A NOTE ON ANALYTICALLY IRREDUCIBLE DOMAINS
ROSWITHA RISSNER
Abstract. This is a survey on the following result: If D is one-dimensional, Noetherian, local
domain, it is well-known that D is analytically irreducible if and only if D is unibranched and
the integral closure D0 of D is finitely generated as D-module. However, the proof is split in
pieces and spread over the literature. This survey collects the pieces and assembles them to
a complete proof. Next to several results on integral extensions and completions of modules,
we use Cohen’s structure theorem for complete, Noetherian, local domains to prove the main
result. The purpose of this survey is to make this characterization of analytically irreducible
domains more accessible.
1. Introduction
A Noetherian, local domain (D, m) is said to be analytically irreducible if the m-adic completion
b of D is a domain and we say D is analytically unramified if D
b is a reduced ring. It is wellD
b has zero-divisors or nilpotents is strongly connected to
known that the question of whether D
certain properties of the integral closure D0 of D.
We say D is unibranched if D0 is local. This paper is a survey on a characterization of analytically irreducible domains that depicts the connection between the existence of zero-divisors in
b and properties of the integral closure D0 . To be more precise, we give a proof of the following
D
well-known theorem.
Theorem 1. Let (D, m) be a one-dimensional, Noetherian, local domain with integral closure D0 .
Then the following assertions are equivalent:
(1) D is analytically irreducible.
(2) D is unibranched and analytically unramified.
(3) D is unibranched and D0 is finitely generated as a D-module.
(4) D is unibranched and if m0 denotes the maximal ideal of the integral closure, then the
m-adic topology on D coincides with subspace topology induced by m0 .
c0 be the m0 -adic
Assume that D is unibranched, let m0 be the maximal ideal of D0 and D
0
0
completion of D . We can consider D as a topological subspace of D . The completion of D with
c0 . On the other hand, we
respect to this subspace topology is the topological closure D of D in D
b
also have the m-adic completion D of D, see Figure 1.
c0
D
b
D
D
D0
D
Figure 1. Completions of D with respect to m and m0
2010 Mathematics Subject Classification. 13-02; 13B22, 13B35 13J05, 3J10,13H05.
Key words and phrases. unibranched, analytically irreducible, one-dimensional, Noetherian, local.
R. Rissner is supported by the Austrian Science Fund (FWF): P 27816-N26.
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2
ROSWITHA RISSNER
b = D which, in turn, is the case
By Theorem 1, D is analytically irreducible is equivalent to D
if and only if D0 is a finitely generated D-module.
Theorem 1 is well-known but its proof is split into pieces and has to be assembled from several
sources. This survey simply collects known results from different references in order to present a
complete proof. For the implication (2) ⇒ (3) we follow the approach of Nagata’s textbook [6,
(32.2)]. For the remaining implications, we pursue the suggestions of [2, Theorem III.5.2]. For
the implication (4) ⇒ (1) one can also refer to [7, Theorem 8].
In order to make this survey more self-contained, we give a short introduction to integral ring
extensions and completions in Sections 2 and 3, respectively. Then, in Section 4 we discuss Cohen’s
structure theorem which allows us to prove that the integral closure of a complete, Noetherian,
local, reduced ring R is a finitely generated R-module in Section 5. Finally, we give a proof of
Theorem 1 in Section 6.
2. Integral ring extensions
In this section we recall some facts on integral ring extensions which we use throughout this
paper.
Fact 2.1 (cf. [1, Proposition 5.1, Corollaries 5.3, 5.4]). Let R ⊆ S be a ring extension. We call
s ∈ S integral over R if the following equivalent assertions are satisfied:
(1) There exists a monic polynomial f ∈ R[X] such that f (s) = 0.
(2) R[s] is finitely generated as R-module.
(3) There exists a ring R[s] ⊆ T which is T is finitely generated as R-module.
Let intS (R) = {s ∈ S | s integral over R} denote the set of elements of S which are integral
over R. Then R ⊆ intS (R) is a ring extension.
We call intS (R) the integral closure of R in S and if R = intS (R) we say R is integrally closed
in S. If S = intS (R), we say R ⊆ S is an integral extension.
If R ⊆ T ⊆ S is an intermediate ring such that both R ⊆ T and T ⊆ S are integral extensions,
then R ⊆ S is an integral extension. In particular, intS (R) = intS (intS (R)).
In case S is the total ring of quotients of R, we simplify and say R0 := intS (R) is the integral
closure of R and R is integrally closed if R = R0 .
Lemma 2.2. Let D be a Noetherian domain with quotient field K. Then x ∈ K is integral over
D if and only if there exists an element 0 6= d ∈ D with dxt ∈ D for all t ∈ N.
Proof. According to Fact 2.1, x is integral over D if and only if D[x] is a finitely generated Dmodule. If D[x] is finitely generated, then there exists an element 0 6= d ∈ D with dD[x] ⊆ D and
hence dxt ∈ D for all t ∈ N
Reversely, let 0 6= d ∈ D be an element with dxt ∈ D for all t ∈ N. Then d−1 D ⊆ K is a finitely
generated D-module and hence Noetherian. Since xt = d−1 dxt ∈ d−1 D holds for all t ∈ N, it
follows that D[x] is a D-submodule of d−1 D and hence finitely generated.
Proposition 2.3. If R is a integrally closed Noetherian ring, then RJXK is integrally closed.
In particular, the ring RJX1 , . . . , Xn K is integrally closed.
Proof. Let K be the quotient field of R.
