αi)) = 0 and π∗α1, · · · , π∗αn are distinct. Therefore, Φ := π∗ :
q∗C(X)[α1, · · · , αn] → p∗C(X)[α01, · · · , α 0
n] is an isomorphism which carries q
p∗C(X) and {α1, · · · , αn} to {α01, · · · , α 0
n}. As a result, we obtain an isomorphism Ψ : Gf →
Autp∗C(X)p∗C(X)[α0
1, · · · , α 0
n] which is defined by Ψ(φ)(g) = (π
∗)φ((π∗)−1g).
Example 4.3. Let fx(z) = zn − x for x ∈ S1 where n ∈ N. Then f is a Weierstrass
polynomial, and its solution space E is an n-fold covering of S1.
Let p : R → X be defined by p(s) = e2πsi which is the universal covering space of
S1. (p∗f )s(z) = zn− e2πsi, where s ∈ R. It is easy to see that roots of p∗f are αj(s) =
e2πi(s+j−1)n , j = 1, · · · , n. Note that for j = 1, · · · , n − 1, e 2πi
n αj = αj+1, and the constant function e2πin is an element in R = p∗C(S1). Therefore, for φ ∈ Gf,
e2πin φ(αj) = φ(e 2πi
n αj) = φ(αj+1).
Hence φ is uniquely determined by φ(α1). Let σ : R[α1, · · · , αn] → R[α1, · · · , αn] be defined
by σ(˜γ)(s) := ˜γ(s + 1). Then for p∗γ ∈ R, σ(p∗γ)(s) = σ(γ ◦ p)(s) = γ(p(s + 1)) = γ(p(s)) = (p∗γ)(s). Hence σ|R= idR. For j = 1, · · · , n − 1, σ(αj) = αj+1, σ(αn) = α1.
Therefore σ ∈ Gf. Furthermore, for j = 0, 1, · · · , n − 1, σj(α1) = α1+j, and σn= id. Thus,
from the above observations, we have
Gf ∼=< σ >∼= Zn.
Proposition 4.4. (Functoriality) Suppose that λ : Y → X is a covering map and f1 is a
Weierstrass polynomial of degree n in X. Let f2 = λ∗f1.
1. There is a covering map eλ : Ef2 → Ef1 such that the following diagram commutes: Ef2 e λ // q Ef1 p Y λ //X
2. If α1, · · · , αn : Ef1 → C are all the roots of p
∗f
1, then eλ∗α1, · · · , eλ∗αn are all the roots
of q∗(f2).
3. eλ∗ : p∗C(X)[α1, · · · , αn] → q∗C(Y )[eλ∗α1, · · · , eλ∗αn] is injective.
4. The map eλ induces a group monomorphism b
defined by
b
λ(φ)(α) = (eλ∗)−1φ(α ◦ eλ) Proof. 1. Since f2 splits in Ef2 and q
∗f
2 = q∗λ∗f1 = (λ ◦ q)∗f1, hence f1 also splits in Ef2. By the definition of splitting covering, there is a covering map eλ : Ef2 → Ef1.
2. This follows from a direct computation: (q∗(f2))y((eλ∗αj)(y)) = ((λ◦q)∗f1)y(αj(eλ(y))) =
((p ◦ eλ)∗f1)y(αj(eλ(y))) = (p∗f1)eλ(y)(αj(eλ(y))) = 0.
3. Since eλ is surjective, so eλ∗ : C(Ef1) → C(Ef2) is injective. For g ∈ C(X), eλ
∗(p∗g) = (p ◦
e
λ)∗g = (λ ◦ q)∗g = q∗(λ∗g) ∈ q∗C(Y ). Thus the restriction of eλ∗ to p∗C(X)[α1, · · · , αn]
is also injective.
4. Observe that for φ ∈ Gf2, φ fixes eλ
∗p∗C(X) ⊂ q∗C(X); hence, φ(eλ∗(p∗C(X)[α
1, · · · , αn])) ⊂
e
λ∗(p∗C(X)[α1, · · · , αn]). Therefore, bλ is well defined. It is a direct checking that bλ is
a group homomorphism. Suppose that bλ(φ1) = bλ(φ2). Then for any αj, φ1(eλ∗αj) =
φ1(αj ◦ eλ) = eλ∗bλ(φ1)(αj) = eλ∗λ(φb 2)(αj) = φ2(eλ∗αj). So φ1 = φ2 and hence bλ is injective.
Proposition 4.5. Let f be a Weierstrass polynomial of degree n in X and split in Y where Y → X is a covering. Let αq 1, · · · , αn be the roots of q∗f in Y . Suppose that T is a subring
of q∗C(X) and G = AutTT [α1, · · · , αn]. Then we have the following group homomorphism
ωY,T : A(Y /X) → G defined by
ωY,T(Φ)(β)(y) := (Φ−1)∗(β)(y) := β(Φ−1(y)).
