Resolving method for 4-fold covers

4.2 Resolving method for 4-fold covers

Letπ:X →Y be a 4-fold cover over a smooth surface. Our strategy is similar to the canonical resolution for double and triple covers. So letσ⁽¹⁾ :Y⁽¹⁾→Y be the blowing-up at a singularity of ∆(π). Then we obtain the following diagram of normalizations ofY⁽¹⁾.

Y⁽¹⁾

X₁⁽¹⁾ X₂⁽¹⁾ Xe⁽¹⁾

X^{(1) π˜(1) ψ(1)2}

ψ⁽¹⁾₃

ψ⁽¹⁾₁ π⁽¹⁾

φ⁽¹⁾

By the succession of this process, the branch locus is modified to smooth for double and triple covers. However, we see, from the following example, that it is not true for 4-fold covers.

Example 4.2.1. The question is local onY.So letY =C², andC(Y) =C(x, y), the rational function field inxandy.LetK=C(Y)(z) such thatf :=z⁴−xy is the minimal polynomial ofz overC(Y),andπ:X→Y theK-normalization of Y. Let E⁽ⁱ⁾ be the exceptional divisor of σ⁽ⁱ⁾. Note that, by Remark 3.3.1, E⁽ⁱ⁾is not in ∆(π⁽ⁱ⁾) if and only ifv_E^Y_(i)⁽ⁱ⁾(xy)≡0 (mod 4).

Letσ⁽ⁱ⁾(i= 1,2,3) be the blowing-ups atP⁽ⁱ⁾∈Y⁽ⁱ⁻¹⁾ as in the following diagram.

1 1

P⁽¹⁾

1 1

2 P⁽²⁾

1 3

2 1 P⁽³⁾

1 5≡1 (4)

1 2

3 P⁽⁴⁾

σ⁽¹⁾ σ⁽²⁾ σ⁽³⁾

Here each number is the valuation at each component. Then P⁽²⁾ ∈Y⁽¹⁾ and P⁽⁴⁾ ∈ Y⁽³⁾ are intersections of divisors with the valuation 2 and 1 (mod 4), and P⁽⁴⁾ is an infinitely near point of P⁽²⁾. Hence the branch locus is never modified to smooth by the above method.

Lemma 4.2.2. Let π : X → Y be a 4-fold cover. We put ∆_• := ∆_•(π) for

•=A, B, C, D.Assume the branch locus∆(π) is an SNC divisor. Then (i) (∆A+ ∆B)∩∆C=∅,and

(ii) ∆D∩∆C=∅.

Proof. If Gal(K/e C(Y)) is isomorphic to eitherZ4, D8 orV4,then ∆C = 0 by Remark 3.3.7, hence there is nothing to prove.

(i) If Gal(K/e C(Y)) ∼= A4, then ∆A = ∆B = 0, thus there is nothing to prove. Hence we suppose Gal(K/e C(Y))∼=S₄.Sinceψ₁◦ψ₂ is anS₃-cover, we

34 CHAPTER 4. RESOLUTION OF4-FOLD COVERS OF SURFACES can regard ∆_A+ ∆_B+ ∆_C as the branch locus of a triple cover that is simply ramified along ∆_A+ ∆_B,and is totally ramified along ∆_C.So (i) is done by the proof of [22, Theorem 1.4].

(ii) We show the assertion in only the case Gal(K/e C(Y)) ∼= S4 since the assertion in the other case can be also showed by the same way. Suppose that

∆D∩∆C 6= ∅. Let P ∈ ∆D∩∆C, and Pe ∈ ˜π⁻¹(P). Since ψ1 is unramified over P, X1, ψ^∗₁∆C and ψ^∗₁∆D are smooth over P. So X2 is smooth at ψ3(Pe).

Moreover, since ψ^∗₁∆D intersects transversally withψ₁^∗∆C overP, ψ₂^∗ψ₁^∗∆D is smooth over P. Hence Xe is smooth at P .e Note that ψ₂◦ψ₃ is an A4-cover.

