SIMULTANEOUS RESOLUTION

7

transform

V'

of

V

such that the normalization

W'

of

V'

in

L

is a nonsingular

projective model of

Ljk.

(6.2) If char(K) = 2 then, given any nonsingular projective model

V

of

K/k

such that the residue field of any point of it is algebraically closed and given any

algebraic field extension

L/ K

of degree 2, there exists an iterated quadratic trans-

form

V'

of

V

such that the normalization

W'

of

V'

in

L

is a nonsingular projective

model of

Ljk.

(6.3) If char(K) = 2 with algebraically closed ground field

k

such that

K/k

has

a minimal model then there exists a Galois extension

L/ K,

with Galois group a

direct sum of 2 copies of a cyclic group of order 2, such that there does not exist

any nonsingular projective model

V'

of

K/k

whose normalization

W'

in

L

is a

nonsingular projective model of

Ljk.

PROOF. In case of (6.1) and (6.2) let V be the given nonsingular projective

model of

K/k,

and in case of (6.3) let

V

be the minimal model of

K/k.

In a moment

we shall construct a sequence of nonsingular projective models (Vi)i=o,

1,2 , ...

of

K / k,

with V0 = V, such that

Vi

is an iterated quadratic transform of Vi-

1

for all i 0.

In case of (6.1) and (6.2) let

L/ K

be the given Galois extension, and in case of

(6.3) let

L/ K

be the Galois extension with Galois group a direct sum of 2 copies

of a cyclic group of order 2, which is to be constructed.

In all the cases let (Wi)i=o,

1,2 , ...

be the sequence of projective models of

L/k

such that Wi is the normalization of

Vi

in L for all i ;:::: 0. Let H(Wi) be the set of

all singular points of Wi. Let

G(Vi)

be the set of all those points

R

of

Vi

for which

there is a point S of H(Wi) such that S dominates R. For each i ;:::: 0, since Wi is

a "normal surface," H(Wi) is a finite set, and hence so is

G(Vi).

For each i 0,

we decree that

Vi

be the quadratic transform of Vi-

1

obtained by quadratically

blowing up G(Vi-1). Note that for each i 0, clearly

G(Vi)

is contained in the

inverse image of

G(Vi-d

under the domination map

Vi

----+

Vi-

1

.

What we need to

show is that, in case of (6.1) and (6.2),

G(Vi)

is empty for all large enough i and,

in case of (6.3),

G(Vi)

is nonempty for all i.

If

G(Vi)

is nonempty for all i then we can take a point Ri in

G(Vi)

such that

Ri is a two dimensional quadratic transform of Ri-

1

for all i 0. In case of (6.1)

and (6.2) this is impossible by (2.1) and (2.2) respectively.

In case of (6.3) let R be any point of V, let X, Y be any generators of M(R),

let F

1

and F2 be as in Lemma (1.3), let

L

be a splitting field of F 1F

2

over

K,

and fori= 0, 1, 2, ... let Ri = R[X/Yi]Pi where Pi is the prime ideal in R[X/Yi]

generated by X/Yi andY. Then by (2.3) we know that

L/K

is Galois with Galois

group Z

2

ffi Z

2

and, for all i, the integral closure Si of Ri in L is a two dimensional

nonregular local domain for every i. It follows that, for all i, the point Ri belongs

to

G(Vi)

and hence G(Vi) is nonempty.

4. Problems

PROBLEM 7. In (6.2), how far can you remove the assumption that the residue

fields are algebraically closed.

PROBLEM 8. In (6.3), how far can you remove the assumption that

K/k

has

a minimal model.