In particular, we have a group homomorphism ωf = ωEf,q∗C(X) : A(Ef/X) → Gf where Ef
q
→ X is the splitting covering of f .
Proof. For Φ ∈ A(Y /X), it is easy to check that (Φ−1)∗ : T [α1, ..., αn] → T [α1, ..., αn] is a
ring automorphism. Since (Φ−1)∗|T = idT, (Φ−1)∗ ∈ AutTT [α1, ..., αn].
5. Galois correspondence
5.1. Correspondences between commutative rings and groups
For a Weierstrass polynomial f in X, we have two groups associated to f : A(Ef/X) and
Gf. We will show that these two groups are actually isomorphic. In order to do that, we use
the Galois theory of commutative rings developed by Chase-Harrison-Rosenberg ([2], [4]). All rings are supposed to be commutative rings with identity and connected, that is, have no idempotent other than 0 and 1 unless otherwise stated.
Definition 5.1. 1. A commutative R-algebra S is separable if S is a projective Se-module where Se = S ⊗
2. Let R be a subring of S and G be a finite subgroup of AutR(S). S is said to be
G-Galois over R if R = SG := {x ∈ S : σ(x) = x, ∀σ ∈ G}, and there exist elements
x1, · · · , xn; y1, · · · , yn of S such that n
X
i=1
xiσ(yi) = δe,σ, ∀σ ∈ G,
where e is the identity in G and δe,σ =
1 , if σ = e 0 , if σ 6= e.
Theorem 5.2. ([2, Theorem 2.3]) Let S be G-Galois over R. Then there is a one-to-one lattice-inverting correspondence between subgroups of G and separable R-subalgebras of S. If T is a separable R-subalgebra of S, then the corresponding subgroup is HT = {σ ∈ G :
σ(t) = t, ∀t ∈ T }. If H is a subgroup of G, then the corresponding separable R-subalgebra is SH = {x ∈ S : σ(x) = x, ∀σ ∈ H}.
Definition 5.3. Let α1, · · · , αnbe some symbols. For σ ∈ Sn, we denote
P i1,··· ,inα i1 σ(1)α i2 σ(2)· · · α in σ(n) by σ(P i1,··· ,inα i1 1 α i2 2 · · · αnin). For (i1, · · · , in−1), (j1, · · · , jn−1) ∈ Nn−1, we define (i1, · · · , in−1) ≺ (j1, · · · , jn−1) ⇔ in−k = jn−k, in−l < jn−l, k = 1, · · · , l − 1, f or some l. We denote α(i1,··· ,in−1) := αi1 1α i2 2 · · · α in−1
n−1 for ik = 0, · · · , n − k. Then from the above “≺”,
we give an ordering to A = {(i1, · · · , in−1) : ik = 0, · · · , n − k, k = 1, · · · , n − 1}. We list
all elements of A according to the ordering:
I1 = (0, 0, ..., 0), I2 = (1, 0, ..., 0), I3 = (2, 0, ..., 0), ..., In! = (n − 1, n − 2, ..., 1)
and define
xj = αIj, j = 1, 2, ..., n!
Let σi := (i i + 1 · · · n) ∈ Sn (note that the order of σi is n − i + 1.) and σ(i1,··· ,in−1) :=
σi1
1 σ i2
2 · · · σ in−1
n−1 for ik = 0, · · · , n − k. Similarly, we define φi = σIi for i = 1, 2, ..., n! Finally,
we define Vn to be the following n! × n! matrix:
Vn:= (φi(xj))i,j=1,··· ,n!.
Example 5.4. For n = 2, V2 =
1 α1
1 α2
, and note that det(V2) = α2− α1.
Example 5.5. For n = 3, x1 = 1, x2 = α1, x3 = α21, x4 = α2, x5 = α1α2, x6 = α21α2,
φ6 = σ12σ2 = (1 3), and V3 = 1 α1 α21 α2 α1α2 α21α2 1 α2 α22 α3 α2α3 α22α3 1 α3 α23 α1 α3α1 α23α1 1 α1 α21 α3 α1α3 α21α3 1 α2 α22 α1 α2α1 α22α1 1 α3 α23 α2 α3α2 α23α2 . Note that det(V3) = det 1 α1 α21 α2 α1α2 α21α2 1 α2 α22 α3 α2α3 α22α3 1 α3 α23 α1 α3α1 α23α1 1 α1 α21 α3 α1α3 α21α3 1 α2 α22 α1 α2α1 α22α1 1 α3 α23 α2 α3α2 α23α2 = det 1 α1 α21 α2 α1α2 α21α2 1 α2 α22 α3 α2α3 α22α3 1 α3 α23 α1 α3α1 α23α1 0 0 0 (α3− α2) α1(α3 − α2) α21(α3− α2) 0 0 0 (α1− α3) α2(α1 − α3) α22(α1− α3) 0 0 0 (α2− α1) α3(α2 − α1) α23(α2− α1) = −[(α2− α1)(α3− α1)(α3− α2)]3. Lemma 5.6. Sn = {σ(i1,··· ,in−1) : ik= 0, · · · , n − k}.
