REGO, C. J.

Rego, C. J.

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Remarks on differential operators on algebraic varieties

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Rego, C.J.
Osaka J. Math.
14 (1977) 481-486
REMARKS ON DIFFERENTIAL OPERATORS ON
ALGEBRAIC VARIETIES
C. J.REGO
(Received August 9, 1976)
We will work over an algebraically closed field of characteristic zero. Let
A be a (commutative) A-algebra and IdA®kA the 'diagonal ideaP, i.e., the ideal
generated by z® 1 — 1®#, z^A. We consider Aξζ>kA as an ^4-module via
Recall that the ^-module of n-jets (over k\ P\A\k}M= A®kA/Γ+1 is of finite
type when A is a localization of a Λ-algebra of finite type. Note that there is a
splitting
P\A I k) = A®kA/In+ί ^ A®kA/I « A
so that Pn(A I k)**I/In+1®A. We write P^ 1 k) for 7//*+1. The β-homomor-
phism d
n
\ A-*Pn(A\k) defined by d
n
(x)=x®\ — \®x is called the 'universal'
differential operator of order n on A,
Now suppose A is a local ring obtained from a β-algebra of finite type with
residue field k. We wish to pose the question:
Q(*) : When does Pζ(A \ k) have a free direct summand for n sufficiently large.
REMARK. For A regular Pξ is a free A module so we are interested in A
non-regular.
To understand the above question recall the definition of a differential
operator on A. Say B^QX^ '••X
s
](χlt...Xs) and A=B/^ί. The differential
operators on B of order <# are the obvious ones, i.e., elements of HomA(.B. B)
where the d/dX/s are the standard derivations of B. They form a jB-module
denoted by ΌΊSn(B \ k). An n-th order differential operator on A is a D as above
with D(Sl)cSl. We identify twτo operators whose difference sends B into 31, to
get an A module, Όifitt(A \ k). By the definition we get DiίPcDiff1*1. We can
identify DifF with the A -dual of Ptt via
482 c.J. REGO
φ^HomA(P", A), φ ι-» φodn(=Όίftn(A \k) .
η
In particular, DifP is a reflexive ^l-module-a fact we shall need. Note that
DifF^φDer^yl) and PJ«Ωi,
Λ
, the module of Kaehler differentials. For
generalities on DiίF see [3].
Now our question comes down to asking whether there exists a differential
operator D without constant term of some order for which there is an x^A with
D(x)=l. The main observation of this note is the
Theorem. Let A be a reduced, local one dimensional k-algebra with residue
field k. Then
(i): if A the completion of A, is a domain, £)(*) is true for A,
(ii): if A is not a domain with maximal ideal as conductor then Q(#) is
negative for A.
Notice that we should not ask for D to be a derivation because of the
Lemma (Zariski). Let D be a derivation of A and x^A with D(x)=l.
Then A^A
λ
\[X]\, i.e., A is analytically a product.
The composition of two differential operators is a differential operator of
higher order. One may ask if DifF is generated by the composites of elements
of DifP. Y. Nakai conjectured that if Diff n is generated by Diff * for all n^ 0 then
A is regular. It is more or less immediate (and checked below) that if Q(*)
has an affirmative answer for a class of rings then the Nakai conjecture is true
for that class. We should remark that the Nakai conjecture has been settled
in dimension one in [2] : it is also verified below in the dimension one analytic-
ally reducible case (which does not follow from the theorem). Finally we
observe (Prop. 2) that an affirmative solution of the Nakai conjecture would
settle the Zariski-Lipman problem which states that if Όeτk(A) is a free A-
module then A is regular.
Proposition 1. An affirmative solution of Q(*) for a class of rings, $>y implies
the Nakai conjecture for j3.
Proof. Let Diff1^) generate DifF(^4), w>0. We need to show that A or
equivalently, A is regular. Writing A as a quotient of &[[Fj, ••• Y
m
]] the above
definition of Όifί*(A) can be carried over to A in an obvious manner once we
define the differential operators on £[[3 ,^ •••¥',»]] to be generated by the
Then ΌΊffn(A\k) consists of express ions of the type 2 a<*^γ~i — *
v
 ,-„ > a
which preserve the ideal defining A. Hence Όiΰn(A\k)=ΐ)ifin(A\k)®AΆ.