P Then the field K((X)) of formal Laurent series contains
the quotient field of RJXK. Let f = i≥z fi X i ∈ K((X)) with z ∈ Z be integral over RJXK.
P
According to Lemma 2.2, there exists g = j≥n gj X j ∈ RJXK with n ∈ N0 such that rf t ∈ RJXK
for all t ∈ N. Without restriction we assume fz 6= 0 and gn 6= 0. It follows from
t
gf t = gn fm
X n+tm + X n+tm+1 (· · · ) ∈ RJXK,
that n ≥ −tm holds for all t ∈ N and hence m ≥ 0 and f ∈ KJXK.
t
Moreover, gn fm
∈ R for all t ≥ 0 which implies that fm ∈ K is integral over R according to
Lemma 2.2. However, R is integrally closed by assumption and hence fm ∈ R. Thus fm X m ∈
RJXK and f − fm X m ∈ KJXK is integral over RJXK. We can repeat the last argument to conclude
that fi ∈ R for all i ≥ m.
The second statement follows since RJXK is Noetherian if R is Noetherian.
A NOTE ON ANALYTICALLY IRREDUCIBLE DOMAINS
3
Fact 2.4 (Cohen-Seidenberg, cf. [1, Corollary 5.9, Theorem 5.11]). Let R ⊆ S be an integral
extension. Then the following assertions hold:
(1) If Q1 ⊆ Q2 are prime ideals of S such that Q1 ∩ R = Q2 ∩ R, then Q1 = Q2 .
(2) If P1 , P2 ∈ spec(R) with P1 ⊆ P2 and Q1 ∈ spec(S) with Q1 ∩ R = P1 , then there exists
Q2 ∈ spec(S) such that Q1 ⊆ Q2 and Q2 ∩ R = P2
(3) dim(R) = dim(S) and max(S) = {P ∈ spec(S) | P ∩ R ∈ max(R)}.
We are particularly interested in the one-dimensional case. The next result is a special case of
the Krull-Akizuki theorem cf. [5, Theorem 11.7]. However, we prove Proposition 2.5 directly as
we do not need the full result in the remainder of this paper.
Proposition 2.5. Let D be a one-dimensional, Noetherian domain.
Then the integral closure D0 of D is a Dedekind domain.
Proof. Since 1 = dim(D) = dim(D0 ) by Fact 2.4 and D0 is integrally closed, we only need to prove
that D0 is Noetherian.
If 0 6= I be an ideal of D0 , then there exists an element 0 6= a ∈ I ∩ D. It follows that D/aD is a
zero-dimensional, Noetherian ring and thus Artinian. Since In = (an D0 ) ∩ D + aD for n ∈ N form
a descending chain of ideals of D/aD, there exists an m ∈ N such that Im = In for all n ≥ m.
It suffices to prove that am D0 ⊆ am+1 D0 + D. Then
D0 /aD0 ' am D0 /am+1 D0 ⊆ (am+1 D0 + D)/am+1 D0 ' D/(D ∩ am+1 D0 ).
Then D0 /aD0 is a submodule of the Noetherian module D/(D ∩ am+1 D0 ) and hence a finitely
generated D-module. This further implies that D0 /aD0 is Noetherian and the submodule I/aD0
is finitely generated. Consequently I is a finitely generated ideal of D0 .
It remains to prove that am D0 ⊆ am+1 D0 + D. We can localize at each maximal ideal of D and
prove the inclusion locally. So assume that D is local with maximal ideal m.
If a ∈
/ m, then a is a unit in D and therefore am D0 = D0 = am+1 D0 + D. Now assume a ∈ m
and let x = cb ∈ D0 . Then 0 6= cD is an ideal with radical m and therefore there exists n ≥ m
with mn+1 ⊆ cD. It follows that
an+1 x ∈ (an+1 D0 ) ∩ D ⊆ In+1 = In+2 = (an+2 D0 ) ∩ D + aD
and hence an x ∈ an+1 D0 + D. If n > m, then
an x ∈ (an+1 D0 + D) ∩ an D0 = an+1 D0 + D
an D}0 ⊆ an+1 D0 + aD
| ∩{z
⊆In =In+1
n−1
and therefore a
n
0
x ∈ a D + D. Repeating this argument completes the proof.
If D is a local (one-dimensional, Noetherian) domain with maximal ideal m, then Fact 2.4
implies that the maximal ideals of D0 are the minimal primes of mD and therefore D0 is always
a semilocal Dedekind domain.
Corollary 2.6. Let D be a one-dimensional, Noetherian, local domain.
Then the integral closure D0 is a discrete valuation domain.
3. Completions
In this section we recall the necessary facts on topologies on rings and modules which are
induced by ideals. Let R be a Noetherian ring, I an ideal of R and M an R-module. Then the
submodules (I n M )n∈N are a filtration on M which induce a linear topology on M , that is, the
sets m + I n M for m ∈ M and n ∈ N form a basis of this topology. We call this the I-adic topology
on M .
Addition, subtraction and scalar multiplication are continuous with respect to this topology. If
M is a ring extension of R, then
S multiplication in M is continuous too.
Moreover, M \ m + I n M = y y + I n M where the union runs over all y ∈ M with m − y ∈
/ I nM
n
and hence each m + I M is both open and closed.
