C OMPOSITIO M ATHEMATICA
P ETER R USSELL
Simple birational extensions of two dimensional affine rational domains
Compositio Mathematica, tome 33, n
o2 (1976), p. 197-208
<http://www.numdam.org/item?id=CM_1976__33_2_197_0>
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197
SIMPLE BIRATIONAL EXTENSIONS OF TWO DIMENSIONAL AFFINE RATIONAL DOMAINS
Peter Russell Noordhoff International Publishing
Printed in the Netherlands
Let k be a field. For any
ring
R let R(n) denote thepolynomial ring
in n variables over R. In this paper we
investigate
affine k-domains A with theproperty
thatA[a/b] = k(2)
forsome a, b
E A.By
our main result(see 1.3),
if A is aunique
factorization domain(UFD),
thenA ~
k(2)
under variousfairly
mild additionalassumptions.
Acorollary (see 1.4)
is thefollowing
littlepiece
of information on the "can- cellationproblem"
fork(2) (see [3]):
Let k beperfect,
A ak-algebra
and t transcendental over A. Assume that
A[t] xi
k(3). If a variable ink(3)
is linear as apolynomial in t,
then A Ioek(2).
This work was
inspired by [9],
where thefollowing
is shown: Let k be of characteristic0,
A =k(2), a, b
E A with b ~ 0 and H = bw - a EA[w]
=k(3)
such thatA[w]/HA[w]
=A[a/b]
isisomorphic
tok(2).
Then there exist
F, G E A[w] ]
such thatk[F,G,H] = A[w].
We ex- tend this result to fields ofarbitrary
characteristic(see 2.3).
The
proof
of 1.3 runs like this: An irreducible factor x of b ink (2)
contracts to a maximal ideal in A
and,
sincek (2) is generated by
oneelement over
A,
defines a line ink (2) (i.e. k(2)/xk(2) ~ kil’).
The crucialproblem
lies inshowing
that there exists yE k(2)
such thatk[x, y] ]=
k(2).
If char k =0,
this is assuredby [1].
Under suitable restrictions onA
(not involving
chark), however,
we can also reach this conclusionexploiting
further the fact that x contractsbirationally
to a maximalideal in A. One then shows A =
k[x, by]
without muchdifficulty.
1 would like to express my thanks here to W. Heinzer. Numerous conversations with him were instrumental in
getting
this research off theground.
1
We
begin
with anelementary
result.1.1. PROPOSITION: Let A be a UFD and a,
a’, b,
b’ E A withGCD(a, b)
= 1 =GCD(a’, b’). Suppose A[alb]
= B =A[a’/b’].
Then(i)
b and b’ have the same irreduciblefactors, (ii)
a’ =aob’
+ caa =
a’ob
+ c’ a’with ao,
a’o,
c, c’ E A and cc’ ~ 1 modGcD(b, b’),
(iii) if
b’ =qb
with q EA,
then a and a’ are units mod q and q is aunit in B.
PROOF : We have
with ai ~ A. Since
GCD(a ,b’) = 1, b’l bn. Similarly, b 1 b’-
for somem. This proves
(i).
From(1)
we obtainwhere
Since ca E A and
GCD(a, b)
=1,
c E A.Similarly,
with
a’o ,
c’ E A. Now( 1- cc’)a’
=aob’
+ca’ob
and henceGCD(b, b’)|1 2014
cc’. This proves(ii).
Suppose
b’= qb
withq E A. By (ii),
c is a unitmod q,
and weobtain from
(3)
with d E A. Hence a is a unit mod q, and so a’ is a unit mod q
by (2).
Write
03B1a’ = 1 + 03B2q
with a,03B2 ~ A.
Thenb 03B1(a’/b’) = 1/q + 03B2
andhence
1/q
E B. This proves(iii).
1.2. COROLLARY: Let p be an irreducible
factor of GCD(b, b’ )
andsuppose the
p-orders of
b and b’ aredifferent.
Then a and a’ are unitsmod p in
Ah,
where h is theproduct of
theprime factors of GCD(b, b’)
different from
p.PROOF:
Replace
Aby
Ah andapply (ii ).
1.3. THEOREM: Let A be a UFD
finitely generated
over k and a,b E A with
GCD(a,b) = 1.
LetB = A[a/b]
and supposeB===k(2).