(When A is complete we must remember that Όifin(A) has a different meaning
than when A is of geometric type). We assume now that A is complete.
DIFFERENTIAL OPERATORS ON ALGEBRAIC VARIETIES 483
Suppose there is a D in Derk(A) and an x<^A with D(x)= 1
Zariski's lemma. We claim that the hypothesis of Nakai's conjecture is satisfied
b y A
γ
. Write A l a n d l e t A = Σ «.-,• - - » ^ ye^ι be generators
U JL
of Der^j). Then Der ^4 is generated by 3/3* and the D,, where d/dx\A1 and
Df(x) are zero. Take a differential operator .2) on Al and represent it
We may assume it has no constant term. Put 3)(X)=Q to extend it to an
operator 3) on ^ [[J?]]. By hypothesis we can express 3) in terms of the Dt and
3/3* and since 3/3 F, and 3/3* commute and ai ^Al (so 1^=0) 3) maybe3
 \ dx '
written as ΣV3o»*^ where b^A^X]] and Q
Λ
 is a monomial in the D/s.
Since --^ , acts as the zero operator on ^C^pΓ]] for />0 putting 5)= )^0(/-Λ.
^ι=Σ*ίβί we find ΦΛA^i). Writing 0^0^
notice that^i^jQ^ restricted to A
λ
 represents 3).
As Qξ is a monomial in the derivations of Al we have verified the hypothesis
of Nakai's conjecture for A19
By induction on dimension of the ring, A1 is regular and we are through.
So let D(A)amA, the maximal ideal of A, for every D in DerA(^4). By hypo-
thesis every JζfeDifF, «>0 is a composite of derivations (plus a constant term)
so if H has no constant term, H(A)ayflA. This contradicts the truth of £)(#)
and proves the proposition.
Theorem. Let A be a reduced, local one dimensional k-algebra of geometric
type with residue field k. Then
(i): if A is a domain Q(*) is affirmative for A,
(ii) : if A is reducible with maximal ideal as conductor then Q(*) is negative
for A.
Proof. Let A be the normalization of A, I the diagonal ideal of A®kA
and / the diagonal ideal of A®kA. The picture
Ψ> "ψs homomorphisms, -v/r induced by ψ*
ylr +
->A
shows that the differential operators of A which send A into A are differential
484 C.J.REGO
operators of A. Note that we do not claim that every differential operator of A
is a restriction of one on ^4-which is in fact false. (However, by the theorem of
Seidenberg every derivation on A comes from one on A). As in the proof of the
proposition, remembering Όifin(Ά)=ΌΊfίn(A)®
 AΆ we suppose A complete.
( i ) : Suppose A is a domain so that ^4«&[[Z]]. We identify A as a subring
of k[[Z]]. Let (Zc) A be the conductor of A in A, i.e., Z'&Afec. Write D
for 9/9Z, the generator of Der^( A). We will construct
H = aJ>+alD°+l-J( ----- \-acD2c,
satisfying
so, in particular, H^ΌΊfί2c(A) and satisfies the requirements of the theorem. Put
a—h Z* where h{ are constants to be chosen presently. For Q^ζj^c write
D*+>(Z'+i) = \dijz'~' ***J> dϊj integersv
 ' l o i<j
suppose we want
H(ZC) = 1, H(Zc+j) = 0, 0<j<c (**)
for which it suffices to solve the linear equations
<*oA = 1
•+<*«*, = o.
Since these equation are clearly solvable we can choose α, so that H satisfies (**).
Further #(Z')-0, i<c and for;>0, H(Z*+i)=Zeb(Z)<=A as ZcΆdA. Hence
.?/ is our required operator. This proves (i).
