arXiv:1403.4200v1 [] 17 Mar 2014

A family of quotients of the Rees algebra
V. Barucci ∗, M. D’Anna †, F. Strazzanti

‡

Dedicated to Marco Fontana on occasion of his 65-th birthday

Abstract
A family of quotient rings of the Rees algebra associated to a commutative ring is studied. This family generalizes both the classical
concept of idealization by Nagata and a more recent concept, the
amalgamated duplication of a ring. It is shown that several properties
of the rings of this family do not depend on the particular member.
MSC: 20M14; 13H10; 13A30.

Introduction
Let R be a commutative ring and let M be an R-module; the idealization,
also called trivial extension, is a classical construction introduced by Nagata
(see [15, page 2], [11, Chapter VI, Section 25] and [8]) that produces a new
ring containing an ideal isomorphic to M. Recently, D’Anna and Fontana
introduced the so-called amalgamated duplication (see [4], [2], studied also in,
e.g., [5], [14] and [1]), that, starting with a ring R and an ideal I, produces a
new ring that, if M = I, has many properties coinciding with the idealization
(e.g., they have the same Krull dimension and if I is a canonical ideal of a
local Cohen-Macaulay ring R, both of them give a Gorenstein ring). On the
other hand, while the idealization is never reduced, the duplication can be
∗

email: barucci@mat.uniroma1.it
email: mdanna@dmi.unict.it
‡
email: strazzanti@mail.dm.unipi.it
†

1

reduced, but is never an integral domain. Looking for a unified approach to
these two constructions, D’Anna and Re in [6] observed that it is possible
to present both of them as quotients of the Rees algebra modulo particular
ideals. This observation leaded to the subject of this paper, where we study
a more general construction, that produces a ring which, in some cases, is an
integral domain.
More precisely, given a monic polynomial t2 + at + b ∈ R[t] and denoting
with R+ the Rees
associated to the ring R with respect to the ideal
L algebra
n n
2 2
I, i.e. R+ =
n≥0 I t , we study the quotient ring R+ /(I (t + at + b)),
2 2
where (I (t + at + b)) is the contraction to R+ of the ideal generated by
t2 + at + b in R[t]. We denote such ring by R(I)a,b .
In the first section we introduce the family of rings R(I)a,b , show that
idealization and duplication are particular cases of them (cf. Proposition
1.4) and study several general properties such as Krull dimension, total ring
of fractions, integral closure, Noetherianity and spectrum. In Section 2 we
assume that R is local; in this case we prove that the rings R(I)a,b have the
same Hilbert function and that they are Cohen-Macaulay if and only if I
is a maximal Cohen-Macaulay R-module. We conclude this section proving
that, if R is a Noetherian integral domain of positive dimension, there exist
infinitely many choices of b such that the ring R(I)0,−b is an integral domain.
Finally in the last section we study the one-dimensional case. If R is local,
Noetherian and I a regular ideal we find a formula for the CM type of R(I)a,b
(cf. Theorem 3.2) and prove that it is Gorenstein if and only if I is a canonical
ideal of R. Moreover, we show the connection of the numerical duplication
of a numerical semigroup (see [7]) with R(I)0,−b , where R is a numerical
semigroup ring or an algebroid branch and b has odd valuation (see Theorems
3.4 and 3.6).

1

Basic properties

Let R be a commutative ring with unity and I a proper ideal of R; let t be
an indeterminate. The Rees algebra (also called Blow-up algebra) associated
to R and I is defined as the following graded subring of R[t]:
M
R+ =
I n tn ⊆ R[t].
n≥0