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ROSWITHA RISSNER
c of M is the inverse limit of the inverse system M/I n M together with the
The completion M
canonical projections M/I n M −→ M/I m M for n ≥ m, that is,
)
(
Y
n
n
n
n
c = lim M/I M = (an + I M )n ∈
M
M/I M an+1 ≡ an (mod I M )
←−
n∈N
A sequence (xk )k in M is an I-adic Cauchy sequence, if for each n there exists kn such that
xkn − xkn +m ∈ I n M for all m ∈ N. As usual, we say two Cauchy sequences (xk )k , (yk )k are
equivalent if (xk − yk )k converges to 0. In particular, (xk ) is equivalent to (xkn )n . Hence each
equivalence class of Cauchy sequences in M contains a so-called coherent sequence (an )n which
c is isomorphic to the set of equivalence classes
satisfies an+1 ≡ an (mod I n M ) for all n. Hence M
of Cauchy sequences.
T
c via m 7−→ (m)n .
If n∈N I n M = 0, then M is I-adically separated and we can embed M into M
c.
We say that M is complete if M ' M
Example 3.1. Let V be a discrete valuation domain with maximal ideal (t) and valuation v. By
Vb we denote the (t)-adic completion of V .
Moreover, let (an )n be a (t)-adic Cauchy sequence with limit a ∈ Vb . If a = 0, then for each
k ∈ N there exists n0 ∈ N such that an ∈ (tk ) for all n ≥ n0 and hence lim v(an ) = ∞.
If a 6= 0, then there exist k, m0 ∈ N such that an ∈
/ (tk ) for n ≥ m0 . However, the sequence
(an+1 − an )n converges to 0 and hence there exists m1 ∈ N such that an+1 − an ∈ (tk ) for all
n ≥ m1 . If m = max{m0 , m1 }, then for all n ≥ m,
v(an+1 ) = v(an+1 − an + an ) ≥ min{v(an+1 − an ), v(an )} = v(an ) < k
holds and the sequence (v(an ))n stabilizes at v(am ).
For a ∈ Vb , we set v(a) = lim v(an ) = v(am ). This extends v to a discrete valuation on Vb .
We are in particular interested in the case where M is a finitely generated module over a
Noetherian ring R.
Fact 3.2 (Artin-Rees, cf. [5, Theorem 8.5]). Let R be a Noetherian ring, I an ideal of R and M
a finitely generated R-module.
If N is an R-submodule of M , then there exists an integer r such that for all k ≥ 0
I r+k M ∩ N = I k (I r M ∩ N ).
Corollary 3.3 (cf. [5, Theorem 8.9, Theorem 8.10]). Let R be a Noetherian ring, I an ideal of R
and M a finitely generated R-module.
T
(1) If N = n∈N I n M , then there exists an a ∈ R with aN = 0 and 1 − a ∈ I.
(2) If I ⊆ Jac(R), then M is I-adically separated and every submodule of M is I-adically
closed.
Proof. (1): According to Fact 3.2, there exists r ∈ N such that (I r+1 M ) ∩ N = I(I r M ∩ N ) ⊆ IN
and hence
\
IN ⊂ N =
I n M ⊆ (I r+1 M ) ∩ N ⊆ IN.
n∈N
Consequently, N = IN and it follows from Nakayama’s lemma that there exists an a ∈ R with
1T− a ∈ I and aN = 0. (2): Since I ⊆ Jac(R), the element a from (1) is a unit of R. Therefore
n
if P is a submodule of M , it follows that
Tn∈N I n M = 0 and M is separated. Consequently,
T
n
I
M/P
=
0
=
P
M/P
and
therefore
P
+
I
M = P.
n∈N
n∈N
Fact 3.4 (cf. [5, Theorem 8.7]). Let R be a Noetherian ring, I an ideal and M a finitely generated
b M
c be the I-adic completions of R and M , respectively.
module. Further, let R,
b ⊗R M ' M
c via (lim rn , m) 7→ lim rn m. In particular, if R is I-adically complete, so
Then R
M is I-adically complete.
A NOTE ON ANALYTICALLY IRREDUCIBLE DOMAINS
5
Proposition 3.5 (cf. [5, Theorem 8.4]). Let R be a complete ring with respect to an ideal I of R
and M an I-adically separated R-module.
If M/IM is a finitely generated R/I-module, then M is a finitely generated R-module.
Proof. Let m1 , . . . , mt ∈ M such that their projections modulo IM generate M/IM as R/IPt
module. Then M = i=1 Rmi + IM and for x ∈ M , there exist r0,i ∈ R and x1 ∈ IM such that
Pt
x = i=1 r0,i mi + x1 . Then again, for x1 ∈ IM , there exist r1,i ∈ R and x2 ∈ I 2 M such that
Pt
x1 = i=1 r1,i mi + x2 . For j > 2, we successively choose rj−1,i ∈ I j−1 and xj ∈ I j M such that
Pn
Pt
xj−1 = i=1 rj−1,i mi + xj . Then ( j=0 rj,i )n∈N is a Cauchy sequence in R which has a limit
ri ∈ R. Moreover,
t
X
\
x−
ri mi ∈
I nM = 0
i=1
n∈N
and therefore M is generated by m1 , . . . , mt .
b Let V be a discrete valuation domain with valuation v and Vb its
3.1. Zero-divisors in R.
completion. If 0 6= a, b ∈ Vb , then v(a), v(b) < ∞ according to Example 3.1. Consequently,
v(ab) = v(a)v(b) < ∞ which implies that Vb is a domain.
However, in general the completion of a domain may not be a domain.
Proposition 3.6 (cf. [2, Lemma III.3.4]). Let R be a Noetherian domain and I an ideal of R.
