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arXiv:1109.4921v1 [math.AC] 22 Sep 2011
ARE COMPLETE INTERSECTIONS COMPLETE
INTERSECTIONS?
RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
Abstract. A commutative local ring is generally defined to be a complete
intersection if its completion is isomorphic to the quotient of a regular local
ring by an ideal generated by a regular sequence. It has not previously been
determined whether or not such a ring is necessarily itself the quotient of a
regular ring by an ideal generated by a regular sequence. In this article, it is
shown that if a complete intersection is a one dimensional integral domain, then
it is such a quotient. However, an example is produced of a three dimensional
complete intersection domain which is not a homomorphic image of a regular
local ring, and so the property does not hold in general.
Introduction
Let R be a commutative local Noetherian ring, and let?R denote the completion
of R with respect to the topology defined by the maximal ideal of R. Let us say that
the absolute definition of R being a complete intersection is that R is isomorphic to
the quotient of a regular local ring by an ideal generated by a regular sequence, and
let us say that the formal definition of R being a complete intersection is that?R
is a complete intersection in the absolute sense. The formal definition of complete
intersection was given in 1967 by Grothendieck in [G, 19.3.1]. Since completion
with respect to the maximal ideal of a Noetherian local ring defines a faithfully flat
functor, this definition works well for most applications and has been predominantly
adopted in recent years as the definition of complete intersection in commutative
ring theory. Nevertheless, the absolute definition also occurs frequently in the
literature. In spite of one’s preference of definition, it has remained a bane that it
is unknown whether the absolute and formal notions of complete intersection are
in fact equivalent. In this paper we give answers which shed light on this question.
In Section 1 below we show that the two notions are the same when R is a one di-
mensional integral domain (of arbitrary codimension). On the other hand, we show
in Section 2 by way of example that there are commutative local Noetherian rings
R whose completion is isomorphic to the quotient of a regular local ring modulo an
ideal generated by a regular sequence, but the ring itself is not the homomorphic
image of a regular local ring. We note that it is well-known (see, for example, [G,
19.3.2]) that if R is a complete intersection in the formal sense, then it being the
homomorphic image of a regular local ring is equivalent to it being a complete in-
tersection in the absolute sense. In the example,?R = R[[x,y,z,w]]/(x2+ y2), and
Date: September 23, 2011.
2000 Mathematics Subject Classification. 13J10, 13C40, 14M10.
Key words and phrases. Complete intersection, completion.
The first author was partially supported by NSF grant DMS-0856124. The second author was
partially supported by NSA grant H98230-10-0197.
1
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2RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
so R is an integral domain of dimension three. The main question thus remains
unanswered when R is of dimension 2, or dimension 1 and not an integral domain.
The proofs in both Sections 1 and 2 rely on the following basic theorem.
Theorem. Consider a diagram of commutative local ring homomorphisms
T
π
????
R
⊆
???R
where (T,m) is a complete regular local ring with dimension equaling the embedding
dimension of R, and π is surjective with kernel generated by a regular sequence
contained in m2. Then R is isomorphic to the quotient of a regular local ring by
an ideal generated by a regular sequence if one can complete this diagram to a
commutative diagram of local ring homomorphisms
S
⊆
??
π|S????
T
π
????
R
⊆
???R
where S is a regular local ring containing a generating set of kerπ, whose completion
is naturally isomorphic to T, and π|S is a surjection. The converse holds if R
contains a field of characteristic zero.
Proof. Of course only the last statement of the theorem is not obvious. To see the
last statement, assume that R is isomorphic to S′/I′where S′is a regular local ring,
and I′is an ideal of S′generated by a regular sequence. Without loss of generality
we can assume that the embedding dimensions of R and S′are the same. Let T′
denote?S′. By Cohen’s structure theorem for complete local rings of characteristic
zero, T′ ∼= F[[X1,...,Xn]] where F∼= T/M is a coefficient field. We can harmlessly
identify T′with its isomorphic copy and also write?R = F[[x1,...,xn]] where the
surjection T′−→?R is the obvious map taking Xito xi. We claim there exists an
isomorphism ϕ : T′−→ T such that the surjection T′−→?R factors through π.
Assuming the claim, the subring S = ϕ(S′) of T then completes the diagram in the
desired fashion.
To see the claim, we first construct a field E such that E ⊆ F is generated over
Q by a set A of algebraically independent elements and such that F is algebraic
over E. Let B ⊂ T be chosen so that π|B : B −→ A is a bijection. Then B
is algebraically independent over Q and so K = Q(B) is a field contained in T
satisfying π|K: K −→ E is an isomorphism. Now let L be the algebraic closure of
K in the integral domain T. Then L is certainly a field. Since T −→ T/M factors
through π, T/M is algebraic over K and so L is necessarily a maximal subfield of
T. Recalling the proof of Cohen’s Theorem, this forces L to be a coefficient field
of T and so we have T = L[[Y1,...,Yn]] for some Y1,...,Yn ∈ T. We can even
choose Y1,...,Ynsuch that π(Yi) = xifor each i. The last key step is to see that
π(L) = F; for this, it is enough that π(L) ⊆ F. Consider any e ∈ L. Then, using
separability, e satisfies an irreducible monic polynomial g(x) ∈ K[Z] and we can
factor g(Z) = (Z − e)h(Z) with h(Z) ∈ L[Z] and h(e) ∈ L a unit. We have an
injection ? π : L −→?R given by the composition L −→?R −→ T/M −→ F −→
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ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?3
?R and we let g(Z),h(Z),g(Z),h(Z) ∈?R[Z] denote the images of the respective
polynomials under π and ? π respectively.
we will not know h(Z) = h(Z) until we have actually proved our claim. Now
0 = g(π(e)) = g(π(e)) = (π(e) − ? π(e))h(π(e)). As h(π(e)) − h(? π(e)) is divisible by
the non-unit π(e) − ? π(e), h(π(e)) is a unit and it follows that π(e) − ? π(e) = 0 and
indeed π(e) ∈ F. Finally there is a unique homomorphism ϕ : T′−→ T such that
ϕ|F is the inverse of π|Land ϕ(Xi) = Yiand this map is what we need to complete
the proof of the claim.
It is obvious that g(Z) = g(Z), but
?
1. One Dimensional complete intersection domains are complete
intersection domains
In this section we prove the diagram of the theorem in the introduction can al-
ways be completed provided R is a one dimensional integral domain. The following
result from [H2] will help us find such an S.
Proposition 1.1. ([H2, Proposition 1]) Let (S,m∩S) be a quasi-local subring of a
complete local ring (T,m). Then S is Noetherian and the natural map?S −→ T is
an isomorphism if and only if S −→ T/m2is onto and IT ∩S = I for every finitely
generated ideal I of R.
We will refer to the condition that IT ∩ S = I for every finitely generated ideal
I of R by saying that finitely generated ideals of R are closed (with respect to T).
Lemma 1.2. Let (T,m) be a local ring and let f1,...,fk,s1,...,sn be a regular
sequence in T. Set K = (f1,...,fk)T and B = T/Ki+1. Let {σj} be the set of
all distinct monomials of degree i in the generators of K, and let σj,sidenote the
respective images in B. If we have an equation?ajσj+?bisi = 0 in B, for
aj,bi ∈ T, then for each j, there exists αj ∈ T with αj ∈ (si...,sn)B such that
αjσj= ajσj.
Proof. The equation?ajσj+?bisi = 0 yields?ajσj+?bisi ∈ Ki+1. As
Ki+1= ({σj}T)K, we get an equation?(aj+ cj)σj+?bisi= 0 with each cj∈ K.
Now fix j and write σj =?fei
ideal (fe1+1
1
,...,fek+1
k
As f1,...,fk,s1,...,sn is regular, it is a straightforward demonstration to see
that aj + cj ∈ (f1,...,fk,s1,...,sn)T.
(f1,...,fk,s1,...,sn)T = K + (s1,...,sn)T. Since (K/Ki+1)σj = 0 in B, the
conclusion follows.
i. Every other σi is necessarily contained in the
)T and so (?fei
i)(aj+cj) ∈ (fe1+1
1
,...,fek+1
k
,s1,...,sn)T.
Since cj ∈ K, this gives that aj ∈
?
Lemma 1.3. Let (T,m) be a Cohen-Macaulay complete local ring, let f1,...,fkbe
a regular sequence contained in m2, and set K = (f1,...,fk)T. Suppose that A is
a local subring of T/Kiwhose completion is naturally isomorphic to T/Ki, and R
is a local subring of T/K whose completion is naturally isomorphic to T/K such
that the natural map T/Ki−→ T/K induces a surjection A −→ R. Assume that
B is a quasi-local subring of T/Ki+1such that the following hold:
(1) The natural map T/Ki+1−→ T/Kiinduces a surjection B −→ A,
(2) f1+ Ki+1,...,fk+ Ki+1∈ B, and
(3) Ker(B −→ T/K) = (f1+ Ki+1,...,fk+ Ki+1)B.
Then B is a local Noetherian ring whose completion is naturally isomorphic to
T/Ki+1.
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4RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
Proof. Let fjdenote fj+Ki+1for 1 ≤ j ≤ k. We first want to show that condition
(3) above implies the condition
(3′) Ker(B −→ T/Ki) = (f1...,fk)iB.
The statement is trivial if i = 1. For i > 1, let x ∈ Ker(B −→ T/Ki). Since
x ∈ Ker(B −→ T/K) = (f1,...,fk)B, we can write x =?bjfjfor bj ∈ B. We
then have?bjfj∈ K2/Ki+1and it follows that bj ∈ K/Ki+1for each j. So
bj∈ Ker(B −→ T/K) = (f1,...,fk)B. Let {σ(m)
distinct monomials of degree m in the f1,...,fk. Then we can write x =?b2,jσ(2)
where b2,j ∈ B for all j, finishing the proof of the condition if i = 2. For i > 2,
?b2,jσ(2)
j
allows us to write x =?b3,jσ(3)
j
where b3,j∈ B for all j. Continuing in this way
we arrive at x =?bi,jσ(i)
j
claimed.
Thus for the rest of the proof we replace condition (3) by condition (3′), and
continue to let fjdenote fj+Ki+1for 1 ≤ j ≤ k. Since T/Ki+1modulo the square
of its maximal ideal is naturally isomorphic to T/m2, by Proposition 1.1 we just
need to show that the map B −→ T/m2is onto and that finitely generated ideals
of B are closed in T/Ki+1. Since T/Kiis naturally isomorphic to the completion
of A, Proposition 1.1 yields that the map A −→ T/(m2+ Ki) = T/m2is onto. It
then follows from assumption (1) that the map B −→ T/m2is also onto.
We will now establish that finitely generated ideals of B are closed in T/Ki+1.
We first show that all ideals of B which contain (f1,...,fk)iB are closed. Suppose
I is an ideal of B containing (f1,...,fk)iB and x ∈ I(T/Ki+1) ∩ B. Since ideals
in A are closed, we have x + (Ki∩ B) ∈ I + (Ki∩ B)/(Ki∩ B). As B −→ A is
surjective, there exists y ∈ I such that y + (Ki∩ B) = x + (Ki∩ B). Next, as
x = (x − y) + y, we may reduce to the case x ∈ Ki∩ B. Since (f1,...,fk)iB ⊆ I,
Condition (3’) tells us that x ∈ I, and we are done.
Next we show that if s1,...,snare parameters in B, then I = (s1,...,sn)B is
closed. Suppose x ∈ I(T/Ki+1)∩B. Let J = (f1,...,fk)iB. Since J +I is closed,
we have x = y + z with y ∈ J and z ∈ I. It suffices to show that y ∈ I, and
we may reduce to the case where z = 0. Write x =?amsm for am ∈ T/Ki+1,
and y =?bjσ(i)
j
?bjσ(i)
j
every j, and therefore reduce to the case where x ∈ JI(T/Ki+1). This allows us
to write x =?cjσ(i)
j
write x =
j
dj−cj∈ K/Ki+1for each j. For fixed j we have cj+ K ∈ R because dj∈ B, and
it is also in the closure of (s1+ K,...,sn+ K)R. However, ideals in R are closed
and so we have elements e1,...,en∈ R with cj+ K =?em(sm+ K). We next
choose a preimage e♯
each j, we have x ∈ I.