2

Lemma 1.1. Let f (t) ∈ R[t] be a monic polynomial of degree k. Then
f (t)R[t] ∩ R+ = {f (t)g(t); g(t) ∈ I k R+ }
P
Proof. Observe first that I k R+ = { ni=0 bi ti ; bi ∈ I k+i }. It is trivial that
each element of the form f (t)g(t), with g(t) ∈ I k R+ , is in f (t)R[t] ∩ R+ .
Conversely, if g(t) ∈ R[t] and if f (t)g(t) ∈ R+ , we prove by induction on
the degree of g(t), that g(t) ∈ I k R+ . If the degree of g(t) is zero, i.e.
g(t) = r ∈ R, and if f (t)r ∈ R+ , then the leading term of f (t)r is rtk and
r ∈ I k ⊂ I k R+ . The inductive step: suppose that the leading term of g(t)
is hn tn ; thus the leading term of f (t)g(t) is hn tk+n . If f (t)g(t) ∈ R+ , then
hn ∈ I k+n and so f (t)hn tn ∈ R+ . It follows that, if f (t)g(t) ∈ R+ , then
f (t)g(t) − f (t)hn tn = f (t)ḡ(t) ∈ R+ , where deg(ḡ(t)) < n = deg(g(t)). By
inductive hypothesis ḡ(t) ∈ I k R+ , hence g(t) = ḡ(t) + hn tn ∈ I k R+ .
We denote the ideal of the previous lemma by (I k f (t)).
Lemma 1.2. Let f (t) ∈ R[t] be a monic polynomial of degree k > 0. Then
each element of the factor ring R+ /(I k f (t)) is represented by a unique polynomial of R+ of degree < k.
Proof. The euclidean division of an element g(t) of R+ by the monic polynomial f (t) is always possible and gives g(t) = f (t)q(t) + r(t), with deg(r(t)) <
k. Moreover, an easy calculation shows that q(t) ∈ I k R+ and r(t) ∈ R+ .
Thus g(t) ≡ r(t) (mod (I k f (t))). Finally, if r1 (t) and r2 (t) are distinct polynomials of R+ of degree < k, also deg (r1 (t) − r2 (t)) < k and they represent
different classes.
It follows from Lemma 1.2 that the ring R is a subring of R+ /(I k f (t)).
Proposition 1.3. The ring extensions R ⊆ R+ /(I k f (t)) ⊆ R[t]/(f (t)) are
both integral and the three rings have the same Krull dimension.
Proof. By the two lemmas above we have the two inclusions. Moreover, the
class of t in R[t]/(f (t)) is integral over R and over R+ /(I k f (t)) as well. It
follows that all the extensions are integral. By a well known theorem on
integral extensions, we get that the three rings have the same dimension.
We observe now that, for particular choices of the polynomial f (t) above,
we get known concepts.
3

Recall that the Nagata’s idealization, or simply idealization, of R with
respect to an ideal I of R (that could be defined for any R-module M) is
defined as the R-module R ⊕ I endowed with the multiplication (r, i)(s, j) =
(rs, rj + si) and it is denoted by R ⋉ I.
The duplication of R with respect to I is defined as follows:
R ✶ I = {(r, r + i) | r ∈ R, i ∈ I} ⊂ R × R;
note that R ✶ I ∼
= R ⊕ I endowed with the multiplication (r, i)(s, j) =
(rs, rj + si + ij).
Proposition 1.4. We have the following isomorphisms of rings:
1) R+ /(I 2 t2 ) ∼
=R⋉I
2 2
2) R+ /(I (t − t)) ∼
= R✶I
Proof. 1) For each residue class modulo (I 2 t2 ), let r + it ∈ R+ , with r ∈ R
and i ∈ I, be its unique representative; the map
α : R+ /(I 2 t2 ) → R ⋉ I
defined setting α(r + it + (I 2 t2 )) = (r, i) is an isomorphism of rings: as a
matter of fact, α preserves sums and, if r, s ∈ R, i, j ∈ I, we have α((r + it +
(I 2 t2 ))(s + jt + (I 2 t2 ))) = α(rs + (rj + si)t + ijt2 + (I 2 t2 )) = α(rs + (rj +
si)t + (I 2 t2 )) = (rs, rj + si) = (r, i)(s, j).
2) Similarly to 1), the map
β : R+ /(I 2 (t2 − t)) → R ✶ I
defined setting β(r+it+(I 2(t2 −t))) = (r, r+i) is an isomorphism of rings. As
for the product, we have β((r +it+(I 2(t2 −t)))(s+jt+(I 2 (t2 −t)))) = β(rs+
(rj +si)t+ijt2 +(I 2 (t2 −t))) = β(rs+(rj +si+ij)t+ij(t2 −t)+(I 2 (t2 −t))) =
β(rs+(rj+si+ij)t+(I 2(t2 −t))) = (rs, rs+rj+si+ij) = (r, r+i)(s, s+j).
The previous proposition makes natural to consider the family R(I)a,b =
R+ /(I 2 (t2 + at + b)), where a, b ∈ R. As R-module R(I)a,b ∼
= R ⊕ I and
the natural injection R ֒→ R(I)a,b is a ring homomorphism; however 0 ⊕ I
in general (if b 6= 0) is not an ideal of R(I)a,b , although this happens for
idealization and duplication.
Both idealization and duplication can be realized in other cases.
4