If there exist ideals J1 and J2 of R such that I = J1 ∩ J2 and R = J1 + J2 , then the I-adic
b of R is not a domain.
completion R
Proof. Since J1 and J2 are coprime there exist a1 ∈ J1 and a2 ∈ J2 such that 1 = a1 + a2 . If k ≥ 1
is an integer, then
2k X
2k i 2k−i
a1 a2
∈ J1k + J2k
1 = (a1 + a2 )2k =
i
i=0
which implies that J1k and J2k are coprime ideals. Hence there exist bk , ck ∈ R such that
bk ≡ 0 mod J1k ,
bk ≡ 1 mod J2k
ck ≡ 1 mod J1k ,
ck ≡ 0 mod J2k
for all k ∈ N. Since bk+1 − bk ≡ ck+1 − ck ≡ 0 mod J1k ∩ J2k = J1k J2k = I k , the sequences (bk )k
b and c = lim ck ∈ R
b their I-adic limits. By
and (ck )k converge I-adically. Let b = lim bk ∈ R
k
k
construction, b 6= 0 and c 6= 0. However, since bk ck ≡ 0 mod J1 ∩ J2 = 0 for all k, it follows that
bc = 0. Hence b and c are non-zero zero-divisors.
It follows from Proposition 3.6 that the completion of a domain may contain zero-divisors,
cf. Proposition 3.8. However, constant sequences behave well as the next lemma states.
b the I-adic completion of
Lemma 3.7. Let R be a Noetherian ring, I a proper ideal of R and R
b
R. If 0 6= d ∈ R is not a zero-divisor in R, then d is not a zero-divisor in R.
Proof. By Fact 3.2, there exists r ∈ N such that for all n ∈ N
(1)
I n+r ∩ dR = I n (I r ∩ dD) ⊆ dI n
b such that dx = 0. Then (dxn )n is an I-adic Cauchy sequence with limit 0.
Let x = limn xn ∈ R
Hence for each n ∈ N there exists t0 ∈ N such that dxt ∈ I n+r for all t ≥ t0 . Then dxt ∈ dI n by
Equation (1) and therefore xt ∈ I k . Hence 0 = limn xn = x.
Proposition 3.8. Let (D, m) be a Noetherian, local domain with quotient field K and D ⊂ R ⊂ K
be an intermediate ring such that R is finitely generated as D-module.
Then the following assertions hold:
(1) R is semilocal,
(2) the m-adic topology on D coincides with subspace topology induced by the Jac(R)-adic
topology on R.
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ROSWITHA RISSNER
b of D is not a domain.
(3) If R is not local, then the m-adic completion D
Proof. R is finitely generated as module over the Noetherian domain D and hence a Noetherian
domain. Moreover, the ring extension D ⊆ R is integral which implies that all prime ideals of R
which lie over m are maximal by Fact 2.4 and are therefore minimal prime ideals of mR. Hence
there are only√finitely many maximal ideals N1 , . . . , Nn in R which proves (1).
Tn
(2): Since mR = i=1 Ni = Jac(R), there exists ` ∈ N such that Jac(R)` ⊆ mR. Then
Jac(R)`k ⊆ (mR)k ⊆ Jac(R)k
and hence the the Jac(R)-adic and the mR-adic topology coincide on R. Thus it suffices to prove
that the m-adic topology on D coincides with subspace topology induced by the mR-adic topology
on R. Clearly, mk ⊆ mk R ∩ D holds for all k. On the other hand, Fact 3.2 implies that there
exists an integer r such that
mk+r R ∩ D = mk (mr R ∩ D) ⊆ mk
for all k ≥ 0 and hence the topologies coincide.
(3): Let M1 , . . . , Mn be the maximal ideals of R with n > 1. We set J1 = M1 · · · Mn−1 =
M1 ∩ · · · ∩ Mn−1 and J2 = Mn . Then J1 ∩ J2 = Jac(R) and J1 + J2 = R. By Theorem 3.6, the
b of R is not a domain.
Jac(R)-adic completion R
b
b Since R
b is not a domain, there exist 0 6= b, c ∈ R
b
By (2), the D is a topological subspace of R.
with bc = 0. However, R is a finitely generated D-module and therefore there exists 0 6= d ∈ D
b ⊆ D.
b Hence (db)(dc) = 0 with db, dc ∈ D.
b By Lemma 3.7, d is not a zero-divisor in D
b
with dR
b
and therefore 0 6= db is a zero-divisor in D.
4. Structure theorem for complete, local domains
In this section we discuss the structure theorem for complete, Noetherian, local domains. As
Proposition 4.7 states, a complete, Noetherian, local domain (D, m) contains a regular local
subring S such that D is finitely generated as S-module. Moreover, S ' RJX1 , . . . , Xm K where R
is either a field which is isomorphic to D/m or a complete discrete valuation domain whose residue
field is isomorphic to D/m. This result allows us in the next section to reduce the investigation
to power series rings of this form.
For a proper ideal I of a ring R, let ht(I) denote the height of I, that is,
ht(I) = max{dim(AP ) | P minimal prime ideal of I}
where dim(AP ) denotes the Krull dimension of the localization AP of A at the prime ideal P . For
Noetherian rings, the height of an ideal is bounded by the number of generated according to the
Krull’s generalized principal ideal theorem:
Fact 4.1 (cf. [3, Theorem 10.2]). Let R be a Noetherian ring and x1 , . . . , xm ∈ R. If I =
(x1 , . . . , xm ) ( R is a proper ideal, then ht(I) ≤ m.
Proposition 4.2 (cf. [6, (9.5)]). Let (D, m) be a Noetherian local ring.