Next we show that any ideal I of B which is primary to the maximal ideal of B
is closed. Since I is primary to the maximal ideal, it necessarily contains an ideal J
which is generated by a complete system of parameters. We have just seen that J is
j
} be the set of images in B of the
j
∈ K3/Ki+1and it follows that b2,j∈ Ker(B −→ T/K) for each j. This
where bi,j∈ B for all j. That is x ∈ (f1,...,fk)iB, as
with bj∈ T/Ki+1. In T/Ki+1we have the equation?amsm−
= 0. Using Lemma 1.2, we may assume bj ∈ (s1,...,sn)(T/Ki+1) for
with each cj ∈ I(T/Ki+1). As J is closed, we also can
with dj ∈ B. Hence?(dj− cj)σ(i)
?djσ(i)
j
= 0. It follows that
m∈ B for each em. Since σ(i)
jcj = σ(i)
j
?e♯
m(sm+ Ki+1) for
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ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?5
closed. Since B −→ T/m2is onto, so is the map B −→ T/(mj+Ki+1) for all j ≥ 1
(see, for example, the proof of Proposition 1 of [H2]). Let J♯denote an ideal of T
generated by a set of preimages in T of a generating set for J. Now for some j we
have mj⊆ (Ki+1+J♯)T, so that the map B −→ T/(Ki+1+J♯)T is also onto. Thus
T/Ki+1= J(T/Ki+1)+B, and we have I(T/Ki+1)∩B = (IJ(T/Ki+1)+I)∩B ⊆
J(T/Ki+1) ∩ B + I = J + I = I as desired.
According to the remark following Lemma 21 in [H2], if the finitely generated
ideals in B are not all closed, then either there exists an ideal I of B which is
not closed and whose closure is primary to the maximal ideal, or there exists an
infinitely generated prime ideal P ∩ B for P ∈ SpecT/Ki+1. To see that neither
of these situations can occur, first note that since both of the maps B −→ A and
T/Ki+1−→ T/Kihave nilpotent kernels, there are natural bijections SpecB ↔
SpecA and SpecT/Ki+1↔ SpecT/Ki. Suppose that I is an ideal of B which
is not primary to the maximal ideal of B. Thus I is contained in a non-maximal
prime ideal of B, and one of our bijections tells us the same is true for the image I
of I in A. As T/Kiis the completion of A, this means that I(T/Ki) is contained
in a non-maximal prime ideal of T/Ki. Then we invoke the other bijection to see
that I(T/Ki+1) is not primary to the maximal ideal of T/Ki+1and so the closure
I(T/Ki+1) ∩ B of I is not primary to the maximal ideal of B. Thus if I is an
ideal of B whose closure is primary to the maximal ideal of B, then I is primary
to the maximal ideal of B, and hence is closed by the proof above. For the other
case, P ∩ B is necessarily the closure of a finitely generated ideal J of B, since all
ideals of T/Ki+1are finitely generated. As each fj becomes nilpotent in T/Ki+1,
we see that (f1,...,fk) ⊆ P ∩B. Therefore P ∩B is also the closure of the finitely
generated ideal J + (f1,...,fk), which is closed by the proof above. This forces
P ∩ B to be finitely generated.
?
Lemma 1.4. Let (T,m) be a Cohen-Macaulay complete local ring, f1,...,fk be
a T-regular sequence contained in m2, and K = (f1,...,fk)T. Further suppose
that T/K has dimension one. Let Ri and R be local subrings of T/Kiand T/K,
respectively, with i ≥ 1 such that
(1) R is an integral domain,
(2) T/Kiand T/K are naturally isomorphic to the completions of Ri and R
respectively,
(3) f1+ Ki,...,fk+ Kiare in Ri, and
(4) We have the following commutative diagram with surjective vertical maps
T/Ki+1
????
Ri
????
⊆
??T/Ki
????
R
⊆
??T/K
Then there exists a local subring Ri+1 of T/Ki+1such that T/Ki+1is naturally
isomorphic to the completion of Ri+1, f1+ Ki+1,...,fk+ Ki+1are in Ri+1, and
the commutative diagram above may be completed to one with surjective vertical
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6RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
maps
Ri+1
????
⊆??T/Ki+1
????
Ri
????
⊆
??T/Ki
????
R
⊆
??T/K
Proof. If we construct Ri+1containing f1+Ki+1,...,fk+Ki+1such that the upper
square is commutative with surjective vertical maps and Ker(Ri+1 −→ T/K) =
(f1+Ki+1,...,fk+Ki+1)Ri+1, then the previous lemma completes the proof. We
let fjdenote fj+ Ki+1for 1 ≤ j ≤ k.
Consider the set of subrings B of T/Ki+1satisfying the following conditions:
(1) (a) f1,...,fk∈ B
(b) In addition, if Char(T/K) = p > 0, Cp⊆ B where C is the full
preimage of Riin T/Ki+1.
(2) The image of B under the map T/Ki+1−→ T/Kiis contained in Ri.
(3) Ker(B −→ T/K) ⊆ (f1,...,fk)π−1(Ri), where π : T/Ki+1−→ T/Kiis
the natural projection.
This set can be ordered by inclusion. We claim that this set contains a maximal ele-
ment, a claim we will prove by Zorn’s Lemma. To see the claim, first we must show
that the set is nonempty. In the characteristic zero case, let B0= Z[f1,...,fk]. Ob-
viously B0satisfies Conditions (1) and (2). As Ker(B0−→ T/K) = (f1,...,fk)B0,
Condition (3) also holds.
In the characteristic p case, let B0 = Cp[f1,...,fk]. Conditions (1) and (2)
are clear for B0. Suppose α ∈ Ker(B0 −→ T/K). Then α =?cjσj with each
cj ∈ Cpand {σj} ranging over a set of distinct monomials in the fj. To show
Condition (3) for B0, it suffices to show cjσj ∈ (f1,...,fk)π−1(Ri) for each j.
The statement is obvious unless σj = 1, so we may assume α ∈ Cp. Suppose
a + Ki+1∈ C is such that ap+ Ki+1∈ K/Ki+1. Then the induced map C −→ R
takes a + Ki+1to a nilpotent element of the integral domain R. So a ∈ K. Hence
a+Ki= (a1+Ki)(f1+Ki)+···+(ak+Ki)(fk+Ki) with each ai∈ T. As T/Ki
is the completion of Ri, we may actually choose a1,...,ak so that each aj+ Ki
is in Ri; so aj+ Ki+1∈ π−1(Ri). Let e =?ajfj and note that a + Ki+1=
(e + Ki+1) + (d + Ki+1) with d ∈ Ki. Finally, since d2∈ Ki+1and pd ∈ Ki+1,
(a + Ki+1)p= (e + Ki+1)p∈ (f1,...,fk)π−1(Ri).
We have thus shown that in any characteristic, the set is nonempty. Next we
consider the union of an ascending chain of elements in the set. The union obviously
satisfies Condition (1) and the other two conditions can be viewed as elementwise
conditions. Since these hold for every set in the union, they must hold for the
union. Thus, by Zorn’s Lemma, the set contains a maximal member B.
Next we claim the map B −→ Riis surjective. If not, let Ai= Image(B −→ Ri)
and let A = Image(B −→ R). We choose r♯∈ Ri− Ai and let r be the image
of r♯in R. If we lift r♯to a preimage ? r ∈ T/Ki+1, we have natural surjections
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ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?7
B[? r] −→ Ai[r♯] −→ A[r] fitting into the diagram
B
?
??
????
B[? r]
????
⊆
??T/Ki+1
????
Ai
???
????
Ai[r♯]
????
⊆
??Ri
????
⊆
??T/Ki
????
A
⊆
??A[r]
⊆
??R
⊆
??T/K
where since Ai[r♯] properly contains Ai, B[? r] properly contains B. We will show
that ? r can be chosen in such a way that B[? r] satisfies the three conditions. This will
contradict the maximality of B and so prove the claim. As the first two conditions
hold for an arbitrary choice of ? r, we need only consider Condition (3).
Note that Ker(B[? r] −→ T/K) = Ker(B[? r] −→ A[r]) since R injects into T/K.
We have a natural presentation
A[X]/I
∼
=
− → A[r]
For h(X) ∈ B[X] we let h(X) denote the polynomial in Ai[X] obtained by reducing
the coefficients of h(X) modulo Ki, and by h(X) the polynomial in A[X] obtained
by reducing the coefficients of h(X) modulo K.
If h(X) ∈ B[X], then h(? r) ∈ Ker(B[? r] −→ T/K) precisely if h(X) ∈ I. So
the proof reduces to showing that h(? r) ∈ (f1,...,fk)π−1(Ri) whenever h(X) ∈ I.
First we consider the case h(X) = 0. By Condition (3) for B, all of the coefficients
of h(X) are in (f1,...,fk)π−1(Ri) and so h(? r) ∈ (f1,...,fk)π−1(Ri) regardless of
which lifting ? r we choose. In particular, we note that Condition (3) holds for B[? r]
if I = (0).
Assume I ?= (0). Suppose g(X) ∈ B[X] is such that 0 ?= g(X) ∈ I and g(X)
is of minimal degree among all such polynomials. As B −→ A is surjective, every
element of I has the form h(X) and so g(X) also has minimal degree among the
set of nonzero polynomials in I. Since R is an integral domain, g(X) is a (not
necessarily monic) minimal polynomial satisfied by r over the quotient field of A.
Claim. We may choose g(X) and ? r so that g(? r) ∈ (f1,...,fk)π−1(Ri).
Either g′(r) = 0 or g′(r) ?= 0. The first case will occur precisely when A ⊆
A[r] is not a separable extension, something which can happen only if T/K has
characteristic p and g(X) has degree at least p. In this case (? r)p∈ B, so we may
choose g(X) = Xp− (? r)pand we see that g(? r) = 0 ∈ (f1,...,fk)π−1(Ri). The
claim actually holds for any choice of ? r.
In the second case, we choose our minimal polynomial g(X) arbitrarily but we
must select ? r carefully. Since g(? r) ∈ K/Ki+1, g(r♯) ∈ (π(f1),...,π(fk))(T/Ki).
Further, as T/Kiis the completion of Ri, g(r♯) ∈ (π(f1),...,π(fk))Ri. So we
may choose elements γj ∈ π−1(Ri) such that g(r♯) =?π(fj)π(γj). Let {σ(m)
be the images in B of the distinct monomials of degree m in the f1,...,fk. Then
we have elements αj∈ T/Ki+1such that g(? r) =?fjγj+?σ(i)
R/(g′(r))R is zero-dimensional and so Ri/(g′(r♯))Riis also zero-dimensional. Since
j
}
jαj. As g′(r) ?= 0,
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8RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
zero-dimensional local rings are complete, the induced map
Ri/(g′(r♯))Ri−→ (T/Ki)/(g′(r♯)(T/Ki))
is an isomorphism.
preimages we see T/Ki+1= π−1(Ri) + g′(? r)(T/Ki+1). There exists for each j,
βj ∈ π−1(Ri),tj ∈ T/Ki+1such that αj = βj+ g′(? r)tj. We now claim that the
desired condition holds if we choose the lifting ? r2= ? r−?σ(i)
series expansion of g(X) about ? r we have
g(? r2) = g(? r) − g′(? r)
=
fjγj+σ(i)
?
fjγj+σ(i)
It follows that T/Ki= Ri+ g′(r♯)(T/Ki) and by taking
jtj. Taking the Taylor
?
?
?
σ(i)
jtj
?
j(αj− tjg′(? r))
jβj∈ (f1,...,fk)π−1(Ri).
=
This completes the proof of the claim.
To derive our contradiction, it only remains to show that, for the choice of ? r given
by this claim, if h(X) ∈ B[X] with h(X) ∈ I, then h(? r) ∈ (f1,...,fk)π−1(Ri).
Choosing b ∈ B−K/Ki+1to be a sufficiently high power of the leading coefficient of
g(X), we can write bh(X) = h1(X)g(X) + h2(X) where h1(X),h2(X) ∈ B[X] and
h2(X) is a polynomial of lower degree than g(X). Since h(X) ∈ I and g(X) ∈ I,
we have h2(X) ∈ I. By degree considerations h2(X) = 0 and, as we noted above,
this yields h2(? r) ∈ (f1,...,fk)π−1(Ri). Further, as h1(? r) ∈ π−1(Ri) and g(? r) ∈
(f1,...,fk)π−1(Ri), we actually get bh(? r) ∈ (f1,...,fk)π−1(Ri). Additionally, we
note that since b maps to a nonzero element of R and every nonzero element of R
is regular on the completion of R, i.e., T/K, b = (e + Ki+1) where f1,...,fk,e is
a regular sequence in T. It follows that b is a regular element on T/Kmfor every
m ≤ i + 1.
