203, 579]589 Ž1998.
JA977350
JOURNAL OF ALGEBRA
ARTICLE NO.
A Computation of Tight Closure in Diagonal
Hypersurfaces
Anurag K. Singh
Department of Mathematics, Uni¨ ersity of Michigan, East Hall, 525 East Uni¨ ersity A¨ enue,
Ann Arbor, Michigan 48109-1109
Communicated by Craig Huneke
Received August 26, 1997
1. INTRODUCTION
The aim of this paper is to settle a question about the tight closure of
the ideal Ž x 2 , y 2 , z 2 . in the ring R s K w X, Y, Z xrŽ X 3 q Y 3 q Z 3 . where
K is a field of prime characteristic p / 3. ŽLower case letters denote the
images of the corresponding variables.. M. McDermott has studied the
tight closure of various irreducible ideals in R and has established that
xyz g Ž x 2 , y 2 , z 2 .* when p - 200, see wMcx. The general case however
existed as a classic example of the difficulty involved in tight closure
computations, see also wHu, Example 1.2x. We show that xyz g Ž x 2 , y 2 , z 2 .*
in arbitrary prime characteristic p, and furthermore establish that xyz g
Ž x 2 , y 2 , z 2 . F whenever R is not F-pure, i.e., when p ' 2 mod 3. We move
on to generalize these results to the diagonal hypersurfaces R s
K w X 1 , . . . , X n xrŽ X 1n q ??? qX nn ..
These issues relate to the question whether the tight closure I* of an
ideal I agrees with its plus closure, Iqs IRql R, where R is a domain
over a field of characteristic p and Rq is the integral closure of R in an
algebraic closure of its fraction field. In this setting, we may think of the
Frobenius closure of I as I F s IR` l R where R` is the extension of R
obtained by adjoining p e th roots of all nonzero elements of R for e g N.
It is not difficult to see that Iq: I*, and equality in general is a formidable open question. It should be mentioned that in the case when I is an
ideal generated by part of a system of parameters, the equality is a result
E-mail: binku@math.lsa.umich.edu
579
0021-8693r98 $25.00
Copyright Q 1998 by Academic Press
All rights of reproduction in any form reserved.
580
ANURAG K. SINGH
of K. Smith, see wSmx. In the above ring R s K w X, Y, Z xrŽ X 3 q Y 3 q Z 3 .
where K is a field of characteristic p ' 2 mod 3, if one could show that
I* s I F for an ideal I, a consequence of this would be I F : Iq: I* s I F,
by which Iqs I*. McDermott does show that I* s I F for large families of
irreducible ideals and our result xyz g Ž x 2 , y 2 , z 2 . F , we believe, fills in an
interesting remaining case.
2. DEFINITIONS
Our main reference for the theory of tight closure is wHHx. We next
recall some basic definitions.
Let R be a Noetherian ring of characteristic p ) 0. We shall always use
the letter e to denote a variable nonnegative integer, and q to denote the
eth power of p, i.e., q s p e. We shall denote by F, the Frobenius
endomorphism of R, and by F e, its eth iteration, i.e., F e Ž r . s r q. For an
ideal I s Ž x 1 , . . . , x n . : R, we let I w q x s Ž x 1q, . . . , x nq .. Note that F e Ž I . R s
I w q x, where q s p e, as always.
We shall denote by R8 the complement of the union of the minimal
primes of R.
DEFINITION 2.1. A ring R is said to be F-pure if the Frobenius
homomorphism F : M ª M m R F Ž R . is injective for all R-modules M.
For an element x of R and an ideal I, we say that x g I F, the Frobenius
closure of I, if there exists q s p e such that x q g I w q x. A normal domain R
is F-pure if and only if for all ideals I of R, we have I F s I.
We say that x g I*, the tight closure of I, if there exists c g R8 such
that cx q g I w q x for all q s p e 4 0.
It is easily verified that I : I F : I*. Furthermore, I* is always contained in the integral closure of I and is frequently much smaller.
3. PRELIMINARY COMPUTATIONS
We record some determinant computations we shall find useful. Note
that for integers n and m where m G 1, we shall use the notation
ž mn / s
Ž n . Ž n y 1 . ??? Ž n y m q 1 .
