- Text
- Generic,
- Subgroup,
- Fundamental,
- Element,
- Proposition,
- Generated,
- Homomorphism,
- Finite,
- Quotient,
- Projection,
- Galois,
- Closures,
- Projections

On Fundamental Groups of Galois Closures of Generic Projections

Proposition 4.2 We can split the short exact sequence (∗) using inertia groups.With respect to this splitting there are the following isomorphisms(Gal(M/L)/C) / (Gal(M/L)/C) Sn∼ = Gal(M ∩ k nr /k)(Gal(M/L)/C) / (Gal(M/L)/C) (1) S∼= Gal(M ∩ K nr /K)n−1where the notations are the ones introduced in Section 3.1If Question 2.14 has an affirmative answer for the finite étale cover Y → X galthen the group C is trivial.PROOF. For every transposition (1 k) we choose a prime ideal Q (1 k),i and denoteby r k the non-trivial element **of** its inertia group. We denote by ¯r k the image **of**r k inside Gal(M/k)/C. The elements ¯r k fulfill ¯r 2 k = 1 and map to (1 k) underthe induced surjection onto Gal(L/k). Since we took the quotient by C also thefollowing relations hold true:(¯r i ¯r i+1 ) 3 = 1 and (¯r i ¯r j ) 2 = 1 for |i − j| ≥ 2.These are precisely the Coxeter relations for S n (cf. Section 5.6) and hence the¯r k define a group isomorphic to a quotient **of** S n . Since there is a surjectivemap from this group onto S n it must be equal to S n . This defines a splittings : Gal(L/k) → Gal(M/k)/C.From Lemma 4.1 we know that C is a subgroup **of** N. So we see that the mapfrom Gal(M/k) onto Gal(M ∩ k nr /k) factors over Gal(M/k)/C. The kernel**of** the map from Gal(M/k)/C onto Gal(M ∩ k nr /k) clearly is the image ¯N **of**N inside Gal(M/k)/C. The group ¯N is generated by the images **of** the inertiagroups.From Lemma 3.1 we know that (Gal(M/L)/C) Sn is generated by the commutators[g, s(τ)]’s where g runs through Gal(M/L) and τ runs through the transpositions**of** S n . The element gs(τ)g −1 is the non-trivial element **of** the inertiagroup **of** some prime ideal lying above P τ . With this said it is easy to concludethe equalitiesN = (Gal(M/L)/C) Sn · s(S n )and N ∩ (Gal(M/L)/C) Sn = (Gal(M/L)/C) SnApplying the second isomorphism theorem **of** group theory we obtainGal(M/L)/C(Gal(M/L)/C) Sn==Gal(M/L)/CN ∩ Gal(M/L)/C = N · Gal(M/L)/CNGal(M/k)/CN∼ =Gal(M ∩ k nr /k).28

Hence the induced homomorphism from Gal(L/K)/C to Gal(M ∩ k nr /k) is surjectivewith kernel (Gal(L/K)/C) Sn .The assertion about the quotient **of** Gal(M/L)/C by (Gal(M/L)/C) (1) Sisn−1proved similarly and left to the reader.Now suppose that the curves corresponding to the Q τ,i ’s fulfill the connectivityproperties **of** Question 2.14.For two disjoint transpositions τ 1 and τ 2 we choose two prime ideals Q τ1 ,i andQ τ2 ,j and let r 1 and r 2 be the non-trivial elements **of** their inertia groups. Sincethe curves corresponding to the two prime ideals intersect there is a maximal idealcontaining both **of** them. The inertia group **of** this maximal ideal is isomorphic to¡ 2 × ¡ 2 and is generated by r 1 and r 2 . Hence these two elements commute andc(r 1 , r 2 ) = 1.If τ 1 and τ 2 have exactly one index in common then there is a maximal idealwith inertia group S 3 that is generated by r 1 and r 2 . So there is a triple commutatorrelation between r 1 and r 2 and so also c(r 1 , r 2 ) = 1 holds true in thiscase.So if Question 2.14 has an affirmative answer for Yc(r 1 , r 2 )’s are equal to 1 and so C is trivial.→ X gal then all the□We now pass to the limit **of** all finite étale covers **of** X gal and keep track **of** theinduced homomorphisms between the corresponding field extensions and their**Galois** groups. We will denote the limit **of** the subgroups C by C proj . UsingProposition 4.2 we arrive at surjective homomorphismsGal(L nr /L)/C proj ↠ Gal(L nr ∩ k nr /k)Gal(L nr /L)/C proj ↠ Gal(L nr ∩ K nr /K).Taking the compositum **of** L with K nr we get a subfield **of** Ω that corresponds toa limit **of** étale extensions X gal . Hence this compositum must be contained in L nrand hence already K nr was contained in L nr . So K nr ∩ L nr is equal to K nr andthe second surjective homomorphism above takes the formGal(L nr /L)/C proj ↠ Gal(K nr /K).Its kernel is (Gal(L nr /L)/C proj ) (1) S.n−1Up to now have actually never needed that k is the function field **of** the projectiveplane over the complex numbers. This means that everything done in thissection works equally well in the affine situation. We denote by L nr,aff the compositum**of** all fields corresponding to finite étale extensions **of** Xgal aff inside Ω. Wethen define C aff to be the subgroup **of** Gal(L nr,aff /k) normally generated by thec(r 1 , r 2 )’s where the r i ’s run through inertia groups in this extension.29