Proof. Let Bl = {σ(0,··· ,0,in−l+1,··· ,in−1) : ik = 0, · · · , n − k, f or k = n − l + 1, · · · , n − 1}, for
l = 2, · · · , n. We claim that
Bl= permu{n − l + 1, · · · , n} := {σ ∈ Sn: σ(j) = j, ∀j = 1, · · · , n − l}.
The cases l = 1, 2, 3 are obvious. Assume Bl−1 = permu{n − l + 2, · · · , n}, l 6 n. Then
Bl−1 6 Sn, and we have a partition Sn/Bl−1. Recall that for τ, ρ ∈ Sn,
τ Bl−1 = ρBl−1 ⇔ τ ρ−1 ∈ Bl−1.
Also note that if 0 6 p, q 6 l − 1, p 6= q, then (σn−l+1p )(σ q n−l+1)
−1 = σp−q
n−l+1 is a cycle of
length l, where σn−l+1 is defined as in Definition 5.3. Thus
Hence Bl=`l−1p=0σn−l+1p Bl−1. Therefore,
|Bl| = l|Bl−1| = l|permu{n − l + 2, · · · , n}| = l!.
Moreover, just from the definition, we have Bl ⊆ permu{n − l + 1, · · · , n}, and |permu{n −
l + 1, · · · , n}| = l! = |Bl|. This implies Bl = permu{n − l + 1, · · · , n}. Thus, by induction,
we proved what we claimed. In particular,
Sn= Bn= {σ(i1,··· ,in−1) : ik = 0, · · · , n − k, k = 1, · · · , n}.
Lemma 5.7. Let T be an integral domain. Suppose α1, · · · , αn ∈ T are distinct. Then for
each n ∈ N, det(Vn) 6= 0.
Proof. Let Ck = Ck,0= (φi(xj))i,j=1,··· ,(n−k)!n! , and Ck,l = (φi(xj))i=1,··· ,(n−k)!n! ,j=l(n−k)!n! +1,··· ,(l+1)(n−k)!n! ,
for l = 0, · · · , (n − k − 1). Moreover, for σ ∈ Sn , we define
Ck,lσ := (φi(σ(xj)))i=1,··· , n! (n−k)!,j=l n! (n−k)!+1,··· ,(l+1) n! (n−k)!. Then for ik+1 = 0, · · · , n − k − 1, Ck,lσ ik+1 k+1 = (φi(σ ik+1 k+1(xj)))i=1,··· ,(n−k)!n! ,j=l(n−k)!n! +1,··· ,(l+1)(n−k)!n! = ((σ(i1,··· ,ik,0,··· ,0)◦ σik+1 k+1)(xj))Ii=(i1,··· ,ik),i=1,··· ,(n−k)!n! ,j=l(n−k)!n! +1,··· ,(l+1)(n−k)!n! = (σ(i1,··· ,ik,ik+1,0,··· ,0)(x j))Ii=(i1,··· ,ik),i=1,··· ,(n−k)!n! ,j=l n! (n−k)!+1,··· ,(l+1) n! (n−k)! = (φi(xj))i=(ik+1)(n−k)!n! +1,...,(ik+1+1)(n−k)!n! ,j=l n! (n−k)!+1,··· ,(l+1) n! (n−k)!
Hence, by definition, we can divide Ck+1 into several blocks in the following way:
Ck+1 = Ck,0 Ck,1 . . . Ck,(n−k−1) Ck,0σk+1 Ck,1σk+1 . . . Ck,(n−k−1)σk+1 .. . ... . .. ... Ck,0σk+1n−k−1 Ck,1σk+1n−k−1 . . . Ck,(n−k−1)σk+1n−k−1 .
We want to express det(Ck) in term of det(Ck+1) and then use induction. For j =
l(n−k)!n! + 1, · · · , (l + 1)(n−k)!n! and xj = α(j1,··· ,jn−1), we have jk+2 = · · · = jn−1 = 0 and
xj = αj11· · · α jk k α l k+1. Therefore, for p = 0, 1, · · · , n − k − 1, σpk+1(xj) = αj11· · · α jk k α l k+p+1.
If l 6= 0, Ck,lσpk+1− Ck,l = (φi(σk+1p (xj) − xj))i=1,··· ,(n−k)!n! ,j=l(n−k)!n! +1,··· ,(l+1)(n−k)!n! = (φi(αj11· · · α jk k )φi(αlk+p+1− α l k+1))i=1,··· , n! (n−k)!,j=l n! (n−k)!+1,··· ,(l+1) n! (n−k)! = (φi(α j1 1 · · · α jk k )φi( l−1 X q=0 αk+1l−1−qαqk+p+1)φi(αk+p+1− αk+1))i=1,··· ,(n−k)!n! ,j=l(n−k)!n! +1,··· ,(l+1)(n−k)!n! . If l = 0, Ck,0σk+1p − Ck,0= 0.
