Page 1

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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2 RAYMOND 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

Page 11

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