Proposition 1.5. 1) If t2 +at+b = (t−α)2 , with α ∈ R, then R(I)a,b ∼
= R⋉I.
2
2) If t + at + b = (t − α)(t − β), with (t − α) and (t − β) comaximal ideals
of R[t], then R(I)a,b ∼
= R ✶ I.
Proof. 1) It is enough to consider the automorphism of R[t], induced by
t 7→ t − α.
2) By Chinese Remainder Theorem, the map Φ : R[t]/(t2 + at + b) → R × R,
defined by Φ(r + st) = (r + αs, r + βs) is an isomorphism of rings, as well as
the map Ψ : R[t]/(t2 − t) → R × R, defined by Ψ(r + st) = (r, r + s), so that
Ψ−1 ◦ Φ : R[t]/(t2 + at + b) → R[t]/(t2 − t),
where (Ψ−1 ◦ Φ)(r + st) = (r + αs) + (β − α)st is also an isomorphism. If
we fix an ideal I of R and if we restrict Ψ−1 ◦ Φ to the subring R(I)a,b , i.e.
to the elements r + it, with r ∈ R and i ∈ I, we get
R(I)a,b ∼
={(r + αi) + (β − α)it; r ∈ R, i ∈ I}
={r ′ + (β − α)it; r ′ ∈ R, i ∈ I};
the last ring is R ✶ J, where J = (β − α)I. To finish the proof we show that
β − α is invertible. In the authomorphism of R[t] induced by t → t + β, the
ideal (t − α, t − β) corresponds to (t − α + β, t) = (β − α, t) and this last ideal
is R[t] if and only if β − α is invertible.
Example 1.6. Let R = Z and t2 + at + b = t2 − 5t + 6 = (t − 2)(t − 3). Then
for each ideal I of Z, Z(I)−5,6 ∼
= Z ✶ I.
In this paper we study the family of rings of the form R(I)a,b , showing
that many relevant properties are independent by the member of the family.
From now on, we denote each element of R(I)a,b simply by r + it (r ∈ R,
i ∈ I).
Proposition 1.7. Let Q be the total ring of fractions of R(I)a,b . Then each
element of Q is of the form r+it
, where u is a regular element of R.
u
Proof. Assume that (s + jt) is a regular element of R(I)a,b and that
(r + it)/(s + jt) ∈ Q. Since (s + jt) is regular, then x(s + jt) 6= 0, for every
x ∈ R \ {0}. Hence, xj = 0 implies xs 6= 0.
Consider, now, the element (ja − s + jt). To prove the Proposition, it is
enough to show that:
5

i) the product u = (s + jt)(ja − s + jt) is a regular element of R,
ii) (ja − s + jt) is a regular element of R(I)a,b .
In fact in this case we can write (r + it)/(s + jt) = (r + it)(ja − s − jt)/u.
Observing that −at − t2 = b ∈ R, we have u = s(ja − s) − j 2 b ∈ R.
If x(ja − s + jt) = 0 (for some x ∈ R \ {0}), then xj = 0, that implies
x(ja − s + jt) = xs = 0, a contradiction. Hence (ja − s + jt) is not killed by
any non zero element of R; it follows that u is regular in R, otherwise there
would exist x ∈ R \ {0} such that ux = 0 that implies (s + jt) not regular in
R(I)a,b , since it is killed by (ja − s + jt)x 6= 0. Thus i) is proved.
ii): if (ja − s + jt) is not regular in R(I)a,b , there exists (h + kt) 6= 0
such that (ja − s + jt)(h + kt) = 0. Hence u(h + kt) = (s + jt)(ja − s +
jt)(h + kt) = 0, so u is not regular in R(I)a,b . But then it is not regular in
R; contradiction.
Corollary 1.8. Assume that I is a regular ideal; then the rings R(I)a,b and
R[t]/(t2 + at + b) have the same total ring of fractions and the same integral
closure.
Proof. Each element of R[t]/(t2 + at + b), let’s say r + r1 t with r, r1 ∈ R,
is in Q, in fact if i is an element of I regular in R, i is also regular in
R[t]/(t2 + at + b) and r + r1 t = (ir + ir1 t)/i ∈ Q. Moreover, if r + r1 t is
regular in R[t]/(t2 + at + b), it is also regular in Q. In fact, according to
Proposition 1.7, an element of Q is of the form (s + jt)/u (s ∈ R, j ∈ I,
u ∈ R and regular); if (r + r1 t)(s + jt)/u = 0, then (s + jt)/u = 0. It follows
that, if (r + r1 t)/(s + s1 t) is an element of Q′ , the total ring of fractions of
R[t]/(t2 + at + b), then r + r1 t and s + s1 t belong to Q and s + s1 t is regular
in Q, so (r + r1 t)/(s + s1 t) ∈ Q. On the other hand, if (r + it)/u ∈ Q,
with u ∈ R and regular in R, then u is also regular in R[t]/(t2 + at + b) and
(r + it)/u ∈ Q′ .
By Corollary 1.8, it follows that the integral closure of R(I)a,b contains
R[t]/(t2 + at + b), where R is the integral closure of R, but it may be strictly
larger. For example, for R = Z and t2 + at + b = t2 + 4, we have that
Z[t]/(t2 + 4) is not integrally closed, in fact (t/2)2 + 1 = 0.
Using the chain of inclusions R ⊆ R(I)a,b ⊆ R[t]/(t2 + at + b) and the fact
that these extensions are integral, we can get information on Spec(R(I)a,b )
with respect to Spec(R).