Then n = dim(D) is finite. Moreover, if y1 ∈ D is an element with ht(y1 ) = 1, then there exist
elements y2 , . . . , yn such that ht(y1 , . . . , yi ) = i for 1 ≤ i ≤ n. In particular, (y1 , . . . , yn ) is an
m-primary ideal.
Proof. Since D is Noetherian, m is finitely generated and therefore ht(m) is finite according to
Fact 4.1. Hence n = dim(D) = ht(m) is finite.
Assume we know y2 , . . . , yj ∈ D such that ht(y1 , . . . , yi ) = i for 1 ≤ i ≤ j. Let P1 , . . . , Pr be
the minimal primes of (y1 , . . . , yj ). Then ht(Pt ) =
Srj < n = ht(m) for 1 ≤ t ≤ r. Hence Pt ( m
for all t and there exists an element yj+1 ∈ m \ t=1 Pt . Therefore ht(y1 , . . . , yj+1 ) = j + 1 by
Fact 4.1. Repeating this argument gives us the desired elements y2 , . . . , yn .
Finally, since ht(y1 , . . . , yn ) = n, m is the only minimal prime ideal of (y1 , . . . , yn ) and therefore
this is an m-primary ideal.
Remark. Fact 4.1 implies that every m-primary ideal has height n and hence at least n generators.
A NOTE ON ANALYTICALLY IRREDUCIBLE DOMAINS
7
Definition 4.3. Let (R, m) be a Noetherian, local ring.
(1) We say y1 , . . . , yn ∈ R form a system of parameters of R if they generate an m-primary
ideal.
(2) We say R is regular local, if m is generated by dim(R) elements.
Example 4.4.
(1) Let S = kJX1 , . . . , Xm K be the ring of formal power series in m variables over a field k.
Then m = (X1 , . . . , Xm ) is a maximal ideal since S/m ' k is a field. Moreover S \ m = k
consists of units and hence S is local. By Fact 4.1 it follows that dim(S) ≤ m. On the other
hand, S/(X1 , . . . , Xi ) ' kJXi+1 , . . . , Xm K is a domain for all 1 ≤ i ≤ m and therefore
0 ( (X1 ) ( (X1 , X2 ) ( · · · ( (X1 , X2 , . . . , Xm )
is a chain of prime ideals of S of length m. Hence dim(S) = m and S is a regular local
ring.
(2) Let R be a discrete valuation domain with maximal ideal (t) and S = RJX1 , . . . , Xm K.
With the same arguments as above, one can show that m = (t, X1 , . . . , Xm ) is the only
maximal ideal of S and that dim(S) = m + 1, so S is a regular local ring.
We are now ready to discuss Cohen’s structure theorem.
Definition 4.5. Let (D, m) be a complete, Noetherian, local domain. We say
(1) D is of equal characteristic, if char(D) = char(D/m) and
(2) D is of unequal characteristic, if char(D) 6= char(D/m).
If char(D/m) = 0, it follows that char(D) = 0 and therefore Z ⊆ D. However, Z ∩ m = 0 which
implies that every integer is invertible in D and thus Q ⊆ D. Similarly, if char(D) = p > 0, then
Z/pZ is contained in D and char(D/m) = p. Hence, a domain of equal characteristic contains a
field. On the other hand, if D contains a field k, then char(k) = char(D). Let π : D −→ D/m
denote the canonical projection. Then π(k) is a subfield of D/m and since char(π(k)) = char(k) it
follows that D is of equal characteristic. Indeed, it is possible to show that a domain D of equal
characteristic contains a field k with π(k) = D/m, cf. Fact 4.6.
If D is a domain of unequal characteristic, then char(D) = 0 and char(D/m) = p > 0. In this
case it is possible to show that D contains a complete discrete valuation ring (R, pR) such that
the residue fields of R and D are isomorphic. We summarize these results in the next proposition.
However, the proof goes beyond the scope of this paper. We refer to Matsumura’s textbook [5]
for details.
Fact 4.6 (cf. [5, Theorem 28.3, Theorem 29.3]). Let (D, m) be a complete, Noetherian, local
domain.
(1) If D is of equal characteristic, then D contains a field k which is isomorphic to D/m via
d 7−→ d + m. We say k is a coefficient field of D.
(2) If D is of unequal characteristic and char(D/m) = p, then D contains a complete discrete
valuation domain (R, pR) such that R/pR is isomorphic D/m via r + pR 7−→ r + m. We
say R is a coefficient ring of D.
The existence of a coefficient field or coefficient ring, respectively, is crucial for the proof of the
structure theorem which we state in the next proposition.
Proposition 4.7 (cf. [5, Theorem 29.4.(iii)]). Let D be a complete, Noetherian, local domain of
Krull dimension n.
There exists a complete regular local subring S ⊆ D such that D is finitely generated as Smodule. Moreover,
(1) if D is of equal characteristic, then S ' kJX1 , . . . , Xn K where k is a coefficient field of D.
(2) if D is of unequal characteristic, then S ' RJX2 , . . . , Xn K where R is a coefficient ring of
D.
8
ROSWITHA RISSNER
Proof. Let m be the maximal ideal of D. First, we consider the case where D is of equal characteristic. According to Fact 4.6, D contains a field k with k ' D/m. Moreover, D has a system
of parameters y1 , . . . , yn by Proposition 4.2 since ht(y1 ) = 1 for a non-unit y1 6= 0. Let X1 , . . . ,
Xn be variables and set T = kJX1 , . . . Xn K. We define the k-homomorphism ϕ : T −→ D by
ϕ(Xi ) = yi for
Pn1 ≤ i ≤ n. Let S = kJy1 , . . . , yn K denote the image of T under ϕ.