Multiplying the first row of Ck+1 by (−1) and added to each other row, we have
det(Ck+1) = det Ck,0 Ck,1 . . . Ck,(n−k−1) 0 Ck,1σk+1− Ck,1 . . . Ck,(n−k−1)σk+1− Ck,(n−k−1) .. . ... . .. ... 0 Ck,1σn−k−1k+1 − Ck,1 . . . Ck,(n−k−1)σn−k−1k+1 − Ck,(n−k−1)
Note that φi(αk+p+1 − αk+1) is a common factor of entries of (p + 1)-row. Take all of them
out and let the remaining matrix be D, we get
det(Ck+1) = det(Ck)det(D)( n−k−1 Y p=1 n! (n−k)! Y i=1 φi(αk+p+1− αk+1)),
Let the matrix
Er,sm = (φi(xj)φi( X q0+···+qm=s−1 αq1 k+1α q2 k+2· · · α qm k+mα q0 k+r+m))i,j=1,..., n! (n−k)!
and the matrix
Dm = (Er,sm)r,s=1,...,n−k−m
Note that D1 = D and Er,1m = Ck.
Then det(Dm) = det E1,1m E1,2m . . . E1,n−k−mm Em 2,1 E2,2m . . . E2,n−k−mm .. . ... . .. ... Em n−k−m,1 En−k−m,2m . . . En−k−m,n−k−mm = det Ck E1,2m . . . E1,n−k−mm 0 E2,2m − Em 1,2 . . . E2,n−k−mm − E m 1,n−k−m .. . ... . .. ... 0 Em n−k−m,2− E1,2m . . . En−k−m,n−k−mm − E1,n−k−mm
Note that Er,sm − Em 1,s = (φi(x)φi( X q1+···+qm=s−1 αq1 k+1α q2 k+2· · · α qm k+m(α q0 k+r+m− α q0 k+1+m)) = (φi(x)φi( X q0+···+qm=s−1 αq1 k+1α q2 k+2· · · α qm k+m( X a+b=q0−1 αak+m+1αbk+r+m)φi(αk+r+m− αk+1+m))) = (φi(x)φi( X q0+···+qm+1=s−2 αq1 k+1α q2 k+2· · · α qm k+mα qm+1 k+m+1α q0 k+(r−1)+(m+1))φi(αk+r+m− αk+1+m)) = (Er−1,s−1m+1 φi(αk+r+m− αk+1+m)) So we have
det(Dm) = det(Ck)det(Dm+1)( n−k−m Y r=2 n! (n−k)! Y i=1 φi(αk+r+m− αk+m+1)).
Since det(Dn−k−1) = det(Ck), inductively we get det(D1) 6= 0. Therefore det(Ck+1) 6= 0.
Because of det(C1) = Q16i<j6n(αj − αi) 6= 0, by induction, det(Ck) 6= 0 for k ∈ N. In
particular,
det(Vn) = det(Cn−1) 6= 0.
Definition 5.8. Suppose that f is a Weierstrass polynomial in X and p : Y → X is a covering where f splits with roots α1, ..., αn. Define V : Y → Mn(C) by
V (y) := (φi(xj(y)))i,j=1,...,n!
and ∆ : Y → C by
∆(y) := det(V (y)) where φi, xj are defined in Definition 5.3.
Lemma 5.9. Let Y → X be a Galois covering. Suppose λ : Y → C is a continuous mapp satisfying
λ ◦ Φ = λ, ∀Φ ∈ A(Y /X). Then λ ∈ p∗C(X).
Proof. Since p : Y → X is a quotient map, it suffices to show that for each x0 ∈ X,
x, y ∈ p−1(x0), λ(x) = λ(y). The group A(Y /X) acts on p−1(x0) transitively, so there is
Φ ∈ A(Y /X) such that Φ(x) = y. Hence
Lemma 5.10. Let f be a Weierstrass polynomial in X and split in Y → X with rootsq α1, · · · , αn. Let R = q∗C(X), and let T be a subring of R containing coefficients of f such
that
1. T [α1, · · · , αn] ∩ R = T
2. 1
∆ ∈ T [α1, · · · , αn]. Then
1. T [α1, · · · , αn] is G-Galois over T where G = AutTT [α1, · · · , αn].
2. Then the group homomorphism ωY,T : A(Y /X) → G = AutTT [α1, · · · , αn] is surjective.
Proof. 1. We have
T ⊂ T [α1, · · · , αn]G⊂ T [α1, · · · , αn]ωY,T(A(Y /X)).