Since the stabilizer ofPebyS₄includes (1 2)(3 4) and (1 2 3),that isA₄.Hence there is a faithful representation ρ : A₄ → GL(2,C) via the action of A₄ on the tangent space atP .e SinceA4 is non-commutative,ρis irreducible. On the other hand, it is known by representation theory that there is no irreducible representation ofA4 of dimension 2,which is a contradiction.

We prove Proposition 1.1.1.

Proof of Proposition 1.1.1. Here we show the assertion in only the case where Gal(K/e C(Y))∼=S4 since that in other cases be shown by the same way.

By the well-known embedded resolution of the singularities of a curve in a smooth surface, there are a finite number of blowing-ups σ⁰ : Y^(r0) → Y such that (σ⁰)^∗(∆(π)) is an SNC divisor. So we may assume that ∆_A(π) +

∆_B(π) + ∆_C(π) + ∆_D(π) is an SNC divisor. Then there is no intersection of

∆_A(π) + ∆_B(π) + ∆_D(π) and ∆_C(π) by Lemma 4.2.2. Hence we can regard

∆_C(π) as the branch locus of the cyclic triple coverψ₂,and its nodes are resolved by the canonical resolution for triple covers. Similarly, since ∆A(π) + ∆B(π) is the branch locus ofψ1,its nodes are resolved by the canonical resolution for double covers. Therefore there are a finite number of blowing-upsσ⁰⁰:Y^(r00)→ Y such that ∆A(π^(r00)) + ∆B(π^(r00)) + ∆C(π^(r00)) is smooth.

Finally, we assume that π satisfies (i) and (ii) of Proposition 1.1.1. Let σ^(P) : Y^(P) → Y be the blowing-up at a point P ∈ Y, E^(P) the exceptional divisor, andπ^(P):X^(P)→Y^(P)the induced 4-fold cover. Let{P1, . . . , Pr} ⊂Y be the set of nodes of ∆D(π) such that E^(Pi) is not in ∆D(π^(Pi)). To obtain (iii), we putσ^(r):Y^(r)→Y the blowing-ups atP1, . . . , Pr.

Remark 4.2.3. Ifπ:X→Y is a 4-fold cover of good type and Gal(K/e C(Y))∼= Z4,then Sing(∆_D(π)) =∅sinceψ₃is a double cover.

Let σ = σ⁽¹⁾◦ · · · ◦σ^(r) be a finite number of blowing-ups such that π^(r) is a 4-fold cover of good type (Definition 1.1.2). Then we have the morphism ν^(r):X^(r)→X such thatσ◦π^(r)=π◦ν^(r) by the Stein factorization.

From now, we assume that π : X → Y is a 4-fold cover of good type.

Then the set of singularities of ∆(π) is (∆A(π)∩∆D(π))∪(∆B(π)∩∆D(π))∪ Sing(∆D(π)).

Lemma 4.2.4. Let P ∈ ∆A(π)∩∆D(π). Then π⁻¹(P) consists of a smooth point and anA₁-singularity.

4.2. RESOLVING METHOD FOR4-FOLD COVERS 35

point coset stabilizer α1 α2

Pe₁ G_Pe

1 {id, (1 2), (3 4), (1 2)(3 4)} (1 2) (3 4) Pe2 (1 3)G_Pe

1 {id, (2 3), (1 4), (1 4)(2 3)} (2 3) (1 4) Pe3 (2 3)G_Pe

1 {id, (1 3), (2 4), (1 3)(2 4)} (1 3) (2 4) Pe4 (2 4)G_Pe

1 {id, (1 4), (2 3), (1 4)(2 3)} (1 4) (2 3) Pe5 (1 2 4 3)G_Pe

1 {id, (2 4), (1 3), (1 3)(2 4)} (2 4) (1 3) Pe6 (1 3)(2 4)G_Pe

1 {id, (3 4), (1 2), (1 2)(3 4)} (3 4) (1 2) Table 4.1: Cosets and stabilizers of thePei.