Set I =
i=1 yi S. Then ID ⊆ m and S/I ' k ' D/m and therefore D/m is a finitely
generated S/I-module. However, D/m ' (D/ID)/(m/ID) and m is a finitely generated ideal
which implies that D/ID is a finitely generated S/I-module too.
Moreover, D/ID is I-adically discrete and hence separated and S is I-adically complete. We can
apply Proposition 3.5 to conclude that D is finitely generated as S-module. Hence the extension
S ⊆ D is integral and dim(S) = dim(D) = n by Fact 2.4. However, since dim(kJX1 , . . . , Xn K) = n
and S ' kJX1 , . . . , Xn K/ ker(ϕ) it follows that ker(ϕ) = 0. Hence S ' kJX1 , . . . , Xn K is a power
series ring over a field k and hence regular local (see Example 4.4).
For the case of unequal characteristic, we need to adapt the proof as follows. Let char(D) = 0
and char(D/m) = p > 0. It follows from Fact 4.6 that D contains a complete discrete valuation
domain R with maximal ideal pR such that R/pR ' D/m. According to Proposition 4.2, we can
choose a system of parameter y1 = p, y2 , . . . , yn ∈ D.
We set T = RJX2 , . . . , Xn K and define ϕ : T −→ D by ϕ(Xi ) = yi for 2 ≤ i ≤ n. Now
S = ϕ(T ) = RJy2 , . . . , yn K is the image of T under ϕ.
Pn
Pn
As in the previous case dim(T ) = n, see Example 4.4. Moreover, I = i=1 yi S = pS + i=2 yi S
and hence S/I ' R/pR ' D/m. With the same argumentation as above we can conclude that
n = dim(S) = dim(T ). Hence ker(ϕ) = 0 and S ' T is a power series ring over a discrete valuation
domain which is regular local.
5. Finiteness of integral closure
Let D be a complete, Noetherian, local domain with quotient field K and K ⊆ L a finite field
extension. The goal of this section is to prove that the integral closure intL (D) of D in L is finitely
generated as D-module, see Proposition 5.3. This allows us to conclude in Corollary 5.4 that
the integral closure R0 of a complete, Noetherian, local, reduced ring R is finitely generated as
R-module. The latter result is essential in the proof of Theorem 1 in the next section.
Following Nagata’s textbook [6], we exploit the structure of complete, Noetherian, local domains, see Proposition 4.7. If K ⊆ L is a finite field extension and intL (D) the integral closure of
D in L. According to Proposition 4.7, D contains a regular local subring S such that D is finitely
generated as S-module, see Figure 2. If F is the quotient field of S, then the extension F ⊆ L
is finite. Moreover, the extension S ⊆ D is integral and hence intL (D) = intL (S) is the integral
closure of S in L.
L
intL (D)
K
D
f.g.
F
=
S
RJX1 , . . . , Xm K
Figure 2. R contains a regular local subring S
If we show that intL (S) is finitely generated as S-module, then it follows that intL (D) is a
finitely generated D-module. Therefore, Proposition 4.7 allows us to reduce the investigation to
domains of the form S = RJX1 , . . . , Xm K where R is either a field or a complete discrete valuation
domain. Note that in both cases S is integrally closed according to Proposition 2.3.
A NOTE ON ANALYTICALLY IRREDUCIBLE DOMAINS
9
To prove that the integral closure of S in L is finitely generated, we distinguish between two
cases, either the field extension F ⊂ L is separable or it is inseparable.
Proposition 5.1 (cf. [6, (10.16)]). Let S be an integrally closed, Noetherian domain with quotient
field F and F ⊆ L a finite field extension.
If F ⊆ L is separable, then the integral closure intL (S) of S in L is finitely generated as Smodule.
Proof. Since F ⊆ L is a finite separable field extension there exists an element α ∈ L such that
L = F (α) (cf. [4, Theorem V.4.6]). Without restriction we can assume α ∈ intL (S).
Then L = F (α) is the quotient field of S[α] and since S[α] ⊆ intL (S) is an integral extension it
follows that intL (S) = intL (S[α]) is the integral closure of S[α].
Since intL (S) ⊆ F (α) = F [α], it follows that for every b ∈ intL (S) there exist x0 , . . . , xn−1 ∈ F
Pn−1
such that b = j=0 xj αj where n is the degree of the minimal polynomial f ∈ F [X] of α.
Let F ⊆ F be a splitting field of f and a1 = α, a2 , . . . , an ∈ F be the (different) roots of f .
Then
ϕi : F [α] −→ F
α 7−→ ai
is an F -homomorphism. Let A = (aji )i,j be an n × n-matrix, x = (xi )i b = (ϕi (b))i . Then Ax = b
Pn−1
since ϕi (b) = j=0 xj aji . If A# denotes the adjugate matrix of A, then
det(A)x = A# Ax = A# b
(2)
However, α ∈ intL (S) is integral over S and therefore there exists a monic polynomial g ∈ S[X]
such that g(α) = 0. Hence f | g in F [X] which in turn implies that g(ai ) = 0 for all 1 ≤ i ≤ n.
Consequently, a1 , . . . , an are elements of the integral closure intF (S) of S in F . Therefore both
A and A# are matrices with entries in intF (S). Moreover, if h ∈ S[X] is a monic polynomial
with h(b) = 0, then 0 = ϕi (h(b)) = h(ϕi (b)) which implies that ϕi (b) ∈ intF (S) for 1 ≤ i ≤ n.
