SIMULTANEOUS RESOLUTION
5
To prove (1.2) it suffices to note that L is a splitting ofF{ over K.
To prove (1.3), since A1 E
M(R)
with B1 =
Y
E
M(R) \
M(R)
2,
we see
that S1 is a two dimensional regular local domain with M(S1)
=
(X, H)S1

Also
y
=
H 2
+
xya H
and substituting this in
we get
where
and hence
F~(Z)
=
F2(Z +H)= Z
2
+ XYa EZ + (Y + H
2
+ xya EH)
F~(Z)
=
Z
2
+
XYa EZ
+
XYaH(1 +E)
{
ya
= (H2
+
XYaH)a
=
(H2 +X[H2 +XYaH]aH)a
=
(H2 + X[H2 +
X(H2
+ XYaH)aH]H)a
=
H
2aD*
with
D*
E
R \ M(R)
F~(Z)
= Z
2
+
XH2aED*Z
+
XH2a+1D**
with
D**
E
R \ M(R)
and therefore
F~'(Z)
=
H- 2 aF~(zHa)
=
Z
2
+
XHaED*Z
+
XHD**.
Consequently "the irreducible surface
F~'
( Z)
= 0 is devoid of singular curves" and
hence, by the following Normality Theorem 3, we see that
S
=
S1[J]
where
I
is a
root
ofF~'
in L. Since the coefficients of Z
1
and
Z
0
in
F~'(Z)
belong to
M(R)
and
M(R)
2
respectively, it follows that Sis a two dimensional nonregular local domain.
Since
F1(Z)
and
F~'
( Z)
are irreducible, we also see that
L / K
is Galois with Galois
group Z2 EB Z2.
THEOREM 2. Let (Ri)i=0,1,2, ... be a two dimensional quadratic sequence with
R
0
=
R,
and let
Si
be the integral closure of
Ri
in
L.
Then we have the following.
(2.1) If char(K)
-I
2 and
R
is pseudogeometric, then
Si
is a two dimensional
semilocal regular domain for infinitely many i.
(2.2) If
n
=
1 with
R
pseudogeometric and
R/M(R)
algebraically closed, then
si
is a two dimensional semilocal regular domain for infinitely many
i.
(2.3) If char(K)
=
2
=
n
with F1 and F2 as in Lemma (1.3) and for i
=
0, 1, 2, ...
we have
Ri
=
R[X/Yi]P;
where
Pi
is the prime ideal in
R[X/Yi]
gen-
erated by
XjYi
andY, then
Si
is a two dimensional nonregular local domain for
every i, and Lj K is Galois with galois group Z2 EB Z2.
PROOF. To prove (2.1), by applying the following Total Embedded Curve
Resolution Theorem 4 to "the plane curve
C
=
0," for all sufficiently large i we
can write
C
=
DX[Y/,
with
DE Ri \ M(Ri)
and nonnegative integers
r, s,
where
M(Ri)
=
(Xi, Yi)Ri.
This amounts to writing
Cj
=
DjX?Y;si
for all j
E
J with
Dj
E
Ri \ M(Ri)
and nonnegative integers
Tj, Sj.
Now we are done by Lemma
(1.1).
To prove (2.2), in view of Lemma (1.2) and Theorem (2.1), it suffices to show
that, assuming char(K)
=
2 with
R
pseudogeometric and
R/M(R)
algebraically
closed, given any irreducible
F(Z)
=
Z
2
+
AZ
+
B
E
R[Z], for infinitely many i
there exists 0
"I
o:i
E
K and f3i
E
K such that for Ff ( Z)
=
o:;
2
F (
O:i
Z
+
f3i) we have
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