(ii): Assuming A complete as before we can write A=k[[Zl]]@ ®
k[[Z
r
]]. Suppose there is an H with H(A)dA and #(#)=!, some x in A. Let
us derive a contradiction. Note that H operates on the total quotient ring of A
which is the same as that of A. So H is a meromorphic operator on A and can
oy
be written as a row (H^ H2, —Hr) each AΓf of the form 2 hij*τj> hij^k((Zj))-j <j£ i
Writing x=(χly > xr] we have assumed H^x^l for all ί. Observe that for
λf eΛ, \=(\!, •••\r)and<y=(jy1, •• jyr)e^, we have //(λ^)=λ/ί(y). In particular
\H(x)=H(\x)=(\ly •• λr). Now if λ. Φλj. for some (ι, j) then (\)&A since ^4
being local its elements have same Value' at each maximal ideal of A. Hence
for λ not in the diagonal of Λ© φft. To get a contradiction it
DIFFERENTIAL OPERATORS ON ALGEBRAIC VARIETIES 485
suffices to verify that \x^A and since the maximal ideal is in the conductor we
need only choose x a non unit. If x is a unit we can write #=#<,+#!, x0 a con-
stant and as H has no constant term H(xQ)=0 so H(x1)= 1. Replace x by xλ to
get the contradiction and complete the proof of the theorem.
REMARK. The statement (i) is much stronger than what is required to prove
the Nakai conjecture in the one dimensional case. It can be verified as follows.
Put AdA=®k[[Zi]] and recall that by Seidenberg's result the derivations of A
extend to A, so they are of the form (^(ZΛ — ~- Y Clearly all the -ψ»f (Zf )'s are\ 9z f J
/
 9 \
not units in Λ[[ZJ] — for otherwise D=i^lri(Zi)-^-J would send (X Z,-), X, eΛ to
(X ψv) which is not in A for suitable X,-. We may assume that there is one / for
r\
which all derivations of A have z-th entry φ(Z, W=- , φ(Zt ) a non-unit in Λ[[Z, ]] —oZi
say /—I. We will now restrict our attention to the first coordinate so let
A=k[[Z]], i.e., r=l. Write D—W^Z)^ for the derivations of A, W{(Z) of
order m^l. By hypothesis the operators of A are composites of the Z)/s at least
for large order. Notice that
r* d
implies that in a monomial Π /V''= Σ ^Λ tne order of ό
rt is not less than a.
* ΘZΛ
gc+i
From this we find that Zc——— is not a combination of D^s for any £>0. But
o£
for ί:>0 eg. so that ZcAdA we note that Zc—^^(Ά)ClA which contradicts
c/ ί^
the hypothesis and proves the conjecture. See [2] for another proof.
Hopefully Q(*) has an affirmative answer in the case when A is normal
thus solving the Nakai and Zariski-Lipman problems because of the
Proposition 2. The truth of the Naked conjecture implies that if A has
Derk(A) a free module then A is regular.
Proof. Let A be as above. By [4] we know A is normal so we may sup-
pose dim A^2 and A is nonsingular in codimension one. The composities of
the derivations Dly D2, " DmJ w—dim A, {D{} a free basis of Der^(^4) generate
(together with constants) submodules Diff* cDifF. It suffices to verify equality
for tt>0. Put J^=Spec A, UdX the set of regular points of X and Y= X-U.
The monomials in the Z)/s form a free basis of DifΓ at points in U since the
higher order operators on a regular local ring are generated by derivations and
486 CJ. REGO
Diff" c Diff ". In particular Diff" \ U= Diff11 | U. Let R be a relation among the
monomials in the Z)/s. Since at points in U the monomials are independent the
coefficients of R are zero on U. Since A is a domain an element vanishing on
an open subset of Spec A is zero. Hence R has all coefficients zero.
Hence Diff" is a free extension of Diff" | U. As X is of depth > 2 at Y and
DifF (being reflexive) has F-depth>2 the local cohomology groups Hγ(Difίn)=
//y(DifF)— 0 [see [1]]. The exact sequence of -..local cohomology says that
Γ (U, DifΓ I U)=Γ(X, Diff"). For the same reason Γ (X, Diffn)=
Γ(C7, D i f f n \ U ) = Γ ( U , Όiff \U)=Diff"9 as Z)#f is of F depth>2. Hence
and we are done.
Acknowledgement: Thanks are due to J. Becker for many interesting
conversations on the topics treated in this article.
TATA INSTITUTE OF FUNDAMENTAL RESEARCH
References
[1] A. Grothendiek: Local cohomology, Springer Lecture Notes.
[2] K.R. Mount and O.K. Villamayor: On a question of Y. Nakai, Osaka J. Math.
10 (1973), 325-327.
[3] Y. Nakai: High order derivation I, Osaka J. Math. 7 (1970) 1-27.
[4] J. Lipman: Free derivation modules, Amer. J. Math. 87 (1965), 875-898.


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