Assume also that one
of
thefollowing
conditions holds :(i)
A contains afield
generator, i.e. there existsf
E A such thatqtA
=k(f, q) for
some q EqtA (qtA
=field of quotients of A), (ii)
char k =0,
(iii) k is perfect
and A isregular.
Then A ~
k(2).
Moreprecisely,
there exist x, y E B such that B =k[x, y], b
Ek[x]
and A= k[x, by].
PROOF: We assume
b ~ k
and claim(see [9], proof
of Lemma3):
(*)
Letb1, . . ., br.
be the irreducible factors of b in B. Then(bi, b;)B
=B for i ~
j, biB
n A =Mi
is a maximalideal, Mi ~ Mj
fori ~ j
andBlbiB c--- (AIMi)(1).
If c E B is irreducible and cB n A a maximalideal,
then cB =
biB
for some i.In
fact,
ifbi
is an irreducible factor ofb,
thenMi = biB
n A ~(a, b)A,
and sinceGCD(a, b)
=1, Mi
is maximal. HencemiB n A = M,
andB/MiB
=AIMi[z],
where z is theimage
ofalb
modMiB.
NowMiB
CbiB,
so z is transcendental overA/Mi
and MiB isprime.
Hence MiB =biB, BIbiB
~(A/Mi)(1)
and Mi ~Mi,
and so(bi, bj)B
=B,
for i ~j.
The last assertion follows from Ab = Bb.Under each of the conditions
(i), (ii), (iii)
we will showby
différentmethods:
(**)
Thereexist x,
y E B such that B =k[x, y] ]
and b Ek[x].
Suppose
this has been done. Then x isintegral
over A and hencex E A. Since
Ab
=Bb, b m y
E A for some m and there exists b’ Ek[x] ]
of smallestdegree
such that v =b’ y
E A. Then b and b’ have thesame irreducible factors
(note
that these are the same whether taken ink[x],
A orB)
andk[x,v] b’ = Ab’ = k [x, y] b’ . Suppose
there is an irreducible c E A such thatcA ~ k[x,u] ]
is maximal. Then c is anirreducible factor of b and
v = b’y ~ cA.
Henceb "y E A,
whereb" =
b/c
Ek[x]
is of smallerdegree
thanb’,
and this isimpossible.
Sono
height
oneprime
in A contracts to a maximal ideal ink[x, v] and
the birationalmorphism Spec
A ~Spec k[x, v]
has finite fibres.By
Zariski’s Main Theorem(see [6,
Cor.2,
p.42]),
it is an open immer- sion. Sincek [x, 03C5]
is aUFD,
A is a localization ofk [x, v ],
and since the units of A are constant,k[x,v] = A.
NowA [v/b’] = k[x,y] = A[a/b],
and b = b’ follows from 1.2.It remains to establish
(**).
Case 1:
f E A
is a fieldgenerator.
Wekeep
the notation of(*).
There exist monic
polynomials
Pi with coefficients in k such thatbi|Pi (f)
in B(the
minimalpolynomials
off mod Ma,
forinstance).
Now
f
is a fieldgenerator
in B ~k(2)
as well as inA,
andby [8,
3.7 and4.5]
we can findx, y E B
such that B= k[x, y] ]
and(a )
thedegree
form of
f
is a monomial in x and y,(03B2) f
is nottangent
to the line atinfinity
ofk[x, y]. (Equivalently, f
=xmy"
+ g wheredeg
g m + n,degx g ~
m,degy
g ~n.)
Theoperations
off orming
apolynomial (with
coefficients in
k)
and oftaking
a factor preserve theseproperties
andhence
each bi
satisfies(03B1)
and(03B2).
On the otherhand,
sinceB/biB
is apolynomial ring
over afield,
thedegree
formof bi
is a monomial in x alone or y alone andhence bi
is apolynomial
in either x or y.(This argument slightly generalizes [8, 4.8].)
Since(bi, b;)B
= B fori ~j, x
and y cannot appear
both,
and we may assume thateach bi,
andtherefore b,
is apolynomial
in x.By
Lemma 1.6below,
we can assume that k isalgebraically
closedin
verifying (**)
under conditions(ii)
and(iii). (Unique
factorization will not be usedagain,
and A remainsregular
over analgebraic
closure of k if we assume
(iii).)
