.--A

 

H'EGHER DERWATW OF A
PLANE ALGEBRAIO CURVE OVER A FIELD
OF PRIME mm

mm for the Degree of m. a.
mum STATE UNIVERSITY
ANNE LARIMER women
1975

 

x _‘.,'C'. “
g s.)

This is to certify that the

thesis entitled
HIGHER DERIVATIONS OF A PLANE ALGEBRAIC

CURVE OVER A FIELD OF PRIME CHARACTERISTIC

presented by
Anne Larimer Ludington

has been accepted towards fulfillment
of the requirements for

Ph.D. degree in Mathematics

 

 

Major professor

Date 20 June 1275

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ABSTRACT

HIGHER DERIVATIONS OF A PLANE ALGEBRAIC
CURVE OVER A FIELD OF PRIME CHARACTERISTIC

BY

Anne Larimer Ludington

Let T be a plane irreducible algebraic curve defined
over an algebraically closed field k. Let P be a point
on the curve and let R be the local ring of f at P.

Der:(R) is the R-module of all n-th order k-deriva-
tions of R to R. Thus w 6 Der:(R) if and only if

-, r E R we have

m E Homk(R,R) and for all r n

0'0.

n v
- 8-1 V o o o o o o o o o o o o
w(ro°°°rn)- §3(-l) . M _ ri ri cp(rO ri ri rn).
s—l 11(---<1S 1 s l 5

Let Der(R) = kJDer:(R). If k has characteristic
zero, we define der(R) to be the subalgebra of Der(R)
generated by composites of lst order derivations. If the
characteristic of k is p #'0, ‘we say Der(R) is gener—
ated by pi-th order derivations if the following condition
is satisfied:

Let I E Der(R) and let n be the smallest integer

such that x € Der:(R;. Let the p-adic expans1on of

n be given by n = 2‘ oip1 . Then there exist
i=0

Anne Larimer Ludington

i
li'H'C’INTmi 6 ”3": ‘1‘"
- _ ... _ o
l '— 0' ’ Np SUCh that A (TlN oooooTlo

a
an 0) e Derfi'1(R).

pl-th order derivations r
+ooo+

TmN °°"°Tm0

Here Der:(R) = 0. Thus Der(R) is generated by pi—th
order derivations if every n—th order derivation is a sum of
composites of pi-th order derivations. If Der(R) is gen-
erated by pi—th order derivations, we write Der(R) = der(R).
Chapter 1 is devoted to theorems which characterize
Der:(R) when it is a free R—module. We show that if it is
free, it must be free on n generators. Further, if Der:(Rl
is free for all n, then there exist derivations
l1,---, In,°-- and elements x1,--o, xn,--- of R such

that xi 6 Der:(R): Der:(R) is generated by l X

1,000. n;

_ r0 j<1i
and xi(xj) — 11 '==i .

In Chapter 2 we consider the following example:
R is the local ring at the origin of F: f(X,Y) = X2 - Y3
over a field of characteristic 2. Since (0,0) is a
singular point of F, R is not a regular local ring. We
show that Der£(R) is a free Remodule for all n and that
Der(R) = der(R). Thus this example shows that over a field
of characteristic p # O the following two conjectures are
false:

(I) Der:(R) is a free R-module for all n if and

only if R is a regular local ring.

Anne Larimer Ludington

(II) Der(R) = der(R) if and only if R is a regular local
ring.
The first conjecture is a generalization of a conjecture by
Lipman: the second is Nakai's conjecture.
The main theorem of Chapter 3 is that if Der(R)==der(R)
and Der:(R) is a free R—module for all n, then R is
analytically irreducible, that is, R, the completion of R,

is an integral domain. Geometrically this means that P has

only one branch at P.

HIGHER DERIVATIONS OF A PLANE ALGEBRAIC
CURVE OVER A FIELD OF PRIME CHARACTERISTIC

BY

Anne Larimer Ludington

A DISSERTATION

Submitted to
Michigan State University
in partial fulfillment of the requirements
for the degree of

DOCTOR OF PHILOSOPHY

Department of Mathematics

1975

TO PAUL

ii

ACKNOWLEDGEMENTS

I am grateful to my thesis advisor, Dr. William C.
Brown for suggesting this topic and for the guidance and
help that he gave me during the last two years. I am
particularly appreciative of the time and effort that he
spent while I was writing the rough draft. I also wish
to thank my husband, Paul, for being a source of inspira—

tion and encouragement.

iii

CHAPTER 1:

CHAPTER II:

CHAPTER III:

TABLE OF CONTENTS

INTRODUCTION . . . .

CHARACTERIZATION OF DER:(R) As A FREE

R—MODULE
AN EXAMPLE

A
THE COMPLETION R

BIBLIOGRAPHY . .

iv

16

28

74

INTRODUCTION

Throughout this entire paper we shall assume that T
is a plane irreducible algebraic curve, defined over an
algebraically closed field k. Let P be a point on the
curve, and let R be the local ring at P. Without loss
of generality, we may assume that P is (0,0). We shall
denote the quotient field of R by K. If F is given by
f(X,Y) = 0, then R = (k[x'Y])(x,y) where
kfX.y]

and f(0,0) = O.

k[X,Y]/(f(X,Y)), f(X,Y) is irreducible over k,

For each n = 1,2,---, we let Der:(R,M) denote the
R-module of all n—th order derivations of R to an R—module
M which vanish on k. Thus, m 6 Der:(R,M) if and only if

r 6 R we have

m E Homk(R,M) and for all rO,---, n

(1)

n
-1 v
m(r °--r )== Z)(-l)s Z: r. -~-r. m(r '°°r. ---r. ...r )

When M = R, we write Der:(R) instead of Der:(R,R).
d:(R) will denote the R-module of all n-th order differ—
entials. DerE(R) is the dual module of fl:(R) so

Der:(R) = HomR(O:(R),R).

Let Der(R) = LJDer:(R). If R has characteristic
n

zero, we define der(R) to be the subalgebra of Der(R)
generated by composites of lst order derivations. If the
characteristic of k is p #’O, we say Der(R) is
generated by pi-th order derivations if the following

condition is satisfied:

Let A E.Der(R) and let n the smallest integer

N .
such that A 6 Der:(R). Let n = Z) OL.pl be the
. 1
i=0
p-adic expansion of n. Then there exist pl-th
i

order derivations T ---,T . E Derp (R), i==O,o--,N,

li' m1 k
such that
a o o a
N O N O n-l
l - (TlN o,,,° r10 +---+ TmN °,,,o rmo) E Derk (R).

Here Derg(R) = 0. Thus by induction we see that Der(R)
is generated by pi-th order derivations if every n-th
order derivation is a sum of composites of pi-th order
derivations. If Der(R) is generated by pi-th order der-
ivations, we shall write Der(R) = der(R). For
K, Der(K) = der(K), [Pr0p. 18:7]. Further, if R is a
regular local ring, then Der(R) = der(R) [Theorem 4.3;4].
Well-known prOperties of Der:(R), Q:(R), and Der(R)
may be found in Nakai's papers [6 and 7].
The starting point of this thesis concerns two conjec—
tures which are known to hold for plane curves when k has

characteristic zero:

(I) Lipman's conjecture: Deri(R) is a free
R-module if and only if R is a regular local

ring [Theorem 1:2].

(II) Nakai's conjecture: Der(R) = der(R) if

and only if R is regular [3].

It is easily shown that (I) is false when the charac-
teristic of k is p #'O; in fact, the example given in
Chapter 2 is a counterexample. That (I) is false is per—
haps to be expected, since when k has characteristic
zero, Der(K) is generated by composities of l-st order
derivations [Pr0p. 18: 7]. However, when the character-
istic is p, Der(K) is generated by composities of pi-th
order derivations, i = 0,1,--- [Pr0p. 18: 7]. Thus, the

following conjecture arises when k has characteristic p:

(III) DerE(R) is a free R-module for all
n = 1,2, --- if and only if R is a regular

local ring.

It is known that if R is regular and k has character-
istic p #'0, then Der:(R) is a free R-module for all

n [Theorem l6.ll.2: l]. The example which we shall give
in Chapter 2 will show that the converses of (II) and (III)
are false. That is, we shall construct a ring R over a
field k of characteristic p which is not regular, but
suCh that Der(R) = der(R) and Der:(R) is a free

R-module for all n.

In order to show (II) and (III) are false, some
results about Der:(R) are needed. Thus, Chapter 1 is
devoted to theorems which characterize Der:(R) when it
is a free Ramodule.

The main result of Chapter 3 is that if
Der(R) = der(R) and Der:(R) is a free R-module for all
n, then R is analytically irreducible; that is, R,
the completion of R, is an integral domain. Geometri—

cally this means that F has only one branch centered at P.

CHAPTER I
CHARACTERIZATION OF DER;1 (R) AS A FREE R-MODULE

The first lemma of this chapter shows that we may
assume K is a separable algebraic extension over k(x).
The remainder of the chapter gives necessary and sufficient
conditions for Der:(R) to be a free R-module. We shall
show that if Der£(R) is free, it must be free on n gen—

erators. Moreover, there exists a set of generators

n .
11, , In of Derk(R) and monomials x1, , xn of
O j<1
k[x,y] such that A.(x.) = { , .
1 J 1 j==l

Lemma 1.1: K is a separable algebraic extension of either

 

k(x) or k(y).

23923: The only case requiring proof is when k has char—
acteristic p #’0. Let fX and fY denote the partial
derivatives of f(X,Y) with respect to X and Y respec-
tively. Suppose fx(x,y) = O and fY(x,y) = O. Pulling
back to k[X,Y] gives fX(X,Y) = h(X,Y)f(X,Y). Viewing
these as polynomials in X with coefficients in k[Y]
gives deg fX < deg f‘g deg fh, which is a contradiction
unless fx(X,Y) = 0. So fX(X,Y) = O and fY(X,Y) = O.

This implies that f(X,Y) = g(XP,YP) = (h(X,Y))p.

contradicting the assumption that f is irreducible.

Hence, either fx(x,y) #’O or fY(x,y) #'0. Thus K is

a separable algebraic extension of k(x) or k(y). QED
Henceforth, we shall assume fY(x,y) # 0: thus K is

a separable algebraic extension of k(x).

 

Theorem 1.2: Der:(K) is a free K-module; it is free on
n generators. If Der£(R) is a free R-module, then it is

free on n generators.

Proof: Since K is a separable algebraic extension of

 

n :_ n . . n
k(x), flk(K) ._ Ok(k [x]) ®k[x]K [p.26,7]. Since Qk(k[x])p
is a free k[x]-module of rank n [11, Prop. 2: 6], we
have that Q:(K) is free of rank n over K. Now,
n _ n n . _
Derk(K) — HomK(flk(K) ,K), so [Derk(K). K] — n.
n _ n m n
Now Derk(R) ®RK - I-lomR(Qk (R) ,R) ®RK — HomK(Qk(R) ®RK,K) .
Since Q:(R) ®RK & Q:(K) [11, Theorem 9; 6],
n a n . n .
Derk(R) ®RK _ Derk(K). Thus, if Derk(R) IS a free

R-module, it must be free on n generators. QED

Since fY(x,y) #’O, f(X,Y) must have a term involVing
Y. Further, since f(X,Y) is irreducible, there must be a
term of the form OLYN , o. E k: otherwise, X would divide

f(X,Y). Hence subdngf(0,Y) > 0.

Lemma 1.3: Let subdngf(O,Y) = N. Then yN-l/x { R.

 

nggf: .As shown above, N > 0. write f(X,Y) = Xh(X,Y) +
YNg(Y), ‘where g(Y) begins with a non-zero constant term.
‘ Suppose yN-l/x 6 R. Then yN'l/x = r(x,y)/s(x,y),
where r(x,y), s(x,y) E k[x,y] and s(x,y) ¢ (x,y). Then

yN-ls(x,y) - xr(x,y) = O. Pulling back to k[X,Y] we

have
(2) YN-ls(X,Y) - r(X,Y)X = t(X,Y)[Xh(X,Y) + YNg(Y)].
Evaluating (2) at x = 0 gives
(3) YN'1s(o,Y) == t(O,Y)YNg(Y).

Thus, s(O,Y) = t(O,Y)Yg(Y). Since s(X,Y) has a constant
term, s(O,Y) # 0. Thus, (3) implies that Y divides

s(O,Y), which is a contradiction. Hence, yN-l/x { R.
QED

In the theorems which follow, we shall often use the
fact that an n-th order derivation is also an (n+1)-st
order derivation [1, Prop. 4: 6]. We shall also use the
result that if A E Der:(R), then A E Der:(K) [I,Theorem
15: 6]. Hence if l1,'--, In are a free basis for Der£(R),
these derivations must also be a free basis for Der:(K).

If Al,°'°, 1n are a free basis for Der:(R), then
we shall write Der:(R) = < ll,'°°, In >.

Before Der:(R) is considered for arbitrary n,
Der%(R) is studied. Special attention should be paid to
the method of proof, since the same technique will be used

‘when n > 1.

