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Year: 1983
A peculiar unmixed domain
Brodmann, M; Rotthaus, C
Abstract: We construct a local noetherian domain R containing a nonzero prime element U and satisfying:
(i) The completion R̂ of R is a domain. (ii) U R̂ has an embedded prime divisor.
DOI: 10.2307/2043342
Posted at the Zurich Open Repository and Archive, University of Zurich
ZORA URL: http://doi.org/10.5167/uzh-23085
Originally published at:
Brodmann, M; Rotthaus, C (1983). A peculiar unmixed domain. Proceedings of the American Mathematical Society, 87(4):596-600. DOI: 10.2307/2043342
A Peculiar Unmixed Domain
Author(s): Markus Brodmann and Christel Rotthaus
Source: Proceedings of the American Mathematical Society, Vol. 87, No. 4 (Apr., 1983), pp. 596
-600
Published by: American Mathematical Society
Stable URL: http://www.jstor.org/stable/2043342
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PROCEEDINGS OF THE
AMERICAN MATHEMATICAL SOCIETY
Volume 87, Number 4, Apnl 1983
A PECULIAR UNMIXED DOMAIN
MARKUS BRODMANN
AND CHRISTEL ROTTHAUS
ABSTRACT. We construct a local noetherian domain R containing a nonzero prime
element U and satisfying:
(i) The completion R of R is a domain.
(ii) UR has an embedded prime divisor.
In [7] Nagata asked the following question: Let (R, m) be a local noetherian
domain and let p C Spec(R). Assume that R is unmixed (which means that
dim(R/f ) = dim(R) for all D E Ass(R)). Is R/I unmixed again?
Clearly, an example as mentioned in the abstract, gives a negative answer to this
question.
The method we propose to obtain R follows partly the method used in [6].
Originally we used a construction based on a similar principle presented in [2]. We
are grateful to the referee for having pointed out to us that modifying the approach
of [6] is more effective in getting the type of ring in question. We also thank him for
his helpful comments, in particular the use of (9), which essentially simplified some
of our original arguments.
We want to point out the essential difference between the rings constructed in [2]
and the ring which will be constructed in this paper. The main result of [2] states:
(1) There is a countable field K (which may be chosen of any characteristic) such
that-with a set X, Y, Vl,. . ., Vmof indeterminates- the following holds: Let a C
K[VI,..., VmIbe an ideal consisting of polynomials without constant term. Then
there is a local domain (R, m) such that:
(i) R=K[X, Y, VI,-.., VMI/A)O.
(ii) q
RnR= (0).
(Vo,. ., Vm)Ris prime and satisfies n
(iii) If p C Spec(R - ((O)}), R/I is essentially of finite type over K.
So (1) allows us to construct examples with arbitrarily"bad" generic formal fibers
((i) and (ii)) and with "good" formal fibers at nonzero primes (iii). On the other
hand, to deny Nagata's question we need a ring of the contrary shape. Its generic
formal fiber has to be better than the formal fiber in some nonzero prime.
We start our construction with a preliminary remark. Let K be a field and let
U, V, W, T be indeterminates. Let %0C K[U, V, W, T] be the prime ideal generated
P4 = W2UP3 WV-UT,
2, P2=WU2-V2,
TV. This is
by Pi -=T'
exactly the prime treated as an example in [5]. We know for example that there is a
Received by the editors October 5, 1981 and, in revised form, July 29, 1982.
1980 Mathematics Subject Classification. Primary 13C15, 13H10.
Key wordsand phrases. Local domain, completion, embedded prime, unmixedness.
?)1983 American Mathematical Society
0002-9939/82/0000- 11 22/$02.00
596
A PECULIAR UNMIXED DOMAIN
597
morphism A' - Spec(K[U, V, W, T]/ ), which is a (Zariski-) homeomorphism,
and an isomorphism apart from the origin. So Spec(K[U, V, W, T]/10) has a unique
singularity-namely the origin (from which we moreover know that it is of proper
Buchsbaum-type[41).In particular(K[U, V, W, T]/%o)(u v,w,T) is a local domainof
dimension 2 and of depth 1.
Now, to perform our construction we start according to [61(with the modification
of introducing an additional variable U). So let k be an infinite-countable
field and let Al,,, A2n, A3, (0 < n < xo) be indeterminates. Put K,,,
3 be additional variak(A 1J, A21, A31/j ? m) and K = U K,. Let X, U, Z I
bles. Put Qj = K[X, Z,, Z21 Z3, U], ro = (X, Zil Z2 Z3, U), Q = (QO)r). r = r0Q.
Now we find a set 4 of prime elements in QOwith the following properties:
(2)(a) XEE .
(b) p, q E $, p # q =*pQ 7# qQ.
(c) If P E Spec(Q - {UQ}) is a prime ideal of height one, p is generated by an
element p GE$.
(d) If q E U Q is a prime element, then q ( 43.