To get h(? r) ∈ (f1,...,fk)π−1(Ri), we will actually prove a more general state-
ment which allows the coefficients of h(X) to be arbitrary elements of T/Ki+1.
Using reverse induction on m, for m = 1,...,i, we will show that if h(X) ∈
(T/Ki+1)[X], b ∈ B − K/Ki+1, and bh(? r) ∈ (f1,...,fk)mπ−1(Ri), then h(? r) ∈
each dj ∈ π−1(Ri). Also, as b is regular on T/Ki, we have h(? r) =?σ(i)
each cj∈ T/Ki+1. It follows that?σ(i)
Thus π(dj) ∈ (π(b),π(f1),...,π(fk))(T/Ki). As T/Kiis the completion of Ri
and so ideals are closed, we get π(dj) ∈ (π(b),π(f1),...,π(fk))Ri. Thus there
exists aj ∈ π−1(Ri) such that π(dj− baj) ∈ K/Ki. Of course, we then have
dj− baj ∈ K/Ki+1and so σ(i)
h(? r) =?σ(i)
have already demonstrated the m + 1 case. As before, we have bh(? r) =?σ(m)
with each dj∈ π−1(Ri) and h(? r) =?σ(m)
gives?σ(m)
j
element aj∈ π−1(Ri) such that π(dj−baj) ∈ (π(f1),...,π(fk))Ri. Thus dj−baj∈
(f1,...,fk)π−1(Ri) + Ki/Ki+1. Then σ(m)
Finally we define k(X) = h(X)−?σ(m)
j
(f1,...,fk)mπ−1(Ri). First consider m = i. Here we have bh(? r) =?σ(i)
jdj with
jcj with
j(bcj− dj) = 0 and so bcj− dj∈ K/Ki+1.
j(dj− baj) = 0.So bh(? r) =
?σ(i)
jbaj, giving
jaj ∈ (f1,...,fk)iπ−1(Ri). Next assume that m ≥ 1 and that we
j
dj
j
cj with each cj∈ T/Ki+1. Again this
(bcj− dj) = 0 and so bcj− dj∈ K/Ki+1. As before, this gives us an
j
(dj− baj) ∈ (f1,...,fk)m+1π−1(Ri).
aj. Then bk(? r) =?σ(m)
j
dj−b?σ(m)
j
aj=
Page 9
ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?9
?σ(m)
(f1,...,fk)m+1π−1(Ri) and so h(? r) ∈ (f1,...,fk)mπ−1(Ri) as desired. The m = 1
case is actually the result we need, the final piece in the demonstration of our con-
tradiction.
We have shown that we can choose B satisfying the three conditions such that
B −→ Riis surjective. Let Ri+1= B. Condition (3) gives Ker(Ri+1−→ T/K) ⊆
(f1,...,fk)π−1(Ri). As Ri+1 −→ Ri is surjective, we now have Ker(Ri+1 −→
T/K) ⊆ (f1,...,fk)(Ri+1+ Ker(T/Ki+1−→ T/Ki)) = (f1,...,fk)Ri+1.
j
(dj− baj) ∈ (f1,...,fk)m+1π−1(Ri). By the induction assumption, k(? r) ∈
?
Theorem 1.5. Let (T,m) be a Cohen-Macaulay complete local ring, f1,...,fkbe a
T-regular sequence contained in m2, and K = (f1,...,fk)T. Further suppose that
T/K has dimension one. Let R be an integral domain which is a local subring of
T/K such that T/K is naturally isomorphic to the completion of R. Then there
exists a local subring S of T with f1,...,fk∈ S such that T is naturally isomorphic
to the completion of S and we have the commutative diagram with surjective vertical
maps
S
⊆
??
π|S????
T
π
????
R
⊆
??T/K
Proof. We apply the previous lemma a countable number of times to obtain a
sequence of surjections
··· −→ Ri+1
ρi+1
− − − → Ri
ρi
− → Ri−1−→ ··· −→ R1
=
− → R
where πi : T/Ki+1−→ T/Kiis the natural map and ρi = πi|Ri. This sequence
of maps forms an inverse system and the inclusion maps Ri −→ T/Kidefine a
morphism of inverse systems. Let S = lim←Ri be the inverse limit of the Ri.
Elements of s ∈ S have the form s = (si+Ki) ∈?Riwith si∈ T and si+1+Ki=
si+ Kifor all i ≥ 1. Since T is complete in the K-adic topology, the natural
map T −→ lim←T/Kitaking t to (t + Ki) is an isomorphism. We will therefore
identify the element x of T with (x + Ki) ∈ lim←T/Ki. It follows that we get a
commutative diagram with surjective vertical maps
S
ψ
??
π|S????
T
π
????
R
⊆
??T/K
First we show that the induced map on direct limits ψ is injective. Suppose that
ψ(s) = 0 in T for s = (si+ Ki) ∈ S. This means that si+ Ki= Kifor all i, and
so (si+ Ki) = (Ki) which is the zero element in S.
Since fj+ Ki∈ Rifor all i, and 1 ≤ j ≤ k. We have fj = (fj+ Ki) ∈ S for
1 ≤ j ≤ k. It follows that we also get all monomials comprised of the fjin S.
To show that S is quasi-local with maximal ideal m ∩ S, we show that every
element of S −m has an inverse in S. Let x = (xi+Ki) ∈ S −m. Since each Riis
quasi-local with maximal ideal m/Ki∩Ri, and xi+Ki∈ Ri−(m/Ki∩Ri), we have
inverses yi+Ki∈ Ri−(m/Ki∩Ri) of xi+Kiin Rifor each i. We just need to show
that yi+Ki−1= yi−1+Ki−1for all i ≥ 1, for then the element (yi+Ki) will be the
Page 10
10RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
inverse of x in S−m. We have xiyi−1 ∈ Ki, and it follows that xi−1yi−1 ∈ Ki−1.
By uniqueness of inverses in Ri−1we see that yi+ Ki−1= yi−1+ Ki−1.
If we show that S −→ T/m2is onto and IT∩S = I for all finitely generated ideals
I of S, then we may apply Proposition 1 to complete the proof. As R −→ T/m2is
onto, certainly S −→ T/m2is onto.
Now let y1,...,ynbe elements of S, with yj= (yj,i+ Ki), and I the ideal of S
they generate. Choose x = (xi+ Ki) ∈ IT ∩ S. Then xi+ Ki∈ I(T/Ki) ∩ Rifor
all i ≥ 0.
Claim. There exists a positive integer N and {tj,i|i ≥ 0,1 ≤ j ≤ n} ⊆ T such that
(1) tj,i+ KN+i+1∈ RN+i+1,
(2) tj,i+1+ Ki= tj,i+ Ki, and
(3) xN+i+ KN+i=?tj,iyj,N+i+ KN+iin RN+i
for each j and for all i ≥ 0. Assuming the claim, conditions (1) and (2) imply that
(tj,i+ Ki) ∈ S. Condition (3) implies xi+ Ki=?tj,iyj,i+ Kifor i ≥ 0, which
means that (xi+ Ki) =?(tj,i+ Ki)(yj,i+ Ki) ∈ I, and thus I is closed.
Proof of the claim. By Artin-Rees there exist an integer N such that IT ∩KN+i=
Ki(KN∩ IT) for all i ≥ 0. We prove the existence of the tj,i by induction on i,
beginning with i = 0.
Since finitely generated ideals in RN+1are closed we have I(T/KN+1)∩RN+1=
IRN+1. Therefore there exists {tj,0} ⊆ T such that
?
xN+1+ KN+1=tj,0yj,N+1+ KN+1
with tj,0+ KN+1∈ RN+1. Of course then,
xN+ KN=
?
tj,0yj,N+ KN.
Now suppose we have successfully chosen {tj,i|1 ≤ j ≤ n} and we want to find
{tj,i+1|1 ≤ j ≤ n}. Since finitely generated ideals of RN+i+1are closed, there exists
{uj,i} ⊆ T such that
?
xN+i+1+ KN+i+1=uj,iyj,N+i+1+ KN+i+1
and uj,i+ KN+i+1∈ RN+i+1. This equation together with
xN+i+1+ KN+i=
?
tj,iyj,N+i+1+ KN+i
yields
?
(uj,i− tj,i)yj,N+i+1∈ KN+i∩ (IT + KN+i+1) = (KN+i∩ IT) + KN+i+1
= Ki(KN∩ IT) + KN+i+1
⊆ KiIT + KN+i+1
Therefore we have?(uj,i− tj,i)yj,N+i+1+ KN+i+1∈ KiI(T/KN+i+1) ∩ RN+i+1.
Since finitely generated ideals of RN+i+1 are closed, there exist {vj,i} ⊆ T such
that vj,i+ KN+i+1∈ Ki(T/KN+i+1) ∩ RN+i+1and
?
(uj,i− tj,i)yj,N+i+1+ KN+i+1=
?
vj,iyj,N+i+1+ KN+i+1
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ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS? 11
Finally, since RN+i+2−→ RN+i+1is onto, we choose tj,i+1+ KN+i+2∈ RN+i+2
such that tj,i+1+ KN+i+1= tj,i+ vj,i+ KN+i+1. It is routine to check that the
elements {tj,i+1} do the job and the induction is complete.
?
Assume that?R is isomorphic to T/K. Upon identifying R with its image in
T/K, Theorem 1.5 proves the following.
Corollary 1.6. Suppose that R is a local Noetherian integral domain of dimension
1. Then?R is isomorphic to the quotient of a regular local ring by an ideal generated
by a regular sequence if and only if R is the quotient of a regular local ring by an
ideal generated by a regular sequence.
2. A 3-dimensional complete intersection that is not a complete
intersection
This section is devoted solely to the construction of an example. The ring R will
be a three dimensional local domain whose completion is T = R[[x,y,z,w]]/(x2+
y2). The example is far from excellent; in fact, while we do not know whether or not
there are excellent examples, the non-excellence is a critical aspect of this example.
The generic point of the singular locus of T, the prime ideal (x,y)T, will intersect
trivially with R, but R will have uncountably many height two prime ideals Pλ
which are contractions of singular points and in fact singular themselves. Showing
that R cannot be lifted to a regular local ring will largely consist of showing that
there is not a simultaneous compatible lifting of the R/Pλ’s. This particular trick
certainly requires both non-excellence and dimension at least three. On the other
hand, there is no reason to believe that the choice of coefficient field was of any
importance. The construction is easier if one knows the cardinality of the coefficient
field and choosing R allowed us to use the very elementary polynomial x2+ y2.
While the construction is rather intricate, the conception behind it is not. We
build an example with the property that if R can be lifted to a regular local ring S
contained in R[[x,y,z,w]], then S must contain an element Θ = f(1 + zω + z2h +
xa+yb) where f = x2+y2and ω ∈ R is known, but we have no information about
h,a,b. We also equip R with a large collection of prime ideals whose extension to T
contains (x,y)T and the construction of each of these prime ideals guarantees the
existence of an element in the lifting which is congruent to f modulo that particular
prime ideal P. This new element and Θ are unit multiples of each other and so their
quotient will be in S. So 1+zω+z2h ∈ R/P and it follows that z(ω+zh) ∈ R/P.
We will construct R so that z ∈ R and so ω + zh is in the quotient field of R/P.
We can’t control h but we do know that, for some fixed value of h, ω +zh must be
in the quotient field of R/P for every one of the primes we construct. So, for every
possible value of h, we construct a prime ideal Phsuch that ω+zh is transcendental
over R/Phand so certainly not in the quotient field. This contradiction rules out
the possibility of a lifting.
We begin with a lemma. It is a minor variant of Lemma 3 from [H1] and the
proof is substantively the same.
Lemma 2.1. Let (T,m) be a local ring which contains an uncountable field F, let
C ⊂ SpecT, and let D ⊂ T. Let I be an ideal of T with I ? P for all P ∈ C. If
|C × D| < |F|, then I ??{r + P|r ∈ D,P ∈ C}.