Case 2: k
algebraically closed,
char k = 0. Let x == b1 be an ir- reducible factor of b. ThenB/xB
~k(1),
andby
the main result of[1],
there exists y E B such that B =k[x, y].
Ifbi
is any other irreducible factor ofb, then bi
= yix + Si with y;, Si E k since(b,, bi)B
= B(see [9,
Lemma
1]).
Hence b Ek[x].
Case 3: k
algebraically closed,
Aregular.
Let x = b1. As in case2, B/xB = k(1), bi
= yix +,Si with yi, Si Ek,
and b Ek[x].
Hence x E A.Let X and Y be
complete non-singular
surfacescontaining
respec-tively Spec
B andSpec
A as dense opensubsets,
with X =P2k.
Thebirational
morphism Spec B-SpecA
induces a birational map ç : X ~ Y and(see [12, part II]
or[10,
Ch.IV, § 3]
for basic facts from thetheory
of birationalcorrespondences
of surfaces usedbelow)
there exists anonsingular
surface Z and birationalmorphisms
~1 : Z ~X, ~2 :Z ~ Y
such that ~ ° ~1 = ~ and ~1, CP2 arecomposites
oflocally quadratic
transformations.(The
centres of these we call thefundamental
points
of çi and ’P2respectively.) Replacing Z,
if neces- sary,by
a surface Z* dominatedby
Z we may assume that(ai)
no irreducibleexceptional
curve E of the first kind on Z(this
means E =
Pl k
and(E, E) = - 1,
where(-, -) denotes
the intersectionpairing)
shrinks to apoint
on both X and Y.For any curve C on X or Y let C’ denote its proper transform on
Z. For À E
k,
letC,
be the curve on X whose ideal in B is(x - 03BB)B.
Put d =
deg Cx
and L = X -Spec
B. The curvesC, together
with dLform a linear
pencil
A =039B (x) (see [8, 1.2]).
Let p ESpec
A C Y bethe closed
point
with idealMi
= xB n A in A.By (*), Cl
Ccp 2 -’(p).
LetE be an irreducible
component
of~-12(p)
such that(E, E) = 2013 1.
Then E ;;é L ’. Infact,
(a2) ~2(L’)
C Y -Spec
A.Otherwise
~2(D’)
CSpec
A for almost all lines D C X p2 andSpec
A carries acomplete
curve, which isimpossible. Also,
E doesnot contract to a
point
on Xby (03B11)
and hence E =C’,
where C C X is an irreducible curve such that C nSpec
B 40. By (*),
C =Co
and hence(a3) (Cl, Cl)
= - 1 andCl
is theonly
irreduciblecomponent
of’P21(p)
with this
property.
For
general
À Ek, ~2(C’03BB)
is a curve on Y whose ideal in A is Ix =(x - 03BB)B
rl A. Since x E Mi CA,
we have x - À E Ix andI03BB ~ Mi.
Hence
p ~ ~2(C’03BB)
and(a4)
no basepoint
of A(see [8, 2.5])
is onC’o.
Let q1, . . ., qs be the base
points
ofA,
let vi =03BC (039B, qi) (see [8, 2.5],
this is the
multiplicity at qi
of the proper transform of ageneral
member of
039B)
and 03BCi =03BC(Co, qi) (see [8, 2.3],
this is themultiplicity
atqi of the proper transform of
Co). Note vi
~ 1and i£i
~ 0.By (a4),
all qi with gi > 0 are fundamentalpoints
of ’P 1.By (a3)’
and since(Co, Co) = d2, ~ 03BC2i ~ d2 +
1.Intersecting Co
with ageneral
memberC,
of A we find Y- iiivi =d2. Also,
since A is apencil, 1 V2 = d2 (see [8, 2.10]). By
Schwarz’s
inequality, 2 03BC2i ~ d2.
If yIl 2= d2 + 1,
then 1vi (03BCi 2014 vi)
= 0 and 1(03BCi 2014 Pi)’
=1,
which cannot be satisfied inintegers
iii, vi with-vi 1. Hence ~ 03BC2i = d2
and 03BCi = vi for all i. SinceC’o ~ Pl,
anymultiple point
ofCo
is a fundamentalpoint
of ~1 and hence one of the qi. For if not,(C’o,C’o) ~ d2 - ~03BC2i - 4 ~ - 4
in contradiction to(a3).