Now we may choose an enumeration N - 4 of 4 such that (with 8(n)-PO)
= X.
(3)(a)p,
(b)p,7 E Kn 2[X, Zi, Z21 Z3, U] (n > 1).
Next, define the following elements and ideals.
(4)(a)qn =
n
I pj(n E N),
A,1qf (i= 1,2,3),
(b) ,o0:= Z7Itin = Z7 +
(C) ,J
=
(41n7 42n1s43n, U)Q (n a_ 0).
Observe that gIn, 42n' t3n, U make part of a regular system of parameters for Q.
So p, is a prime ideal of height 4.
(5) PROPOSITION.
Pn ( Pn,
Vn E N.
PROOF.See [6] (the arguments are not affected by the presence of the additional
variable U).
Now, put A - K[U, V, W, TI(U.v,w,T)and (using the above notation) =OA,
t (U, V, W, T)A. Next we define
(6)(a) rt
(b) Zn
:=
=
As U, ,In
Q = KI[U,tin,
1,. ... ,4),
P(U, (Ins 42n' 430)(
(77lnI
..
Isqr4n)Q-
42n' i3n X are a regular system of parameters for Q,
42n' p3n,X], where denotes the completion. So there is a
we have
canonical
isomorphism
f3,,
= KffX, U, V, W, TI - Q
7) ACX]
with
,{3n(X)
/,n(V)
=
= X,
3l, n(W)
U) = U,
Bn(
= 42n /n(T)
= 43n-
,8nsends AI[X] to ClnQ. As the integral closure of A/0 is a (regular) local domain
and as A is of finite type over K, B is analytically irreducible. So BA[ X] is a prime
ideal (of height 2). Thus Z nQ, hence in particular Z n I is a prime ideal of height 2.
598
MARKUS BRODMANN AND CHRISTEL ROTTfHAUS
Let L be the quotient field of Q. As
obviously have the relation
( 8)
q I(
)tz
In particular it follows Q[q,;L''jR'= ,tQ[q-,
as a subring of L.
& (i/,,,
Er,-
I
(use (4) and (6)) in L we
q,t,)Q
1
q, Q[ qs
JQ[q,'1J.
This allows us to define
(9) LEMMA.Let a e Q such that q,7 E aQ. Then it holds that aR' n Q = (a, ?1't)Q.
PROOF.Z_ C q,tR' implies (a, zn)Q C aR', thus the inclusion " D ". To verify
the reverse inclusion it suffices to show that aQ[q,7 ?-rnj]n Q C (a, ?.t)Q for all
It follows that
m >>n. So let r = a(a() +
+q -ni a.) ( Q (a, E ?Z)
E
(r
qAn
aao)
Moreover we have q,, i .t, To see this assume the opposite. Then there is a least
natural number s s m with p5 E v, By (4) it follows that 4,(, --1) ( t), (i
1, 2, 3),
thus , -I LC , These primes both have height 4, so they are equal, which induces
the contradictionps E p1 =- p v (see (5)).
Now L CP implies q M Z, Li,,* being prime, it follows that (r - aao) ) SZi,,
thusr e (a, Li,,)Q.By (4) we have(,, - ,) E q7Q, thus(7r1,- iTl &
qG'qQ C aQ.
=
(
From this we get (a,
Zrn)
(a, Li,), thus r G ( a,
?,1)Q.
(10) COROLLARY.R'/p,R' = Q/(lp,p, i,1).
PROOF.(8) (applied for all m > n) and the observation that R' = U
Q qn ?]
show that R' - Q + pnR'. Thus the map Q - R'/p"R' is onto. By (9) its kernel is
P,? ? )Q'
(pi
( 11) COROLLARY.r R'-
(X, Z1, Z2 Z3, U)R' is a maximal ideal of R' and R'/r
R'
K.
r.
PROOF.Apply (10) with p
X, making use of p,
=
Next we put R RrR, m = rR and consider the following elements in Q =
KIIX,Z1, Z2, Z35UDJ
00
(a)
(12)
(b)
,Z,Z+
? 3)
A,,qJ
(As:i
I
'Tj= PJ(U- (1, 42 (4)
(
j -,- 4).
(13) PROPOSITION.R is a 3-dimensionallocal domain and UR is a prime ideal.