- Page 1: On Fundamental GroupsofGalois Closu
- Page 5 and 6: ContentsIntroductioniii1 A short re
- Page 7 and 8: IntroductionSchon winkt der Wein im
- Page 9 and 10: is defined by a line bundle L on X.
- Page 11 and 12: depends only on G and n and not on
- Page 13 and 14: 1 A short reminder on fundamental g
- Page 15 and 16: Miyaoka [Mi] gave a construction of
- Page 17 and 18: 2 Generic projections and their Gal
- Page 19 and 20: 2.2 Galois closures of generic proj
- Page 21 and 22: Proposition 2.12 Let L be a suffici
- Page 23 and 24: 3 Semidirect products by symmetric
- Page 25 and 26: Remark 3.2 The assumption n ≥ 3 i
- Page 27 and 28: Corollary 3.5 Let n ≥ 2.1. If G i
- Page 29 and 30: The assertions about the torsion an
- Page 31 and 32: For j ≠ i, j ≥ 2 the group X τ
- Page 33 and 34: since the product over all componen
- Page 35 and 36: 4 A first quotient of π 1 (X gal )
- Page 37 and 38: 4.2 The quotient for the étale fun
- Page 39: of K. The normalisation of X inside
- Page 43 and 44: zero. Hence the proof also works al
- Page 45 and 46: an isomorphism between these two gr
- Page 47 and 48: In particular, for H = 1 we obtain
- Page 49 and 50: For an element g of the inertia gro
- Page 51 and 52: eThis automorphism has order e i an
- Page 53 and 54: Hence the homomorphisms from π top
- Page 55 and 56: 5 A generalised symmetric group5.1
- Page 57 and 58: Theorem 5.3 For n ≥ 5 there exist
- Page 59 and 60: Proposition 5.6 (Rowen, Teicher, Vi
- Page 61 and 62: First, assume that i = n. ThenNow a
- Page 63 and 64: We note that for affine subgroups n
- Page 65 and 66: Every element of H 2 (G) maps to an
- Page 67 and 68: PROOF. We let F d /N ∼ = G and F
- Page 69 and 70: 5.4 ExamplesWe now compute ˜K(−,
- Page 71 and 72: Theorem 5.22 For a K(G, 1)-complex
- Page 73 and 74: Example 5.24 The homomorphism ψ ma
- Page 75 and 76: 6 ConclusionJetzt nehmt den Wein! J
- Page 77 and 78: This can be also formulated as foll
- Page 79 and 80: Severi claimed that a curve with on
- Page 81 and 82: where ϕ denotes the splitting of
- Page 83 and 84: Corollary 6.3 Under the assumptions
- Page 85 and 86: Proposition 6.5 Under the isomorphi
- Page 87 and 88: Remark 6.8 Proposition 6.5 shows us
- Page 89 and 90: We remark that the group on the lef
- Page 91:
E(π top1 (Z), n). By Corollary 3.3

- Page 94 and 95:
7.3 Surfaces in 3Let X m be a smoot

- Page 96 and 97:
If we denote by Π g the fundamenta

- Page 98 and 99:
NotationsVarieties and morphismsf :

- Page 100 and 101:
[GR1][GR2][GH][SGA1]H. Grauert, R.