Theorem 1.4: Deri(R) is a free R-module if and only if

there exists I E Deri(R) and z 6 R such that k(z) = 1.

Proof: Let Deri(R) be generated by Y as a free

R-module. Suppose y(r)€(x,y) for all r E R. Then y(r)

may be written as y(r) = xrx + yry With rx, ry E R. As
in Lemma 1.3, write f(X,Y) = Xh(X,Y) + YNg(Y). Now con-
sider (yN-l/x)y which is certainly a derivation from K

to K. For r 6 R,

(yN-l/E)Y(X) = (yN'l/karx + yry)

= YN'er + (yN/xny
_ N-l
— y rx - (h(x.y)/s(y))ry.

Since g(y) is a unit in R, (yN'l/x)y: R 4 R. Thus
N-l 1 . . . N-l _
(y /x)Y E Derk(R) 'Wthh implies that (y /x)y — ty
for some t 6 R. Thus (yN-l/x - t)y = 0 on R, hence on
K. Thus yN—l/x - t = O or yN’l/x = t E R which is a

contradiction. Thus there exists 2 6 R such that y(z)

is a unit in R. Let A = y/y(z). Then Der;(R) < A >

and k(z) = 1.
Conversely, suppose such a l and z exist.
Deri(R) C Deri(K) and I may be regarded as the generator

of the K—module, Der:(K). Now let 6 e Deri(R). Then since

6 6 Deri(K), 5 tx ‘where t E K. Evaluating at 2 gives
6(2) = tX(z) = t. Thus t = 6(2) e R.’ Hence Deri(R) is

a free R-module. QED

Theorem 1.4 yields an easy proof of (I), Lipman's
conjecture, for a plane curve I defined over a field of
characteristic zero. For if Deri(R) is free on I,
then by Theorem 1.4 there exists 2 6 R such that I(z)==1.
Thus by Zariski's Lemma [Theorem 2: 2], R 2 B[[z]] where
z is analytically independent over. B. By Chevalley's
Theorem [Theorem 31, p. 320:9], R has no nilpotent ele—
ments. Thus, the dimension of R is l which implies
that B is a reduced, zero-dimensional local ring. There-
fore, B is a field and R is regular. Hence, R is

regular.

Theorem 1.5: Der:(R) is a free R-module if and only if

 

there exist n-th order derivations I1,"', In and
distinct elements xl,---, xn 6 R such that

0 ji<i
I.(xj)= 1 j=l°

n — 000
Proof: Assume Derk(R) — < 51, , 6n >. Suppose 61(r) E

 

(x,y) for all r E R. Then, as is the proof of Theorem

1 4 ( N"1/x)6 E Dern(R) So (yN—l/x)6 = E) t 6
° ' Y 1 k ' ' 1 i=1 i i
N 1 n
with t. e R. Thus, (t — y ' /x)6 + Z t.5. = 0 on
i l 1 i=2 i i
R, hence, on K. This implies yN_1/x = t1 6 R which is
a contradiction. So, there exists an x1 6 R such that

61(x1) is a unit. Let I1 = 61/61(x1). Then

6 >.

n — ...

10

we now proceed by induction. Suppose we have found

derivations I ---, I and elements x1,---, x E R

l m-l m—l
n — .0. 0..
such that Derk(R) - < I1. . xm-l'fim' , 6n > and
O j‘Ci . .
Ii(xj)={1 j=i for l_<_i_<_m—l and igggm—i.

Define Im as follows:

_ m-l
Im = 6m - 1E: riIi where r1 = 6m(xl)
i-l
and ri = qm(xi) - jg) rjIj(xi).

Computing Im(x£), L = l,°°‘, m - 1, gives

_ m—l
1mm) = smog) - . , HEW)
i=1
1—1
= 6m(X£) ‘ E7 rixih‘z) ’ ‘33
1—1
= o.
n — CO. - .0.
Also, Derk(R) — < I1, , Im_1, Im, , 6n >. As before,

there is an element x 6 R such that Im(xm) is a unit

m
in R. Finally, let Im = Im/Im(xm). Therefore

n
Derk(R) — < Il,---, Im,5m+l,---, 6n > and
o j<i
Xi(xj)={1j=i for l_<_i__<_m and l_<_j_<_m.

Thus, by induction, Der:(R) = < I1,---, In > and

there are elements x1,---, xn of R such that
{ 0 j<<i
11(ij " 1 j=i'

11

Conversely, suppose such elements and derivations

exist. Consider

(4) aI +---+aI =0

1 1 n n

with a1 6 R C K. Evaluating (4) at x1 gives a1 = 0
since Ii(x1) = 0 for i > 1. Suppose ai equals zero
for l g i _<_ m-l. Then evaluating (4) at xm gives
am = 0. Hence ai = 0 for 1.3 i.g n.

Thus, Der:(K) = < I1,---, I > . Let Y 6 Der:(R).

Then y may be written as y =

ll 13:!
PM

t.I. with 12.1 E K.

i i i

It suffices to show t.1 E R. Now y(x1) = t1 6 R. Induc- ‘

tively assume that t1.°°-. tm_1 6 R. Then
m-l
y(xm) = 1E1 tiIi(xm) + tm Since again Ii(xm) = 0 for
m—
m4-l g i S_n. So tm = y(xm) — iEI tiIi(xm) E R. Hence,
n _ ...
Derk(R) — < I1, , In >. QED

Theorem 1.5 does not require that Der;(R) be a free
R-module for i < n. With this added assumption we get a

much stronger result.

Theorem 1.6: Suppose Der2—1(R) is a free R-module with

 

generators I1,°--, I where Ii €_Der:(R) for

n-l
1 g_i g n-l. Further suppose that there exist distinct

elements of R, x .°°°, x , such that
l n-l
0 j<<i n
Ii(xj) = { 1 j==i . If Derk(R) is a free R-module, then

the following hold:

12

(a) Der:(R) = Der:'1(R) e Ry where y €.Der:(R).

(b) There exists an n-th order derivation In and an
element xn E R such that xn # xi, 1 = 1,°°°,n-l:In(xi)==0,

i = 1,-°°, n-1: and In(xn) = 1.

Proof: Suppose Der:(R) = < 61,---, 6n >. Since an

(n-1)-st order derivation is an n—th order derivation,

n
n — —
In_1 e Derk(R) and I — Z) ri6i where ri e R. Eval

“’1 i=1 n
uating In_l at Xn-l gives 1 = 1E; ri6i(xn_l). So, for
some I, 1 3.1 g'n, rI and 6I(xn_1) are units in R. we‘

reorder, if necessary, so that I = n-l. Thus,

 

 

 

1 n—2 ri rn
6n-l _ r xn-l - Z) r 6i - r 6n
n-l i=1 n—1 n-1

n _ 0..
And so, Derk(R) — < 61, , 6n_2,In_1,6n > .

Inductively we assume

n — 0..
Derk(R) - < 61, , 6 I

m' °, xn-l'bn > , and we show

m+1'..
that 6m may be replaced by Im, after relabeling the

. . n .
61's if need be. Since Im E Derk(R), Im may be written

a 8
IE: nil
(5) I = t.6. + > t.I. + t 6 .
m i=1 1 1 j=m+1 n n

Evaluating (5) at xm gives

13

m n-l
—_— . . 6
1 iii ti6i(xm) + j=§i1 tJIJ(xm) + tn n(xm)

m
= .23 ti6i(xm) + tn6n(xm)

i=1
since Ij(xm) = 0 for j = ma+1,--~, n-—l. For some I,
tI and 6I(xm) are units in R, I = l,'°°, m,n. Re-

arrange the 61, if necessary, so that I = m. Thus,

6m-?xm--Z?-5i— 4‘3 th-E‘bn-
m 1=1 m j=m 1 m m
n — 00. .0.
Hence, Derk(R) — < 61, , 6m-l'xm’ , In_1,6n > .

. . n
Therefore by induction, Derk(R) - < I1, -, In_l,6n >.

This completes the proof of (a).
To prove (b), we assume by (a) that Der:(R) =
< I1,:°°, In_1,6n > . We define a new n-th order deriva—

tion as follows:

n-l
(6) y = 6n + '53 riIi where r1 = — 6n(x1)
1- i-l
and ri = - 6n(xi)". : rjIj(xi).
J=1
Then since Ii(xm) = 0 for i = m4—1,°°°, n-1, evalua—
ting (6) at xm gives
n-l
y(xm) = 6n(xm) + _ZZ rixi(xm)
i=1
m—l
= 6n(xm) + 1:: riIi(xm) + r1“

= O.

Thus y(xm) = 0 for 1 g_m < n-l. Moreover,

n _ ...
Der-k(R) -' < All 0 Ari-lav >-
As in Theorems 1.4 and 1.5, there must exist an
element xn E R such that y(xn) is an unit in R.

Finally then, let In = y/y(xn). The theorem is now

proved. QED

We shall use the following lemma to show that the xi
given in Theorems 1.5 and 1.6 may be assumed to be

monomials in k[x,y] c R.

Lemma 1.7: If I is an n-th order derivation and I(r)==1

 

for some r 6 R, then I(xlyj) is a unit in R for some

x1 J 6 k[x,y] c R.

Proof: ‘Write r = s(x,y)/t(x,y) where s(x,y),t(xgy)€kIX.y]

and t(x,y) t (x,y). By [I,Theorem 5:6]

1 = I(r)
= I(s/t)
n
= (4)“ 23 (-1)‘“ < “1:1 > tmI(tn—ms)/tn+1.
m=0 ‘

For some m, I(tn.ms) is a unit in R. write tn-ms =

ZloijxlyJ E k[x,y]. So, I(xlyj) is a unit for some
xlyj. QED
The following theorem summarizes the results of this

chapter.

15

 

Theorem 1.8: If Der:(R) is a free R—module for all n,

then there exist derivations I1,'°°, In and monomials
xl,---, xn 6 k[x,y] such that
(a) x e Deri(R)

i k

n

(b) Derk(R) - < I1. . In >
() I() {Oj<i
c . x. =

1 3 1 j==i

2529;: The only part of the theorem that requires proof
is that the xi may be choosen to be monomials. We show
this by induction on n. If n==1, then by Theorem 1.4
there exists a derivation I such that Deri(R) = < I >
and I(z) = 1 for some z 6 R. By Lemma 1.7 there exists
a monomial x1 6 k[x,y] such that I(xl) is a unit in R.
Let I1 = I/I(x1). Then Deri(R) = < x1 > and I1(xi) = 1

where x1 is a monomial in k[x,y].

n-l - 0..
We now assume that Derk (R) — < I1, , In_1 > and

0 j<ii .
Ii(xj) — { 1 j==i for monomials xi 6 k[x,y]. By
Theorem 1.6 there exists a derivation In and an element
- n — o o o - - =
xn E R such that Derk(R) — < I1, , In_1.In2>.In(Xi) 0

for i < n, and I (

n xn) = 1. Again by Lemma 1.7, there

exists a monomial xn 6 k[x,y] such that In(xn) is a
unit in R. Now let In = In/In(xn). Then

n -— so. =3 .
Derk(R) - < I1, , In >, In(xi) 0 for i < n, and

In(xn) = 1 where xn is a monomial in k[x,y]. QED

CHAPTER II
AN EXAMPLE

In this chapter we give an example of a curve F
defined over a field of characteristic 2. I is given by
f(X,Y) = X2 — Y3. Since (0,0) is a singular point of F,
the local ring, R, at (0,0) is not regular. By con-
structing i-th order derivations I1 and monomials
x.1 6 k[x,y] which satisfy the conditions of Theorem 1.5,

we shall show that Der§(R) is a free R-module for all n.
We shall also show Der(R) = der(R). Thus, this example
shows that both (II) and (III) are false.

The following lemma will be used repeatedly in the
example. The results hold for any characteristic p #’0.
Thus, we shall prove the lemma in this more general setting
even though in the example k has characteristic 2.

n
Lemma 2.1: If I 6 DerE (R) and r,s E R, then

 

n n n
(a) I(rp s) = rp I(s) + sI(rp )

n+i n+i
(b) I(rp s) = rp I(s) for i 21.

Proof: The proof of (a) follows immediately from the de—

finition of a pn—th order derivation: this is equation (1).

16

17

For (b), the previous part gives

i

n+ i n
I(rp s) I((rp )9 s)

n+i n+i
rp I(s) + sI(rp ).

i n+i

rp I(s).

n+i n+
Since I(rp ) = 0 [I,PrOp. 10:6], I(rp s)

QED

Example 2.2: Let R be the local ring at (0,0) of

F:f(X,Y) = X2 — Y3 over a field k of characteristic 2.

3
R = (k[X'Y])(x,y) and x2 = y . Then Der:(R) is a free

 

R-module for all n and Der(R) = der(R).