In conclusion, the entries of the vector on the right-hand side of Equation (2) are integral over S
which implies that det(A)xi ∈ intF (S) for 1 ≤ i ≤ n.
Q
However, A is a Vandermonde matrix, so 0 6= d = det(A)2 = i<j (ai − aj )2 is the discriminant
of f . Hence d ∈ F and therefore dxi ∈ intF (S) ∩ F = S since S is integrally closed. It follows that
db =
n−1
X
dxj αj ∈ S[α].
j=0
and therefore 0 6= d intL (S) ⊆ S[α]. However, S[α] is a finitely generated S-module and hence
Noetherian. Therefore intL (S) ' d intL (S) is finitely generated too.
Proposition 5.2. Let S = kJX1 , . . . , Xn K be a power series ring in n variables over a field k with
quotient field F and L a finite purely inseparable field extension of F .
Then the integral closure intL (S) of S in L is finitely generated as S-module.
Proof. Let p > 0 be the characteristic of the field F and q = pe = [L : F ] < ∞ be the degree of
the field extension F ⊆ L. Since the extension is purely inseparable, every element a ∈ L is a q-th
root of an element in F . In particular, if a ∈ intL (S) is an integral element, then aq ∈ intL (S) ∩ F .
However,
2.3, S is integrally closed. Hence aq ∈ S and there exists cν ∈ k with
P by Proposition
ν1
q
νn
a = ν∈Nn cν X1 · · · Xn where ν = (ν1 , . . . , νn ) ∈ Nn0 .
Let F be an algebraically closed extension of F that contains L. Then F contains elements dν
and Yi such that cν = dqν for all ν and Xi = Yiq for all 1 ≤ i ≤ n. Hence
!q
X
X
ν1
ν1
q
νn
νn
a =
cν X1 · · · Xn =
dν Y1 · · · Yn
ν∈Nn
ν∈Nn
10
ROSWITHA RISSNER
holds and since the Frobenius homomorphism b 7→ bp is injective, it follows that
X
a=
dν Y1ν1 · · · Ynνn .
ν∈Nn
Thus
1
intL (S) ⊆ k q JY1 , . . . , Yn K
1
where k q = {d ∈ F | dq ∈ k} denotes the subfield of F consisting of the q-th roots of elements in
k.
1
Let J = (Y1 , . . . , Yn ) be the ideal of k q JY1 , . . . , Yn K be the ideal generated by Y1 , . . . , Yn . Then
for f ∈ intL (S) there exists a minimal i ∈ N0 such that f ∈ J i . We set h(f ) to be the homogeneous
summand of f of degree i.
Claim. If f1 , . . . , fr ∈ intL (S) are elements such that h(f1 ), . . . , h(fr ) are linearly independent
over k[X1 , . . . , Xn ], then f1 , . . . , fr are S-linearly independent.
To prove the claim, let h1 , . . . , hr ∈ S with hi 6= 0 for some i. Then d = min{j ≥ 0 |
h(hi ) h(fi ) ∈ J j for some 0 ≤ i ≤ r} is finite and we set I = {0 ≤ i ≤ r | h(hi ) h(fi ) ∈ J d }.
h(fr ) are linearly independent over k[X1 , . . . , Xn ] which implies that
P By assumption, h(f1 ), . . . ,P
r
d
i∈I h(hi ) h(fi ) 6= 0. Hence
i=1 hi fi ∈ J and
!
r
X
X
h
hi fi =
h(hi ) h(fi ).
i=1
i∈I
Pr
In particular, i=1 hi fi 6= 0 which proves the claim.
Now let w1 , . . . , wq be a F -basis of L. Without restriction, we can assume that wi ∈ intL (S)
for all 1 ≤ i ≤ q. For each f ∈ intL (S) there exist h1 , . . . , hq ∈ S and v1 , . . . , vn ∈ N such that
X1ν1
· · · Xnνn f
=
q
X
hi wi .
i=1
According to the claim above, this implies that h(f ), h(w1 ), . . . , h(wq ) are linearly dependent
over k[X1 , . . . , Xn ]. This further implies that each coefficient of h(f ) is k-linearly dependent on
the coefficients c1 , . . . , ct of h(w1 ), . . . , h(wq ). Hence h(f ) ∈ k(c1 , . . . , ct )[Y1 , . . . , Yn ].
1
Moreover, k(c1 , . . . , ct ) is a finite dimensional extension of k since c1 , . . . , ct ∈ k q are algebraic
over k. If d1 , . . . , dr is a k-basis of k(c1 , . . . , ct ), then
k(c1 , . . . , ct )[Y1 , . . . , Yn ] =
r
X
i=1
kdi [Y1 , . . . , Yn ] =
r
X
X
Sdi Y1ν1 · · · Ynνn
i=1 0≤νi ≤q−1
1≤i≤n
is a finitely generated S-module and hence Noetherian. Thus the S-submodule
M = {h(f ) | f ∈ intL (S)}
is finitely generated as well.
Let b1 , . . . , bm ∈ intL (S) such that h(b1 ), . . . , h(bm ) generate M . Since b1 , . . . , bm are integral
over S, it follows that the ring T = S[b1 , . . . , bm ] is finitely generated as R-module. Hence, S is a
topological subspace of T by Proposition 3.8 and T is complete with respect to the maximal ideal
m = (X1 , . . . , Xn ) of S by Fact 3.4.
To finish the proof, we show that every element in intL (S) is the limit of an mT -adic sequence
in T . Then T ⊆ intL (S) ⊆ Tb = T which implies that intL (S) = T is a finitely generated S-module.