6

Proposition 1.9. For each prime ideal P of R, there are at most two prime
ideals of R(I)a,b lying over P . Moreover if t2 + at + b is irreducible on R/m
for any maximal ideal m of R, then there is exactly one prime ideal of R(I)a,b
lying over P .
Proof. Every prime ideal of R(I)a,b , lying over P has to be the contraction
of a prime ideal of R[t]/(t2 + at + b). It is well known (see e.g. [9, Chapter
6]) that for every prime ideal P of R, P [t] is a prime of R[t] lying over P
and there exist infinitely many other primes in R[t], lying over P , all of them
containing P [t] and with no inclusions among them. In particular, there is a
bijection between these ideals and the nonzero prime ideals of (Q(R/P ))[t]
(here Q(R/P ) denotes the field of fractions of R/P ); therefore the image of all
these prime ideals J in (Q(R/P ))[t] is of the form (f (t)), for some irreducible
polynomial f (t); hence J = ϕ−1
P ((f (t))), where ϕP is the composition of the
canonical homomorphisms R[t] → (R/P )[t] ֒→ (Q(R/P ))[t]. Thus the prime
ideals of R[t]/(t2 + at + b) lying over P are of the form J/(t2 + at + b), with
J ⊇ (t2 + at + b). This means that the polynomial f (t), corresponding to
J, divides the image of t2 + at + b in Q(R/P )[t]. Hence, if t2 + at + b is
irreducible in Q(R/P )[t], there is only one prime of R[t]/(t2 + at + b) lying
over P ; on the other hand, if t2 + at + b has two distinct irreducible factors in
(Q(R/P ))[t], there exist exactly two prime ideals in R[t]/(t2 + at + b) lying
over P . Hence there are at most two primes in R(I)a,b lying over P and the
first part of the proposition is proved.
Suppose that J/(t2 + at + b)) ∈ Spec(R[t]/(t2 + at + b)) and J/(t2 + at +
b)) ∩ R = P . We know that J = ϕ−1
P ((f (t))), where f (t) is an irreducible
factor of t2 + at + b in Q(R/P )[t]. If P ′ ∈ Spec(R), P ′ ⊂ P , then the prime
ideals of R[t]/(t2 + at + b) lying over P ′ correspond to the irreducible factors
of t2 + at + b in Q(R/P ′ )[t]; since the factorization of t2 + at + b in Q(R/P ′)[t]
induces a factorization in Q(R/P )[t], f (t) is irreducible also in Q(R/P ′)[t]
and we have a prime ideal of R[t]/(t2 + at + b) lying over P ′ of the form
J ′ /(t2 + at + b), with J ′ = ϕ−1
P ′ ((f (t))) ⊂ J. In particular, if m is a maximal
ideal of R containing P and t2 + at + b is irreducible on R/m, then there
is one and only one prime ideal of R[t]/(t2 + at + b) lying over P and the
same happens for R(I)a,b because the extension R(I)a,b ⊆ R[t]/(t2 + at + b)
is integral.
Remark 1.10. 1) Notice that, for particular a and b, the factorization of
t2 + at + b in Q(R/P )[t] may not depend on P . For example, in the case of
the idealization, the equality t2 = t · t, implies that there is only one prime
7