Proof: Let A = k[x,y] = k[X,Y]/(XZ-Y3). 50

m
— - O O O . 2
R _ (A)(x,y)' For m — 0,1, , define Yzm E Derk (k[Y])
as follows:
. o i<2m
7) Y (Y1) = { m
( 2m 1 i==2

Thus, y m is a Zm-th order derivation of k[y] to
2 GI a0
k[y]. Define Yi = y I °,,,° Y1 where the aj's are the

2

coefficients in the 2—adic expansion of i: that is

I . .
i = Z: c.23. It is easily shown that y.(yj) = { . ..
j=O J l 1 j==i

O j<Ii

We now define a 2n—th order derivation on k[x] to

k[x,y], n = 0,l,--°, . For n = 0, define

18

n
I n E Deri (k[x], k[x,y]) as follows:
2
(8a) I n(x) = 0
2
2j _ 3j . _ n—l
(8b) I n(x ) — Y n—1(Y ) J — 1.°--. 2
2 2
(BC) I n0:23”) = XI (x23) j = 1,---, 2n'1-1.
2 2n

For each n, I n extends uniquely to a 2n-th order
2

derivation of k(x) to k(x,y) [I,Theorem 15:6]. We

call this extension I n' Since k(x,y) is separably
2

algebraic over k(x),I n extends in a unique way to a
2

2n~th order derivation of k(x,y) to k(x,y) [Theorem
17:7]. This extension is also called I n'
2
2n
We shall show that I n 6 Derk (A). First however
2

we shall show that (8b) and (8c) hold for all values of j.

From the definition of I , we have that

2n
2n 2n
I n(x ) = y . For, by Lemma 2.1,
2
n
n n-1 n-1 n-1+2
2 2 2 3 2 2
I m )=I HX) )=Y (W) )=w (y )
2n 2n 2n—l 2n—l
2n 2n-l 2n

= y Y2n_l(Y ) = y

We now show by induction that for all j we have

23' _ 33'
(sh) 12n<x ) - Y2n_1(y ).

19

First, suppose 2j = 2n + 2i where i = l,'°°, 2n-1
Evaluating both sides of equation (8b) gives
. n .
I n(x23) = I n(x2 x21)
2 2
n . n
= X2 I n(X21) + X21 I n(X2 )
2 2
_ 2n ( 3i) + 21 2n
- X Y n-l Y X Y
2
and
. n-l
3 2+1 2 +'
Y n_1(y 3) = Y n_l(y( )( 1))
2 2
2n “'1 3i
— Y n_l(y y y )
n n-l .
_ 2 2 3i
— y Y2n_1(Y y )
2n 2’“1 3i 2n 31
= y y Y2n_1(Y ) + y Y
n . n .
2 3i 2 21
- x Y2n_l(Y ) + y X -
Thus, (8b) is valid for 2jIg 2n+1.
. n+k
Now suppose that (8b) holds for 23.3 2 where
k > 0. We show (8b) also holds for 2j = 2n+k + 2i where
i = 1,---, 2n+k-1 . Again we compute I n(x23) and
2

3j .
Y2n_l(Y ).

20

2n+k 2n+k-l

33' _
Y2n_1(y ) " Y2n—l (y y Y )

2n+k 2n+k—l

= Y Y Y
2n

1(y )

2n+k

31
= X Y n_1(y )
2

Thus (8b) holds for 2j g 2n+k+l and so by induction holds

for all values of j.

Now we show that (8c) is valid for all values of j.

That is, I (x23+1) = XI (x23). First suppose
2n 2n
2j + 1 = 2n + 2i + 1 where i = o,---, 2‘"1 — 1. Then
2'+1 2n 2i+l
I n(x j ) = I n(x x )
2 2
n . . n
= x2 I n(x21+l) x21+1 I n(X2 )
2 2
n . . n
= x2 xI (x21) + x21+1 I (x2 )
n n
2 2
n . n
= x(x2 I (X21) + 21 A (X2 ))
2n 2n
n .
= xI n(x2 x21)
2
= xI n(x23).

21

Now suppose that (8c) holds for 2j + 1 g_2n+k-1
where k > 0. We show (8c) holds for 2j + 1 = 2n+k4-2i_.1
where i = l,-~-, 2n+k-1-_1.

. n+k .
I n(X23+1) = I n(X2 X21+1)
2 2
n+k .
= X2 A n(X21+l)
2
n+k .
= X2 XI n(X21)
2
= xI n(x23).
2
n+k+1

Thus, (8c) is valid for 2j + 1.3 2 and hence, by

induction (8c) holds for all values of j.

n
We now show that I n 6 Der: (A). In order to show
. 2 .
this, we compute I n(yl) and I n(xyl) and show that
2 2
(9a) 1 (yi> = v (yi>
2n 2n—l
i _ i
(9b) I n(XY ) - XY n_l(y )
2 2
where i = l,2,°°° and n = 1,2,---

To show (9a), there are several cases depending on

0,1,2 (mod 3).

whether i

Case 1: i E 0 (mod 3)
Let i = 3L. Then from (8b), I (yl) = I (Y3!) =
2n 2n
21 32 i
I(x)=Y (Y)=Y (y).
2n 2n-l 2n—l

22

1 (mod 3) and 2n 2 1 (mod 3).

Case 2: i

2 (mod 3) and 2n 5 2 (mod 3).

H)

Case 3: 1

These cases are considered together for in both

i + 2n+1 e 0 (mod 3). Let i + 2n+1 = 32. Then
1 2n+1 2n+1 i
I (y y ) = y I (y ). 0n the other hand
2n 2n
3 2
I n<y ‘) = I n<x ‘)
2 2
32
= Y ( )
2n-l
- n+1
i 2
= Yn_1(yy )
2
_ 2n+1 ( i)
—Y Yn—l Y °
2
n+1 . n+1 . . .
2 i _ 2 i i __ 1
So. y I n(y ) — y Y n_1(y ) or I n(y )-Y n_1(y ).
2 2 2 2
Case 4: i I 2 (mod 3) and 2n 5 1 (mod 3).
Case 5: i E 1 (mod 3) and 2n 5 2 (mod 3).
. n+2 _
In both of these cases, i + 2 = 0 (mod 3). Say,
. n+2 n+2 .
i + 2n+2 = 32. Then, I (yly2 ) = y2 I (yl). 0n
n n
2 2
the other hand
32 22
I (Y ) = I (x )
2n 2n
_ 32
— Y n_1(y )

23

. n+2
i 2
2
n+2 i
= y2 Y2n_1(Y )
2’”2 i 2n+2 i i i
80. y I n(y ) = y Y n_1(y ) or I n(y )==Y n_1(y )-
2 2 2 2
Now we show that I (xyl) = xY (yl). The cases
2n 2n-l

are the same as above.

Case 1: i E 0 (mod 3).

Let i = 32. The result follows immediately from

(BC):
I n<xy1) = I n<xx2‘)
2 2
= xI n(x2£)
2
3
= XY n_l(y L)
2
= XY n_1(Y1) 0
2
Case 2: i 2 1 (mod 3) and 2n 2 1 (mod 3).
Case 3: i E 2 (mod 3) and 2n 5 2 (mod 3).
. . n+1 _ . n+1
Again 1 + 2 = 0 (mod 3). Say, 1 + 2 = 32.
. n+1 n+1 .
Then, I n(xyly2 ) = y2 I n(xyl).
2 2

Also

24

NOW

32 22
I (XY ) = X (XX )
2n 2n
= xI n(x22
2
3
= XY n_1(y z)
2
2n+1
= XY n-1(Y y )
2
2n+1 i
= xy Y n_1(y ).
2
n+1 . n+1 . . .
2
Thus, y I n(xy1)==xy2 y n_]_(y1) or I n(xyl)==xY n_1(yl).
2 2 2 2
Case 4: i E 2 (mod 3) and 2n 5 1 (mod 3).
Case 5: i a 1 (mod 3) and 2n 5 2 (mod 3).
Here i + 2n+2 E 0 (mod 3). Let i + 2n+2 = 32.
1 2n+2 2n+2 l
I n(xy y ) = y I (xy ) 0n the other hand
2 2n
32 22
I (xy ) = I (xx )
2n 2n
= xI n(le’)
2
3
- XY n_1(Y L)
2
i 2n+2
= XY n_1(y y )
2
2n+2 i
= xy Y n_1(y )-

25

2n+2 i 2n+2
So again we have y I n(xy ) = xy Y

2 2

I (xyl) = xY 1(yl). Thus, we have completed the proofs
2n 2n‘

i
n_1(y ) or

of (9a) and (9b).

. . . . i
Since every monomial in A can be written as xjy

with j = 0 or 1 and 0.3 i, (9a) and (9b) imply that
I n: A 4 A for n = 1,2,°°° . To show that Il:A 4 A we
2

compute Il(y). Since

0 = 11(x2) = I1(y3) = y2 11(y).

6 Deri(A). Hence, I n:A 4 A for

I1(Y) = 0. Thus, I
2

l
n = 0,1,-°°

By taking composites, we define an m—th order deriva-

tion for m = l,2,--- . Write m in its 2-adic expansion
M .
as m = Z) (1.21 ‘where o. = 0 or 1. Define
. i 1
i=0
a a
_ M 0 m

We now make some observations about Im. We first

oL a1 L 1
consider I = I °,,,° I where 22 = 23 o.2 .
22 L 2 ._ i
2 1-1
Equation (9a) shows that I n when restricted to k[y]
2
equals Y 1. Thus, if I is restricted to k[y], we
2n— 22
have
a a
_ L ,, 1 _
X2,¢|k[y’] — Y L-l ° ° Y1 - Y2°

2

26

i i r 0 i<<2
By the definition of Y1, I2z(y ) = Y2(Y ) = 1 1 i==2.

Also I2 (x) = 0 since I (x) = 0 for n > 1. And
2 2n -—

. i _ i
finally, I2£(xy ) — xI2£(y ) by (9b).

Now consider I = I I

. 22+1 21° 13
I1(y1) = o, and I1(xy1) = Y1, the following hold:

Since I1(x) = 1,

(11a) I2£+l(x) = 0
i . _
(11b) X22+1(y ) — 0 1 - 1.2.
i _ i _
(11c) I2£+1(xy ) - I21(y ) 1 — 1.2.

Thus, we have defined derivations In, n = 1,2,

from A to A. Since R = it follows that

(A)(X.y)'
In e Der:(R) [1, Theorem 15:6].

Summarizing the above, we have that the I satisfy

the following:
(12) I1(X) = 1

L - i — . —’ ...
I2£(y ) — l and I21(y ) — 0 i - 1, , 2
I (xyi) = o i = 0 ~--. 2
22 '
I (xyz) = 1 and I (yi) = 0 i = 1 --° 2
22+1 22+1 ' '

i . ...
X22+1(XY ) - 0 i — 0, o L

where 2 = 1,2,°

27

We now define xi as follows:

_ L — .00
(13) x22+1 — xy 2 — 0,1,

Thus using the notation of (13), equation (12) says
j<i

0
that Ii(xj) = { 1 Thus, the conditions of

j==i
Theorem 1.5 are satisfied and Der:(R) is a free R-module
for all n. And.by construction, Der(R) = der(R).

Since I has a singular point at the origin, R is
not regular. Thus, R is an example of a local ring
which is not regular but Der:(R) is a free Remodule for
all n and Der(R) = der(R). Hence, we have shown that

(II) and (III) are false.

CHAPTER III

A
THE COMPLETION R

In this chapter we shall prove the following theorem:

If Der:(R) is a free R-module for all n and

A
Der(R) = der(R), then R is an integral domain.

If the characteristic of k is zero, then Deri(R) being
free implies that R is a regular local ring [Theorem 1:2].
Hence R is regular and thus an integral domain. Thus,
‘we shall assume throughout this chapter that the charac-
teristic of k is p #’0. Further, if P is a simple
point of I, then R is regular. So, R is an integral
domain and the theorem is trivial in this case. Thus, we
also assume P is a singular point of F: that is,
subdeg f 2.2.

As in Chapter 1, we shall assume in this chapter that
fy(x,y) #'0. Thus, K is a separable algebraic extension
of k(x).

We shall use the following notation in this chapter.
Let 6i denote the i—th order derivation of k[x] to
j<i

. j 0
k[x] defined by 6i(x ) = { 1 j==i°

The following

results hold for 61:

28

29

(14) Der2(k[x]) < 61,'°°, 6n > as a k[x]—module.

< 6

Der:(k(x)) ', 6n > as a k(x)-module

1'..
[pp. 26,27: 7].