Page 12
12RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
Moreover, if t ∈ I −?
such that tu / ∈?{r +P|r ∈ D,P ∈ C}. Further, if D′⊂ T is such that |D′| < |F|,
then we may additionally choose u / ∈?{r + m|r ∈ D′}.
Proof. We first prove I ??{P|P ∈ C}. It suffices to prove I ??{P +mI|P ∈ C}.
By Nakayama’s Lemma, I ? P + mI for any P ∈ C. Letting V = I/mI, we have
reduced the problem to showing that if a finite-dimensional vector space V over
T/m is the union of |C| subspaces where |C| < |T/m|, then one subspace must be
all of V . This is surely well known and the proof is even written out as part of the
proof of Lemma 3 in [H1].
To prove the lemma, it suffices to prove the second paragraph. We choose an
element t ∈ I −?{P|P ∈ C}. Then, for any (r,P) ∈ D × C, we have tu ∈ r + P
if and only if u ≡ t−1r modulo P. If r + P / ∈ (t + P)(T/P), then tu / ∈ r + P.
Otherwise, t−1r ≡ s modulo P and we can obtain tu / ∈ r+P by choosing u / ∈ s+P.
For each such pair (r,P), we choose a coset representative s and we label the full
set of coset representatives D∗. Then |D∗∪ D′| ≤ |C × D| + |D′| < |F| and so
we can choose u ∈ F such that u ?≡ s modulo m for any s ∈ D∗∪ D′. Clearly
u / ∈?{s+P|s ∈ D∗,P ∈ C}?{r +m|r ∈ D′} and so tu / ∈?{r +P|r ∈ D,P ∈ C}
and u / ∈?{r + m|r ∈ D′}.
P∈CP (and such a t must exist), then there exists u ∈ F
?
We repeat a definition from [H1].
Definition 2.2. Let (T,m) be a complete local ring and let (R,m∩R) be a quasi-
local unique factorization domain contained in T satisfying:
(1) |R| ≤ sup(ℵ0,|T/m|) with equality only if |T/m| is countable,
(2) Q ∩ R = (0) for all Q ∈ Ass(T), and
(3) if t ∈ T is regular and P ∈ Ass(T/tT), then Height(P ∩ R) ≤ 1.
Then R is called an N-subring of T.
In the present circumstances, we will let T = R[[x,y,z,w]]/(x2+ y2). Here
condition (2) of N-subring is vacuous and condition (1) is just |R| < |R|.
Definition 2.3. (R,Γ) is called an N-pair if R is an N-subring of T and Γ is a set
of pairs {(Pλ,Vλ)|λ ∈ Γ} with {Pλ} a distinct set of primes such that, for every λ,
(1) z,w ∈ R
(2) (x,y)T ∩ R = (0)
(3) Pλ= (x,y,z + uλw)T for some uλ∈ R ∩ R
(4) Vλ⊂ T and |Vλ| ≤ |R|
(5) the images in T/Pλof the elements of Vλare algebraically independent over
R/Pλ∩ R
(6) if Qλ = Pλ∩ R, then QλRQλ= (x + αλtλ,y + βλtλ,tλ)RQλfor some
αλ,βλ∈ R where tλ= z + uλw.
Note that the requirement that the elements of the set {Pλ} be distinct and of the
specified form forces |Γ| ≤ |R|. In practice, Vλwill always be a singleton set, but
the additional generality does not require any extra effort.
Lemma 2.4. Suppose R is an N-subring of T, v ∈ T, Λ = {(Pλ,Vλ)|λ ∈ Λ} where
each Pλ∈ SpecT and each Vλ is a subset of T, H is a finite subset of the set of
Page 13
ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?13
height one prime ideals of T which have the form rT for r ∈ R, and I is an ideal
of T. Let G = {P ∈ Ass(T/rT)|r ∈ R} and ∆ = {Pλ|λ ∈ Λ}. Assume
(1) I ? P for every P ∈ ∆ ∪ (G − H)
(2) if P′∈ H, then P′? P for every P ∈ ∆
(3) |Λ| ≤ |R| and |Vλ| ≤ |R| for every λ
Then there exists an element d ∈ I such that if S is the intersection of T with the
quotient field of R[v+d], S is an N-subring of T such that, for each λ ∈ Λ, the image
of v + d in T/Pλis transcendental over (R/Pλ∩ R)[Vλ] and for each P ∈ G − H,
the image of v + d in T/P is transcendental over (R/P ∩ R). Moreover, if I = tT
is a principal ideal, we may choose d = tu with u ∈ R and u / ∈?{r + m|r ∈ D′}
for any prescribed set D′with |D′| < |R|.
Proof. Let C = ∆ ∪ (G − H). For each P ∈ C, we define a subring A(P) of T/P.
If P / ∈ ∆, A(P) = R/P ∩ R. If P = Pλ∈ ∆, A(P) = (R/Pλ∩ R)[Vλ]. In either
case, A(P) has cardinality at most |R| and so the same is true for the algebraic
closure of A(P) in T/P. Let DP be a complete set of coset representatives modulo
P of {t ∈ T|v + t is algebraic over A(P)}. Therefore |DP| ≤ |R|. Now |C| ≤ |R|
and so also, if D =?
Hence we may apply Lemma 2.1 to choose d ∈ I so that the image of v + d in
T/P, is transcendental over A(P) for every P ∈ C. We use this choice of d to
define S. The construction immediately forces the image of v + d in T/Pλ to be
transcendental over (R/Pλ∩ R)[Vλ] for each λ ∈ Λ and the image of v +d in T/P
to be transcendental over (R/P ∩R) for each P ∈ G−H. Additionally, in the case
where I = tT is principal, Lemma 2.1 allows us to choose d = tu so that the final
statement of the lemma holds.
It only remains to show that S is an N-subring. Suppose P ∈ Ass(T/tT) for some
nonzero t ∈ T. If P∩R = (0) and Q = P∩S, then R(0)[v+d] ⊂ SQ⊆ R[v+d](0)and
SQis either a discrete valuation ring or a field. In either case, Height(P ∩ S) ≤ 1.
If P ∩ R ?= (0) then P ∈ G. If P ∈ G − H and Q = P ∩ S, the fact that v + d is
transcendental over R/(P ∩R) tells us that SQis the localization of RP∩R[v+d] at
the unique prime ideal minimal over P ∩ R. Thus Q is a height one prime ideal of
S principally generated by the generator of P ∩ R. Finally, if P ∈ H, there exists
r ∈ R such that P = rT. Then Q = P ∩S ⊂ rT and since rT ∩S = rS, we see that
Q is also a principal height one prime ideal of S. This shows that condition (3) of
N-subring is satisfied and that is the only one of the three enumerated conditions
that needs to be checked as |S| = |R|. We also note that the intersection of a local
domain with a subfield of its quotient field is always quasi-local. To prove S is a
UFD, we may first invert any known principal prime elements and so reduce to the
case where our ring is a localization of R(0)[v +d] and so is of course a UFD. Thus
S is an N-subring of T.
P∈CDP then |D| ≤ |R|, and finally |C × D| ≤ |R| < |R|.
?
Remark 2.5. As we will repeatedly employ Lemma 2.4, it seems useful to give a
quick summary of how it is used. We start with a ring R (which is always obvious
from the context) and must specify v,Λ,H,I. The lemma then gives us an element
d and a ring S satisfying the properties we need.
Remark 2.6. The reader may have noted that the proof did not require H to be
finite, nor did it require hypothesis (2). However, as any nonzero ideal I can be
contained in at most finitely many members of G, none of which are contained in
Page 14
14RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
∆, these hypotheses place no restriction on the situations in which Lemma 2.4 can
be used. The hypotheses are critical in the next two lemmas.
It will be useful if we can describe S more precisely.
Lemma 2.7. Let S be the N-subring obtained in Lemma 2.4 and let η = v+d. Let
g ∈ m∩R[η] be a prime element of R[η]. If g ∈ R, then g is a prime element of S.
If g / ∈ R, then there is an element r ∈ R, possibly r = 1, such that every height one
prime ideal of T which contains r is in H, and g/r is either a prime element or
a unit of S. Consequently, if H = {r1T,...,rnT} with each ri∈ R and t =?ri,
then there exists a domain L such that R[η] ⊆ L ⊂ R[η,t−1] and S = Lm∩L. In
particular, if H = ∅, we have S = R[η]m∩R[η].
Proof. The fact that prime elements of R remain prime in S was noted in the proof
of Lemma 2.4. So suppose g / ∈ R. As g is prime in R[η], no height one prime ideal
of R contains all of the coefficients of g. By construction, η is transcendental over
R/P ∩ R for every P ∈ G − H and so g / ∈ P. As H is finite, it follows that we can
find r ∈ R such that g/r ∈ S, g/r / ∈ P for every P ∈ G and every height one prime
ideal of T which contains r is in H. Now, to determine the prime factorization of
g/r, we can safely invert all nonzero elements of R. This localization of S must be
an overring of the principal ideal domain K[η], where K is the quotient field of R,
and so a localization of K[η]. As g/r is a prime element of K[η], it is either a prime
element or a unit in S. Now let L = R[η,t−1] ∩ S. Let c ∈ S. As S is contained in
the quotient field of R[η], we can write c = a/b with a,b ∈ R[η] and express both
as a product of prime elements of R[η]. We then alter the factorizations slightly,
replacing each g ∈ m ∩ R[η] by the corresponding g/r, which is an element of L.
(We actually write g = (g/r)r, so a,b do not change.) After canceling common
factors, b is necessarily a unit in S and since a,b ∈ L, S is a localization of L. The
final statement is now obvious.
?
Definition 2.8. If (R1,Γ1) and (R2,Γ2) are N-pairs such that R1⊆ R2, Γ1⊆ Γ2,
and |R1| = |R2|, we say (R2,Γ2) is an A-extension of (R1,Γ1).
Lemma 2.9. If (R,Γ) is an N-pair, then any application of Lemma 2.4 with Γ ∪
{((x,y)T,∅)} ⊆ Λ will produce an N-subring S such that (S,Γ) is an A-extension
of (R,Γ).
Proof. We must show (S,Γ) is an N-pair; certainly |S| = |R|. To do this, we need
only check the six conditions and conditions (1)-(4) are obvious. Consider any
λ ∈ Γ. By lemma 2.7, S is a localization of R[η,t−1] ∩ S and since t / ∈ Pλ, SPλ∩S
is a localization of R[η,t−1] and so a localization of RPλ∩R[η]. Let Kλdenote the
quotient field of R/Pλ∩ R. As η is transcendental over Kλ, it follows that the
quotient field of S/Pλ∩ S is just Kλ(η). The fact that η is transcendental over
Kλ[Vλ] tells us that Vλis algebraically independent over Kλ(η) and so condition
(5) holds. Finally, since SPλ∩S is the localization of RPλ∩R[η] at the extension of
the prime ideal (Pλ∩ R)RPλ∩R, condition (6) is also satisfied.
?
Lemma 2.10. Let (R,Γ) be an N-pair and suppose Q ∈ SpecT is a height two
prime ideal. If Q ?= Pλ for all λ ∈ Γ, then there exists S, R ⊆ S ⊆ T, such that
(S,Γ) is an A-extension of (R,Γ) and Q ∩ R ? Pλfor any λ ∈ Γ.
Proof. Let Λ = Γ ∪ {((x,y)T,∅)}, let v = 0, and let I = Q. Now apply Lemma 2.4
with H = ∅ to find an N-subring S = R[d]m∩R[d]which clearly satisfies the conclu-
sion as d ∈ Q −?Pλ.
?
Page 15
ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?15
Lemma 2.11. Let (R,Γ) be an N-pair and v ∈ T. Then there exists S, R ⊆ S ⊆ T,
such that (S,Γ) is an A-extension of (R,Γ) and there exists c ∈ S with v −c ∈ m2.
Proof. Let Λ = Γ∪{((x,y)T,∅)} and let I = m2. Now apply Lemma 2.4 with H = ∅
to find an N-subring S which contains c = v + d and the result is immediate.
?
Lemma 2.12. Let (R,Γ) be an N-pair, c ∈ R, I a finitely generated ideal of R and
c ∈ IT. Further suppose that I ? Pλ for all (Pλ,Vλ) ∈ Γ. Then there exists S,
R ⊆ S ⊆ T, such that (S,Γ) is an A-extension of (R,Γ) and c ∈ IS.