.
Ž m . Ž m y 1 . ??? Ž 1 .
581
A COMPUTATION OF TIGHT CLOSURE
LEMMA 3.1.
ž
det
ž
n
aqk
n
aqky1
/ ž
/ ž
n
aqky1
n
aqk
ž na /
ž
n
aq1
???
/
???
/
ž
ž
n
a q 2k
/
n
a q 2k y 1
/
???
ž
ž
n
aqk
/
n
nq1
nqk
???
aqk aqk
aqk
.
aqk aqkq1
a q 2k
???
aqk
aqk
aqk
ž
s
???
/
/ž
/ ž /
/ ž /
/ž
Proof. This is evaluated in wMu, p. 682x as well as wRox.
Let F Ž n, a, k . denote the determinant of the matrix
LEMMA 3.2.
¡ žn/
a
ž
M Ž n, a, k . s
nq2
aq1
ž
ž
/
n
aq1
nq2
aq2
???
/
/
???
ž
ž
n
aqk
nq2
aqkq1
???
¢ž
n q 2k
aqk
/ ž
n q 2k
aqkq1
/
???
¦
/
ž
n q 2k
a q 2k
/
.
/§
Then for k G 1 we ha¨ e
F Ž n, a, k .
s n
a
F Ž n q 2, a q 2, k y 1 .
k
k
s Ž s q 2 a y n.
ž / Ł Ł Ž a q r.Ž n y a q r. .
ss1 rs1
Hence
ž na / ž na qq 22 / ??? ž na qq 22 kk /
F Ž n, a, k . s
ž
aqk
k
/ž
aqkq1
a q 2k y 1
???
ky1
1
/ ž
/
/ ž
/ ž
2a y n q k 2a y n q k q 1
2a y n q 2k y 1
???
k
ky1
1
?
.
nyaqk nyaqky1
nyaq1
???
k
ky1
1
ž
/ž
ž
/ž
/
/
582
ANURAG K. SINGH
Proof. We shall perform row operations on M Ž n, a, k . in order to get
zero entries in the first columns from the second row onwards, starting
with the last row and moving up. More precisely, from the Ž r q 1.st row,
subtract the r th row multiplied by Ž naqq2rr .rŽ naqq2rryy12 . starting with r s k,
and continuing until r s 2. The Ž r q 1, s q 1.st entry of the new matrix,
for r G 1, is
ž
n q 2r y
aqrqs
/
s
ž naqq2rr /
ž
n q 2r y 2
aqry1
s Ž s q 2 a y n.
/ž
n q 2r y 2
aqry1qs
/
n q 2r .
Ž a q r.Ž n y a q r. a q r q s
ž
/
We have only one nonzero entry in the first column, namely Ž na . and so we
examine the matrix obtained by deleting the first row and column. Factoring out sŽ s q 2 a y n. from each column for s s 1, . . . , k and 1rŽ a q r .Ž n
y a q r . from each row for r s 1, . . . , k, we see that
det M Ž n, a, k . s n
a
k
k
s Ž s q 2 a y n.
ž / Ł Ł Ž a q r.Ž n y a q r.
ss1 rs1
=det M Ž n q 2, a q 2, k y 1 . .
The required result immediately follows.
LEMMA 3.3. Consider the polynomial ring T s K w A1 , . . . , A m x where Ir, i
denotes the ideal Ir, i s Ž A1i , . . . , Air .T for r F m. Then
a
b
Ž A1 ??? A ry1 . Ž A1 q ??? qA ry1 . g Iry1, aq g q Ž A1 q ??? qA ry1 .
aq g
T
for positi¨ e integers a , b , and g implies
a
b q g y1
g Ir , aq g q Ž A1 q ??? qA r .
Ž A1 ??? A r . Ž A1 q ??? qA r .
aq g
T.