(15) Since K is a separable algebraic extension of
k(x), 6n 6 Der:(K) [Theorem 17: 7]. Hence,

n — o o o

n
a
5 N 0‘0
N 61
(16) 6n = 527-0...° 6r7- where the p—adic expansion of
N' O'

N .
n is given by n = Z) dip1 [Pr0p. 18: 7].
i=0

 

 

i
P
Proof: 6Pi is a pl+1-order derivation. Hence,
P
pi+1
6pi - .2: r 6 It is easily verified that 6pi(x3) - 0
p 3:1 3 P
for j = l.°°°. p“1 80: rj = 0. j = 1.--°. pi+1o
(18) 5n05m=5m05n
P P P P
Proof: ono5m=omoon+[5n.6m] where [én'ém]
P P P P P P P P

is a derivation of order pn + pm-1 [1, Cor. 6.2: 6].
So, 6 ° 6 = 6

pn+ Ill-1
n m m o 5 n + $1 r161. But
P P P P 1'-

3O

3' _ 9 j __
6 "6 m(x ) — 0 and 6 m 6 n(x ) - 0 for
P P P P

j=lo°°°v Pn+Pm-l. Hence, 5 05 =5 05
P P P P

A
Now consider the completion of R, denoted by R,
with respect to its maximal ideal m = (x,y). We note

that if I 6 DerE(R), then I extends to an N-th order

A A
k-derivation from R to R. Let I 6 Der§(R). Define

..A A ..
I:R 4 R 'by I(r) = lim I(rn) where r = 1im rn with

{rn} a Cauchy sequence in R. We show I is continuous.
n

For any n we must find n such that I((m)n)c m 0

0
the ideal m. Let n = ng+1.

using the definition of an N-th order derivation, equation
n _

(1), that I(mn) c m 0. Hence, I is continuous. That

A
I is an N-th order derivation on R follows from the

for

Then it is easily checked

fact that I is an N-th order derivation on R. Hence-

forth, we shall denote I by I.

The next theorem relates derivations and zero

divisors.

Theorem 3.1: Let A be a commutative reduced k—algebra

 

where k is a field of characteristic p. Let
D = {0}LJ{zero divisors in A]. Then D is closed under

every derivation I.

Proof: Suppose I is a derivation of order n. Choose
N such that pN > n. Then I may be viewed as a deriva—

tion of order pN

31

Now let r 6 D, r #’o, and find 5 # 0 such that
N N
rs = 0. Then rsp = 0 and 3p #'0. Then

N

N N N
0 = I(rsP ) = sp I(r) + rI(sP ) = sp I(r)

N
since I(sp ) = 0 [I,Pr0p. 10:6]. Thus, I(r) 6,D. So,

D is closed under I. QED

Theorem 3.1 assumes that A is reduced: that is,

A has no nilpotent elements. We can apply this theorem
A A

to R because R has no nilpotent elements if k is

perfect [Theorem 31,p. 320:9]. This is our case.

A A
R = k[[x,y]]: that is, every element in R may be
written as ZiaijxlyJ ‘with aij 6'k. This representation,

however, is not necessarily unique.

Theorem 3.2: Suppose Der:(R) is a free Remodule for all

n and Der(R) = der(R). Let [Ii] be any set of gener—

1 n - .0.
ators such that Ii 6 De k(R) and Derk(R) — < I1. , In >.

Define derivations Ym as follows:

I M 0‘o
M I1 M i
(19) Ym = 527-0...° EIT' where m = Z) dip .
M' 0' i=0

n "" O O 0
Then Derk(R) — < Y1, , Yn >.
Before we prove this theorem we note that Theorem 1.8

implies the existence of such Ii's.

ProOf: We first show that the following three relation—

ships hold:

32

i 3'
(20a) (I .°rI .) - (rI iOI j) 6Derfi +p -1(R) for r 6R.

P P P P
n
(20b) r I . o...o r I . - ( H r.)(I . o...° I . ) 6
1 11 n in j=l j 11 in
P P P
n i.
Dart-I(R) where m = 23 p 3.
j='l
i l
p 1 p n n
(20C) (3.231 Slej) o-o-o (jgl SnjAj) "" (jg-:15. ij))\ .110...o A in
JP P P
n i.
Deri-1(R) where m = 23 p 3.
j=1
For (a),
I1°rxj=rIon +[AiorAj]
P P P P P P

= r(I o 0A . + [A ”I 1]) + [A i'r)‘ j]-
P P P P P P

Since [I i,rI .] and [I j,I i] are derivations of order

P P:J P P

p1 + p3-l [I,Cor. 6.2:6], the result follows.

Result (b) follows from (a) by induction. And (c)
follows immediately from (b).
We prove the theorem by induction. For n = 1,

Y1 = I1 and Deri(R) = < I1 > = < Y1 > . Assume that
Darn-I(R) = < °°' > To show Dern(R)==< --- \
k Y1' ' Yn—i ° k Y1' ' Yn .

n
it suffices to show that I = Z} r.Y. for some r. 6 R.
n i=1 i i 1

Since Der:(R) = DerE-1(R) ® RIn, we see that n is the

33

smallest integer such that In 6 Der:(R). Since

Der(R) = der(R), we have

a a o a
— N o... O ... N 0
(21) In - TIN o T10 + + TmN o...o TmO + o
where o 6 Dern'1(R) and T 6 Derpj(R) H r
k ij k ' e e
N j
n = Z) a.P is the p—adic expansion of n. By absorbing

j=0
any bogus terms in with O, we may assume p3 is the

j
smallest power of p such that Tij 6 Derfi (R), j==0,---,N.

N a.
Thus, Z) ord 7.? = n for i = 1,'°°, m.
i=0 ‘3
0‘N 0‘0
Now consider one of the summands TiN o...o r10 in

j
(21). Since Derfi (R) = < I I . > we have by (20c)

1'...'
P3
N
a a p a a
N o _ N o
(22) TiN o...o T10 — (‘2; siN2 IL) o...o (siolIl)
o
= s(IaN o o I O) + 0'
pN 1
where 0‘ 6 Dern-1(R) and s 6 R.

k
Thus I has the form I = r(IOLN o... IaO)-+o”
n n pN ° 1

II n-l _ . . . .
where r 6 R and o 6 Derk (R) - < Y1, , Yn-l ,. Now
0. C1. N
Y = aI N °°°°o I 0 where o = l/ H o.! is a non-zero
n N 1 . 1
p i=0
constant in k. Thus In = (r/a)Yn + c” or

In 6 < Y1."’. Yn > . QED

34

Corollary 3.3: Under the assumptions of the previous

n-l
theorem and using the notation of it, In = rYni- Z} riYi
i=1
n-1
and Y = sI + Z} s.I. where r and s are units in IL
n n i=1 i 1

Proof: The only part requiring proof is that r and s

n-l
are units. Substituting for Yi in In = rYn + i=1riYi
n-l
gives In = rsIn + iEI tiIi. Thus, rs = 1 and Since
r,s 6 R, they must be units. QED

Corollary 3.4: Again assume the conditions and notation
n-l
of Theorem 3.2. For Y = Y + Z} r.Y. 6 Dern(R) for

n i=1 i i k

some r1 6 R, there exists an element r 6 R such that

 

Y(r) is a unit in R.

Proof: As in Lemma 1.3, write f(X,Y) = Xh(X,Y) + YNg(Y).
Suppose Y(r) 6 (x,y) for all r 6 R. Then
N-l n . n _ ...
(y /k)Y E Derk(R). Slnce Derk(R) — < Y1. . Yn > .
N-l n .
(y /x)Y = Z) s.Y. With 5. 6 R. Also
i=1 1 1. 1

n-l
(YN-l/k)Y = (EN-l/X)Yn + .2; (yN-l/x)riYi. But this
1:

implies that yN-l/x = 3n 6 R: this is a contradiction.
Therefore, there exists some r 6 R such that Y(r) is a

unit. QED

Proposition 3.5: Let S = (k[x,z])(x z) where k has

characteristic p, k[x,z] = k[x,z]/(h(X,Z)), and h(X,Z)

35

is an irreducible polynomial such that hz(x,z) #'0 and

h(0,0) = 0. Suppose 6 .1(2) = 0 for all i less than

N
g(Xp ,Z) and 6 i2 = 26 i
P P

P
some fixed N. Then h(X,Z)
for i < N.

Proof: Since hz(x,z) # 0, 6 1 extends uniquely to
P

k(x,z). We shall View all calculations which we make in

 

the proof of this prOposition as taking place in k(x,z).

Now hx(x,z)6l(x) + hz(x,z) 61(2) = 0. So, 61(2) =
-hx(x,z)/hz(x,z) and since 61(2) = 0, hx(x,z) = 0. As
in Lemma 1.1, this implies that hx(X,Z) = 0. Thus,
h(X.Z) = 91(xp.2).

i
Inductively, we suppose that h(X,Z) = gi(Xp ,Z) with

i
i < N. Now 6 i is a lst order derivation on k[xp ]
P

i
[1, Theorem 14:6]. Since (gi)z(xp ,z) = hZ(x,z) # 0,6 i
P

1
extends uniquely to a lst order derivation on k(xp ,z).

i i
Thus. (gi)z(xp .2)6 '1(2) + (91) 16 i(xp ) = 0. SO.

P XP P

pi i
6 1(2) = - (91) i(x .2)/(gi)z(xp .2). Slnce 6 i(2)==0.
P XP P

i i
(gi) i(xp ,z) = 0. And therefore (gi) i(Xp ,Z) = 0.
xp XP
i+1
Hence, h(X,Z) = gi+l(Xp ,Z). By induction then
n

P

h(X.Z) = g(X .Z).

We now ShOW’by induction that 6 i2 = 26 i < N.

P P
For i = 0, 61(zr) = r61(z) + 261(r) = 261(r), for any

il

r 6 S. Thus 612 = 261.

For 1 < i < N:

36

consider the polynomials

9,29

I

‘0

1

.(2p )

The

i . .
ZP ’19. write ng, j = 0,°°°, pl—l, as
i .
. p N i N
(23) ng = Z) r. (XP ,ZP )zn + r.(Xp )
=1 Jn J
N i N i
where r. (Xp ,ZP ) = Z) a. (Xp )k(Zp )m. Since
3n k jnkm
,m
6 i(zjg) = 0, evaluating (23) gives
P
(24)
P
i . i .
N i p N i
= I: r. (xp ,zP )5 .(zn)+ )3 z“: .(r. (xP ,zP ))+o
n=1 3n Pi n=1 pi 3n
1 i .
P 11 P n N k 1
_ p m
-- n§1 rjn6 1(2 ) +n§1 Z Opi&?m ajnkmucp ) (Z ) )
i i .
P N i
= $3 rjn6 i(zn) + 2 2n 2 a'nkmxkp 6 i(sz )
i i .
P N i
= Z r'n6 .(zn) + EX 2n 2 a. ka m(zp )m-16
3 i jnkm 1
~ n=1 P n=1 kom p
i .
P n l p1
= Z“ rjnO 1(2 )+rj6 i(z )
n=1 p p
Pi
- n
= r. . z
nZ-Pl Jn6p1( )
' _ . i “ _ l
where rjn — rjn if n < p and r. i — r. i + rj.
JP JP

equations in (24) yield the following matrix equation:

37
6 .(z) 0

jn .
l

6 J-_(zp) o

P

Suppose that the determinant of (Ejn) is zero. Then
.).

there exists a non-zero solution, say (t1,---, t

_ N Npl
Since (rjn) 6 [k(xp ,Z)] i' we have tj 6k(xP ,2),
P
j==l:"': p1. By replacing the tj with some multiple, if

N
necessary, we may assume tj 6'k[xp ,2].

Now consider the short exact sequence:

N i W i
o .. (g(xp .z)) -v k[xp .z] 4 k[xp ,z] a o.

i .
Let T]. 6k[XP ,2] such that 1T(Tj) = tj, j=1,---. p1.

We define 6 i: k[x] 4 k[x] to be the canonical pl—th

p i
. . . - j j<P
order derivation: that is, 6 .(X ) = { . 1 Now we
pl 1 J==P
_ i i

define I 6 Derfi (k[xP ,Z]) as follows:
I(z)) = T3. 3 = l, - , p1
_ i _ i .
I((xP )‘) = 6 1((XP )‘) z = 1.- . p1

P
_ i . i ._ i .
I((xpflzl) =(xP)‘Tj+zjai((xP)‘) o<z+j_<_p1

P

i
extend I to monomials of k[Xp ,Z] using (1).

38

_ - i
We first Observe that I = 6 i on k[Xp ]. This is

. P i
obvious since both are pl—th order derivations on k[Xp ]

‘ p.1 2 _ ‘ p1 2 _ i
and I((X ) ) — 6 i((X ) ) for 2 - 1,°°°, p .
P

We next show
_ i . i . . _ i
(26) I((xp )‘23) = (xp )‘ I(zj) + 23 I((xp )‘)

for all 2 and j. By the definition of I, (26) holds
for 2 + j S_p1. We first prove (26) for 2 and j when
2 < p1 and j < p1. The proof is by induction. Suppose

2 + j = pli-l. Using (1) and the fact that

Z: (-1)m (2) = 0, we get
m=0 m

_ i . j 2 . i _ i .
I((xp )‘zJ)== 23 z:(—1)n+m-1(;>(g>(xp >mz“I((xP )“mz3’")

n=0 m=0,
lsn+m5p1
j g . i i _ .
= E Z (-1)n+‘“'1(,fi)(g) (Xp )mznuxp )"’“I(23’“)
n=0 m=0.
lin+m$p1

. _ i
+ 23‘“ I((xp )“m)}

j 2 . i _ .
= Z? z:(_i>n+m-1(;>(g)[<xp )‘ z“I(z3‘n)
n=0 m=0.
1Sn+msp1

. i m _ i 2
+ z3 (xP ) I((xp ) ‘m)}

j 2 . i - .
= z: 2: (-1>n+m-1<;)(g>((xp )‘ z“I(23’“)
n=0 m=0

. i m _ i 1 m
+ z3(xp ) I((xp ) ' )1

39

. i _ . . _ i
+ <3)(g>[(xp )‘ I(zj) + 23 I((xp )‘)J

j . _ . i 2
= z: (-1)“'1(g)z“ I(zj‘“)(xp )‘ z>(-1)m(‘)
n=0 =0 m
j . 2 i ._ i
+ 2: (-1)“(r31> E (-1)"“'1(‘)(xP )mZJquP )L‘m)
n=0 m=0

i _ . . _ i
+ (xp )‘ I(ZJ) + 23 I((xp )‘)

i _ . i
= (xp )‘ I(ZJ) + 23 I((Xp )‘) .