Hence 1
03BCi(03BCi 2014 1)
=(d - 1)(d 2014 2) = 2 vi(vi - 1)
and thegeneric
member
039Bn
of A is a curve of genus zero overk(x) (see [8,
2.8 and2.11]).
Since k isalgebraically closed, 039Bn
is a rational curve, i.e.qtB,
the function field of039Bn,
ispurely
transcendental overk(x). (There
is aconic in
P2k(x) birationally equivalent
to039Bn (see [2,
Ch.II, §6]). By
Tsen’s
Theorem, A,
has aplace
rational overk(x) (see [11,
Ch.II,
3.2 and3.3])
and hence is rational(see [2,
Ch.II, §3]).) Equivalently,
x isa field generator in
B,
andby [8, 4.8]
there exists y E B such that B =k[x, y].
1.4. COROLLARY: Let k be
perfect,
A ak-algebra,
t transcendentalover A and assume
A[t]
=k[x,
y,z] ~ k(3).
Assume also that z = bt - awith a, b
E A. Then A ~k(2).
PROOF: If b =
0,
the result is proven in[3,
Theorem4.1].
If b ~0,
thehomomorphism 03C3 : A ~ A[t] ~ A[t]/zA[t]
= B isinjective. Clearly
A is a
regular
UFDfinitely generated
over k andGCD(a, b)
= 1.Identifying A
withu(A)
we have B ~A[alb]. By
1.3(iii),
A =k(2).
1.5. LEMMA: Let
k’/k
be aseparable algebraic
extension. Let x EB ~k(2)
such thatB ~kk’ = k[x, y’] for
somey’
EB Ok k’.
Then thereexists y E B such that B
=k[x, y].
PROOF: Let t be transcendental over k and
put
R =B ~kk(t)/(x - t)B ~k k(t).
ThenR ~k(t)k’(t) ~k’(t)(1),
and sincek’ (t)1 k(t)
isseparable, R~ k(t)(1),
as is well known(see [7, 1.1]).
Hence x is a field generator
(see [8, 1.3])
and we canapply [8, 4.8].
1.6. LEMMA: Let
k’/k be a separable algebraic
extension. Let b EB ~ k(2)
such that there existx’, y’
E BOk
k’ with BOk
k’ =k’ [x’, y’]
and b E
k’[x’].
Then there exist x, y E B such that B =k[x, y ]
and b Ek[x].
PROOF: We can find a,
(03B2 ~ k’, a ~ 0,
such that03B1x’ + 03B2 ~ B.
Theproof
is the same as theproof
ofCorollary
1 of[9].
Put x = ax’ +03B2.
Then
k’[x,y’] = k’[x’,y’ ]
and b~k’[x] ~ B
=k[x].
The existence of y such that B= k [x, y] follows
from 1.5.2
In this section we extend the result of
[9]
on "linearplanes"
tofields of
arbitrary
characteristic. It ispossible
to dothis,
once 1.3 isestablished, by referring
to details in theproof
of[9].
It may beworthwhile, nevertheless,
to write down an argument moredirectly adapted
to our line ofreasoning. Also,
2.2 below(which,
more orless,
can be found hidden in
[9]), giving
a construction for somewhat unfamiliar(since
ingeneral
not"tame") automorphisms
of thepolynomial ring R(2), where
R is any commutativering,
deserves to bementioned
explicitly.
We record the
following
well known facts(see
also the remark after Lemma 5 in[9]):
2.1. Let S be a commutative
ring
and a ES[T] ~ S(1),
a = ~aiT’
iwith ai E S. Then
(i)
a isnilpotent ~
ai isnilpotent
for alli,
(ii)
a is aunit~03B1o
is a unit and ai isnilpotent
for i ~1, (iii) S [a ] = S[T]~03B11 1 is
a unit and ai isnilpotent
for i ~ 2.2.2. PROPOSITION: Let R be a commutative
ring, b E R,
03B1 = Y-aiv
i ~ R[v ] ~
R(1) with aiE R,
and H = bw + a E R[v,
w] c-- R(2).
Assume that al is a unit mod bR and that ai is
nilpotent
mod bRfor
i ~ 2. Then there exists
cp(T)
ER[T]
such thatcp(a)
~ v modbR[v].
For any such cp,
with G E
R [v, w]. Moreover,
PROOF: The existence of cp follows from
2.1(iii) applied
to S =R/ bR.