Proof. Consider the binomial expansion of Ž A 1 ??? A r . a Ž A 1
q ??? qA r . bq g y 1 into terms of the form Ž A 1 ??? A ry 1 . a Ž A 1
q ??? qA ry1 . bq gy1yj A ar qj. Such an element is clearly in Ir, aq g whenever
583
A COMPUTATION OF TIGHT CLOSURE
j G g , and so assume g ) j. Now
a
b q g y1yj
Ž A1 ??? A ry1 . Aar qj Ž A1 q ??? qA ry1 .
g Ir , aq g q Aar qj Ž A1 q ??? qA ry1 .
: Ir , aq g q Ž A1 q ??? qA ry1 , A r .
: Ir , aq g q Ž A1 q ??? qA r .
aq g
aq2 g y1yj
2 a q2 g y1
T
T
T.
4. TIGHT CLOSURE
We now prove the main theorem.
THEOREM 4.1. Let R s K w X 1 , . . . , X n xrŽ X 1n q ??? qX nn . where n G 3
and K is a field of prime characteristic p where p ¦ n. Then
ny 2
g Ž x 1ny1 , . . . , x nny1 . *.
Ž x 1 ??? x n .
Note that there are infinitely many e g N such that p e s q ' 1 mod n.
By wHH, Lemma 8.16x, it suffices to work with powers of p of this form,
and show that for all such q we have
Ž x 1 ??? x n .
Ž ny2 . qq1
g Ž x 1Ž ny1. q , . . . , x Žnny1. q . .
Letting q s nk q 1, it suffices to show
Ž x 1 ??? x n .
Ž ny2 . n k
g Ž x 1Ž ny1. n k , . . . , x Žnny1. n k . .
Let A1 s x 1n, . . . , A n s x nn and note that A1 q ??? qA n s 0. In this notation, we aim to show
Ž A1 ??? A n .
Ž ny2 . k
g Ž AŽ1ny1. k , . . . , AŽnny1. k . .
Our task is then effectively reduced to working in the polynomial ring
K w A1 , . . . , A ny1 x ( K w A1 , . . . , A n xrŽ A1 q ??? qA n . where we need to
show Ž A1 ??? A ny1Ž A1 q ??? qA ny1 ..Ž ny2. k g 2 Iny1, Ž ny1. k q Ž A1 q ??? q
A ny 1 .Ž ny1. k . By repeated use of Lemma 3.3, it suffices to show
Ž A1 A 2 .
Ž ny2 . k
Ž
.
k
Ž A1 q A 2 . g ž AŽ1ny1. k , AŽ2ny1. k , Ž A1 q A 2 . ny1 k / .
We have now reduced our problem to a statement about a polynomial ring
in two variables. The required result follows from the next lemma.
584
ANURAG K. SINGH
LEMMA 4.2. Let K w A, B x be a polynomial ring o¨ er a field K of characteristic p ) 0 and e be a positi¨ e integer such that q s p e ' 1 mod n. If
q s nk q 1, we ha¨ e
Ž A, B . 2 ny3 k : I s Ž AŽ ny1. k , B Ž ny1. k , Ž A q B .
Ž
.
Ž ny1 . k
..
In particular, Ž AB .Ž ny2. k Ž A q B . k g I.
Proof. Note that I contains the following elements: Ž A q B .Ž ny1. k =
A B Ž ny3. k , Ž A q B .Ž ny1. kAky1 B Ž ny3. kq1 , . . . , Ž A q B .Ž ny1. k B Ž ny2. k . We
take the binomial expansions of these elements and consider them modulo
the ideal Ž AŽ ny1. k , B Ž ny1. k .. This shows that the following elements are in
I:
k
žŽ
žŽ
n y 1. k
Ž n y 1 . k AŽ ny2. k B Ž ny1. k ,
AŽ ny1. k B Ž ny2. k q ??? q
k
2k
žŽ
n y 1. k
Ž n y 1 . k AŽ ny2. k B Ž ny1. k .
AŽ ny1. k B Ž ny2. k q ??? q
0
k
/
/
n y 1. k
AŽ ny1. k B Ž ny2. k q ??? q
ky1
???