Thus, (26) holds for 2+-j==p4-1 when 2 < pi and j<<pi.
Suppose (26) holds for 2' < p1 and j' < p1 whenever

2I + j' < 2 + j, 2 < pi, j < pi. We now show (26) is valid
for 2 + j. Since 2 < p1 and 2 + j 2_pi + 2, j 2_3.
Write j = j1 + j2 where j1 = p -2. Then

1 . i j j i 1 j
(xP )‘z3 = (xp )‘ z 1(z 2) = xp ---XP z---Z(z 2).

Equation (1) gives

- i . - i J J
I((Xp )‘zj)= I((Xp )‘ z 1(z 2))

j1 1 1 j i j

23 2 t2 (-1>“+m+t’1(,fi)(,})(xp )mznm 2)t
n=O m=0 =0

1Sn+m+tsp1

- i _ j -n j

I((xP >‘ m z 1 (z 2)1't)
jl jl 2 —m Zj —n+j

= Z E(-1)“*‘“' 1(‘H ,3pr in)mz I((xp’ 1 2)
n=0 m=0
l£n+m 3
1 2 j n+i
+ Z? Z? (-1)n+m(nf)(n1)(xp )m z 2
n=0 m=0

n+msp1

_ i j -n
I((Xp )"m z 1 )
jl n+m— 1 2 jl m- '-n
= z: 23(- 1) (m )( F)(xp ) mz ni[(xP ) mI(zj )
n=0 m=0
lSn+m
+ZjnI((xpi") m)J
jl j1 n+i i - j —n
+ 2: z:(-1)m+n(‘)( §)(xp )mz 2[(xp )"mI(z 1 )
n=0 m=0
n+m$p1
j —n_ i
+ z 1 I((Xp >“m)]
jl .
= z: 23(-1)“+m 1(‘)(jv>(xp ) ”z I(zj'n)
=0 m=0
jl2 j . i_ i
+ z: Z‘(-1)“+m'1(;)(13)z3(xp )mI((XP )"m)
n=0 m= 0

. . _ i
+ ( 31(V>1<xp > I(z)) + 23 I((xp )‘>]

jl 2 n+j _ j -n
+ 2: 23(-1)“+m(m)(:1)(xp W)‘ 2 I(z 1 )
n=0 m- -0
j1 j i . _ i
+ z: 23(-1)n+m(m)(,3)(xp )mzJ I((XP )“m)
n=0 m= 0

p.1 L ' j j ‘ pi L
(x ) I(z ) + z I((X ) ).

Thus, we have shown that (26) holds for all 2 and j

when 2 < p1 and j < p1. We now prove (26) for arbitrary

2 and j. Write 2 = 21 + 22p1 + 23p1+1 with

i ._ . . i . i+l
0.3 21 < p , 0322 < p, o_g 23 and j-314-32p 4-j3p
with 0 g,jl < pl, 0.3 j2 < p, 0.3 j3. Then using

Lemma 2.1, we have

41

i ‘ p1+1j j pi j p1+1
3 Z 12 2 Z 3

1 l p1
2 (XP I

_ i . _ i z '
I((xp #23):qup I 1(xP I )

i+1 . 1+1 . i . .
z p 3 p - 1 L i L p 3 3 p
J<xp ) 3 z 3 I((xp I 1(xP I 2 z 1z 2

pi £3p1 j3p pi £1 11- pi fizpl jzp1
=(X) Z [(X ) z I((X) Z I

1 z p1 j p1 i z j
+(xp ) 2 z 2 I((Xp I 1z 1)]

i 1 p1 j p i L 31 i 2 pi j p.
-(xp I 3 z 3 [(xp I 121((xp I 2 I(z 2 I

Zj pi i 1 pi i L p1 j pi i L
+ 2 I(xp I 2 I+<xP I 2 z 2 ((xP I 1I(zj1 I

1 L

j _
+zl)\((xp)1

))J

j p' j _ j p' j p' - j
(XP ”)1 3 (z 1 I(z 2 )+-z 2 I(z 1))

+1 .
1p . 1 £1 lip
+ (x9 I 3p 23((xp I 1I((xp I
1 zzpi

' 1 L
+ (Xp )

{((xp ) 1

))
. . . . i
1 J P _ 3 +3 P

i+1
13p - 1 11+‘2Pi

+ (xP I zj k((XP I I

p.1 z ‘ j j ' pi L
(x ) I(z ) + z I((x ) I.

Hence, (26) holds for arbitrary L and j.

We want to show

_ i i i i _ _ i i
(27) I(s(xp ,zP )r)==s(Xp ,zp )I(r) + r I(s(xP .29 I)

42

i
for any r 6 k[xP ,2]. To do this we first show

i i .
n(zp )mi((xp )‘zJ)

_ i i i . '1
k((xP )n(zp )m(xp )‘23) = (xp )
i . i i
... n m
+ (x9 >‘23x(<xp ) (Z? > ).
Using (26) and Lemma 2.1, we have
_ i i i . i i_ i. . i__ i i
Mxnp zmp sz zj) = an sz Hz]mp 23) +23]sz )4an sz )
n i L i m i_ . ._ m i
= x p x p [z P x(23) + zJX(z p )]
. i _ i L i
+zjzmp[6 i(xnp xp)]
P
i i i_ . ._ i
= an zmp [x‘p M23) + 23“le )]
L i . n i_ m i i_ n i
+xpz:'[xp)\(z“-’)+z’“p MXPH
i i_ L i . L i ._ i i
= an zmp Mx P 23) +x 9 zJMXnp zmp ).
' pfL n p1 m p1 23'
Now we consider X((X ) (Z ) r) where r = 23a£j(x ) Z :

_ i i _ i i i .
Mxnp zmp r) =)\(xnp zmp Z) a£j(xp )LZJ)

i i_ Li. Li._ imi
Za£j[xnp zmp k(xp 23)+XP z3Mxnp z p )]

i i _ i i
= xnp zmp Mr) + ruxnp zmp )

i i i n pi m
Now we prove (27) with s(XP‘,Zp ) = Zlanm(xp ) (Z ) :

43
.. i i _ i i
MS(XP .2? )r) = mzanmmp )n(zP )mm
= Ea X(anlzmplr)
nm

nimi_ ‘nimi
=Zanm(xpzp k(r)+r)\(Xpr))

i i _ .. i i
np mp nP mp
Zanmx Z Mr) + rk(2anmx Z )
i i _ _ i i
= s(xp ,zP )x(r) + rX(s(Xp .Zp ).

Thus, we have proven (27) . We use (27) to show
_. i+1 i+1 i+1 i+1 _
<28) x(s(xP .zP )r) = s(xp .zP )x(r)

i
for r €k[Xp ,Z]. By (27) we have

_ i+1 i+1 i+1 i+1 .. _ i+1 i+1
k(s(Xp ,zP )r)==s(xP ,zP )x(r)+-rx(s(xp ,zP ).

Now

__ i+1 i+1 i+1 i+1
x(s(xP .zP ))

II
>» I
M
39
3?
’U
:3
13
"O
VB

II
M
9
>’ I
35
U
'U
H.
+
N
3
"U

_ i+1 i+1 i+1 i+1 _
Thus, x(s(xP , zp )r) = s(xp ,zp ) k(r).

44

We now use (27) and (28) to show that i((g)) s: (g).

i i i N
Suppose f(Xp ,Z) 6 (g) . Then f(Xp , z) =k(xp ,z)g(xp ,z) =

p -1 p-l i+1 i+1 i z m N
Z Z- smmp ,Zp )(xp) 2 g(xp ,2). Thus,
m=0 t:
. i+1 .
_ l _ p -1 p-l i
Mf(xp ,Z)) = M Z 23 smx‘P 2mg)
m=0 i=0
i+1 .
p -l p-l _ 1
= Z Z k(sm‘X‘!’p ng)
m=0 £=O
i+1 .
p -1 p-l _ 1.
23 Z smux‘p ng)
m=0 i=0
i+1. i+1
by(28) since s)"m E k[Xp ,ZP ]. Thus, it suffices to

+1

_ i .
show k(xl'p 2mg) 6 (9) ‘where O g_z < p, O g.m g.p1 —1.

Let m = npli-j ‘where 0 g.n < p and j < pl.
.11 _‘ini. Lini-' ._‘ini
k(xp zmg)=x(xPzPZJg)=xPzpk(ZJg)+zjgx(xpzp)

by (27). So, it suffices to show X(ng) E (g) for

'Then

0$j<p1-

i .
. p N 1. N
Now by (23), 239 = 2 rjn(xp ,zp )2n + rj(xp ).
n=1

Thus, i
- - P ._ N i _ N
M239) = 23 Mr. (xp ,2p )2“) + >\(r.(xP ))
n=1 3n 3
"5 i n "5 n- - PN
"’ n=1 rjn (Z ) + =1 Z k(rjn) + 6pi(rj (X ))

by (27). Using the notation of (24), we have

45

i i .
_ . _ p _ N 1
X(ng) = 23rjnx(zn) + Z} znx( Z) ajnkmxkp zmp ) + 0
n=1 n=1 k,m
i i .
_ p N_ 1
= %rjnuzn) + )3 Zn 2‘ ajnkmka Mzmp )
n=1 n=1 k,m
i i . .
p _ N 1 _ 1
= ernuzn) + 22 Zn 2 ajnkmxkp m(zp )m’lxwp )
n=1 n=1 k,m
i .
1
_. % " n I " p
— r. Z + . Z
n=1 3n1( ) rJ X( )
i
= 2);. k(zn)
n=1 3n
g-
= r. T
n=1 3n n
1 i
P .. _ N
Now rr( ernTn) = % rjnmp ,z)tn = 0. Thus,
n=1 n=1

g(zjg) E ker w = (9). Thus, we have shown that {(9) g (9).

Now i induces a pl-th order derivation x on the

1
ring k[xP ,2] as follows:
Ma) = Trim) where 1rA =a

or k(A + (9)) = X(A) + (9) where A e k[xpl,z]. If

A + (g) = B + (9) then A43 6 (9): Km) -X(B) =i(A—B) 6(9)
since X(g) : (9). Thus, 1(A + (9)) = 1(3 + (9)) and

so 1 is well defined.

We observe that k(zj) = tj:

sz) = M23 + (9)) = MD) + (g) = T]. + (g) = tj

46

. i i
for j = 1,---, p1. Also, we have k((xP )‘)==6 i((xp )L)
P

i z i z - z
for all i. For, k((xp ) )==x((xP ) +(g))==x((xp ) )-+(g)==

i z 1 z
aiuxp))+ (g)=ai((xp) ).
P P .
. 1
Thus, l is a pl—th order derivation from k[xp ,z)

i 1 i
to k[xp ,2]. 30, X E Derp (k(xp ,z)) and A agrees

k

with 6 i on k(xp ). This is a contradiction since 6 i

P i P
has a unique extension to k(xp ,2). Hence, the deter-

minant in (25), |(E. )I, is non-zero.

3n
Multiplying (23) by Xm, m = 1,"', p1-1, gives
i .
. N 1 . N
XmZJg = $3 rjn(XP ,Zp )XmZJ + erj(xP ). These give rise

n=1

to the following equations in k(x,z)z

(29)

o = 6 i(xmzjg)

P
Pi pi N
= ngar'bn 1(anz)+ I%'_)xmnzai(r.n)~I~r.(xp){5j_(Xm)
-1 J p1 n-l p 3 J p
i pi
= %r.n5 1(szn) + xm Z znb i(r.n) + 0
n=1 3 p n=1 p 3
= 33L!) imnmz)+xmr.‘ (zpi)
n=1 jn6p j 6p1
p1 __ m p1 pi m __ m p1
Now “(x z ) — x 6 1(2 ) + z 6 i(x ) — x 6 1(2 ):
51P P P P
6 i(xm) = 0 since m < p1. So, (29) can be written as

P

47

i

P
O = Z} rjnb i(xmzn) + r56
=1 P p

m pi
i(x Z )

g;

= r. 6 .(x z ).
n 1

=1 3 p

Since |(Ejn)| #’o, 6 i(xmzn) = O for O g_m < pl.