lying over P , both in R[t]/(t2 ) and in the idealization. As for the case of the
duplication, the equality t2 − t = t · (t − 1), implies that there are two primes
in R[t]/(t2 − t) lying over P , namely (P, t) and (P, t − 1). Contracting these
primes to the duplication we get the same prime if and only if P ⊇ I (see,
e.g., [4]).
2) By the proof of Proposition 1.9 we see that the extension R ⊆ R[t]/(t2 +
at + b) and the extension R ⊆ R(I)a,b as well fulfill the going down property.
In particular a minimal prime of R(I)a,b lies over a minimal prime P of R.
3) The proof of the previous proposition also implies that a sufficient
condition for R(I)a,b to be an integral domain is that R is an integral domain
and t2 + at + b is irreducible in Q(R)[t]. We will see in the next section
that, under particular assumptions on R, we can prove the existence of such
polynomials.
We conclude this section characterizing the rings R(I)a,b which are Noetherian.
Proposition 1.11. The following conditions are equivalent:
(i) R is a Noetherian ring;
(ii) R(I)a,b is a Noetherian ring for all a, b ∈ R;
(ii) R(I)a,b is a Noetherian ring for some a, b ∈ R.
Proof. If R is Noetherian, also the Rees algebra R+ is Noetherian; hence
it is straightforward that R(I)a,b is Noetherian for every a, b ∈ R, being a
quotient of a Noetherian ring.
Since the condition (iii) is a particular case of (ii), we need to prove only
that (iii) implies (i). Assume by contradiction that R is not a Noetherian ring;
then there exists an ideal J = (f1 , f2 , . . . ) of R that is not finitely generated
and we can assume that fi+1 ∈
/ (f1 , . . . fi ) for any i. Consider the ideal
JR(I)a,b of R(I)a,b ; by hypothesis, it is finitely generated and its generators
can be chosen from those of J (regarded as elements of R(I)P
a,b ). Hence we can
s
assume that JR(I)a,b = (f1 , . . . , fs ). This implies fP
=
s+1
k=1 fk (rk + ik t),
s
for some rk ∈ R and ik ∈ I, and therefore fs+1 = k=1 fk rk ; contradiction.

2

The local case

Assume that R is local, with maximal ideal m. Then it is known that both
R ✶ I and R ⋉ I are local with maximal ideals m ⊕ I (in the first case under
8

the isomorphism R ✶ I ∼
= R ⊕ I). More generally:
Proposition 2.1. R is local if and only if R(I)a,b is local. In this case the
maximal ideal of R(I)a,b is m ⊕ I (as R-module).
Proof. Let R be local; we claim that all the elements r + it with r ∈
/ m
are invertible in R(I)a,b . As a matter of fact, looking for s + jt such that
(r + it)(s + jt) = 1, we obtain the linear system
(
rs − ibj = 1
is + (r − ia)j = 0
which has determinant δ = r 2 − iar + i2 b ∈ r 2 + m. Thus δ is invertible in R;
moreover, it is easy to check that if (s, j) is the solution of the system, then
j ∈ I; hence s + jt ∈ R(I)a,b and it is the inverse of r + it.
Conversely, if R(I)a,b is local, R has to be local, since R ⊆ R(I)a,b is an
integral extension (cf. Proposition 1.3).
It is also clear that, if (R, m) is local and if we denote by M the maximal
ideal of R(I)a,b , then k = R/m ∼
= R(I)a,b /M. In the sequel, we will always
denote with k the common residue field of R and R(I)a,b .
Remark 2.2. Since R(I)a,b is an R-algebra, every R(I)a,b -module N is also
an R-module and then λR(I)a,b (N) ≤ λR (N) (where λ( ) denote the length
of a module).
If we consider a R(I)a,b -module N annihilated by M, we have that, as
R-module, N is annihilated by m. Hence it is naturally an R(I)a,b /M-vector
space and an R/m-vector space; in particular, λR(I)a,b (N) = dimk (N) =
λR (N) (where k = R/m ∼
= R(I)a,b /M).
For a Noetherian local ring (R, m), we denote by ν(I) the cardinality
of a minimal set of generators of the ideal I. The embedding dimension
of R, ν(R), is by definition ν(m) and the Hilbert function of R is H(n) =
λR (mn /mn+1 ).
Proposition 2.3. Let (R, m) be a Noetherian local ring. Then, for every
a, b ∈ R the rings R(I)a,b have the same Hilbert function. In particular, they
have the same embedding dimension ν(R(I)a,b ) = ν(R) + ν(I) and the same
multiplicity.
9