/
ž
žŽ
/
/
n y 1. k
AŽ ny2. k B Ž ny1. k ,
2k y 1
ž
/
The coefficients of AŽ ny1. k B Ž ny2. k , AŽ ny1. ky1 B Ž ny2. kq1, . . . , AŽ ny2. k B Ž ny1. k
form the matrix
¡ Ž n y 1. k
ž
žŽ
k
n y 1. k
ky1
/ žŽ
/ žŽ
n y 1. k
kq1
n y 1. k
k
/
/
???
???
žŽ
žŽ
???
¢ž
Ž n y 1. k
0
/ ž
Ž n y 1. k
1
/
???
žŽ
n y 1. k
2k
n y 1. k
2k y 1
?
n y 1. k
k
/¦
/
.
/§
To show that all monomials of degree Ž2 n y 3. k in A and B are in I, it
suffices to show that this matrix is invertible. Since q s nk q 1 we have
Ž Ž n y 1k. k q r . s Žy1. k Ž 2 k ky r . for 0 F r F k, and so by Lemma 3.1, the deter-
585
A COMPUTATION OF TIGHT CLOSURE
minant of this matrix is
žŽ
n y 1. k
k
k
k
n y 1. k q 1
nk
???
k
k
kq1
2k
???
k
k
/žŽ
ž /ž
s Ž y1 .
k Ž kq1 .
/ ž /
/ ž /
ž /ž / ž /
ž /ž / ž /
2k
k
k
k
2k y 1
k
???
k
k
kq1
2k
???
k
k
s 1.
With this we complete the proof that Ž x 1 ??? x n . ny 2 g Ž x 1ny 1 , . . . , x nny1 .*.
5. FROBENIUS CLOSURE
Let R s K w X 1 , . . . , X n xrŽ X 1n q ??? qX nn . as before, where the characteristic of K is p ¦ n.
LEMMA 5.1. Let R s K w X 1 , . . . , X n xrŽ X 1n q ??? qX nn . where K is a field
of characteristic p. Then R is F-pure if and only if p ' 1 mod n.
Proof. This is Proposition 5.21Žc. of wHRx.
The main result of this section is the following theorem.
THEOREM 5.2. Let R s K w X 1 , . . . , X n xrŽ X 1n q ??? qX nn . where K is a
field of characteristic p. Then
ny2
g Ž x 1ny1 , . . . , x nny1 .
Ž x 1 ??? x n .
F
if and only if p k 1 mod n.
One implication follows from Lemma 5.1, and so we need to consider
the case p k 1 mod n.
The case n s 3 seems to be the most difficult, and we handle that first.
Let R s K w X, Y, Z xrŽ X 3 q Y 3 q Z 3 . where p ' 2 mod 3. We need to
show that xyz g Ž x 2 , y 2 , z 2 . F .
Let A s y 3, B s z 3, and so A q B s yx 3. We first show that when
p s 2, we have xyz g Ž x 2 , y 2 , z 2 . F by establishing that Ž xyz . 8 g
Ž x 2 , y 2 , z 2 .w8x. Note that is suffices to show that Ž xyz . 6 g Ž x 15 , y 15, z 15 ., or
in other words that Ž AB Ž A q B .. 2 g Ž A5, B 5, Ž A q B . 5 ., but this is easily
seen to be true.
586
ANURAG K. SINGH
We may now assume p s 6 m q 5 where m G 0. We shall show that in
this case Ž xyz . p g Ž x 2 , y 2 , z 2 .w px, i.e., that
Ž xyz .
6 mq5
g Ž x 12 mq10 , y 12 mq10 , z 12 mq10 . .
Note that to establish this, it suffices to show
Ž xyz .
6 mq3
g Ž x 12 mq9 , y 12 mq9 , z 12 mq9 . ,
i.e., that Ž ABŽ A q B .. 2 mq1 g Ž A4 mq3, B 4 mq3, Ž A q B . 4 mq3 ..
LEMMA 5.3. Let K w A, B x be a polynomial ring o¨ er a field K of characteristic p s 6 m q 5 where m G 0. Then we ha¨ e
Ž AB Ž A q B . .
2 mq1
g I s Ž A4 mq3 , B 4 mq3 , Ž A q B .