. P
1 _<_ n 3 p1.
Now we consider the pl-l order derivation [6 1,2]
P
given by [6 i,z](r) = 6 i(zr) - r6 i(z) - z6 i(r)
P P . P P
[10P0376]° For m = lo°°°o pl-ll [6 1,2](Xm) =
P
5 i(zxm) — xm6 jL(z) — 25 i(xm) = 0. Now
P P P i
1_1 p -1
[6 1.2] €Der£ (k(x.z)) so. [5 ..z] = Z 125. for
1 °— 3 3
P p 3-
some rj E k(x,z). Evaluating [6 .1,2] at x shows that
P
r1 = 0. Similarly, since [6 i,z](xm) = O, rm = O for
. P
m = 1' .0 pl‘lo Thus, [6 1,2] = 00 Hence,
P
0 = [6 i.Z](r)
P
= 6 i(zr) " r6 i(z) - 26 i(r)
P P P
= 6 J-L(zr) - 26 i(r).
P P
Thus, 6 i2 = 26 i for i < N. QED
P P

We now prove a special case of the main theorem.
Namely, if 6n 6 Der:(R) for all n, then a is an

integral domain. If 6n 6 Der:(R) for all n, then

48

Derk(R) - < 510°°°0 6n >. Also Der(R) = der(R) since
at 0‘0
6 = 6 /e 1 o...o 6 /e E where the p—adic expansion of
n pt t 1 O

t .
n is given by n = Z} a.p1. Thus, 6 E Der:(R) for
i=0 1 n ,

all n implies that both hypotheses of the main theorem
are satisfied. That is, Der:(R) is free for all n and

Der(R) = der(R).

A
Theorem 3.6: If 6n e Der:(R) for all n, then R is

 

an integral domain.

Proof: We define a sequence of elements in R as follows:

 

i i
p-l - 6 (Y)x
(3O) 21 = y + Z (-l) l —-];-]--.———
i=l '
i ipn-l
p—l i jpn—l(zn-l)x
2n = zn-l + i2; ('1) i!

Consider zn. Since K is a separable algebraic
extension of k(x) and since zn E K, zn satisfies a
minimal polynomial g(Z) with coefficients in k(x) such
that gZ #’O. By multiplying g by some d(x) €'k[x], 9;
may be assumed to be in k[x][Z]. If g factors over
k[x,z] so that g = 9192' then either g1 or g2 is in
k[x] since degzg = deng1 + degzg2 and g is the

polynomial of minimal Z-degree which zn satisfies.

‘ I.‘\l.lllll.l lulllulll

 

 

49

Thus, 9 may be assumed to be irreducible over k. Thus,
for each zn there exists an irreducible polynomial
g(X,Z) 'w1th g(x,zn) = O and gZ(x,zn) # 0.

We shall show that 6pjzn = zn6pj for J < n. We

first consider 61.

+(—1)P"2511"1(y)xp'2/<p—2) :+ (-1)P'1a§‘1(y)xp’2/(p-2) :

+(-1)p‘15§(y)xp'l/(p-1)!

(-1)P‘16‘1’(y)xp'1/(p-1) :
= 0

since 6§(Y) = 0 by (17). Thus, by Proposition 3.5,

1 2151'

NOW'We assume that 6 jzn_1==zn_l6 j for j==l,---,

P P
n—2. Evaluating 6 n_l(zn), as above, gives that

6 n—l(zn) 0. Also, since 6 j<36 ==5 .
P P P P P
6 j(zn_1) = O for j = l,---, n-2, we have
P
1 ip
p—l i 5 ml6 j(""n--1)X
5 .(z ) = 5 .(zn_1) + .Z}(-1) .2

p3 n p3 1=1

 

i!

= O.

50

Thus, 6 j(2n) = O for j < n. Hence by Proposition 3.5,

P
6 jzn = zn6 j for j < n.
P P n
- - _ P
Now zn is a Cauchy sequence s1nce zn+1 2n 6 m .
Let 2 = lim Zn: 2 E R. Now 6 j is a continuous
P
A
function on R. Thus for r 6 R with r = lim rn.rn ER,
we have
6pj(zr) = 6Pj(11m zn 11m rn)
= 6 j(1im znrn)
P
= lim 6 j(znrn)
P
= 11m (znb j(rn))
P
= 11m zn 11m 6 j(rn)
P
= z 6 j (1im rn)
P
= 26 -(r).
p3
Thus, 6 . z = 26 . for all j and hence, 6.2 = 26.
J J 1 1
P P
for all i.
Now 2 = 21 + (zZ-zl) +°°°+ (Zn-zn-l) +--° : so,
2 = y + Z) ai.xly3. Hence, y = z — Z} ai.x1yj. We
DO 3 i>o 3

substitute for y on the right hand side of the equation;

thus, y = z — Z: a..x1(z — Z) a..x1y3)j. Continue
i>0 13 i>o 13

51

substituting for y. This gives that y may be written

as y = Zjaijxlzj and from this follows that every

A . .
element r E R is of the form r = Z3uijxlzj.

A
Suppose r is a zero divisor in R. Then
= i j . =
Z>aijx z ‘Wlth aij E‘k and GOO 0. By Theorem

3.1, 61(r) is a zero divisor. Now

61(r) = a1 + 23a. 61(x izj). 61(xizj) zj6l(xi) E (x,y)

ii

for i + j 2.2, so, 61(r) is a unit unless alo = O.

= 0. Suppose aiO = O for i < n. Then,

6n(r) = a + Z3aij6n (x123) is a zero divisor. Since

n(x izj) E (x, y), a 0. Hence, we have shown that

no
aio = o for all i. Thus, elements of the form ZaijxlzJ

A
‘with aIO #'O for some I are not zero divisors in R.

Now we suppose that rs = O with r = ZaijxlzJ

and s = 2 E3. .xlzj. Then writing r = 21’ Z a. .xlzJ-l’with
1] jzz 1]
#’o for some I, the above shows that Z) a..xlzj-£

(1.
IL jzz 1]

is not a zero divisor. Note that if r is a zero
divisor, but not equal to zero, then L must be greater

than zero. Likewise = 2m ZIBijxiz zj m ‘with BI m#'0
32m
j— —m . L m__
and 2) Bi .x 1z is not a zero divisor. Hence, 2 Z -0

j=m j
and since R has no nilpotent elements, z==0. Thus,

A
r==0 and R is an integral domain. QED

52

Theorem 3.7: If Der:(R) is a free Remodule for all n

 

A
and if Der(R) = der(R), then R is an integral domain.

 

Proof: Since Deri(R) is a free Remodule, by Theorem

1.8 there exists a derivation Al and a monomial xl

such that l1(x1) = 1. Because Al is a lst order deri-

vation, xl can only be x or y. If fx(x,y) # 0,

then the curve does not distinguish between x and y
since fy(x,y) #'0. Thus, without loss of generality we
may assume x1 = x. If fX(x,y) = 0, then
fx(x,y)11(x) + fy(x,y)kl(y) = 0. Hence, 61(Y) = O and
so, x1 = x. Hence, we shall assume x1 = x. Further,
since l1 and 61 agree on k(x) and since both extend
uniquely to k(x,y), XI = 61.

Now if 6n €.Der:(R) for all n, then by Theorem 3.6

A
R is an integral domain. Thus, the only case we need

consider is if some 6n t Derfi(R). Since

at do t i
5 = 5 /d E o...o 5 /d !, when n = Z) a.p , there

must exist a positive integer N such that 61: R 4 R

N
for i < pN, but 6 N ¢,Der£ (R).
P
If 61(y) = (a+s)/(l+t) where a #‘0, a 67k, and

s,t 6 m, let y' = y - ax. Note that y’ #'0 since R
is not regular. wa 61(y') = (a+s)/(l+t) - a =
(s-at)/(l+t) E‘m. Then since k[x,y] = k[x,y'] and

_ I 0 _ o =
(X.y) - (my ). (k[X.y ])(X'y:) — R. Let g(X.Y )

f(X,Y' + ax). Then g(x,y') = 0. Also,

S3

' _ t _ t —
ng(xoY ) — le(ny + (IX) " fY(XpY)dY/dY -

fY(x,y) d(Y' + aX)/dY' = fY(x,y) #‘o by the chain rule.

Thus gYI(x,y') # 0. Thus we may assume that 61(y) E m.

We have then the following assumptions on R:

(31) Der:(R) is a free R-module for all n
Der(R) = der(R)
Der1(R) = < a >
k l

61(y) €,m.

Now we define zn for n S_N as we did in (30).

 

That is,
i .
P-1 i 61(Y)X1
21 = y + Z (-l) —1—,—-
'_ 1.
1—1
. n—l
1 1p
p-l i 5 n-l(zn 1)x
2n = zn-l + iii (-1) i!
Let 2 = 2N. As in the previous theorem, 61(2) = O for
i < pN. Thus, by PrOposition 3.5, there exists an
N N
irreducible polynomial g(Xp ,Z) such that g(xP ,z)==0

N.
and gZ(xp ,z) #'0. Also, 6iz = 261 for i < pN. We

observe that 21 = y + xr1(x,y) ‘with r1(x,y) €;m since

61(Y) 6 m. Thus,

(32) z = y + xr(x,y) for some r(x,y) G (x,y).

54

We now con31der the ring R1 = (k[x,z])(x'z). we

shall show that R c R. Let s(x,z)/t(x,z) 6 R ‘with

1 1'
s,t E k[x,z] and t t (x,z). Since k[x,z] c k(x,y),
s,t 6 k[x,y]. t t (x,z) implies that t = a + h(x,z)
'with a a non-zero constant in k and h(x,z) E (x,z).
Thus, h(x,z) 6 (x,y) since (x,z) : (x,y) and so, t
is a non-unit in R.‘ Hence, s/t e R. Thus, R1 c R.
We now show that $1 = a. Since 2 = y + xr(x,y),
y = z - xr(x,y). substituting for y on the right hand

side of the equation and continuing this process gives

i j A
that y = 23a..x z for some o.. 67k. Thus, y E R1 and
A A 13 13
so, R1 = R.
Let K1 = k(x,z). K1 is a separable algebraic ex-

N
tension of k(x) since gz(xP ,z) #'0. Thus, 6n extends

uniquely to K1 for all n. Since 612 = 261 for

i < pN, 6i:k[x,z] 4 k[x,z]. Hence, 6i 6 Der;(R for all

1)
i < pN. In fact, Der:(Rl) = < 61,°°°, 6n > for n < p“.

‘We shall show that Der:(R1) is a free Rl-module for

all n. In order to do this, we shall need to know that
there exist derivations Xi and monomials

x. E k[x,z] c R such that Der:(R) = < 11.'°°, 1n > and

1 1
{0 j<i
k.(x.) = , , . Thus, we show this first. For
1 j 1.1 3:1
1 < pN, we let xi = 61 and xi = x1. Then clearly we

have satisfied the desired requirements for ii and xi.

 

55

N N
Now consider Deri (R). By Theorem 1.6 Der: (R)==

N
Derfi -1(R) ® RY ='< 61.-°°, 6 N >19 RY ‘where

p -1

N
Y 6 Derfi (R). Further, there exists an element r E R

A A
such that y(r) = 1. Now r 6 R c R = R1 so,
r = Z) ai.xlzj. Thus, 1 = y(r) = Z) ai.y(xlzj).
O‘i+j 3 131+]. J

This implies that y(szJ) must be a unit in R for some

I and J. Let 1 N = (l/y(szJ))Y. Then

P
pN N
Derk (R) = < 11,°°-, 1 N > where xi = 61 for 1 < p .
P
And we have shown that there is a monomial
x N E k[x,z] : R1 such that 1 N(x N) = l.
P , P P
We now proceed by induction. Suppose Der2'1(R) =
< 11,---, xn—l > and that there are monomials
O j<<i .
x1, , Xn-l 6 R1 such that 11(xj) — { 1 j==i . Again

61 and x1 = x1 for i < pN. Now

we are taking 11
n — 0..

by Theorem 1.6, Derk(R) - < 11, , Xn-l'Y > ‘where

y(xi) = 0 for i < n and y(r) = l for some r E R. As

before r = ZaijxlzJ and so, y(szJ) is an unit in R

for some I and J. Thus there exists a monomial

xn E k[x,z] such that y(xn) is a unit in R. Let

- n - .0.
1n — Y/Y(Xn)- Then Derk(R) - < 11. , An > and
{0 i<n

1 (x.) = , with x a monomial.
n 1 1 1:11 n

56

we want to show that the 11's which have just been
found are in fact derivations from Rl to R1. In order

to do this we shall need to know that R n k(x,z) = R

1.
To see that this is the case we consider the following:
A A A
R Q k(x,z) 0 R c k(x,z) 0 R = k(x,z) n R . Since ® R
1 1 R1 1

A
is exact [Cor. 17.11,p.57:5], k(x,z) n R1 = R1 [Theorem

8.4,p.59:5]. Hence, R = k(x,z) n R.