Proof. Let Λ = Γ ∪ {((x,y)T,∅)}. If I is contained in a height one prime ideal
of T, because R is a UFD, I = bJ with J not contained in a height one prime
ideal of T. It suffices to prove the lemma with J in place of I and b−1c in place
of c, and so we may assume I is not contained in a height one prime ideal of T.
Choose any nonzero y1∈ I. As y1 / ∈ (x,y)T, y1∈ Pλif and only if Pλis minimal
over (x,y,y1)T. Hence y1∈ Pλfor at most finitely many λ. By the usual prime
avoidance lemma, we can find y2 ∈ I such that (y1,y2)R is contained in no Pλ
and in no height one prime ideal of T. We then choose y3,...,yn ∈ I so that
I = (y1,...,yn)R. As c ∈ IT, we may write c =?yitiwith each ti∈ T. We now
apply Lemma 2.4 with v = t3, H = ∅, and (y1,y2)R playing the role of I. Since
d ∈ (y1,y2)T, this allows us to find a new expression for c as a linear combination
of y1,...,ynwith t3replaced by t3+ d and with suitably altered values of t1and
t2. Hence, by enlarging R, we reduce to the case t3∈ R. We proceed similarly with
t4,...,tn. So c − t3y3− ··· − tnyn∈ (y1,y2)T and so it will suffice to prove the
lemma in the case I = (y1,y2)R.
The proof in this case closely resembles the proof of Lemma 2.4. Let ∆ = {Pλ|λ ∈
Λ}, G = {P ∈ Ass(T/rT)|r ∈ R}, and C = ∆ ∪ G. For each P ∈ C, we define a
ring A(P). If P / ∈ ∆, A(P) = R/P ∩ R. If P = Pλ∈ ∆, A(P) = (R/Pλ∩ R)[Vλ].
In each case, A(P) has cardinality at most |R| and so the same is true for the
algebraic closure of A(P) in T/P. If y1 / ∈ P, let DP be a complete set of coset
representatives modulo P of {t ∈ T|t2+ ty1is algebraic over A(P)}. If y1 ∈ P,
then y2/ ∈ P and we let DP be a complete set of coset representatives modulo P of
{t ∈ T|t1−ty2is algebraic over A(P)}. Therefore |DP| ≤ |R|. Now |C| ≤ |R|, and
so also if D =?
we may apply Lemma 2.1 to choose t ∈ T so that either the image of t2+ ty1in
T/P or the image of t1− ty2in T/P is transcendental over A(P) for every P ∈ C.
Let S be the intersection of T with the quotient field of R[t2+ ty1], which is also
the intersection of T with the quotient field of R[t1−ty2] as c = t1y1+t2y2. Clearly
c = y1(t1− ty2) + y2(t2+ ty1) ∈ IS. Verifying that (S,Γ) is an N-pair follows the
same steps performed in the proof of Lemma 2.4 and of course |S| = |R|.
P∈CDP, then |D| ≤ |R|, and finally |C × D| ≤ |R| < |R|. Hence
?
Lemma 2.13. Let R be an N-subring of T with quotient field K. Suppose P ∈
SpecT and Q = P ∩ R. Then RQ= TP∩ K.
Proof. Certainly, as R − Q ⊆ T − P, we have RQ ⊆ TP ∩ K.
containment fails, there exist a,b ∈ R with a/b ∈ TP−RQ. As R is a UFD, we may
express a and b as products of prime elements and, canceling if necessary, we may
assume a,b have no common factors. Further, we may delete any prime elements
not contained in Q as they will not affect membership in either RQor TP. Finally,
let p be any factor of b — one such must exist. As p ∈ Q, p ∈ P, and so there
If the reverse
Page 16
16RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
exists a height one prime ideal P1 of T which is contained in P and contains p.
Now a/b ∈ TP implies a/p ∈ TP ⊆ TP1which implies a ∈ P1∩ R = pR. But p,a
are relatively prime elements; this contradiction completes the proof.
?
Lemma 2.14. Let (R0,Γ0) be an N-pair. Let Ω be a well-ordered set with least
element 0 and no maximal element. Suppose that for every α ∈ Ω, |{β ∈ Ω|β <
α}| < |R|. Let γ(α) = sup{β ∈ Ω|β < α}. Suppose {(Rα,Γα)|α ∈ Ω} is an
ascending collection of pairs such that if γ(α) = α, then Rα=?
?
(S,Γ) = (?Rα,?Γα) satisfies all conditions to be an N-pair except the cardinality
condition. In fact, |S| ≤ sup(|R0|,|Ω|) and so (S,Γ) is an N-pair if |Ω| < |R|.
β<αRβ and Γα=
β<αΓβ, while if γ(α) < α, (Rα,Γα) is an A-extension of (Rγ(α),Γγ(α)). Then
Proof. We replace Ω by Ω′= Ω ∪ {ζ} with ζ > α for all α ∈ Ω. Let (Rζ,Γζ) =
(S,Γ). The lemma then follows if we can show that for each α ∈ Ω′: (Rα,Γα)
satisfies all conditions to be an N-pair with the cardinality condition replaced by
|Rα| ≤ sup(|R0|,|{β ∈ Ω|β < α}|). We do this by transfinite induction, the case
α = 0 being trivial.
First note that since our proof will show that we have N-pairs all along the
way, A-extensions are permissible. Now assume the induction hypothesis holds for
all β < α. If γ(α) ?= α, it holds immediately for (Rα,Γα) by the definition of A-
extension. Therefore assume γ(α) = α. Rαis an N-subring of T by [H1, Lemma 6]).
(Alternatively, the reader may check the straightforward details directly.) Certainly
Γαhas the form {(Pλ,Vλ)|λ ∈ Γ} and so it only remains to check the six conditions.
Conditions (1)-(4) are trivial. For condition (5), note that if the images of the
elements of Vλare not algebraically independent over Rα/Pλ∩Rα, there is a relation
which necessarily involves only finitely many elements of Rα. Hence there exists
β < α such that each of these elements is contained in Rβand (Pλ,Vλ) ∈ Γβ. This
would contradict the fact that the images of the elements of Vλ are algebraically
independent over Rβ/Pλ∩ Rβ and so condition (5) holds. Likewise, for condition
(6), if a ∈ Qλ(Rα)Qλwe can find β < α such that a is in the quotient field of Rβ
and (Pλ,Vλ) ∈ Γβ. Let Qλβ= Pλ∩ Rβ. By lemma 2.13, (Rα)Qλand (Rβ)Qλβare
the intersection of TPλwith their respective quotient fields and so a ∈ Qλβ(Rβ)Qλβ.
Thus we have a ∈ (x+αλtλ,y+βλtλ,tλ)(Rβ)Qλβ⊂ (x+αλtλ,y+βλtλ,tλ)(Rα)Qλ.
Therefore condition (6) holds as well.
?
Lemma 2.15. Let R be an N-subring of T. If ω,h ∈ T are such that ω + zh
is transcendental over R as an element of T/(x,y)T, then there exists a subset
Ψ ⊂ R such that |Ψ| ≤ |R| and whenever u ∈ R−Ψ, ω+zh is transcendental over
R/(x,y,z + uw)T ∩ R as an element of T/(x,y,z + uw)T.
Proof. We choose Ψ to be the set of real numbers that make the element algebraic
and all that remains is to bound the cardinality of Ψ. Now u ∈ Ψ if rn(ω +zh)n+
···+r0∈ (x,y,z +uw)T for some rn,...,r0∈ R, not all of which are contained in
(x,y,z +uw)T. Since the number of finite ordered subsets of R is equal to |R|, we
can restrict our attention to a single ordered subset, that is, we can fix rn,...,r0and
ask when δ = rn(ω +zh)n+···+r0∈ (x,y,z +uw)T. As ω +zh is transcendental
over R as an element of T/(x,y)T, δ / ∈ (x,y)T. Therefore (x,y,z + uw)T must
be minimal over (x,y,δ)T, something that can be true for at most finitely many
values of u. This completes the proof.
?
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ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?17
Lemma 2.16. Let (R,Γ) be an N-pair.
transcendental over R as an element of T/(x,y)T, then there exists u ∈ R and a
ring S = R[u]m∩R[u], such that (x,y,z + uw)T ∩ R = (0), (S,Γ) is an A-extension
of (R,Γ), and ω +zh is transcendental over R as an element of T/(x,y,z +uw)T.
If ω,h ∈ T are such that ω + zh is
Proof. First we use Lemma 2.15 to find a set Ψ1so that u / ∈ Ψ1will give ω +zh is
transcendental over R/(x,y,z +uw)T ∩R as an element of T/(x,y,z +uw)T. Let
Ψ2= {u ∈ R|(x,y,z + uw)T ∩ R ?= (0)}. As (x,y)T ∩ R = (0), no element of R
can be contained in infinitely many prime ideals of the form (x,y,z +uw)T and so
|Ψ2| ≤ |R|. Let Λ = Γ ∪ {((x,y)T,∅)}. Now we use Lemma 2.4 with v = 0, I = R,
and H = ∅ to find the desired S. Since t = 1, we can choose u = d ∈ R−(Ψ1∪Ψ2).
By Lemma 2.7, S has the described form.
?
Lemma 2.17. Let t = z + uw for some u ∈ T and let P = (x,y,t)T = (r,s,t)T
for some r,s ∈ T. Suppose F is a subfield of the quotient field of T, ξ ∈ T, and
E is the intersection of TP with the quotient field of F[r,s,t,ξ]. Further assume
s2∈ F[r,t,ξ]+sF[r,t,ξ], r2+s2∈ tT, and E = LP∩Lwhere L = F[r,t,ξ,s,t−1]∩
T[F]. If there exists σ1,σ2,σ3∈ T with σ1,σ2algebraically independent over F as
elements of T/P such that ξ = σ1r + σ2s + σ3t, then E = F[r,s,t,ξ](r,s,t,ξ).
Proof. We first show that F[r,s,t,ξ] ∩ tT[F] = tF[r,s,t,ξ]. As F[r,s,t,ξ] ⊆ T[F],
one containment is clear. For the reverse, let g ∈ F[r,s,t,ξ] ∩ tT[F]. As s2∈
F[r,t,ξ] + sF[r,t,ξ], we may assume g ∈ F[r,t,ξ] + sF[r,t,ξ] and since the terms
involving t are certainly in tF[r,s,t,ξ], we may assume g ∈ F[r,ξ]+sF[r,ξ]. In this
case, we will show that g = 0. If g ?= 0, we can write g = gn+gn+1+···+gmas a sum
of homogeneous polynomials in r,s,ξ with gnthe nonzero homogeneous summand
of lowest degree (n). We have gn= −(gn+1+ ··· + gm) + g ∈ Pn+1T[F] + tT[F]
and we will see by induction that no nonzero homogeneous polynomial of degree
n is in Pn+1T[F] + tT[F]. This is obvious for n = 0 as in that case gn∈ F. For
n > 0, we can write gn(r,s,ξ) = f1rn+ f2srn−1+ ξh(r,s,ξ) where f1,f2 ∈ F
and h is a homogeneous polynomial of degree n − 1. As ξ − (σ1r + σ2s) ∈ tT.
we can replace ξ by σ1r + σ2s, giving gn(r,s,σ1r + σ2s) = f1rn+ f2srn−1+
(σ1r + σ2s)h(r,s,σ1r + σ2s) ∈ Pn+1T[F]+ tT[F]. Using the relation r2+ s2∈ tT,
gn≡ e1rn+ e2rn−1s modulo tT and necessarily e1,e2∈ PT[F]. However, e1,e2
are polynomials over F in σ1,σ2 with respective constant terms f1,f2. As σ1,σ2
are algebraically independent over F, e1,e2 must both be identically zero and so
f1 = f2 = 0. Therefore gn = hξ.Next we take advantage of the fact that,
modulo tT, ξ(σ1r − σ2s) ≡ σ2
gn(σ1r − σ2s) ∈ (Pn+1+ tT)(r,s)T[F] ⊂ (rn+2,rn+1s,t)T[F]. So hξ(σ1r − σ2s) ≡
h(σ2
we have h ∈ ((rn+2,rn+1s,t)T :T[F]r2) = (rn,rn−1s,t)T[F] and by the induction
assumption, h = 0. This contradiction shows that there can be no nonzero gnand
so F[r,s,t,ξ] ∩ tT[F] = tF[r,s,t,ξ].