4 mq3
..
Proof. To show that Ž ABŽ A q B .. 2 mq1 g I, we shall show that the
following terms grouped together symmetrically from its binomial expansion,
f 1 s Ž AB .
3 mq1
3m
Ž A q B . , f 3 s Ž AB . Ž A3 q B 3 . , . . . ,
f 2 mq1 s Ž AB .
2 mq1
Ž A2 mq1 q B 2 mq1 . ,
are all in the ideal I. Note that I contains the elements Ž AB . m Ž A q
B . 4 mq3 , Ž AB . my 1 Ž A q B . 4 mq5, . . . , Ž AB .Ž A q B . 6 mq1, Ž A q B . 6 mq3. We
consider the binomial expansions of these elements modulo Ž A4 mq3,
B 4 mq3 ., and get the following elements in I:
ž
ž
ž
4m q 3
4m q 3
4m q 3
f q
f q ??? q
f
,
2m q 2 1
2m q 3 3
3m q 2 2 mq1
/
/
ž
ž
/
/
ž
ž
/
/
4m q 5
4m q 5
4m q 5
f q
f q ??? q
f
,
2m q 3 1
2m q 4 3
3m q 3 2 mq1
???
6m q 3
6m q 3
6m q 3
f q
f q ??? q
f
.
3m q 2 1
3m q 3 3
4 m q 2 2 mq1
/
ž
/
ž
/
587
A COMPUTATION OF TIGHT CLOSURE
The coefficients of f 1 , f 3 , . . . , f 2 mq1 arising from these terms form the
matrix
¡ 4m q 3
ž
ž
2m q 2
¢ž
6m q 3
3m q 2
4m q 5
2m q 3
/ ž
/ ž
4m q 3
2m q 3
4m q 5
2m q 4
/
/
???
???
ž
ž
4m q 5
3m q 3
/¦
/
ž
6m q 3
4m q 2
/§
4m q 3
3m q 2
???
/ ž
6m q 3
3m q 3
/
???
.
We need to show that this matrix is invertible, but in the notation of
Lemma 3.2, its determinant is F Ž4 m q 3, 2 m q 2, m. and is easily seen to
be nonzero.
The above lemma completes the case n s 3. We may now assume n G 4
and p s nk q d for 2 F d F n y 1. If k s 0, i.e., 2 F p F n y 1, we have
Ž x 1 ??? x n .
Ž ny2 . p
s y Ž x 1 ??? x ny1 .
Ž ny2 . p
n
x nŽ ny2. pyn Ž x 1n q ??? qx ny1
.
Ž ny1. p
g Ž x 1Ž ny1. p , . . . , x ny1
..
In the remaining case, we have n G 4 and k G 1. To prove that Ž x 1 ???
x n . ny 2 g Ž x 1ny 1, . . . , x nny1 . F , we shall show
Ž x 1 ??? x n .
Ž ny2 . p
g Ž x 1Ž ny1. p , . . . , x nŽ ny1. p . .
This would follow if we could show
Ž x 1 ??? x n .
Ž ny2 . n k
g Ž x 1Ž ny1. n kqn , . . . , x nŽ ny1. n kqn . .
As before, let A1 s x 1n, . . . , A n s x nn. It suffices to show that
Ž A1 ??? A n .
Ž ny2 . k
g Ž AŽ1ny1. kq1 , . . . , AŽnny1. kq1 . .
By Lemma 3.3, this reduces to showing
Ž A1 A 2 .
Ž ny2 . k
k
Ž A1 q A 2 . g I
s AŽ1ny1. kq1 , AŽ2ny1. kq1 , Ž A1 q A 2 .
ž
Ž ny1 . kq1
/.
The only remaining ingredient is the following lemma.
LEMMA 5.4. Let K w A, B x be a polynomial ring o¨ er a field K of characteristic p ) 0 where p s nk q d and where n G 4, k G 1, and 2 F d F n y 1.
588
ANURAG K. SINGH
Then
Ž A, B . 2 ny3 k : I s Ž AŽ ny1. kq1 , B Ž ny1. kq1 , Ž A q B .
Ž
.