Let~(T) = ~i~0 03B2i Ti
iwith l3i
Cz R. Then03B1103B21 2014
1 andthe 03B2i
fori ~ 2 are
nilpotent
mod b. Writewith 6
ER [v].
We havewhere
~(i)(T) = ¿j2d (ji)Tj-i.
Hence~(H)
= v+ b03B2
+ bwP withSo
~(H)
is of the desired form withMoreover, ~(1)(03B1)
is a unit mod b and~(i)(03B1)
isnilpotent
mod b fori ~ 2. Hence P is a unit mod b in
R[v,bw ] (apply 2.1(ii)
toS = R [v]/bR[v])
and there existQ, QI E R[v, bw]
such thatPQ = 1 + bQ1.
Thenw = (G - 03B2)Q - wbQ1. Clearly v = ~(H) - bG ~ R[G,H]
and hencebw = H - 03B1 ~R [G,H].
It follows that(3, Q, Q,
ER [G, H].
Hence w ER [G, H].
2.3. THEOREM: Let k be a
field,
a, b E A =k(2)
with b ~0,
andH = bw - a E
A[w] =
k(3).Suppose A[w]/HA[w] = k(2).
Then there exist u, v E A such that A =k[u, v],
b Ek[u]
and a - v isnilpotent
mod bA.
Moreover,
there exists G EA[w]
=k[u,
v,w]
such thatk[u,
v,w]
=k[u, G, H].
PROOF: Let B =
A[alb].
Then B~A[w]/HA[w] ~ k(2). Clearly GCD(a,b) =
1.Also,
A contains a fieldgenerator,
andby 1.3(i),
wecan find x, y E B such that B
= k [x, y], b
Ek[x]
and A =k[x, by].
Putu = x and
v’ = b y.
ThenB = A[v’/b],
andby 1.1(ii), a = a0b + cv’
where ao, c E A and c is a unit mod b.
By 2.1(ii), c = ~ civ’i
i withci
~ k[u],
co a unit mod b and cinilpotent
mod b for i ~ 1.Finally,
co Y-
ylu
i with yl~ k, i yo ~
0 and co - yonilpotent
mod b. Put v =yov’.
Then a - v isnilpotent
mod b. The existence of G follows from 2.2applied
to R =k[u].
3
The conditions under which we
proved
1.3 may not be the bestpossible. Unique
factorization forA, however,
is essential. For aneasy
example,
considerAlso,
it does nothelp
to assume that A isregular,
as the nextexample
shows
(which
the author learned from W.Heinzer).
where u = x and v = xy.
Here
Spec B - Spec
A is an openimmersion,
and this istypical
in away. For we have
3.3. THEOREM: Let k be
perfect,
A afinitely generated regular
k-domain and a, b E A with b 4 0. Assume B =
A[alb] ~
k(2). Thenthere exist x, y E B and b’ E
k[x] ]
such that B =k[x, y],
A C
k[x, b’y]
= B’ andSpec
B’ -Spec
A is an open immersion.PROOF: Not all irreducible factors of b in B now
necessarily
contract to maximal ideals in
A,
but the claim(*)
we made in theproof
of 1.3 holds for those irreducible factorsb 1, ..., br
that do. If there are none, we are doneby
Zariski’s Main Theorem. Otherwisewe can,
exactly
as under condition(iii)
of1.3,
find x,y E B
withB =
k[x, y]
andbi
Ek[x] for
all i. We claim A Ck[x, biy],
and to finishthe
proof
of the theorem we choose b’ Ek[x]
of maximaldegree
suchthat A C
k[x, b’y].
It will beenough
to establish the claim in case k isalgebraically
closed. Forthen,
ifbi1, . . ., bis
are the(distinct)
ir-reducible factors
of bi
ink’[x,y] = B’,
where k’ is analgebraic
closure of
k,
we havebi1
1 =bi ... bis,
eachbij
contracts to a maximal ideal inA’ = A ~kk’
andA’ ~ k’[x,bijy]
for allj.
HenceA’
~ k’[x,biy]
and A Ck’[x, biy i
nk[x, y]
==k [x, biy].
Let then k be
algebraically
closed. We may set x =bi.