Proof. By Remark 3.3.7, we obtain ∆A(π) = 0 or ∆D(π) = 0 if Gal(K/e C(Y)) is isomorphic toA4, V4orZ4.Hence it is sufficient to show the assertion in the two cases where Gal(K/e C(Y)) is isomorphic toS4 andD8.Here we show that in only the case Gal(K/e C(Y)) ∼=S4 since that in the other case is shown by the same way.

Suppose Gal(K/e C(Y))∼=S4.The question is local onY,so we may assume that ∆(π) = ∆A(π)+∆D(π),and Sing(∆(π)) ={P}.Since ∆(ψ1) = ∆A(π), X1

is smooth. Since ∆A(π) and ∆D(π) intersect transversally atP, D1:=ψ^∗∆D(π) is smooth overP, and intersects transversally with A₁ := (ψ^∗₁∆_A(π))_red. Since ψ₂ is unramified, we obtain three copies of the reduced divisor A₁+D₁, say A_2,i+D_2,i (i = 1,2,3). The bidouble cover ψ₃ is branched at D_2,1+D_2,2+ D2,3.So we have A3,i+D3,i (i= 1, . . . ,6) on X,e where A3,iand D3,i intersect transversally over P for each i = 1, . . . ,6. Let Pei be the intersection of A3,i

andD3,ifori= 1, . . . ,6.As the proof of Lemma 3.2.2, we can regard ϕas the quotient map by S₃ =h(1 2), (1 2 3)i. By some permutation of {Pe_i}⁶i=1, the stabilizerG_Pe

1 ofPe₁ coincide withh(1 2), (3 4)i.

If A_3,1 is in the ramification locus of ϕ, then there is a representation ρ : G_Pe

1→GL(2,C) such thatρ((1 2)) =α1andρ((3 4)) =α2,where α1=

µ1 0 0 −1

¶

, α2=

µ−1 0 0 −1

¶ .

By some permutation of{Pei}⁶i=2,the corresponding cosets and stabilizers are as shown in Table 4.1. Since (j4)6∈S3(j= 1,2,3), A3,1, A3,2andA3,3are in the ramification locus of ϕ,andA_3,4, A_3,5 andA_3,6 are not in that one. However, sinceα2 fixes just one point,Pe4, Pe5andPe6 are in that one. Thereforeπ⁻¹(P) consists of a smooth point and anA1-singularity.

By the above proof, there are two types of ramification points of ϕ over P ∈ ∆A(π)∩∆D(π). We call Pe ∈ π˜⁻¹(P) a curvewise ramification point if ϕ(Pe) is smooth point ofX,and apointwise ramification point otherwise.

We resolve the singularities of X over ∆A(π)∩∆D(π) in the case where Gal(K/e C(Y))∼= S₄. (In the other case, we can resolve the singularity by the

36 CHAPTER 4. RESOLUTION OF4-FOLD COVERS OF SURFACES

X

−1

−1

−1

−2 π

ϕ

˜ π

ψ1

ψ₂ ψ₃

σ3

ϕ⁰⁰

: the branch locus ofψ₁ orψ₃ : the ramification locus ofϕor ϕ⁰⁰

Figure 4.2.1: The resolution of a singularity overA∩D.

same way.) Assume ∆(π) = ∆A(π)+∆D(π) and Sing(∆(π)) ={P}as the above proof. Let σ₃ : Xe⁰⁰ → Xe be the blowing-ups at three pointwise ramification points of ϕ. Then one can check that S₃ acts on Xe⁰⁰, and each irreducible component of the exceptional divisor of σ₃ is fixed by an involution of S₃. Hence the quotient X⁰⁰ := Xe⁰⁰/S₃ is smooth. Since σ₃ is an S₃-equivariant morphism, there is a morphism ν⁰⁰ : X⁰⁰ → X. The resolution is as shown in Figure 4.2.1.