Ž ny1 . kq1
..
In particular, Ž AB .Ž ny2. k Ž A q B . k g I.
Proof. Note that I contains the elements Ž A q B .Ž ny1. kq1Ak B Ž ny3. ky1 ,
Ž A q B .Ž ny1. kq1A ky1 B Ž ny3. k , . . . , Ž A q B .Ž ny1. kq1 B Ž ny2. ky1. We take the
binomial expansions of these elements and consider them modulo the
ideal Ž AŽ ny1. kq1, B Ž ny1. kq1 .. This shows that the following elements are in
I:
žŽ
žŽ
žŽ
n y 1. k q 1
Ž n y 1 . k q 1 AŽ ny2. k B Ž ny1. k ,
AŽ ny1. k B Ž ny2. k q ??? q
kq1
2k q 1
/
/
n y 1. k q 1
AŽ ny1. k B Ž ny2. k q ??? q
k
???
ž
žŽ
/
/
n y 1. k q 1
AŽ ny2. k B Ž ny1. k ,
2k
n y 1. k q 1
Ž n y 1 . k q 1 AŽ ny2. k B Ž ny1. k .
AŽ ny1. k B Ž ny2. k q ??? q
1
kq1
/
ž
/
The coefficients of AŽ ny1. k B Ž ny2. k , AŽ ny1. ky1 B Ž ny2. kq1, . . . , AŽ ny2. k B Ž ny1. k
form the matrix
¡ Ž n y 1. k q 1
ž
žŽ
¢ž Ž
kq1
n y 1. k q 1
k
n y 1. k q 1
1
/ žŽ
/ žŽ
/ žŽ
n y 1. k q 1
kq2
n y 1. k q 1
kq1
n y 1. k q 1
2
/
/
/
???
???
???
???
žŽ
žŽ
žŽ
n y 1. k q 1
2k q 1
n y 1. k q 1
2k
n y 1. k q 1
kq1
/¦
/
/§
.
To show that all monomials of degree Ž2 n y 3. k in A and B are in I, it
suffices to show that this matrix is invertible. The determinant of this
matrix is
žŽ
n y 1. k q 1
kq1
kq1
kq1
ž
n y 1. k q 2
nk q 1
???
kq1
kq1
kq2
2k q 1
???
kq1
kq1
/žŽ
/ž
/ ž
/ ž
/
/
which is easily seen to be nonzero since the characteristic of the field is
p s nk q d where 2 F d F n y 1.
A COMPUTATION OF TIGHT CLOSURE
589
Remark 5.5. It is worth noting that xyz g Ž x 2 , y 2 , z 2 .* in the ring
R s K w X, Y, Z xrŽ X 3 q Y 3 q Z 3 . is, in a certain sense, unexplained. Under mild hypotheses on a ring, tight closure has a ‘‘colon-capturing’’
property: for x 1 , . . . , x n part of a system of parameters for an excellent
local Žor graded. equidimensional ring A, we have Ž x 1 , . . . , x ny1 .: A x n :
Ž x 1 , . . . , x ny1 .* and various instances of elements being in the tight closure
of ideals are easily seen to arise from this colon-capturing property.
To illustrate our point, we recall from wHo, Example 5.7x how z 2 g Ž x, y .*
in the ring R above is seen to arise from colon-capturing. Consider the
Segre product T s RaS where S s K wU, V x. Then the elements x¨ y yu,
xu and y¨ form a system of parameters for the ring T. This ring is not
Cohen]Macaulay as seen from the relation on the parameters
2
2
Ž zu . Ž z¨ . Ž x¨ y yu . s Ž z¨ . Ž xu . y Ž zu . Ž y¨ . .
The colon-capturing property of tight closure shows
Ž zu . Ž z¨ . g Ž xu, y¨ . :T Ž x¨ y yu . : Ž xu, y¨ . *.
There is a retraction R m K S ª R under which U ¬ 1 and V ¬ 1. This
gives us a retraction from T ª R which, when applied to Ž zu.Ž z¨ . g
Ž xu, y¨ .*, shows z 2 g Ž x, y .* in R.
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