Proof. First of all, let us consider M 2 ; we have M 2 = m2 +mIt (and hence, as
R-module, it is isomorphic to m2 ⊕ mI): in fact, if (r + it) and (s + jt) are in
M, then their product rs−bij+(rj+si−aij)t ∈ m2 ⊕mI. Conversely, pick an
element in m2 ⊕ mI of the form rs + uit (with r, s, u ∈ m and i ∈ I); we have
rs+uit = rs+u(it) ∈ M 2 ; since m2 ⊕mI is generated by elements of this form
we have the equality. Arguing similarly for any n ≥ 2, we immediately obtain
that M n = mn + mn−1 I and, as R-module, it is isomorphic to mn ⊕ mn−1 I.
It follows that, as R-modules, M n /M n+1 ∼
= mn /mn+1 ⊕ Imn−1 /Imn . By
the previous remark the length of M n /M n+1 as R(I)a,b -module coincides with
its dimension as k-vector space and with its length as R-module. The thesis
follows immediately.
Remark 2.4. Let R be a Noetherian local ring. By Propositions 1.3 and
2.3 we get
dim R(I)a,b = dim R ≤ ν(R) ≤ ν(R) + ν(I) = ν(R(I)a,b ).
The first inequality is an equality if and only if R is regular and the second
if and only if ν(I) = 0, that is equivalent to I = 0, by Nakayama’s lemma.
This means that R(I)a,b is regular if and only if R is regular and I = 0;
clearly if I = 0 one has R(I)a,b = R.
We want to show that, if R is a local Noetherian integral domain, we can
always find integral domains in the family of rings R(I)a,b . The following
proposition was proved in [6] and we publish it with the permission of the
second author.
Proposition 2.5. Let (R, m) be a local Noetherian integral domain with
dim R ≥ 1 and Q(R) its field of fractions. Then for any integer n > 1,
not multiple of 4, there exist infinitely many elements b ∈ R such that the
polynomial tn − b is irreducible over Q(R).
Proof. We will use the following well-known criterion of irreducibility: if b is
not a p-th power for any prime p|n and b 6∈ −4Q(R)4 if 4|n, then tn − b is
irreducible (see [13, Chapter VI, Theorem 9.1]). In particular, if 4 does not
divide n and b is not a d-th power for any integer d > 1 such that d|n, then
tn − b is irreducible.
Taking a prime ideal P ⊂ R such that ht P = 1 we have dim RP = 1 and,
by the Krull-Akizuki Theorem, its integral closure RP of RP in Q(RP ) =
Q(R) is Noetherian (see, e.g. [12, Theorem 4.9.2]), hence it is a Dedekind
10

ring. So there is at least a discrete valuation v : Q(R)∗ → Z with v((RP )M ) =
N (with M maximal ideal of RP ). Since R ⊆ RP ⊆ (RP )M have the same
field of fractions, it follows that v(R) ⊆ N is a semigroup containing two
consecutive integers; so there exists c > 0 such that any x ∈ N, x ≥ c
belongs to v(R).
In particular, there exist infinitely many elements b ∈ R such that v(b)
is prime to n, so b cannot be a d-th power in Q(R) for any d > 1 such that
d|n. Hence we can find infinitely many b ∈ R such that (tn − b) ⊂ Q(R)[t] is
irreducible.
Corollary 2.6. Let R be a local Noetherian integral domain with dim R ≥ 1,
let Q(R) be its field of fractions and let I ⊂ R be an ideal. Then there exist
infinitely many elements b ∈ R such that R(I)0,−b is an integral domain.
Proof. By Proposition 2.5 we can find b such that (t2 − b) is irreducible in
Q(R)[t]. The thesis now follows by point 3) of Remark 1.10.
Now we want to investigate the Cohen-Macaulayness of R(I)a,b .
Assume that R is a CM ring; we set dim R = depth R = d; moreover,
Ann(R(I)a,b ) = (0), hence the dimension of R(I)a,b as R-module (i.e., since
R(I)a,b is a finite R-module, dim(R/Ann(R(I)a,b ))) equals the Krull dimension of R(I)a,b . We can assume that d ≥ 1, otherwise both R and R(I)a,b are
trivially CM.
Given a regular sequence x = x1 , x2 , . . . , xd of the ring R, it is not difficult
to check that it is an R(I)a,b -regular sequence if and only if its image in
R(I)a,b is a regular sequence of R(I)a,b as a ring. Moreover, since x is a
system of parameters of R, then it is a system of parameters of R(I)a,b (since
R ⊆ R(I)a,b is an integral extensions) and x is a system of parameters for
the R-module R(I)a,b . Hence, arguing exactly as in [2] we have that R(I)a,b
is a CM ring if and only if it is a CM R-module.
Since R(I)a,b ∼
= R ⊕ I as R-module, it follows that depth (R ⊕ I) =
min{depth I, depth R} = depth I and therefore R(I)a,b is a CM R-module if
and only if I is a CM R-module of dimension d (that is if and only if I is a
maximal CM R-module).
Hence we can state the following:
Proposition 2.7. Assume that R is a local CM ring of dimension d. Then
R(I)a,b is CM if and only if I is a CM R-module of dimension d. In particular, the Cohen-Macaulayness of R(I)a,b depends only on the ideal I.
11

Remark 2.8. We notice that if I is a canonical ideal of R, since R(I)a,b ∼
=
R ⊕ I, we can apply a result of Eisenbud (stated and proved in [2]) to get
that R(I)a,b is Gorenstein for every a, b ∈ R. We will see that in the onedimensional case we can determine the CM type of R(I)a,b and deduce that
it is a Gorenstein ring if and only if I is a canonical ideal.
Remark 2.9. In [3, Corollary 5.8], under the assumption that the ring (R, m)
is a local CM ring with infinite residue field, it has been proved the following
formula about the multiplicity of the duplication: e(R ✶I) = e(R)+λR (I/IJ)
(where J is any minimal reduction of m); in particular, if dim R = 1, then
e(R ✶I) = 2e(R).
By Proposition 2.3 we can state that, under the same assumptions, the
same formulas hold for the multiplicity of R(I)a,b , for every a, b ∈ R.