Letbe as in case 3 of the
proof
of1.3,
with theembedding Spec
B C X =p2
so chosen thatCo,
the curve whose ideal in B isxB,
hasdegree
d = 1. Let
(03C3: Z1 ~ X
be thelocally quadratic
transformation with centre p 1 =Co f1 L (L
= X -Spec B), E1 = 03C31-1(p1)
and 03C32 :Z2 ~
Zl thelocally quadratic
transformation with centre P2= Côl’
nEt,
whereCôl’
is the proper transform of
Co
onZi.
The proper transform onZ2
of acurve C on X or
Zl
will be denotedby C(2),
and the proper transformon Z of a curve C
on X, Zi
or Z2by
C’. Put (T == (Tl 0 0-2. Since(C’o, C’o) = - 1 (see (a3)
in theproof
of case3)
and since there are nofundamental
points
of ~1 onSpec B,
pi and P2 areprecisely
thefundamental
points
of y onCo.
It follows that there exists a mor-phism oi :
Z ~Z2
such that cp = 0" 003C8.
Wehave, putting E2 = 03C32-1(p2),
with
Co(2)
andE2, E2
andEi2’, E1(2)
andL (2) meeting normally
in onepoint
and all other intersectionsempty.
There are no fundamental
points of oi
onCo(2)
and hence p == cp 0 o- is amorphism
in aneighbourhood
ofCo(2).
Inparticular,
sinceCo
contracts to a
point
inSpec A,
We claim that there are no fundamental
points of «/1
onE2 - E1(2).
Inf act, F
= Y -Spec
A is connected(see [4,
Ch.II, 6.2])
and hence(b3) ’P2l(F)
is connected.Since ~ is a
morphism
onSpec B, ’P2l(F)
C~1-1(L) =
«/1-l(E2
UE1(2)
UL (2»).
Now L’ C~2-1(F) by (a2),
butE’2 ~ ~2-1(F) by (b2).
It follows from
(bi)
and(b3)
thatif q
EE2 - E1(2)
is a fundamentalpoint
of oi,
then«fJ-1(q)
ncp-1(F) == 0.
Hence~2(03C8-1(q))
CSpec
A and there-fore is a
point,
and this contradicts(ai).
By
what we haveshown,
p is amorphism
onZ2 - (E1(2)
UL (2»).
Let03C4 :Z2 X1
be the contraction ofC0(2). (Note (C0(2) = - 1
andC0(2) ~ Pk1. Xi
isisomorphic
to the ruled surfaceF2.)
Let p== p’ 0 r.
Then
p’
is amorphism
onU = X1 - 03C4(E1(2) ~ L(2) .
Also r o03C3-1(Spec B)
C U. It iseasily
verified that U is affine and that0393(U) = k [x, xy ](0393(U)
=ring
of functions defined onU).
Hence A C k[x, xy].
3.4. REMARK: The
proof given
aboveactually
shows:Let k be
perfect,
A afinitely generated regular k-domain,
A C B ~k(2)
withqtA
=qtB.
Let x E B such that xB fl A is maximal andB/xB ~ k(1).
LetX, Y,
Z be as above. If the proper transform on Z of the curve on X definedby
x isexceptional
of the first kind(briefly,
"xshrinks
first";
one can see that this isindependent
of the choice ofX, Y,
and Z aslong
as Z satisfies(al))
then there exists y E B such that B =k [x, y ] and
A Ck [x, xy ].
Theexample
shows that it is not
enough
to assume that xB fl A is maximal.(Here
yhas to shrink before x can
shrink.)
3.5. REMARK:
Suppose A[t]
=k[x,
y,z]
as in 1.4. One may ask what information our methodsgive if,
say, t = bz - a witha, b
Ek[x, y],
b 0 0. Notmuch, unfortunately.
We havek[x,
y,a/b] = A,
but this does notguarantee
A =k(2)
under the best of conditions forA,
aswe will see. If
f
is an irreducible factor of b inA,
thenA/ fA =
k(1)(assume k
isalgebraically closed),
sok[x,
y,z]/fk[x,
y,z]
=k(2).
Iff
islinear in x, y or z, we can use 2.3 and refer to
[3,4.1],
but there is no reasonwhy
it should.Now
put
a = x and letb ~ k[x,y] = B
be irreducible such that(x, b)B = (x, y)B
= M andB/bB ~ k(1) (for
instance b =XY2 +
y +x2).
Let A
= B[a/b].
It iseasily
checked that A isregular
with constant units.Also,
bA fl B ~(x,b)B
=M,
so bA fl B = M andAlbA ~
k (1).Hence bA is
prime.