By Proposition 4.2, we may assume Y = Ef which is Galois, and hence Lemma 5.9
and the assumptions demonstrate that
T ⊂ T [α1, · · · , αn]G ⊂ T [α1, · · · , αn] ∩ R = T. Since 1 ∆ ∈ T [α1, · · · , αn], we may define y1(t) y2(t) .. . yn!(t) := (V (t))−1 1 0 .. . 0 for t ∈ Y then V (t) y1(t) y2(t) .. . yn!(t) = 1 0 .. . 0 In other words, n! X i=1 σ(xi)yi = δe,σ, σ ∈ Sn
by Lemma 5.6. Since T contains all coefficients of f , G merely permutes α1, · · · , αn.
Therefore,
n!
X
i=1
σ(xi)yi = δe,σ, σ ∈ G
2. Since Y is a Galois covering over X, from Lemma 5.9,
T ⊂ T [α1, · · · , αn]G⊂ T [α1, · · · , αn]ωY,T(A(Y /X))⊂ R ∩ T [α1, · · · , αn] = T
which implies
T [α1, · · · , αn]G= T [α1, · · · , αn]ωY,T(A(Y /X)).
Therefore, by part one and Chase-Harrison-Rosenberg Theorem, ωY,T(A(Y /X)) = G
Theorem 5.11. Let f be a Weierstrass polynomial in X and split in Y → X with rootsq α1, · · · , αn. Let R = q∗C(X). Then R[α1, · · · , αn] contains ∆−1. In particular, R[α1, · · · , αn]
is Gf-Galois over R and the group homomorphisms
ωEf : A(Ef/X) → Gf is surjective.
Proof. Since ωY(A(Y /X)) < Gf, Lemma 5.9 implies
R ⊂ R[α1, · · · , αn]Gf ⊂ R[α1, · · · , αn]ωY(A(Y /X))⊂ R.
Hence R[α1, · · · , αn]Gf = R. By definition, ∆ ∈ R[α1, · · · , αn], and for each σ ∈ Gf which
permutes α1, · · · , αn,
σ(∆) = det((σ ◦ φi(xj))i,j=1,··· ,n!) = sign(σ)∆.
Therefore, ∆2 ∈ R[α1, · · · , αn]Gf = R. By Lemma 5.7, ∆(t) 6= 0, ∀t ∈ Y . Hence ∆12 ∈ R, and 1
∆ =
∆
∆2 ∈ R[α1, · · · , αn]. By Lemma 5.10, R[α1, · · · , αn] is Gf-Galois over R.
Corollary 5.12. (cf. Example 4.3) Let X = S1 = {z ∈ C : |z| = 1} and ˜X = Rp(s)=e 2πsi
−→ X
is a universal covering space of S1. Then G
f is cyclic with a generator φ∗, where φ : R →
R : s 7→ s + 1.
Example 5.13. (cf. Example 4.3) Let X = S1 = {z ∈ C : |z| = 1} and f = (z2−x)(z2−2x),
x ∈ S1. Then f is a Weierstrass polynomial, and its solution space E is homeomorphic to a
disjoint union of two circles and a 4-fold covering of S1.
˜
X = R p(s)=e 2πsi
−→ X is the universal covering space of S1. p∗f = (z2 − e2πsi)(z2 − 2e2πsi),
where s ∈ R. It is easy to see that
α1(s) := eπis, α2(s) := −α1(s), β1(s) :=
√
are all roots. By the previous corollary,
Gf =< φ∗ >∼= Z2.
Here we give another direct proof of , e.g., α1 7→ α2, β1 7→ β1 can’t induce an element in
Gf. In fact, this is immediately from the observations
α1β1 ∈ R = p∗C(X),
but
α1β1 6= α2β1.
5.2. The fundamental theorem of Galois theory
In this section, we will develop a theorem corresponding to the fundamental theorem of Galois theory. More precisely, we will construct an 1-1 correspondences between certain ring extensions and Galois groups, Galois groups and covering transformations, covering transformations and covering spaces.
Definition 5.14. Let Y → X be a covering space and x ∈ X. The cardinality of pp −1(x)
is called the degree of Y over X, denoted by [Y : X]. If H is a subgroup of G, we denote |G/H| by [G : H].
The following result is clear.
Lemma 5.15. If Z → Y and Yq → X are two covering spaces with finite fibres, then Zp → Xpq is a covering and
[Z : X] = [Z : Y ][Y : X].
Lemma 5.16. If Y → X is a Galois covering, then G := A(Y /X) has order [Y : X].p Proof. Since Y is Galois over X, the quotient space π : Y → Y /G is a covering equivalent to Y → X. Hence the number of each fibre is |G|.p
Lemma 5.17. Let (Z, z0) q
→ (Y, y0) and (Y, y0) p
→ (X, x0) be two covering spaces. If Z pq
→ X is Galois, then Z → Y is Galois.q
Proof. Since Z is Galois over X, (pq)∗π1(Z, z0) π1(X, x0). Also note that (pq)∗π1(Z, z0) <
p∗π1(Y, y0). Hence (pq)∗π1(Z, z0) p∗π1(Y, y0) which implies
q∗π1(Z, z0) π1(Y, y0).