PROOF.First we show that UR is a prime ideal. So let f, g & R and assume that
fg E UR. Write fg = Uh (h G R). We may choose an element e E R' - rR' such
that for an appropriate n c N it holds that ef, eg, e2h E Q[q- Z,,]. As e is a unit in
R, it suffices to verify that ef or eg belongs to UR. Observing the equality
(ef )(eg)
U(e2h) we may replacef, g and h respectively by ef, eg and e2h. So we
may assume that f, g and h belong to Q[q Znnl. Choosing v & N appropriately we
have the relations qnVf,qnVg, qnvh & Q, which show that (qnp)2fg e UQ. As UQ is
prime in Q, we may assume without loss of generality that q7nf & UQ. Write
qnVf = Ur. Applying (9) with nv instead of n and qn, - a we get Ur z (qn,, Zi,)Q
A PECULIAR UNMIXED DOMAIN
599
Write Q= Qlcn and denote images modulo en by a bar, . Now, with an
appropriate s E Q, we may write Ur = 4?s. From the proof of (9) we know that
qnV6 S nv thus qn. 4 p nvOn the other side all associated primes of ( Z n, U ) are
contained in P,n. To see this notice that the isomorphism /3n: A'X[
Q maps
(U, V, W, T) A[X] to p n,Qand (t, U)A,[XI to (nv U) Q. Now, our claim
X
follows as Ass( A'IXIJ/( , U)) has only members contained in 13(which follows as
z C A, U E A) and by the faithful flatness of Q - Q.
The above statement shows that qn is regular with respect to Q/( 4nv, U). So 4,, is
regular with respect to Q/(U). It follows that s E UQ. Writing s = Ut we obtain
Ur = 4nn7tU,thus r = 4,nt. Thus induces r E (q7V, ?in)Q. By (9) (applied with
a = q,7) we get r E qn7R'. Consequently qn2f E q,7UR', thus f E UR'. This shows
that UR is prime.
Now we are ready to show that R is noetherian. So let p E Spec(R) - (0)). We
have to show that p is finitely generated. Assume first that p n Q # UQ. As R C L
we have p n Q 7# (0). So there is an n with Pn E p). (10) shows that R'/p,,R' is
noetherian. Thus p is finitely generated. Assume now that p n Q UQ. It suffices
c 1)n
to show p = UR. Assume that this is false. It follows that UQ[q-"
Q[q-n,Z for some n. So we find an element f p n Q[q "?,j -UR. For a
E Q n p= UQ C UR. This induces q,7 E UR
suitable v E N it must hold that qnnVf
n Q =UQ. This contradicts (2)(d).
It remains to show that R is of dimension 3. By (10) we have R/p, R Q/( p,,, ,).
Now the result is clear as R is a domain and as the right-hand ring is of dimension 2.
X]
There is an isomorphism a: A4X1/c AjX
(14) PROPOSITION.
to
U.
Umod(!Au[XI)
R which sends
PROOF.As ,, 2' 3, X, U form a regular system of parameters for Q we may write
2' 43, X, U], these generators being analytically independent over K. So
Q = K[(,,
Q, given by
there is a canonical isomorphism 'yo: A[ XI = KMU,V, W, T, XI
W - 2, T -3.
X - X, U- U, V .1,
By the inclusion map Q R we get a
homomorphism E: Q R, which maps U to U.
We claim that E is surjective. By (11) the maximal ideal mi C R is generated by
(U, Z1l Z2, Z3, X) and it holds that R/m = K. Thus we may write R
Z3)1. This shows the surjectivity of e.
KE( U ), E( Z),...,E(
R is surjective.We claim that y(
So y :=
XII) =0. As y((P,)
y(y: A[XI
0. To see this observe that
=7r(j
1,...,4) we only have to show that 4( 7i)
>
VTI7T11, = qtl + I, with a suitable T,,,E Q for all n 0.
-
In R we thus obtain the relation
E()
~(
j,)
?
qn
E(
q;"It1II
+
qn + I
(qtl , )
)
,1)
C muR
As this holds for all n, our claim follows immediately.
+ E(q
600
MARKUS BRODMANN AND CHRISTELROITHAUS
As already remarked above, BA[X] is a prime of height 2. By (13) we have
dim( R) = 3. By the previous claim and noticing that Al Xi is catenary of dimension
5 it follows that Ker(y) = BAIXI. So y induces an isomorphism a with the
requested properties.
(15) COROLLARY. R is a 3-dimensional local noetherian domain with the following
properties:
(i) R is a domain,
(ii) UR is a prime ideal.
(iii) t := a(U, V, W, T)/1 AllXI belongs to Ass( R/ UR) as an embedded component.
(i) Use (14) and the fact that t AOA
XII is a prime ideal.
(ii) See (13).
(iii) As (U, V, W, T)A E Ass(A/(!, U)A) the claim follows by (14) and observing
that ht (U, V, W, T)Al X]l/> Al XI = 2.
So we have constructed a ring as announced in the abstract.
PROOF.
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niormiial
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DEPARTMENTOF MATHEMATICS,UNIVERSITYOF BASEL, RHEINSPRUNG21. CH-4000 BASEL
DEPARTMENTOF MATHEMATICS,UNIVERSITY OF MUNSTER, EINSTEINSTRASSE62. D-4400 MONSTER.
WEST GERMANY (Currentl address of Christel Rotthaus)
Current uddress (Markus Brodmann): Mathematischcs Institut, UniversIty of Zurich, Raniistrasse 74.
CH-Zurich,
Swltzerland