4
SHREERAM S. ABHYANKAR AND MANISH KUMAR
R' of R we have that the integral closure S' of R' in L is a two dimensional regular
semilocal domain.
(1.2) Assume that
n
= 1 and there exists 0-=/= o: E K together with (3 E K such
that for
F{
=
F{ (
Z) = o:-
2
F1 ( o:Z
+
(3) = Z
2
+
A' Z
+
B'
we have
F{(Z)
E R[Z] with ordRF{(Z) = 1. Then Sis two dimensional regular
semilocal domain.
(1.3) Assume that char(K) = 2 = n and
A1 = xya and A2 = XYaE with B1 =
B2
= Y
where
a
= a positive integer
and
E
= xtyu with nonnegative integers
t, u
at least one of which is positive.
Let H be a root of F
1
in
£
1. Then 8
1
is a two dimensional regular local domain
with M(Sl) = (X, H)S1
,
and S is a two dimensional nonregular local domain.
Moreover,
L/ K
is Galois with Galois group Z
2
EB Z
2
.
PROOF.
To prove
(1.0)
note that L1 = K(H1
).
Also note that (i) follows from
the discriminant theory given in [A03], (i') and (i") are straightforward, and, in
view of (i') (resp: (i")), (ii') (resp: (ii")) follows by taking
{
(Rj, Lj, X, Hi) and (F1, ... , F1-1, FJ+l, ... , Fn)
(resp: (Rz, Lz, Hz, Y) and (F1, ... , Fz-1, Fl+l, ... , Fn))
for (R, K, X, Y) and (F1, ... , Fn) in (i). Likewise, in view of (i'), (iii) follows by
taking
(RJ, Li, X, Hi) and (F1, ... , F1-1, Fi+l' ... , Fn)
for (R, K, X, Y) and (F1, ... , Fn) in (i"). Likewise, in view of (i') (resp: (i")), (iii')
(resp: (iii")) follows by taking
{
(Rj,LJ,X,HJ) and (F1, ... ,Fj-l,FJ+l,···,Fn)
(resp: (Rz,Lz,Hz,Y) and (Fl, ... ,Fz-1,Fl+l,···,Fn))
for (R,K,X,Y) and (F1
, ...
,Fn) in (ii") (resp: (ii')). (iv) follows from (ii'), (ii"),
(iii), (iii'), and (iii"). To prove (v) suppose that
(J'
U J") =
0
and J
111
-=/=
0
and let
R'
is any two dimensional quadratic transform of
R.
Suitably relabelling
X, Y
we
may assume that Y/X E R'. Now ifY/X E M(R') then M(R') =(X, Y/X)R' and
for all j E J we have
C1
=
D1Xrjysj
where
D
1
E R' \ M(R') with
rj
=
r1 + s1
and
sj
=
s1,
and hence
rj
= 0 with
sj
= 0 or 1, and so we are reduced to (iv). Likewise if
Y /X
'f.
M(R') then
M(R') = (X, Y')R' for some Y' E M(R'), and for all j E J we have
C1
=
Djxri
(Y')sj
where
Dj
=
D
1
(Y/X)
8 J
E R' \ M(R') with
rj
=
r1 + s1
with
sj
= 0, and hence
rj + sj
= 0, and so we are again reduced to (iv).
(1.1)
follows from parts (iv) and (v) of
(1.0).
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