Lemma 4.2.5. Let P ∈ ∆B(π)∩∆D(π). Then π⁻¹(P) consists of an A1 -singularity.

Proof. If Gal(K/e C(Y)) is isomorphic toA4orV4,then ∆B(π) = 0.Thus there is nothing to prove. In the case Gal(K/e C(Y)) ∼= Z4, it is easy to show the assertion. We show that in the only case Gal(K/e C(Y))∼=S4 below since that in the case Gal(K/e C(Y))∼=D4 is shown by the same way.

Suppose Gal(K/e C(Y)) ∼= S4. The question is local on Y, so we may as-sume that ∆(π) = ∆_B(π) + ∆_D(π) and Sing(∆(π)) = {P}. As the proof of Lemma 4.2.4, X₂ is smooth, and we have B_2,i +D_2,i (i = 1,2,3). Let Pe ∈ π˜⁻¹(P). If Xe is smooth, then G_Pe includes V4 by Section 4.1. Thus G_Pe ∼=D8. So there is an irreducible representation ρ: D8 → GL(2,C). Since points ofXe over B2,i are fixed by the cyclic subgroup of order 4 in D8, there is a subspace of the vector space C² which is fixed by that subgroup via the representationρ.On the other hand, sinceρis irreducible, the fixed locus inC² of that subgroup is only the origin, which is a contradiction.

Therefore ˜π⁻¹(P) consists of six A1-singularities. By Lemma 3.2.2, ϕ is unramified, soπ⁻¹(P) consists of anA1-singularity.

We resolve the singularities of X over P ∈ ∆_B(π)∩∆_D(π) in the case

4.3. RAMIFICATION STRUCTURE OF S4-COVERS 37

anA1-singularity

∆B(π)

∆D(π)

ψ₁ ψ2

ψ3

ϕ

π

6A1-singularities _ψ0 6

3

ϕ⁰

−1

−1

−1

−2

−2

−2 σ2

Figure 4.2.2: The resolution of a singularity overB∩D.

where Gal(K/e C(Y))∼=S4.(In other cases, it can be resolved by the same way.) Assume that ∆(π) = ∆_B(π)+∆_D(π),and Sing(∆(π)) ={P}.Letσ₂:X₂⁰ →X₂ be the blowing-ups at three points of (ψ₁◦ψ₂)⁻¹(P).ThenS₃ acts onX₂⁰.Let ψ₃⁰ : Xe⁰ → X₂⁰ be the K-normalization ofe X₂⁰. Then ψ⁰₃ is unbranched at the exceptional divisor ofσ2by Corollary 4.1.2. HenceXe⁰ is smooth. LetX⁰be the quotientXe⁰/S3,andϕ⁰:Xe⁰ →X⁰ the quotient map. Thenϕ⁰is unramified, so X⁰ is smooth. Since there is anS4-equivariant morphismXe⁰ →Xe by the Stein factorization, we obtain a morphismν⁰ :X⁰ →X.The resolution is as shown in Figure 4.2.2.

Lemma 4.2.6. Let P∈Sing(∆D(π)).ThenX is smooth overP.

Proof. If Gal(K/e C(Y))∼=Z4,then there is nothing to prove by Remark 4.2.3.

Hence we suppose Gal(K/e C(Y))6∼=Z4,thenψ3 is a bidouble cover.

The question is local on Y, so we may assume that ∆(π) = ∆D(π), and Sing(∆(π)) ={P}.By the definition of a 4-fold cover of good type and Corol-lary 4.1.2,Xe is smooth. By Lemma 3.2.2,ϕis unramified, soX is smooth.

By the above argument, we construct a resolutionν :=ν^(r)◦ν⁰⁰ :X⁰⁰→X for any 4-fold coverπ:X →Y as Figure 1.1.2.