Since Ab = Bb is aUFD,
A is a UFD(see [5]).
Onthe other
hand,
A~k(2), for
otherwiseB/bB = k(1) by
1.3.3.6. REMARK: 1.5 remains true for
purely inseparable
extensionsk’ 1 k
ifB/xB ~ k(1)
is included in theassumptions. (The proof
is morecomplicated.)
Otherwise the conclusion is false ingeneral.
Infact,
letchar k = p > 0
andx = vp + u + 03B1up ~B =k[u,v]
with aE!k-kP.Then
B/xB ~
k(1)(see [7])
andk [x, y] 4
B for all y EB,
butk’ [u, v ] =
k’[x,v + 03B2u ],
where03B2p = 03B1
andk’= k(03B2).
This alsoshows,
with b = x, that 1.6 may fail ifk’ /k
is notseparable.
Moreinteresting
in thepresent
context is the fact that it does nothelp
to assumeBIBB K(1),
where K is a field. For let b = xp - a. If b is a
polynomial
in somew E
k[u, v],
thenclearly w
is a power of yx + 8 withy, 6
Ek,
andagain B/wB ~ k(’). However, letting -
denote residues modb,
we havex = (v + xu)p + u
and henceBlbB
=k[x, v + xu] = k’[v + xu] ~ k’(1).
3.7. REMARK: If
char k = p > 0,
there existsa ~ A = k[u,v] = k(2)
such that
A/aA ~ k(1),
butA ~ k[a,y]
for ally E A,
e.g. a =up2 + v + vrp,
withr > 1, GCD(r,p) = 1 (see
the introduction of[1]).
One deduces
easily
that theassumption b 4
0 in 2.3 cannot bedrop- ped.
We conclude
by raising
twoquestions suggested by
theproof
of1.3.
3.8.
QUESTION:
If A is afinitely generated
k-domain such thatqtA ~ qtk(2),
when does A contain a fieldgenerator?
One shouldassume that the units of A are constant and may want to
impose
additional
conditions,
such as, in somecombination, (i)
A ck[x, y] == k(2)
withqtA
=k(x, y),
(i’) A[alb]
=k[x, y]
for some a, b EA, (ii)
A isregular,
(iii)
A is a UFD.3.9.
QUESTION:
Letf ~ B ~k(2)
be irreducible. When does there exist aregular
k-domain(regular UFD)
A C B withqtA
=qtB
suchthat
fB
contracts to a maximal ideal in A?(Clearly,
if k isalge- braically closed, Spec B/ fB
is anonsingular
rational curve, but whatelse?)
One couldrequire
in addition thatfB
is theonly height
1prime
in B
contracting
to a maximalideal,
or, somewhatweaker,
thatf
hasthe
property
of x in 3.4 of"shrinking
first".REFERENCES
[1] S. S. ABHYANKAR and T.-T. MOH: Embeddings of the line in the plane, J. Reine Angew. Math. 276 (1975) 148-166.
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VI. Am. Math. Soc., New York, 1951.
[3] P. EAKIN and W. HEINZER: A cancellation problem for rings, Conference on
commutative algebra. Lecture Notes in Mathematics 311. Springer, Berlin-Heidel- berg-New York, 1973.
[4] R. HARTSHORNE: Ample subvarieties of algebraic varieties. Lecture Notes in Mathematics 156. Springer, Berlin-Heidelberg-New York, 1970.
[5] M. NAGATA: A remark on the unique factorization theorem. J. Math. Soc. Japan
9 (1957) 143-145.
[6] M. RAYNAUD: Anneaux locaux Henséliens. Lecture Notes in Mathematics 169.
Springer, Berlin-Heidelberg-New York, 1970.
[7] P. RUSSELL: Forms of the affine line and its additive groups. Pacific J. Math. 32, No. 2. 527-239 (1970).
[8] P. RUSSELL: Field generators in two variables. J. Math. Kyoto Univ. 15-3 (1975) 555-571.
[9] A. SATHAYE: On linear planes (in Proc. Amer. Math. Soc.)
[10] J. R. SCHAFAREWITSCH: Grundzüge der algebraischen Geometrie. Vieweg, Braunschweig, 1972.
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(Oblatum 1-IX-1975) Department of Mathematics
McGill University
Montreal P.Q.
Canada