3

One-dimensional case

Assume for all this section that (R, m) is a one-dimensional, Noetherian, and
local ring and I a regular ideal; in this section we determine the CM type of
R(I)a,b .
Since I is regular, it is a maximal Cohen-Macaulay R-module and R is
a CM ring; therefore R(I)a,b is also CM by Proposition 2.7. In this case the
type of R(I)a,b equals the length of (R(I)a,b : M)/R(I)a,b as R(I)a,b -module,
where M is the maximal ideal of R(I)a,b ; so we start studying R(I)a,b : M.
Lemma 3.1. For any a, b ∈ R, the R(I)a,b -module R(I)a,b : M is equal to


r i
i
r
+ t; ∈ I : m, ∈ (I : I) ∩ (R : m)
s s
s
s
Proof. Consider a generic element r/s + (i/s)t of Q(R(I)a,b ), where r, s ∈
R, i ∈ I and s is regular (cf. Proposition 1.7). It is an element of R(I)a,b : M
if and only if
(r/s + (i/s)t)(m + jt) = rm/s + (im/s)t + (rj/s)t + (ij/s)t2
= rm/s − ijb/s + (im/s + rj/s − ija/s)t
is an element of R(I)a,b , for any m ∈ m and for any j ∈ I, that is (rm/s −
ijb/s) ∈ R and (im/s + rj/s − ija/s) ∈ I.

12

Suppose that r/s + (i/s)t ∈ R(I)a,b : M; in particular, if j = 0 we have
rm/s ∈ R and im/s ∈ I, that is r/s ∈ R : m and i/s ∈ I : m. Moreover
since ja ∈ I ⊆ m and i/s ∈ I : m, we have im/s, ija/s ∈ I, hence rj/s ∈ I
for any j ∈ I and then r/s ∈ I : I.
Conversely, suppose that i/s ∈ I : m and r/s ∈ (I : I) ∩ (R : m). Then
rm/s − ijb/s ∈ R + I = R and im/s + rj/s − ija/s ∈ I + I + I = I,
consequently r/s + (i/s)t ∈ R(I)a,b : M.
Theorem 3.2. The CM type of R(I)a,b is




(I : I) ∩ (R : m)
I:m
t(R(I)a,b ) = λR
+ λR
;
R
I
in particular, it does not depend on a and b.
Proof. Consider the homomorphism ϕ of R-modules
(I : I) ∩ (R : m) I : m
×
R
I


r
i
r i
+ t 7→
+ R, + I .
s s
s
s

R(I)a,b : M →

Thanks to the previous lemma, ϕ is well defined and surjective; moreover,
its kernel is given by the elements r/s + (i/s)t with r/s ∈ R and i/s ∈ I,
that is ker ϕ = R(I)a,b ; hence
R(I)a,b : M ∼ (I : I) ∩ (R : m) I : m
×
.
=
R(I)a,b
R
I
Consequently, using Remark 2.2, we have