Therefore, Z → Y is Galois.q
Theorem 5.18. Let f be a Weierstrass polynomial of degree n in X, and α1, ..., αn be the
n roots of f in the splitting covering q : (Ef, e0) → (X, x0). Suppose T is a subring of R
1. T [α1, · · · , αn] ∩ R = T
2. 1
∆ ∈ T [α1, · · · , αn]. Then
1. ω = ωEf,T : A(Ef/X) → G = AutTT [α1, · · · , αn] is an isomorphism.
2. We have the following one-to-one correspondences among (based) covering spaces be-tween (Ef, e0)
q
→ (X, x0), subgroups of A(Ef/X), subgroups of Gf, and separable
sub-rings of T [α1, · · · , αn] over T (Ef, e0) ←→ < e > ←→ < e0 > ←→ T [α1, · · · , αn] ↓ ∧ ∧ ∪ (L, l0) ←→ H ←→ H0 ←→ L0 ↓ ∧ ∧ ∪ (M, m0) ←→ J ←→ J0 ←→ M0 ↓ ∧ ∧ ∪ (X, x0) ←→ A(Ef/X) ←→ G ←→ T
which are given by the theory of covering spaces, ω, and Chase-Harrison-Rosenberg theorem, that is, H = A(Ef/L), H0 = ω(H), L0 = T [α1, · · · , αn]H
0 , and H0 = GL0 = {φ ∈ G | φ|L0 = idL0}. Moreover, [L : M ] = [J : H] = [J0 : H0]. In particular, A(Ef/X) ∼= Gf. Proof. Let Ef q
→ X be the splitting covering constructed in the section of splitting coverings and αi be the projection of the (i + 1)th component, i = 1, · · · , n. Suppose Φ ∈ ker(ω) and
write Φ : Ef → Ef as Φ(x, z1, · · · , zn) = (Φ1(x, z1, · · · , zn), · · · , Φn+1(x, z1, · · · , zn)). Since
(Φ−1)∗(αi) = αi, αi◦ Φ = αi, i = 1, · · · , n. Therefore
Φi+1(x, z1, · · · , zn) = αi(Φ(x, z1, · · · , zn)) = αi(x, z1, · · · , zn) = zi,
and
Φ1(x, z1, · · · , zn) = q(Φ(x, z1, · · · , zn)) = q(x, z1, · · · , zn) = x.
Hence Φ = idEf. In other words, ω is injective. Furthermore, from Theorem 5.10, ω is surjective; therefore, ω is an isomorphism.
For the second part, the correspondences follow part one. Since Ef is Galois over X, it
is also Galois over L. Hence [Ef : L] = |A(Ef/L) = H|. Similarly, [Ef : M ] = |J | Therefore,
Corollary 5.19. Let f be a Weierstrass polynomial of degree n on X with roots α1, · · · , αn:
Ef → C where Ef q
→ X is the splitting covering. Then C(Ef) = q∗C(X)[α1, · · · , αn].
In particular,
A(Ef/X) ∼= Autp∗C(X)C(Ef).
Proof. Clearly, C(Ef) ⊃ q∗C(X)[α1, ..., αn]. Conversely, we consider the map ω : A(Ef/X) →
Autq∗C(X)C(Ef) defined by
ω(Φ) = (Φ−1)∗. By Lemma 5.9,
C(Ef)H = q∗C(X)
where H is the finite group ω(A(Ef/X)). Moreover, from the proof of Lemma 5.10, there
are x1, ..., xn!, y1, ..., yn! ∈ q∗C(X)[α1, ..., αn] ⊂ C(Ef) such that n!
X
i=1
σ(xi)yi = δe,σ, σ ∈ H.
Therefore, C(Ef) is H-Galois over q∗C(X). As a result of Chase-Harrison-Rosenberg Theorem
and Lemma 5.10,
C(Ef) = q∗C(X)[α1, ..., αn]
since C(Ef) ⊃ q∗C(X)[α1, ..., αn] are both separable over q∗C(X) and
Hq∗C(X)[α1,...,αn]= {e} = HC(Ef).
In general not all covering spaces are equivalent to some polynomial covering spaces. The following fact is a classical result.
Theorem 5.20. ([5, Theorem 6.3, pg 110]) Suppose the π1(X) is a free group. Then every
finite covering map onto X is equivalent to a polynomial covering map, that is, there is a Weierstrass polynomial on X such that its solution space is equivalent to that finite covering as a covering space.
It is natural to ask whether any connected Galois covering of finite degree is equivalent to the splitting covering of a Weierstrass polynomial. At this moment we only have some partial results.
Proposition 5.21. Suppose π1(X) is free. Then any finite connected Galois covering of X
Proof. Let Y → X be a finite connected Galois covering of X. Theorem 5.20 demonstratesp that there is a Weierstrass polynomial f on X such that its solution space is equivalent to Y → X. Moreover, by Corollary 3.6, the splitting covering Ep f
q
→ X is equivalent to Y → X.p
Corollary 5.22. Suppose π1(X) is free and Y p
→ X is a finite connected Galois covering of X. Then
A(Y /X) ∼= Autp∗C(X)C(Y ), and C(Y ) is A(Y /X)-Galois over p∗C(X).