1
As we have already seen, 11 = 61:R1 4 R1. Suppose

1nduct1ve1y that X1,---, xn-l: R1 4 R1. Then 1.1:Kl 4 K1

for i < n. Now 1 -, 1n are a basis for Der:(K)

1...

n
n .
and 6n 6 Derk(K) so, 6n - 12; r11i ‘w1th ri E,K and

. n
rn #’O. ‘We show that ri €,K Since 6n €.Derk(K

1' 1"

6n(x1) = r1 6 K1 (in fact, r1 = 0). Suppose

n
G K Then 6n(xj) = Z) riki(xj) =

r '...' r,
i 3’1 i=1

1'

3
Z) rixi(xj) wh1ch implies that r. = 6n(xj) -

i=1 3
j-1 .
iii rixi(xj) 6 K1. Thus, ri 6 K1 for 1 = l,---, n.
n-l
Now An = 6n/rn - iii riAi/rn; 1n: R1 4 R while
n-l
6n/rn - i=1 rixi/rn: R1 4 K1. Thus, 1n:R1 4 R 0 K1 = R1.
Thus, for all i, Xi E Der:(R1). Also, there exist
0 j<:i
monomials xi 6 R1 such that Ai(xj) = { 1 j==i Thus,

by Theorem 1.5, Der:(R1) is a free Rl-module generated

l'°.°' 1n; this holds for all n. Further, we have
i

_ _ - N
that Xi - 6i and xi - x for 1 < p .

57

By assumption 6 N:R-74> R. ‘We want to show the same

P
15 true for R1; that 1s, 6 N:Rl-7L> R1. S1nce
P N
61, , 6 N are a free basis for Derfi (K),
p 11
p -l .

1 = r6 + Z) r.6.. But 1 (x1) = 1 (x.) = o for

N N .

p p 1:1 1 l pN pN l

i < pN, so, ri = O for i < pN. Thus, 1 N = r6 N'

N P P
Now r = 1 N(xp ) E R. Since 6 N: R 7L> R, r 6 (x,y).
P P
Also, 1 = 1 N(x N) = r6 N(x N) so, 6 N(x N) = l/r t R.
P P P P P P
Now x N e R1; 1f 6 N: R1 4 R1, then 1/r = 6 N(x N) e
P P P P
R1 9 R which is a contradiction. Thus, 6 N:R1 74> R1.

P

And further we have

N
N N ‘where r = 1 N(xP )6 (x,z).
P P P

(33) 1 = r6

'Finally for R1, we show that Der(Rl) = der(Rl). It
suffices to show 1n is generated by pl—th order deriva—

tions for all n. Define n“ €1De£:(R) as follows:

0‘11 “o i
(34) yh = 1 M/dM! o...o 11 /'aog where m = .23 dip .
p 1=O
n — O O .
By Theorem 3.2 Derk(R) — < Y1: . Yn >. So,
n
1 = Z: r.y..'with r. E R. On the other hand, it is clear
n i=1 1 1 1
n n
that Yn E Derk(R1) so, Yn = .2) ti).i w1th ti E R1. By

1=l

58

Corollary 3.3 rn and tn are units in R. Thus,

r = l/tn €IK1 n R = R1 and so, tn is a unit in R .

n l
- n-l
So, 1n = yn/tn - 1E) ti1i/tn. Assume by induction that
1

for i < n, 11 = jg) sinj with sij 6 R1. we can do
this Since 11 = Y1“ Then

n—l 1
1n = yn/tn - .E) ti(.§> sijyj)/tn. Hence,

—1 3—1
at al n l
- ' O. O ' -

1n (1/tn)1Pt/nt. o o 11 /oo. 6 Derk (R1). Thus, 1n

is generated by composites of pl-th order derivations.

Hence Der(R = der(R

1) 1) .

We have shown then that the following hold for R1:

n - .0. .
(35) Derk(R1) — < 11, , 1n > where 1i is also

an i-th order derivation of R to R.

O j<<i
There ex1st xi 6 R1 such that 11(xj) = { 1 j==i.

_ _ i . N
1.1 — 6i and xi — x for i < p .

_ - N
112 — z1i for 1 < p .

— . n
Der(Rl) — der(Rl) and in fact Derk(R1)
(I. (I.
-’ M ' 00. 00 o '

< Yl'”°' yn> where Ym

M .
Z dipl.
i=0

m

N
(k[xp .21) N . We still have

(xP .2)
N

N
9(Xp .2) = O and gz(xP ,2) #’O. Just as R1 was

Let R2

59

contained in R so too is R2 : R1. we shall show that

all the assumptions on R. which are given in (31) also

hold for R2. Before we do this however, we Observe that
. A . . . A A A .
if R2 is a domain then so is R = R1. Suppose R is

not a domain. Then there exists r #’o and s #’0 such
A
that rs = 0. By Chevalley's Theorem R has no nil-

N N
potent elements so, rP #’O and sp #'0. But
N

N N N
A A
rP sp = O and rp ,sp 6 R2. Thus, if R is a domain,

2
either r = 0 or s = 0.

An important relationship between the derivations
1n and R2 is that 1i(r) = O and 1ir = r11 when
r 6 R2 and i < pN. Also, if pN X n, then yh(r) = O

for r 6 R2. To see this we write n in its p—adic ex-
t .

pansion as n = Z) a.p1. Since puyl n, some a. # 0
i=0 1 1

for i < N. Let I < pN be the smallest i such that

0"t 0‘1
ol #’0. Then yh = 1 t/ot! o---o 1 I/'aI! . Since
P P
1pI(r) = O for r E R2, yn(r) = 0. Since 11.- = 6i for

i < pN, this relationship also holds for 6n 6 Der:(K).

That is, if pN X n, then 6n(r) = O for r 6 R2.

Before we study R further, we make an observation

2
A A
about zero divisors in R. For any r 6 (x,y)R, we write

r as

N N
P -1 - P -1 -
Z 6.x1 + Z) rix1

(36) r 1
'=1 i=0

60

N N
where Bi €.k and ri 6 (xP ,z)k[[xP ,2]]: ‘we can do

A A
this since R = R1. If r is a zero divisor, then by

Lemma 3.1 6n(r) is also a zero divisor for n < pN.

Since 6n is linear with respect to R2. we have

N N
6 ( ) [3 :pfsl B 6 ( j') P E; 6 ( i)
r = + . x + r- X o
n n i=n+1 1. n i 0 1 n

A
Since all terms but the first are in (x,y)R, 6n must

zero. Thus, 6. O for i = 1,'°°, pN-l. Hence, a

1.:
A
R

zero divisor in has the form
PN'l i

(37) r = Z) rix
i=0

N N
where ri e (xP ,z)k[[xp .21].

We shall show that the derivations Y N' n==1,2,---,
nP
give rise to generators which freely generate Der:(R2)

for all n. we make use of the fact that 6 N is an
nP

N
n-th order derivation on k[xp ] [1, Theorem 14: 6] and

N
that 6 N extends uniquely to k(xp ,z) = K2 and is
nP

an n-th order derivation on K2-

rs8N
Now consider y N; y N = r16i ‘with ri E K.
np np '=1

. n N
For r 6 R2. Y N(r) = E? ri6i(r) = Z) r 6 (r).
np i=1 ' '
n
So, y N E Derk(K2,K).
nP

61

We first show that there exist monomials

N
E k[xp ,2] such that y (y ) is a unit in R .
npN n 1
N
Suppose yN(r) 6 (x,z) for all r €k[xp ,2]. Since
P
y N(k[x,z]) 5t (x,z), there exists some monomial x121
P
such that y N(xlz
P N
N
since N(k[xp ,z]) c (x, 2). Also, p X i, otherwise

yn

3) is a unit in R1. Clearly i > O,

1‘23 6 k[xpN ,2]. Thus, there exists an element of the

form xjs 6 k[x,z] : R1 such that Y N(xjs) is a unit in
P

N
R s E k[xp ,z] and j is as small as possible with

1'

0 < j < pN. Consider the derivation y N =
p +(1:>N -j)

YNoYN. Forany xtr where O_<_t<pN and r€R2
P P-J
we have
YN°YN .(xtr) = YN°6N .(xtr)
P P.) P P-J
=meN .(xtn.
P P-J
N.
Now for tZP-J
5N .(xt) =
P-J
PN __ t-pN+j+s N_._
Sid} )(s Nt>CtpN+j+s-1)( 6N.(xp38)
PN “)8 P-J

=< M

P'J

62

[I, Pr0p. 9:6]. So,

N
P
YN°YN (xr)-
P P-J N
o 0.3 t < p -j
. t . N. .
By assumption Y N(x r) 63m Since t-p +3 < 3. Thus,

P

y N N (s) E'm for all s E R1 ‘which is a contradiction
p+p -j

to Corollary 3.4. Thus, there exists a monomial

N

yl E k[xp ,2] such that y N(yl) is a unit in R Let
P

l'

’ -
P P P

o N
E Der1p (R1)

Suppose we have found derivations k

a
Y

ipN
N

and monomials yi E k[xp ,2] such that the following hold:

i—l
y. N = rY. N + Z; rjy. N with rj 6 R1, r a unit in R
1P 1P j: 3 P

j<i
Y N(ya-H {1. .

1

ip 3‘1
_ n-l ’
for 1 < n. Let y N = Y N - .23 riY N where
np np i=1 ip
jil ‘
r = Y (17). r. = Y (y.) - r.Y (y-). SO.
1 npN 1 j npN j i= 1 ipN j
r.l 6 R1. Then by construction Y N(yi) = O for i < n.

63

N

Suppose (r) 6 (x,z) fbr all r E k[xp ,2]. By

s

npN
Corollary 3.4 there exists s 6 R1 such that § N(s)
nP

is a unit in R1. Again choose j as small as possible

so that § N(XJE) is a unit in R1, 0 < j < pN and
an
E 6 k[xp ,2]. Then as before one shows that
_ n-l
Y 0 6 = Y + Z) r.y :R 4 (x,z)
N N . N N . . . N . 1
up p -3 up +9 -3 1=1 1 1p +9 -3

contradicting Corollary 3.4. Thus, there is a monomial

N
yn E k[xp ,2] such that y N(yn) is a unit in R1. Let
nP

y' N = Y N/Y N(yn). This completes the induction step.
np np np

Thus, we have constructed derivations y' N and

N “P

monomials yn 6 k[xp ,2] ‘which satisfy the following

properties:

y' N is an n-th order derivation from K2 to K and

nP
and npN-th order derivation from R1 to R1.
‘ n-l .
y N = ry N + 'Z) riY. N where ri E R1 and r 13 a
np np i=1 1p
unit in R1.
' O i<<n
anN(Yi) — { 1 i==n

If pN X’m, define v; = vm. So, Y; is defined for all

I

n and Yn € Der:(R1). we now show by induction that

Der:(R1) = < y£,---, Y; >. The first step is clear.

64

- _ _ _ I _ I
Since 61 - l1 — Y1 — Y1, Der1k(Rl) - < Y1 >. Suppose

Derfi 1(R1) = < Y1,---, y 1 >. It suffices to show that

n...
_ N _ I
Yn - 121 s. HY for some 31 E R1. If p X’n, Yn — Yn .
If p N‘ n, so that n = fipN, then Yn = y_ N =
nP
I n—l ngg-ls ’

Y— N/r " 2' riY N/r = Y__ N/r + iYi by the
np i=1 ip np i=1

induction hypothesis. Thus, Der:(Rl) = < y£,---, Y; >.

n — I o o o '
We also hgve that Derk(K1) — < Y1, , Yn >. So,
11% t ’ f
‘ 0 Y. or son‘e to E K
npN i=1 1 i 1

On R we have

5 1° 2

n

6 N = Z) t NY’ N since for pN X i, yf(s) = y.(s) = O
._ . . i

nP 3-1 JP JP

when s 6 R2.

We shall show that y' N: R2 4 R2 for all n. In

nP
order to do this, we shall need to know that thlK2==R -

SO. we prove this first. Suppose r(xp ,z)/s(xp ,z) =
U(X.Z)/V(X.Z) in R1 with r,s,u,v, e k[x,z] and v(x,z)

a unit in R1. Then pulling back to k[x,z] gives
N N N
r(xp ,Z)v(X,Z) -s(xp ,Z)u(X,Z) =h(x,Z)g(xp ,Z).

PN'l P i

Write v(x,z) = Z3 vi(X ,Z)X . Do the same for u and
i=0

h. Then

65

N N N

r(xp ,Z)(v +---+v x? '1).-s(xP ,Z)(u +---+u xp '1)
O O N

P '1 p -l

N N
= (ko+---+k Xp "1)g(xp ,Z)

and so,

N N N N N N
r(xp .z)vo(xp ,Z)-s(xp .Z)uo(xp ,Z) = ko(xp ,Z)g(xp ,2)
since all other terms involve X1, 0 < i < pN. Since
v(x,z) is a unit in R1, v(x,z) begins with a constant

N
term. Thus, vo(xp ,z) is a unit in R2. Hence,

N N pN N
r(xP ,z)/'s(xp ,z) = uo(x ,z)/Vb(xp ,z) E R2. Therefore,
K2 0 R1 = R2.