R(I)a,b : M
R(I)a,b : M
= λR
=
t(R(I)a,b ) =λR(I)a,b
R(I)a,b
R(I)a,b


(I : I) ∩ (R : m) I : m
=λR
=
×
R
I




I:m
(I : I) ∩ (R : m)
+ λR
.
=λR
R
I


13

Corollary 3.3. The ring R(I)a,b is Gorenstein if and only if I is a canonical
ideal of R.
Proof. Recall first that a ring is Gorenstein if and only if it has CM type 1.
Recall also that I is a canonical ideal of a one-dimensional CM local ring R,
i.e. an ideal I such that I : (I : J) = J for each regular ideal J of R, if and
only if λR ((I : m)/I) = 1 (cf. [10], Satz 3.3). Notice that, for any ideal I
regular and proper, λR ((I : m)/I) ≥ 1.
Thus, by the formula of Theorem 3.2 we get: R(I)a,b is Gorenstein if
and only if t(R(I)a,b ) = 1 = 0 + 1; hence, λR ((I : m)/I) = 1, i.e. I is a
canonical ideal. Conversely if I is a canonical ideal, then I : I = R and
λR ((I : m)/I) = 1; by the same formula we get t(R(I)a,b ) = 0 + 1 = 1, i.e.
R(I)a,b is Gorenstein.
We conclude this section studying two particular cases of one dimensional
rings: numerical semigroup rings and algebroid branches; in both cases we
show the connection with the numerical duplication of a numerical semigroup
(see [7]).
Recall that a numerical semigroup is a submonoid of N with finite complement in N and it can be expressed in terms of its minimal set of generators,
S = hn1 , . . . , nν i, with GCD(n1 , . . . , nν ) = 1. A semigroup ideal E is a subset
of S such that S + E ⊆ E. We set 2 · E = {2s| s ∈ E}. According to [7], the
numerical duplication of S with respect to a semigroup ideal E of S and an
odd integer m ∈ S is the numerical semigroup
S ✶mE = 2 · S ∪ (2 · E + m).
A numerical semigroup ring is a ring R of the form k[[S]] = k[[X n1 , . . . , X nν ]],
where k is a field and X an indeterminate. Such a ring is a one-dimensional,
Noetherian, local integral domain; moreover, it is analytically irreducible (i.e.
its integral closure R is a DVR, which is a finite R-module) and in this case
R = k[[X]], the ring of formal power series. The valuation v induced by
k[[X]] on k[[S]] is given by the order of a formal power series and, if I is an
ideal of k[[S]], v(I) = {v(i); i ∈ I, i 6= 0} is a semigroup ideal of S.
Theorem 3.4. Let R = k[[S]] be a numerical semigroup ring, let b = X m ∈
R, with m odd, and let I be a proper ideal of R. Then R(I)0,−b is isomorphic
to the semigroup ring k[[T ]], where T = S ✶mv(I).

14

Proof. If S = hn1 , . . . , nν i, an element of R(I)0,b is of the form
r(X) + i(X)t
where r(X) = r(X n1 , . . . , X nν ) ∈ k[[S]] and i(X) = i(X n1 , . . . , X nν ) ∈ I.
Taking into account that we are factoring out the ideal (I 2 (t2 − X m )), we
can easily check that the map Φ : R(I)0,−b → k[[T ]], defined by
Φ(r(X) + i(X)t) = r(X 2 ) + i(X 2 )X m ,
is an isomorphism of rings.
Example 3.5. If R = k[[X 2 , X 3 ]], b = X 5 and I = X 3 k[[X 2 , X 3 ]], then
R(I)0,b ∼
= k[[X 4 , X 6 , X 11 ]]. According to Corollary 3.3, we get a Gorenstein
ring (in fact the semigroup h4, 6, 11i is symmetric), because the ideal I is a
canonical ideal of R.
We consider now the case of algebroid branches, i.e. local rings (R, m) of
the form k[[X1 , . . . Xn ]]/P , where P is a prime ideal of height n − 1 and k is
algebraically closed. We have that (R, m) is a one-dimensional, Noetherian,
complete, local integral domain; moreover, R is analytically irreducible with
integral closure isomorphic to k[[X]] and k ⊂ R. If we consider the valuation
v induced by k[[X]] on R, we get again that v(R) = {v(r); r ∈ R, r 6= 0} is
a numerical semigroup and that v(I) = {v(i); i ∈ I, i 6= 0} is a semigroup
ideal of v(R).
Theorem 3.6. Let R be an algebroid branch and let I be a proper ideal of
R; let b ∈ R, such that m = v(b) is odd. Then R(I)0,−b an algebroid branch
and its value semigroup is v(R) ✶mv(I).
Proof. Since v(b) is odd, by Proposition 2.5, t2 − b is irreducible in Q(R)[t]
and R(I)0,−b is an integral domain. Moreover, applying the results of the
previous sections, we know that R(I)0,−b is local (we will denote by M its
maximal ideal), Noetherian and one-dimensional. It is not difficult to check
that the m-adic topology on the R-module R(I)0,−b coincide with the M-adic
topology, hence it is complete. Since R(I)0,−b contains its residue field k, by
Cohen structure theorem, it is of the form k[[Y1 , . . . , Yl ]]/Q, for some prime
ideal Q of height l − 1; so it is an algebroid branch.
Let V = k[[Y ]] be the integral closure of R(I)0,−b in its quotient field
Q(R(I)0,−b ) = Q(R)(t) = k((Y )). We denote by v ′ the valuation associated
15

to k[[Y ]]; in particular v ′ (Y ) = 1. Since Q(R) = k((X)), we have k((Y )) =
k((X))(t); moreover, t2 = b implies that 2v ′ (t) = v ′ (b) = mv ′ (X). In order
to obtain v ′ (Y ) = 1 it is necessary that v ′ (t) = m and v ′ (X) = 2.
Now, it is straightforward that v ′ (R(I)0,−b ) = v(R) ✶mv(I).

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