Proof. By Proposition 5.21, there exists a Weierstrass polynomial f on X with splitting covering Ef
q
→ X equivalent to Y → X, that is, there is a covering equivalence Φ : Y → Ep f.
Therefore, Φ∗ gives an isomorphism from C(Ef) to C(Y ) mapping q∗C(X) onto p∗C(X). As
a result of Proposition 5.19,
A(Y /X) ∼= A(Ef/X) ∼= Autp∗C(X)C(Y ), and C(Y ) is A(Y /X)-Galois over p∗C(X).
6. Groups as Galois groups
The inverse Galois problem is a hundred-year old open problem which asks if every finite group can be realized as some Galois group over the field of rational numbers. Hilbert used his irreducibility theorem to realize some groups like Sn and Shafarevich used tools from
number theory to show that every solvable finite group can be realized. Recently rigidity method is used to realize many finite groups but there is no way which may realize all finite groups. Here we provide a new viewpoint from our semi-topological Galois theory toward the inverse Galois problem.
Proposition 6.1. Let G be any finite group. For any compact disc D ⊆ C, there exist some disjoint open discs D1, ..., Dm ⊆ D and an irreducible Weierstrass polynomial defined on
X = D − ∪m
j=1Dj with semi-topological Galois group G.
Proof. Since G is finite, there exists a finite generated free group F and a normal subgroup N of F such that
G = F/N.
If F is generated by m elements, then we take m disjoint open discs D1, ..., Dm in D such
that
π1(X) ∼= F
where X = D−∪m
j=1Dj. By the Galois correspondence of covering spaces, there is a connected
covering space E → X such thatp
and
A(E/X) ∼= F/N ∼= G.
Hence E → X is Galois. By Theorem 5.20, E is equivalent to the solution space Ep 0 of
some irreducible Weierstrass polynomial f defined on X. Since p0 : E0 → X is Galois, by Corollary 3.6, E0 is the splitting covering of f . By Theorem 5.18, the semi-topological Galois group of f is G.
Corollary 6.2. Let G be any finite group and D be a compact disc in C. Then there exist finite disjoint open discs D1, · · · , Dm in D and an irreducible Weierstrass polynomial in
X = D − ∪mj=1Dj with coefficients in Q(i)[x] and semi-topological Galois group G.
Proof. By the above result, there is a Weierstrass polynomial f = zn+ a
n−1zn−1+ · · · + a0 ∈
C(X)[z] such that Gf ∼= G for some X = D − ∪Nj=1Dj. Let a : X → Bn be the continuous
function defined by
a(x) = (a0(x), · · · , an−1(x))
Then a(X) ⊂ Bn is compact where Bn = Cn− Z(δ) and δ is the discriminant polynomial. Since Z(δ) is closed in Cn, the distance between a(X) and Z(δ) is ε = d(a(X), Z(δ)) > 0.
By the Stone-Weierstrass theorem, there are ˜a0, · · · , ˜an−1 ∈ Q(i)[x] such that
k˜aj− ajk < ε/2n, j = 0, · · · , n − 1
where ||˜aj− aj|| = maxx∈X|˜aj(x) − aj(x)|. Hence
||˜a − a|| ≤
n
X
i=1
||˜aj− aj|| < ε/2.
Then for any x ∈ X,
d(˜a(x), Z(δ)) ≥ d(a(x), Z(δ)) − d(a(x), ˜a(x)) > ε − ε/2 = ε/2.
Therefore we have a map ˜a = (˜a0, · · · , ˜an−1) : X → Bn and a Weierstrass polynomial
˜ f = zn+Pn−1 j=0 ˜ajz j. Let H(x, t) := (1 − t)a(x) + t˜a(x)
for t ∈ [0, 1], x ∈ X. Then |a(x) − H(x, t)| = t|a(x) − ˜a(x)| < tε/2 ≤ ε/2, so H : X × I → Bn
is a homotopy between a and ˜a. Hence Corollary 3.10 and Theorem 5.18 imply that Gf˜∼= Gf ∼= G.
In the following, we fix X a path-connected subset of C. Let Ef q
→ X be the splitting covering of a Weierstrass polynomial f , and let α1, · · · , αn : Ef → C be all roots of f. We
define T := {q ∗g q∗h | g, h ∈ Q(i)[x], h(x) 6= 0, ∀x ∈ X} and W := {q ∗g q∗h | g, h ∈ Q(i)[x], h 6= 0} ∼= Q(i)(x). Lemma 6.3. AutTT [α1, · · · , αn] ∼= AutWW [α1, · · · , αn].