It follows that L = F[r,s,t,ξ]. The result is now clear.
1r2− σ2
2s2≡ (σ2
1+ σ2
2)r2and σ2
1+ σ2
2/ ∈ P. Now
1+σ2
2)r2∈ (rn+2,rn+1s,t)T[F]. Finally, as (rn+2,rn+1s,t)T is a primary ideal,
?
Lemma 2.18. Let t = z + uw for some u ∈ T and let P = (x,y,t)T Let F
be a subfield of the quotient field of T and suppose for r,s,ξ1 ∈ T that D =
F[r,s,t,ξ1](r,s,t,ξ1)is the intersection of TP with the quotient field of F[r,s,t,ξ1].
Assume P = (r,s,t)T and r2+ s2∈ tT. Further assume
Page 18
18RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
(1) s2∈ F[r,t] + ξ1F[r,t] + sF[r,t].
(2) δ1,δ2,τ,σ1,σ2 ∈ T are algebraically independent over F as elements of
T/P.
(3) ξ1= δ1r + δ2s + δ3t with δ3∈ T.
(4) δ3= f(δ1,δ2) where f(X1,X2) ∈ (F[t] ∩ T)[X1,X2] is a polynomial in two
indeterminates.
Let d1= δ1+ σ1t, d2= δ2+ σ2t, and d3= δ3− σ1r − σ2s. Let ξ2= f(d1,d2) −
d3. Then if E denotes the intersection of TP with the quotient field of D[d1,d2]
and E = LP∩L where L = F[r,t,ξ1,s,d1,d2,t−1] ∩ T[F[d1,d2]], we have E =
F(d1,d2)[r,t,ξ2,s](r,t,ξ2,s). Moreover we have
(1) s2∈ F(d1,d2)[r,t] + ξ2F(d1,d2)[r,t] + sF(d1,d2)[r,t].
(2) σ1,σ2,τ ∈ T are algebraically independent over F(d1,d2) as elements of
T/P.
(3) ξ2= σ1r + σ2s + σ3t with σ3∈ T.
(4) σ3= g(σ1,σ2) where g(X1,X2) ∈ (F(d1,d2)[t]∩T)[X1,X2] is a polynomial
in two indeterminates.
Proof. As δ1,δ2are algebraically independent over F as elements of T/P and we
have di≡ δimodulo P for each i, it follows that d1,d2are algebraically independent
over F as elements of T/P and so F(d1,d2) ⊂ E. As ξ1= d1r + d2s + d3t, d3is
in the quotient field of E, and so d3∈ E. It follows that ξ2∈ E. By taking the
Taylor expansion of f(d1,d2) about (δ1,δ2), we see that ξ2 = −d3+ f(d1,d2) =
−δ3+ σ1r + σ2s + (f(δ1,δ2) + tσ3) where σ3∈ (F[t] ∩ T)[σ1,σ2,δ1,δ2] =
(F[t]∩T)[σ1,σ2,d1,d2] ⊂ (F(d1,d2)[t]∩T)[σ1,σ2]. Conclusions (2),(3),(4) are now
clear. Also, as ξ1= d1r+d2s+d3t ∈ F[d1,r]+sF[d2]+ξ2F[t]+F[d1,d2,t], we get
s2∈ F[r,t]+ξ1F[r,t]+sF[r,t] ⊂ F(d1,d2)[r,t] +ξ2F(d1,d2)[r,t]+sF(d1,d2)[r,t].
So conclusion (1) also holds.Finally, as d3 = f(d1,d2) − ξ2 and ξ1 = d1r +
d2s + d3t, we observe that E, being a localization of L, is also a localization of
F(d1,d2)[r,s,t,ξ2,t−1] ∩ T[F(d1,d2)]. We now apply Lemma 2.17 to complete the
proof.
?
Lemma 2.19. Let (R,Γ) be an N-pair. Suppose ((x,y,z)T,{ω}) ∈ Γ and also
suppose h ∈ T such that ω+zh is transcendental over R as an element of T/(x,y)T.
Then there exist elements u,α,β,µ,ν ∈ R and r,s,t,A,B,C,q ∈ T, and a ring S,
R ⊆ S ⊆ T, such that the following nine conditions hold.
(1) t = z + uw
(2) r = x + αt
(3) s = y + βt
(4) A = tµ + rq − 2α(1 + tq)
(5) B = tν + sq − 2β(1 + tq)
(6) C = −rµ − sν + (α2+ β2)(1 + tq)
(7) R[u,r,s,t,A,B,C] ⊂ S
(8) q = 0
(9) (S,Γ ∪ ((x,y,t)T,{ω + zh}) is an A-extension of (R,Γ)
Suppose instead that R = Q[z,w](z,w)and Γ = ∅. Here we may set u = 0 and find
such a ring S and an element ω ∈ R satisfying conditions (1)-(7) with the last two
conditions replaced by
(8) q = ω
Page 19
ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?19
(9) (S,((x,y,z)T,{ω})) is an A-extension of (R,Γ)
Proof. To prove this result, we must choose elements α,β,µ,ν ∈ R, as well as
either u or ω, depending on the form of R. Then we define r,s,t,A,B,C,q so that
Assertions (1-6) and Assertion (8) are satisfied. Satisfying Assertion (7) is easy and
so the only real obstacle is Assertion (9). The basic plan is to obtain S =?Ri
with (Ri+1,Γ) an A-extension of (Ri,Γ) for every i. Then, by Lemma 2.14, (S,Γ)
is an A-extension of (R,Γ) and it only remains to adjoin one more element to Γ.
Throughout the proof, P will denote the prime ideal (x,y,t)T and Qiwill denote
P ∩ Ri. To verify that (P,{ω + zh}) satisfies the necessary properties to allow it
to be adjoined to Γ, we really do not need to understand S, only the simpler ring
SP∩S.
We will deal with both cases simultaneously although we must note a few differ-
ences along the way. In the u = 0 case, as |R/(x,y,z)T ∩ R| < |T/(x,y,z)T|, we
may choose ω ∈ R such that ω is transcendental over R/(x,y,z)T∩R as an element
of T/(x,y,z)T. Then, by setting h = 0 in the u = 0 case, the two assertion (9)’s
become identical. For the transcendental case, we begin by applying Lemma 2.16
to get the appropriate element u and an extension R1of R satisfying all conclusions
of that lemma; in particular, u ∈ R1and (R1,Γ) is an N-pair. For the u = 0 case,
we simply let R1= R. Either way, with t = z +uw and P = (x,y,t)T, we see that
Q1= P ∩R1= tR1. Moreover, if F is the quotient field of R in the transcendental
case or Q(w) in the u = 0 case, (R1)Q1= F[t](t). We clearly have ω + zh tran-
scendental over R1/Q1as an element of T/P. Next we claim that tT ? Pλfor all
λ ∈ Γ. Certainly the only possible Pλ which can contain it is (x,y,t)T. In both
cases, there is no such Pλby hypothesis. Hereafter, except for the construction of
R4, the two cases will be handled identically.
Now set Λ = Γ ∪ {((x,y)T,∅)} and ∆ = {Pλ|λ ∈ Λ}. As tT is a principal prime
ideal of T, we can set H = {tT} and then tT ? P′for every P′∈ ∆ ∪ (G − H). In
fact, we will use H = {tT} and this choice of Λ in every application of Lemma 2.4
contained in this proof. Here we apply Lemma 2.4 with v = x and I = tT to find
d = αt so that if R2is the intersection of the quotient field of R1[x + αt] with T,
(R2,Γ) is an N-pair. Further, we may choose α ∈ R so that α is transcendental
over (R1/Q1)[ω + zh] as an element of T/P. Similarly we apply Lemma 2.4 to R2
with v = y and I = tT to find d = βt so that if R3is the intersection of the quotient
field of R2[y +βt] with T, then (R3,Γ) is an N-pair. We choose β ∈ R so that β is
transcendental over (R2/Q2)[ω + zh,α] as an element of T/P and transcendental
over (R2/tR2)[α] as an element of T/tT. Therefore we have u,r,s,t ∈ R3.
We have not yet chosen µ,ν ∈ R, but it is useful at this time to perform a
calculation which will be valid for any choice of µ,ν. We claim that r2+s2+Art+
Brt + Ct2= 0. The verification of the claim is a straightforward calculation. For
future reference, we will actually perform the calculation in R[[x,y,z,w]]. This
makes sense as all of the elements involved are contained in R + xR + yR+ zR +
wR ⊂ T and so have natural liftings to R[[x,y,z,w]]. After noting r2= x2+2αrt−
Page 20
20RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
α2t2and s2= y2+ 2βst − β2t2, we have
r2+ s2+ Art + Bst + Ct2
= (x2+ y2) + rt(A + 2α) + st(B + 2β) + t2(C − (α2+ β2))
= (x2+ y2) + rt(tµ + rq − 2tqα) + st(tν + sq − 2tqβ) + t2(−rµ − sν + tq(α2+ β2))
= (x2+ y2) + rt(rq − 2tqα) + st(sq − 2tqβ) + t2(tq(α2+ β2))
= (x2+ y2) + tq(r2− 2rtα + t2α2+ s2− 2stβ + t2β2)
= (x2+ y2)(1 + tq)
This element is of course zero in T since x2+ y2= 0. For now, what this means
is that r2+ s2∈ tT and so Y3 = −(r2+ s2)/t ∈ R3. We will next establish the
properties of R3that we need.
As observed above, Q1= tR1 and ω + zh is transcendental over R1/Q1as an
element of T/P. Next we claim that r is transcendental over R1/tR1as an element
of T/tT. If not, x is algebraic over R1/tR1 as an element of T/tT. So we have
rnxn+ ··· + r0∈ tT with rn,...,r0∈ R1, rn / ∈ tR1, and n minimal. As x ∈ P,
r0∈ Q1= tR1. Thus the relation continues to hold if we replace r0by 0. However,
now we can divide by x and reduce the degree of the polynomial, contradicting the
minimality of n and so proving the claim. It follows from this and Lemma 2.7 that
R2= R1[r]m∩R1[r]and so we have (R2)Q2= F[r,t](r,t).
We claim that R3 is just a localization of R2[Y3,s]. By Lemma 2.7, R3 is a
localization of a subring of R2[s,t−1] and so of course also a localization of a subring
of R2[Y3,s,t−1].Thus, we need only show R2[Y3,s] ∩ tT = tR2[Y3,s].
(4),(5),(6), which are valid equations in T for any choice of µ,ν, we see that Y3=
Ar+Bs+Ct = (r2+s2)q+(1+tq)(−2αr−2βs+(α2+β2)t) and so, as r2+s2= −tY3,
we have Y3= (1+tq)−1(1+tq)(−2αr−2βs+(α2+β2)t) = −2αr−2βs+(α2+β2)t.
We next show that Y3 is transcendental over R2/tR2 as an element of T/tT. It
suffices to show this with αr + βs in place of Y3. Since r ∈ R2 and s2≡ −r2
modulo tT, s is a nonzero element algebraic over R2/tR2. Hence, if Y3is algebraic
over R2/tR2and so also algebraic over R2/tR2[α], β is algebraic over R2/tR2[α],
contradicting our choice of β. Of course, Y3is also transcendental over R2/tR2[s]
as an element of T/tT. Now suppose g ∈ R2[Y3,s] ∩ tT. As s2∈ R2[Y3], we may
write g = (aks + bk)Yk
As Y3 is transcendental over R2/tR2[s], each ai+ bis ∈ tT. Now we will have
R2[Y3,s]∩tT = tR2[Y3,s] if we can show ai,bi∈ tR2= tT ∩R2. If not, either aior
bior both is not in (rm,t)T for sufficiently large m and so is also not in (rm,t)R2
for sufficiently large m. Choosing m maximal so that ai,bi∈ (rm,t)R2, we may
write ai = cirm+ ? cit and bi = dirm+?dit. As (tT :T rm) = tT, ci+ dis ∈ tT
and ci,diare not both contained in (r,t)R2= Q2. But ci∈ (s,t)T ∩ R2⊂ Q2=
(r,t)R2 and this forces di ∈ ((r,t) :R2s) ⊆ Q2. This contradiction proves each
ai,bi ∈ tR2 and completes the proof of the claim that R3 is just a localization
of R2[Y3,s]. This means then that R3 is just a localization of R1[r,s,Y3] and so
(R3)Q3= F[r,s,t,Y3](r,s,t,Y3).