We now show that Y' NzR2 4 R2. We first consider
n

P

I . 3 _ . . .
Y N' Since Y N — ul N where u is a unit in R1, we
P P P N

I _ _ i P
have by (33) that Y N — r6 N where r — Y N(x ) 6 R1.
P P P
__ l _ . .
We have 1 — Y N(y1) — r6 N(yl)' Since 6 N'KZ 4 K2,
P P P

_. ' . 0 0

r — 1/5 N(yl) 6 K2. Now Y H.122 4 R1 and r5 N.R2 4 K2,
P P P
I 0 =
thus, Y N.R2 4 R1 n K2 R2.
P
Inductively we assume Y' N:R2 4 R2 for i < n.
ip
Recall that 6 N when restricted to R2 may be written
n np
_ I .
as 5 N — .Z’ tiY. N, With ti 6K1 and tn #0.
np i=1 ip

Evaluating 6 N at yi gives ti € K2 Since

np

I

5 :K4K.So,Y -5 /t-

npN 2 2 npN nPN n

n-l I

.5) tiY. N/tn:R2 4 R1 0 K2 = R2. Thus, by Theorem 1.5

i—l ip

Dern(R ) is a free R —module generated by Y' ,---,Y'

R 2 2 N N

P nP

We should again note that Y' N is also an npN-th order

nP
derivation from R to R.

Finally we

show that Der(R2) der(Rz). Let

IaM I00
2 ' .0 t ._ . .
Om YpM /aM. Y1 /a0 where the p adic expanSion
M i
of m is given by m = Z3 dip . By Theorem 3.4
i=0
n — O. =
Derk(R1) — < 01 , , on >. To prove that Der(Rz)
der(RZ), it suffices to show for all n that
' n
(38) Y N = .2 ‘1". N
np i=1 ip
with r.l E R2. To see this suppose that (38) holds. We
first Observe that o N is a j-th order derivation on
ij
. J .
R2. We write j = Z? d.p1. Then ij = Z} a.p1+N.
. . i . 1
i=0 i=0
1a.: 90
_ I I °
Thus, 0. N — Y J+N/dJ' o...o Y N /do. . Since
JP P P
I . I .
Y :R 4 R , O :R 4 R . Further, Since Y . is a
. N 2 2 . N 2 n+i
1P JP 2 P
pl-th order derivation on R2, 0 has order
J jp
Z) dip1 = j. Now let X E Der(RZ) and suppose that n
i=0

is the smallest integer such that A E Der:(R2). write n

67

in its p-adic expansion as n = Z3 a.p . Since

n
n "" ' o o o ' = .
Derk(R2) - < Y N , , Y N >, X .2) t.Y. N With
p np i=1 ip
t.l E R2 and tn # 0. Using (38) and the above remark,
n-l

E Derk

we have that l — tnrno N (R Thus,

nP

2)°

n-l

k (R

a
l — tn rn (Y N+S/as E o...o Y’NO/ool) 6 Der ) where

SP

2

i
Y N+' E Derp (R ) and Z) a. lpi = n. Thus, if (38) holds,
P 1 k 2 i=0

Der(Rz) = der(RZ).

We now show (38). The result is true for n = 1
since 0 N = Y'N. Assume it holds for i < n. Now
p p n
n I
o E Der (R ) so, 0 = Z) u.Y , u. E R . In
npN k 3 npN i=1 i ipN i 2
np
I __ .
R1, Y N — .E) tici With ti 6 R1. By Corollary 3.5 un
np i—l
and t N are units in R1, so, t N = l/un 6K2f1R1==R2.
np nP
. . . I _ _
So, un 15 a unit in R2. Hence, Y N — o N/un
nP nP
n- 1 n-l
Z? u. lY. N/'un = O N/un + Z? ri O. N With ri E R2 and
i= 1. ip npN i=1 ip

(38) is proven.

Thus we have shown that Der: (R2 ) is a free Rz-module

for all n and Der(Rz) = der(RZ).

We now examine Deri(R2) more closely. We know

N pN
92(XP ,z) ¢ 0. ‘We shall also show that g N(xp ,z) ¢ 0.
P
X

68

N N N N
In M):13 .z)g N(xP .2)6 N(xp) + gzhcP .2)6 N(z) = 0
xp p P

N

since 5 N is a first order derivation on k(xp ,2). If

P

N
g N(xp ,z) = 0, then 6 N(2) = 0. Then by Proposition
P P
X

5 N2 = 26 N’ If this were the case then 6 N:R1 4 R1
P P P
N
which does not happen. Thus. 9 N(xp ,z) #'0.
P
X

N
Since y1 is the monomial in k[xp ,z] paired with

the first order derivation Y'N, it must be either 2 or
P

I

’ N
Y = r6 where r = Y (xp ). Since
N N N
P P P

N
5PN:R2‘7L> R2. Y;fi(xp ) is not a unit in R2. Hence, y1

must be 2.

Let 6: represent the canonical derivations on k[z]

23. oj<i N 2
That is, 5.(z ) = { . Since 9 (xp ,z);!0, 6.

1 1 j==i N 1

P
X

extends uniquely to K2. To summarize then, we have the
following:
(39) Der:(R2) is free for all n.

Der(Rz) = der(Rz).

N
g N(xp Oz) #0“
XP

1 _ 2 . . 2 _ ,
Derk(R2) - < 51 >. in fact, 61 — YpN.

69

N N
5i<xp ) 6 (XP 02) I

Thus, all the hypotheses that we originally had for R,

which are listed in (31), hold for R Here 2 plays

2.
N
the role of x and xP plays the role of y.

There are two cases. If bisz * R2 for all i,

A A
then by Theorem 3.8 R2 is a domain. Hence, R is a
domain. The second case is that there exists N(l) such

that si:R 4 R for i < pN(1)

2 2 , but 6PN(1):R2 +> R2.

We now proceed exactly as before and construct rings R3

and R . R3 = (k[u,z])(u'z) and

4
N(1)
R4 = (k[u,zp ]) N(1) . Here u plays the same
(“IZP )
role that z played in R1 and R2. Hence,

N N N N
u = xp + zr(xP ,z) for some r(xp ,z) E (xp ,2); this

is equation (32). Since 2 E (x,y), u E (x,y)z. We also
N(l)

n _ 2 ... 2
have that Derk(R2) — < 61 , , 5n > for n < p and
bir = r6: for i < pN(1) and r E R4. Further,
2 . N(l) . . . .
bi. i < p , may be Viewed as a derivation on R Since
n _ t o O . I I .
Derk(R2) — < Y N , , Y N > and Y_ N.R 4 R.
P np N ip
A
Now any element ri E (xp ,z)R2 may be written as
pN(1)_1 . PN(1)_1 .
(40) ri = 23 6.1.23 + Z ri.zJ
N(1) N(l)

where aij E k and rij E (u,zp )k[[u,zp ]].

70

A
This is just equation (36) with R2 playing the role of
A
R, u playing the role of z, and 2 playing the role
of x.

We now consider r E (x,y)R. Suppose r is a zero

-1 i
divisor. Then by (37) r has the form r = P2) r. 1x1 .
i=0
Using (40) we substitute for rj:
N N
p _1 p (1)_1 i p N_1 pN(1)-l i j
r = Z) .23 fii.x1 zJ + 23 _Z: ri.x z .
i=0 3=1 3 i=0 3=0 3
MD MD A
Here fiij E k and rij E (u,zp )k[[u,zp ]]s:(x,y)2R.

2
we now apply 6 o 6 to r:

62
o 6 (r) = B
6N 1 N -
p ( ) -1 p -1 pN 1.me-l
N
+ Z Z rij°um °°N (“in
i=0 j=0 p -l p —1

Since 62N(l) «:6 N (r) is a zero divisor,

—1 p -1
B N N(l) = 0. If we continue evaluating r at
P-lIP 1
2 2 2 2
6 06 .'°°,606 .6 06 I"'I6I
PN(1)_2 pN-l 1 pN_1 PN(1)_1 pN_2 1

we get Bij = 0 for 0.3 i < pN and O < j < pN(l).

Thus, if r is a zero divisor r E (x,y)2 R

71

we continue this process and construct a sequence

of rings sz such that sz relative to R2(j—l)

satisfies the analogous properties listed in (39). We
23'

1 . There

denote the canonical derivations on sz by 6

are two cases. The first is that there exists j such

that 5?”:R -+ R for all 1. Then by Theorem 3.6,

A 1 2J 2J A

R2J is : domain. This implies that R2(J_1) is a domain.
Hence, R is a domain. The second case is that for all

2:]

j there exists N(j) such that 6 N(j):
P

sz +> R23” In

A
this case we have that for a zero divisor r E R.

A A
r E (x,y)n R for all n. Since fl(x,y)n R = O, r = O.
n

A
Thus, in either case we have that R is an integral

domain. QED

The next theorem gives a geometric interpretation to
A ' .
this result: R is an integral domain only if P is uni-
branched at the origin. Algebraically this means that

f(X,Y) = (ax + BY)n + f +--- + fm.

n+1

Theorem 3.8: Let R be the local ring of the irreducible

 

curve f(X.Y) at (0.0) over an algebraically closed
field k. Suppose f has r distinct branches at (0.0).
Then the integral closure R of R has r maximal ideals

A
and the completion of R is R = R1 ®'°°® Rr'

Proof: Consider the integral closure R of R. It is a

semilocal ring with maximal ideals ml .°-°, mt. Rm is
i

72

A

a discrete rank 1 valuation ring so, Rm. 27k[[t]].
1

Hence, R -—fi> k[[t]]. Consider the image of (x,y) in
k[[t]], say, (g(t),n(t)). Since f(x,y) = O, we have
f(§(t),n(t)) = 0. Thus, mi determines a branch of
f(x,y).

Now suppose mi fi'mj determine the same branch.
6

T en Kai 9.1 [[t]] With (x,y) -——> (§l(t),n1(t)) an

A e

fim. 93 k[[tl] With (x.y) —§-> (§2(t).n2(t)). Since
3 2

mi and mj determine the same branch, there exists a

substitution 0 of order 1 such that

0(§1(t).nl(t)) = (§2(t).n2(t)) [Theorem 12.3: 8]. Let

K be the quotient field of R. Since k is algebrai-
A

cally closed, K is the quotient field of Rm and
i

A

Rm . So, on K ‘we have 0.391 = 92: K 4 k((t)). Now

3

since mi #mj choose r Emi such that r {my Then
91(r) e (t) so, 0°91(r) e (t) but 92(r) t(t); this
is a contradiction. Therefore, distinct maximal ideals
determine distinct branches. So, t.g r.

Consider a branch Yi' This branch determines a
local homomorphism eizR 4 k[[tJ]. 9i extends to K so,
GizK .. k[[t]]. Consider eglm n 1'1; this is prime

hence, maximal in R. Suppose 9;l(t) fl §==6;1(t)!lR==m.

73

Then
9

——i—--> k[[tn

l>
a”

- e...e’.1
e . 1 J

v
k[[tJ]
1

91° 9; is an automorphism since 911(t)I1R==631(t)f1R.

1

Thus, Bi<>97 gives a substitution of order 1 and

3

hence, Yi and Yj are the same branch. Thus, r'g t.

Therefore r = t and the number of maximal ideals
in R equals the number of branches of f at (0,0).
Thus, R = R1 ®--~@ Rr‘ [Theorem 37.9: S].

We have immediately the following corollaries.

A
Corollary 3.9: If R is an integral domain, then f

 

has one distinct branch at (0,0).

Corollary_3.lo: If Der:(R) is a free R-module for all

 

n and if Der(R) = der(R), then f has one distinct

branch at (0,0).

BIBLIOGRAPHY

BIBLIOGRAPHY

Grothendick, A.,"Elements of Geometrie Algebrique,"
Part 4, Publications Mathematiques de L'IHES
No. 32.

Lipman, J.,"Free Derivation Modules on Algebraic
Varieites,"Amer. J. Math., 87 (1965),

Mount, K.R. and Villamayor, O.E.,"On a Conjecture
of Y. Nakai,"Osaka J. Math. to appear.

,"Taylor Series and Higher Derivations,"
Universidad de Buenos Aries, No. 18.

 

Nagata, M., Local Ringg, John Wiley and Sons, Inc.,
New York, 1962.

Nakai, Y.,"High Order Derivations I,"Osaka J. Math.,
7 (1970). pp. 1—27.

Nakai, Y., et. al.,"High Order Derivations II,"
J. Sci. Hiroshima Univ. Sci. A-I, 34 (1970),
pp. 17-27.

Siedenberg, A., Elements of the Theoryiof Algebraic
Curves, AddisonAWesley Publishing Company,
Reading, Massachusetts, 1968.

Zariski, O. and Samuel, P., Commutative Algebra II,
D. Van Nostrand Company, Inc., Princeton, 1960.

 

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