Proof. Since T is a subring of W , we have a restriction map r : AutWW [α1, · · · , αn] →
AutTT [α1, · · · , αn] by sending ϕ to ϕ|T [α1,··· ,αn]. We use that notation α
I := αi1
1 α i2
2 · · · αinn
where I = (i1, i2, ..., in).
For ψ ∈ AutTT [α1, · · · , αn], define eψ : W [α1, · · · , αn] → W [α1, · · · , αn] by
e ψ(X I aI bI αI) = 1 Bψ( X I aIBIαI) where B =Q IbI and BI = bBI. Well defined: IfP I aI bIα I = 0, thenP IaIBIα I = 0. So 1 Bψ( P IaIBIα I) = 0 which implies that eψ(P I aI bIα I) = 0. Homomorphism: eψ((P I aI bIα I)(P J cJ dJα J)) = 1 BDψ(( P IaIBIαI)( P JcJDJαJ)) = [1 Bψ( P IaIBIα I)][1 Dψ( P JcJDJα J)] = eψ(P I aI bIα I) eψ(P J cJ dJα J) where D =Q JcJ and DJ = cD J. Injective: If eψ(P I aI bIα I) = 0, then 1 Bψ( P IaIBIαI) = 0, so ψ( P IaIBIαI) = 0. The map ψ is an automorphism, soP IaIBIα I = 0 and hence P I aI bIα I = 0. Surjective: If P I aI bIα I ∈ W [α 1, ..., αn], then eψ(PI abI I(ψ −1(α))I) = P I aI bIα I. Consequently, eψ ∈ AutWW [α1, · · · , αn].
We obtain a map Φ : AutTT [α1, · · · , αn] → AutWW [α1, · · · , αn] defined by
Φ(ψ) = eψ. Since Φ(ψ1◦ ψ2)( X I aI bI αI) = 1 Bψ1◦ ψ2( X I aIBIαI) = 1 Bψ1( X I aIBIψ2(αI)) = eψ1( X I aI bI ψ2(αI)) = eψ1( 1 Bψ2( X I aIBIαI)) = eψ1◦ eψ2( X I aI bI αI) = Φ(ψ1)Φ(ψ2)( X I aI bI αI), Φ is a group homomorphism. It is clear that Φ ◦ r = id and r ◦ Φ = id, so Φ is a group isomorphism.
Proposition 6.4. Suppose T [α1, · · · , αn] ∩ R = T where R = q∗C(X). Then Gf occurs as
a Galois group of a Galois extension of Q(i).
Proof. Since ∆2 ∈ R and ∆ ∈ T [α1, ..., αn], so ∆2 ∈ T by assumption. So ∆2 = q
∗g
q∗h for some g, h ∈ Q(i)[x] and h(x) 6= 0 for x ∈ X. By Lemma 5.7, ∆12 =
q∗h
q∗g ∈ R, so g(x) 6= 0 for all x ∈ X. This implies ∆12 ∈ T . Then
1
∆ =
∆
∆2 ∈ T [α1, ..., αn]. By Theorem 5.18,
AutT(T [α1, · · · , αn]) ∼= A(E/X) ∼= Gf
Lemma 6.3 implies that
AutW(W [α1, · · · , αn]) ∼= Gf.
Moreover, W [α1, · · · , αn] is the splitting field of f ∈ W [z], and hence, W [α1, · · · , αn] is a
Galois extension of W . Since W ∼= Q(i)(x) and Q(i) is Hilbertian, Gf occurs as a Galois
group of certain Galois extension L of Q(i) as wished (see [11]).
Corollary 6.5. If the degree of T [α1, ..., αn] over T is n, then Gf can be realized as a Galois
group over Q(i).
Proof. Let α ∈ T [α1, ..., αn] be a primitive element over T , i.e., T [α] = T [α1, ..., αn], and
let h ∈ T [z] be the minimal polynomial of α. For ϕ ∈ T [α] ∩ R, ϕ = a0 + a1α + · · · +
an−1αn−1 for some a0, ..., an−1 ∈ T . Let x ∈ X and π−1(x) = {e1, ..., en}. Then ϕ(x) =
a0(x) + a1(x)α(ej) + · · · + an−1(x)α(ej)n−1 for all j = 1, ..., n. Since Ef over X is Galois,
for each ej, there is a unique gj ∈ Gf such that gj(e1) = ej. Applying all gj to ϕ, their
sum is nϕ(x) = na0(x) + a1(x) Pn j=1(gj · α)(e1) + · · · + an−1(x)( Pn j=1gj · αn−1)(e1). Since Pn j=1(gj · αk)(e1) = Pn
j=1αk(ej) and it is well known that
Pn
j=1αk(ej) can be expressed as
a polynomial of coefficients of h, so Pn
j=1(gj · αk) ∈ T which implies that ϕ ∈ T . Hence by
the result about, we have our claim.
Acknowledgements The authors thank the National Center of Theoretical Sciences of Taiwan (Hsinchu) for providing a wonderful working environment.
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