Next, let A3= rq−2α(1+tq), B3= sq−2β(1+tq), and C3= (α2+β2)(1+tq).
Since we chose α,β so that α,β,ω + zh are algebraically independent over F as
elements of T/P, clearly A3,B3,ω + zh are algebraically independent over F as
Using
3+ ··· + (a0s + b0) for some integer k and each ai,bi∈ R2.
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ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?21
elements of T/P. In the q = 0 case, we also have C3= (1/4)(A2
n = 3 and q = 0, with F3= F, we have the following:
3+ B2
3). Thus, for
(1) (Rn,Γ) is an N-pair
(2) (Rn)Qn= Fn[r,s,t,Yn](r,s,t,Yn)
(3) Yn= Anr + Bns + Cnt with An,Bn,Cn∈ T such that An,Bn,ω + zh are
algebraically independent over Fnas elements of T/P
(4) s2∈ Fn[r,t] + YnFn[r,t] + sFn[r,t]
(5) Cn∈ (Fn[t] ∩ T)[An,Bn]
In the q = ω case, (1)-(4) are satisfied.
We will construct an ascending union of N-subrings R3⊂ R4⊂ R5⊂ ···, each
of the same cardinality, such that Rn satisfies conditions (1)-(5) for every n ≥ 4
along with the additional condition that Yn−1∈ (r,s,t)Rn. Note that Lemma 2.13
tells us that (Rn)Qnis the intersection of TP with the quotient field of Rn, a fact
which allows us to freely use Lemmas 2.17 and 2.18 in our construction process.
We now describe the construction of Rn. We wish to adjoin elements? An =
An−1+ tσ1nand? Bn= Bn−1+ tσ2n. To do this, we may apply Lemma 2.4 twice
with I = tT and v = An−1,Bn−1respectively to get an extension Rnwith (Rn,Γ)
an A-extension of (Rn−1,Γ) and? An,?Bn∈ Rn. In defining these extensions, we can
and do impose additional conditions on σ1nand σ2n. We choose σ1n∈ R to be tran-
scendental over (Rn−1/Qn−1)[ω +zh,An−1,Bn−1] as an element of T/P. Likewise
we choose σ2n∈ R to be transcendental over (Rn−1/Qn−1)[ω+zh,An−1,Bn−1,σ1n]
as an element of T/P. Thus ω + zh,An−1,Bn−1,σ1n,σ2nis a set of algebraically
independent elements over Rn−1/Qn−1. Since An−1,Bn−1are algebraically inde-
pendent over Fn−1as elements of T/P, Fn= Fn−1(? An,?Bn) is a field contained in
(Rn)Qn. We also note that if?Cn= Cn−1−rσ1n−sσ2n, Yn−1=? Anr+?Bns+?Cnt and
so?Cnis in the intersection of T with the quotient field of Rn, giving?Cn∈ Rnand
thus Yn−1∈ (r,s,t)Rn. By Lemma 2.7, Rnis a localization of Rn−1[? An,?Bn,t−1]∩T.
So (Rn)Qnis a localization of Fn−1[r,s,t,Yn−1,? An,?Bn,t−1] ∩ T. If Rn−1satisfies
conditions (1)-(5), we get the full hypothesis of Lemma 2.18 when we consider
the situation D = (Rn−1)Qn−1, τ = ω + zh, ξ1 = Yn−1, and δ1,δ2,δ3 equal to
An−1,Bn−1,Cn−1 respectively. It immediately follows that Rn also satisfies con-
ditions (1)-(5). Of course, Ynis the element ξ2specified in the conclusion of that
lemma and An,Bn,Cnequal σ1,σ2,σ3, respectively.
In the case n = 4,q = ω, we must check conditions (2)-(5) directly. We define
Y4=?C4− (1/4)(? A2
B2
4+?B2
4), A4= qα − σ14, B4= qβ − σ24, and C4= (t/4)(A2
4) + (1/2)(? A4A4+?B4B4). Since? A4,?B4∈ F4, condition (5) is immediate.
4+
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22RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
To see condition (3), we simply calculate, first noting that? A4+ 2A4t = rq −
2α − tσ14and?B4+ 2B4t = sq − 2β − tσ24.
Y4− A4r − B4s − C4t
=?C4− (1/4)(? A2
=?C4− (1/4)(? A4(? A4+ 2A4t) +? B4(?B4+ 2B4t)) − A4r − B4s − (t2/4)(A2
= (α2+ β2)(1 + tq) − rσ14− sσ24− (1/4)(rq − 2α(1 + tq) + tσ14)(rq − 2α − tσ14)
− (1/4)(sq − 2β(1 + tq) + tσ24)(sq − 2β − tσ24) − r(qα − σ14) − s(qβ − σ24)
− (t2/4)(σ2
= −(1/4)(r2q2− 4αrq − t2σ2
14− 2αtq(rq − tσ14)
+ s2q2− 4βsq − t2σ2
24− 2βtq(sq − tσ24))
− rqα − sqβ − (t2/4)(σ2
= −(1/4)(r2q2− 2αtq(rq) + s2q2− 2βtq(sq)) − (t2/4)(q2(α2+ β2))
= −(q2/4)((r2+ s2) − 2t(αr + βs) + t2(α2+ β2))
= −(q2/4)(r2+ s2+ tY3) = 0
4+?B2
4) − A4r − B4s − (t2/4)(A2
4+ B2
4) − (1/2)(? A4A4+? B4B4)t
4+ B2
4)
14+ σ2
24− 2q(σ14α + σ24β) + q2(α2+ β2))
14+ σ2
24− 2q(σ14α + σ24β) + q2(α2+ β2))
The rest of condition (3) follows immediately from the construction.
For condition (4), we first note that?C4 = Y4+ (1/4)(? A2
Thus Y3 = ? A4r +?B4s +?C4t ∈ rF4+ sF4+ tF4+ tY4.
tY3 ∈ F4[r,t] + Y4F4[r,t] + sF4[r,t]. We also note Y3 ∈ F4[r,t,Y4,s]. Finally,
by Lemma 2.7, R4is a localization of R3[? A4,?B4,t−1] ∩ T, so (R4)Q4is a localiza-
tion of F3[r,s,t,Y3,? A4,?B4,t−1] ∩ T, which forces it to also be a localization of the
larger ring F4[r,s,t,Y4,t−1] ∩ T[F4]. Now the entire hypothesis of Lemma 2.17 is
satisfied and we get condition (2).
Let S =?Rn; by Lemma 2.14, (S,Γ) is an A-extension of (R,Γ). Note that,
referring to the construction of R4, if we let µ = σ14, ν = σ24, A =? A4, B =
?B4, and C =?C4, we have the desired A,B,C ∈ S. Finally, to see that (S,Γ ∪
{((x,y,t)T,{ω + zh})}) is an N-pair, we let Q = P ∩ S and we need only show
QSQ= (r,s,t)SQand ω+zh is transcendental over R/Q as an element of T/P. As
QSQ= (r,s,t,Y3,Y4,···)SQand each Yn∈ (r,s,t)S, the first is true. As ω +zh is
transcendental over Rn/Qnas an element of T/P for every n, the second is true as
well.
4+?B2
Hence s2= −r2−
4) ∈ Y4+ F4.
?
Lemma 2.20. Let (S,{((x,y,z)T,{ω})}) be an N-pair constructed as in the proof
of Lemma 2.19 and let m be any positive integer. Then there exist dm,em∈ S such
that (x,y,zm)T = (dm,em,zm)T.
Proof. For any m > 0, we will find dm,em∈ S such that x(1 + zω/2) − dm and
y(1 + zω/2)− emare in zmT. That forces dm,em∈ (x,y,zm)T ∩ S and clearly, as
1+zω/2 is a unit, (dm,em,zm)T = (x,y,zm)T. We will show that we can find dm;
the other case is identical.
To find dm, we first observe x = r − zα = r − (z/2)(−A + zµ + rω − 2αzω) =
r −(z/2)(−A+zµ+xω − αzω) and so x(1 +zω/2) = r −(z/2)(−A+z(µ −αω)).
Next recall from the construction that µ = σ14= αω − A4and so x(1 + zω/2) =
Page 23
ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?23
r−(z/2)(−A+z(−A4)) = r+(z/2)(A+zA4). Now, for n ≥ 5, An−1+zσ1n=? An∈ S
and An= σ1n. So A4=? A5−zA5=? A5−z? A6+z2A6and so on. An easy induction
shows A4∈ S+zmT for any m. So x(1+zω/2) = (r+zA/2)+(z2/2)A4∈ S+zmT
for any m.
?
Lemma 2.21. Let (R,Γ) be an N-pair with ((x,y,z)T,{ω}) ∈ Γ, and h,v ∈ T. If
ω + zh is transcendental over R as an element of T/(x,y)T, then there exists an
A-extension (S,Γ′), where Γ′= Γ ∪ {(P,{ω + zh})} for some P ∈ SpecT, and an
element c ∈ S such that
(1) v − c ∈ m2
(2) For every finitely generated ideal I of R such that I ? Pλ for all λ ∈ Γ′,
we have IT ∩ R ⊂ IS.
(3) If P′is a height two prime ideal of T which is minimal over IT for some
finitely generated ideal I of R and P′?= Pλfor all λ ∈ Γ′, then P′∩S ? Pλ
for all λ ∈ Γ′.
If ω+zh is algebraic over R as an element of T/(x,y)T (a case we will later see
cannot occur), there exists an A-extension (S,Γ) and an element c ∈ S such that
conditions (1)-(3) above hold.
Proof. In the transcendental case, we first employ Lemma 2.19 to obtain an A-
extension (R0,Γ′) of (R,Γ) where Γ′= Γ ∪ {(P,{ω + zh})} for some P ∈ SpecT
of the proper form. In the algebraic case, we simply let (R0,Γ′) = (R,Γ) and for
the remainder of the proof, the two cases are identical. Next employ Lemma 2.11
to obtain an A-extension (R1,Γ′) of (R0,Γ′) with c ∈ R1such that v −c ∈ m2. Set
Ω1= {(I,d)|I finitely generated ideal of R such that I ? Pλfor all λ ∈ Γ′and
d ∈ IT ∩ R}. Set Ω2= {P′∈ SpecT|HeightP′= 2,P′?= Pλfor all λ ∈ Γ′, and
P′minimal over IT for some finitely generated ideal I of R}. Set Ω = Ω1∪ Ω2.
Clearly |Ω| = |R|. Well-order Ω, letting 1 denote its initial element, in such
a way that Ω does not have a maximal element; then it satisfies the hypothesis
of Lemma 2.14. Next we recursively define a family {Rα|α ∈ Ω} to satisfy the
hypothesis of Lemma 2.14. We begin with R1. If γ(α) ?= α and γ(α) = (I,d),
then we choose Rαto be an A-extension of Rγ(α)given by Lemma 2.12 such that
d ∈ IRα. If γ(α) ?= α and γ(α) = P′, then we choose Rαto be an A-extension of
Rγ(α)given by Lemma 2.10 such that P′∩ R ? Pλ for all λ ∈ Γ′. If γ(α) = α,
choose Rα =
β<αRβ. Set S =
that |Ω| = |R|, we see that (S,Γ′) is an A-extension of (R1,Γ′). Also, if I is any
finitely generated ideal of R such that I ? Pλfor all λ ∈ Γ′and d ∈ IT ∩ R, then
(I,d) = γ(α) for some α ∈ Ω. So d ∈ IRα⊂ IS. Thus IT ∩ R ⊂ IS. Likewise, if
P′?= Pλis a height two prime ideal of T which is minimal over a finitely generated
ideal of R, P′∩ S is not contained in any Pλ.
??Rα. By Lemma 2.14 and the observation
?
Lemma 2.22. Suppose (R,Γ) is an ascending union of N-pairs. Assume that the
natural map R −→ T/m2is surjective and that for every finitely generated ideal I
of R such that I ? Pλ for all λ ∈ Γ, IT ∩ R = I. Further assume that if P′is a
height two prime ideal of T which is minimal over IT for some finitely generated
ideal I of R and P′?= Pλ for all λ ∈ Γ, P′∩ R ? Pλ for all λ ∈ Γ. Then R is
Noetherian and the natural homomorphism?R −→ T is an isomorphism.
Page 24
24RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
Proof. By Proposition 1 of [H2], it suffices to prove IT ∩ R = I for every finitely
generated ideal I of R in order to get our conclusion. Suppose the statement is
false; we may choose a finitely generated ideal J such that JT ∩ R ?= J and JT
is maximal among such ideals. By our hypothesis, JT ⊆ Pλ for some λ. As no
(m ∩ R)-primary ideal is contained in any Pλ, we have IT ∩ R = I whenever I is
primary to the maximal ideal. Thus we may apply Lemma 21 of [H2] to see that
JT ∩R is infinitely generated. Also, if I is an ideal of R which is infinitely generated
and properly contains JT ∩ R, there is necessarily a finitely generated ideal I0⊂ I
such that I ⊂ I0T. Then I0T ∩R ?= I0, contradicting the maximality of JT. Hence
JT ∩ R is maximal among infinitely generated ideals of R. Cohen’s theorem tells
us that JT ∩ R is a prime ideal Q. It is easy to see that it is the contraction of
one or more of the associated prime ideals of JT. As the contraction of height one
prime ideals are principal, it is the contraction of a height two prime ideal which
is minimal over JT. By the hypothesis, if P′?= Pλfor all λ, P′∩ R cannot be a
height two ideal contained in some Pλ. However, necessarily J, and so also JT, is
contained in some Pλ. So Q = Pλ0∩ R for some λ0∈ Γ.
For notational ease, let P = Pλ0and Q = P ∩ R. Using the notation from the
statement of Lemma 2.19, Q is locally generated by r,s,t and QT is generated
by r,s,t. We will show that Q = (r,s,t)R and so contradict the notion that Q
is infinitely generated, thus completing the proof. Suppose h ∈ Q; we must show
h ∈ (r,s,t)R. There exists d ∈ R−Q,e1,e2,e3∈ R such that dh = e1r+e2s+e3t.
We also have σ1,σ2,σ3 ∈ T such that h = σ1r + σ2s + σ3t. Thus (e1− dσ1)r +
(e2− dσ2)s + (e3− dσ3)t = 0. It follows that e2− dσ2∈ ((r,t)T :T s) = P and so
e2∈ P +dT = (r,s,t,d)T. As (r,s,t,d)R is (m∩R)-primary, e2∈ (r,s,t,d)T ∩R =
(r,s,t,d)R. We can write e2= ρ′
setting ρ1= e1+ sρ′
ρ1,ρ2,ρ3,ρ4∈ R. Now it suffices to show h−ρ3s ∈ (r,s,t)R and so we can reduce to
the case ρ3= 0. Then, as s2∈ (r,t)R, we can write dh = δ1r +δ2t with δ1,δ2∈ R.
Next observe that P is the unique height two prime ideal of T containing (r,t)T
and as d / ∈ P, r,t,d is a system of parameters. Therefore, as T is Cohen-Macaulay,
δ1∈ (d,t)T ∩ R. Now d / ∈ Pλ0and t / ∈ Pλfor λ ?= λ0, giving (d,t)T ∩ R = (d,t)R
and δ1 ∈ (d,t)R. Therefore dh = (δ3d + δ4t)r + δ2t with δ3,δ4 ∈ R. Finally
d(h − δ3r) ∈ tR and so h ∈ (r,t)R as desired.
1r + ρ2s + ρ3d + ρ′
4, we have dh = ρ1r + (ρ2s + ρ3d)s + ρ4t with
4t with ρ′
1,ρ2,ρ3,ρ′
4∈ R and
1,ρ4= e3+ sρ′
?
Theorem 2.23. There exists a local UFD R with completion T such that (R,Γ) =
?(Rα,Γα) is an ascending union of N-pairs which satisfies every defining condition
of N-pair except that |R| = |R|. Moreover, there is an element ω ∈ R such that
for every h ∈ T, there exists an element (P,{ω + zh}) ∈ Γ constructed using
Lemma 2.19. In particular, we have ((x,y,z)T,{ω}) ∈ Γ.
Proof. We start with the N-subring R0= Q[z,w](z,w)and note that (R0,∅) is an
N-pair. Then we apply Lemma 2.19 to obtain an element ω ∈ R and an N-subring
R1such that (R1,{((x,y,z)T,{ω})}) is an N-pair. Next let Ω be the set T×T, well-
ordered so that ∀α ∈ Ω, |{β ∈ Ω|β < α}| < |R|. Let γ(α) = sup{β ∈ Ω|β < α}.
Define an ascending collection of pairs {(Rα,Γα)|α ∈ Ω} so that if γ(α) = α,
then Rα =
β<αRβ and Γα =
an A-extension of (Rγ(α),Γγ(α)) given by Lemma 2.21 so that if γ(α) = (h,v),
then v ∈ Rα+ m2and Γ(α) = Γ(γ(α)) ∪ {(Pα,{ω + zh})} whenever ω + zh is
?
?
β<αΓβ, while if γ(α) < α, then (Rα,Γα) is
Page 25
ARE COMPLETE INTERSECTIONS COMPLETE INTERSECTIONS?25
transcendental over T/(x,y)T. Of course, if ω + zh is algebraic over T/(x,y)T, a
case we will rule out later in the proof, then v ∈ Rα+ m2and Γ(α) = Γ(γ(α)).
Let R =?Rα and Γ =?Γα. By Lemma 2.14, (R,Γ) satisfies all conditions
to be an N-pair except the cardinality condition. We still must show that R is a
Noetherian ring with completion T to complete the proof of the first statement. We
do this by showing that R satisfies the entire hypothesis of Lemma 2.22. We begin
by noting that it is clear from the construction that the natural map R −→ T/m2
is surjective. Next we claim that if I is any finitely generated ideal of R such that
I ? Pλfor all λ ∈ Γ, then IT ∩ R = I. To prove the claim, let I = (y1,...,yn)R
and suppose d ∈ IT ∩ R. As y1,...,yn,d is a finite set, there exists α such that
y1,...,yn,d ∈ Rα. By Lemma 2.21, we get d ∈ (y1,...,yn)Rα+1and so IT ∩R = I,
proving the claim. Finally let P′be a height two prime ideal of T which is not
equal to any Pλ but is minimal over IT for some finitely generated ideal I of R,
and consider any specific λ ∈ Γ. Again we let I = (y1,...,yn)R and note that
there exists α such that y1,...,yn∈ Rαand (Pλ,Vλ) ∈ Γα. By Lemma 2.21, we
get P′∩Rα+1? Pλ. Now Lemma 2.22 tells us that R is Noetherian and the natural
homomorphism?R −→ T is an isomorphism.
Finally we show that for each h ∈ T there exists some (P,{ω + zh}) ∈ Γ.
If not, when we encountered (h,v) in the recursive process, we discovered that
ω+zh ∈ T/(x,y)T was algebraic over the subring Rαin use at that time. However,
as Rα⊂ R, ω +zh ∈ T/(x,y)T is also algebraic over R. We complete the proof by
contradicting the assumption that ω + zh ∈ T/(x,y)T is algebraic over R.
The algebraic assumption gives elements rn,...,r0∈ R, not all zero, such that
rn(ω + zh)n+ ··· + r0∈ (x,y)T. As (x,y)T ∩ R = (0), (rn,...,r0)R ? (x,y,zm)T
for sufficiently large m. Choose m minimal with respect to the property that
(rn,...,r0)R ? (x,y,zm)T.By Lemma 2.20, there exists d,e ∈ R such that
(d,e,zm)T = (x,y,zm)T. As ideals in R are closed, (x,y,zm)T ∩ R = (d,e,zm)R
and (x,y,zm−1)T ∩ R = (d,e,zm−1)R. Let J denote (d,e,zm)T. For each j, as
rj∈ (x,y,zm−1)T, we can write rj= ajd + bje + cjzm−1with aj,bj,cj∈ R. Since
rn(ω+zh)n+···+r0∈ J and d,e ∈ J, cnzm−1(ω+zh)n+···+c0zm−1∈ J. Thus
zm−1(cnωn+ ··· + c0) ∈ (x,y,zm)T and so cnωn+ ··· + c0∈ (x,y,z)T. However,
by the choice of m, at least one cj / ∈ (x,y,z)T. This contradicts the fact that ω is
transcendental over R/Q.
?
Theorem 2.24. Let R be as constructed. Then R is not the homomorphic image
of a regular local ring.
Proof. Assume the theorem is false. By the theorem in the introduction, we have
a commutative diagram
S
⊆
??
π|S????
R[[x,y,z,w]]
π
????
R
⊆??R[[x,y,z,w]]/(x2+ y2)
Let f = x2+y2∈ R[[x,y,z,w]]. Throughout the construction of R, we chose all
elements of interest to be members of the vector space R+xR+yR+zR+wR ⊂ T.
For these elements, there is a natural lifting to R[[x,y,z,w]]. Of course, these
canonical lifts are unlikely to be in S, but we employ them to simplify notation.
Page 26
26RAYMOND C. HEITMANN AND DAVID A. JORGENSEN
So for example, if ALdenotes a lifting of A, we can write AL= A + fA0for some
A0∈ R[[x,y,z,w]].
Consider ((x,y,tλ)T,{ω + zhλ}) ∈ Γ. We have Pλ= (x,y,tλ)T = (rλ,sλ,tλ)T
with tλ= z + uλw, r = x + αλtλ, and s = y + βλtλ. As nearly everything depends
on λ, we will suppress the subscript in our calculations. We begin by lifting the
equation 0 = r2+ s2+ Art + Bst + Ct2to the full power series ring. Let AL
denote the lift of A, etc. and we get an element Θ in the kernel of π|S such that
Θ = (rL)2+ (sL)2+ ALrLtL+ BLsLtL+ CL(tL)2= (r + fρ)2+ (s + fσ)2+
(A + fA0)(r + fρ)(t + fτ) + (B + fB0)(s + fσ)(t + fτ) + (C + fC0)(t + fτ)2=
r2+ s2+ Art + Bst + Ct2+ f(2ρr + 2σs + Arτ + Atρ + A0rt + Bsτ + Btσ +
B0st + 2Ctτ + C0t2) + f2d where ρ,σ,τ,A0,B0,C0 ∈ T and d ∈ T is a sum of
terms which we don’t need to keep track of individually. Recall that in the proof of
Lemma 2.19, we actually did the calculation in R[[x,y,z,w]] and determined that
r2+ s2+ Art + Bst + Ct2= f(1 + tq) and so Θ = f(1 + tq + (2r + At)ρ + (2s +
Bt)σ+(Ar+Bs+2Ct)τ +A0rt+B0st+C0t2+fd). As r = x+αt and s = y+βt,
Θ ∈ f(1+tq+(2α+A)tρ+(2β+B)tσ+(Aα+Bβ+2C)tτ+(x,y,t2)T). Also, from the
defining equations for A,B,C, we see that 2α+A,2β+B,Aα+Bβ+2C ∈ (x,y,t)T
and so Θ ∈ f(1 + tq + (x,y,t2)T).
Now we consider the specific ((x,y,z)T,{ω}) ∈ Γ, which we will refer to as
λ = 0. As q0 = ω, we have Θ0 ∈ f(1 + zω + z2h + (x,y)T) for some h ∈ T.
There is another element in Γ corresponding to h which it is convenient to denote
λ = h. (Admittedly, we need a different name if h happens to equal zero.) This
is (Ph,{ω + zh}). Clearly Θh∈ f(1 + Ph). As both Θ0and Θhare elements of S
and they are unit multiples of each other, we see that Θ0/Θh∈ S. The map S −→
R −→ R/Ph∩R clearly takes Θ0/Θhto 1+zω+z2h. Thus z(ω+zh) ∈ R/Ph∩R.
This contradicts the fact that ω + zh is transcendental over R/Ph∩ R, completing
the proof.
?
References
[G] A. Grothendieck, El´ ements de g´ eometrie alg´ ebrique, Chap. IV. Publ. Math. I.H.E.S. 32,
1967.
[H1] R. C. Heitmann, Characterization of completions of unique factorization domains, Trans.
AMS 337 (1993), 379–387.
[H2] R. C. Heitmann, Completions of local rings with an isolated singularity, J. Algebra 163
(1994), 538–567.
Raymond C. Heitmann, Department of Mathematics, University of Texas at Austin,
Austin, TX, 78712
David A. Jorgensen, Department of Mathematics, University of Texas at Arlington,
Arlington, TX, 76019
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