Let d ∈ intL (S) be an arbitrary element. We recursively construct a sequence (dn )n∈N in T
which converges towards d. Set d0 = 0 and assume that we already determined d1 , . . . , dn−1 such
that d − di ∈ (mT )i .
Pt
Since h(d − dn−1 ) ∈ M it follows that h(d − dn−1 ) = i=1 λi h(bi ) for λ1 , . . . , λm . Without
restriction, we can assume that the λi ∈ S are homogeneous polynomials which implies that
t
X
h(
λi bi ) = h(d − dn−1 ).
i=1
A NOTE ON ANALYTICALLY IRREDUCIBLE DOMAINS
11
Pt
Hence dn = dn−1 + i=1 λi bi ∈ T is an element with d − dn ∈ (mT )n . Thus d = limn→∞ dn is
the limit of a sequence in T .
Proposition 5.3. Let D be a complete, local, Noetherian domain with quotient field K and K ⊆ L
a finite field extension.
Then the integral closure intL (D) of D in L is finitely generated as D-module.
Proof. According to Proposition 4.7, D contains a complete regular local subring S such that D
is finitely generated S-module. Let F denote the quotient field of S. Hence F ⊆ L is a finite field
extension, S ⊆ D is integral and intL (S) = intL (D), cf. Figure 2. Therefore, it suffices to show
that intL (S) is a finitely generated S-module.
Moreover, S ' RJX1 , . . . , Xm K where R is either a field or a complete discrete valuation domain
by Proposition 4.7. In particular, it follows from Proposition 2.3 that S is integrally closed. Consequently, if the field extension F ⊆ L is separable, then the assertion follows from Proposition 5.1.
Let us assume that F ⊆ L is inseparable. Then char(F ) = p > 0 and hence char(S) = char(D) =
p which implies that D is of equal characteristic. Therefore S ' RJX1 , . . . , Xm K where R is a field
and R ' D/m.
Let L(n) be the normal hull of L and E be the fixed field of the automorphism group AutK (L(n) )
of K ⊆ L(n) . Then K ⊆ E is a purely inseparable extension and E ⊆ L(n) is a separable extension,
cf. [4, Proposition V.6.11], see Figure 3.
Moreover, since K ⊆ L is a finite extension, it follows that K ⊆ L(n) is finite which in turn
implies that K ⊆ E is a finite extension too.
L(n)
LE
intLE (S)
separable
L
intL (S)
E
intE (S)
purely inseparable
F
S
Figure 3. Field extensions of F and the corresponding integral closures of S
Hence E ⊆ LE is a finite separable extension and it follows from Proposition 5.1 that intLE (S) =
intLE (intE (S)) is finitely generated as intE (S)-module. In addition, intE (S) is finitely generated
as S-module according to Proposition 5.2.
Consequently, intLE (S) is finitely generated as S-module and therefore a Noetherian S-module.
However, intL (S) is an S-submodule of intLE (S) and thus finitely generated.
We conclude this section with the analogous assertion for complete, Noetherian, local, reduced
rings.
Corollary 5.4. Let R be a complete, Noetherian, local ring.
If R is reduced, then the integral closure R0 of R is finitely generated as R-module.
Proof. Let P1 , . . . , Pn be the minimal prime ideals of R. For 1 ≤ i ≤ n, let Qi be the quotient
field of the Noetherian, local domain R/Pi . Then Q = Q1 × · · · × Qn is the total ring of quotients
of R/P1 × · · · × R/Pn . Moreover,
(3)
intQ (R/P1 × · · · × R/Pn ) = intQ1 (R/P1 ) × · · · × intQn (R/Pn )
The Noetherian, local domain R/Pi is a finitely generated R-module and hence m-adically
complete by Fact 3.4. As the m-adic topology coincides with the m/Pi -adic topology on R/Pi , it
12
ROSWITHA RISSNER
follows that intQi (R/Pi ) is a finitely generated (R/P
Qn 5.3. ToQni )-module according to Proposition
gether with Equation (3), it now follows that intQ ( i=1 R/Pi ) is a finitely generated ( i=1 R/Pi )module.
Qn
Qn
However, i=1 R/Pi is clearly a finitely generated R-module. It follows that intQ ( i=1 R/Pi )
is finitely generated as R-module as well and therefore a Noetherian R-module. Let
ε : R −→ R/P1 × · · · × R/Pn
r 7−→ (r + P1 , . . . , r + Pn )
Tn
ThenQker(ε) = i=1 Pi = nil(R) = 0 since R is reduced byQhypothesis. Hence we can embed R
n
n
into i=1 R/Pi via ε. Similarly, we can embed R0 into intQ ( i=1 R/Pi ) since ε can be canonically
extended to the total ring of quotients T of R, see Figure 4.
Q
i
R/Pi
f.g.
i
intQi (R/Pi )
=
Q
Q
intQ ( i R/Pi )
Q
R0
T
f.g.
R
Figure 4. Embeddings via ε
Qn
Thus R0 is isomorphic to a submodule of the Noetherian R-module intQ ( i=1 R/Pi ) and hence
finitely generated.
6. Proof of the theorem
Finally, we are ready to give a proof of Theorem 1.
b and integral
Definition 6.1. Let (D, m) be a Noetherian, local domain with m-adic completion D
0
closure D . We say
(1) D is unibranched, if D0 is local.
b is a reduced ring.
(2) D is analytically unramified, if D
b
(3) D is analytically irreducible, if D is a domain.
Theorem 1. Let (D, m) be a one-dimensional, Noetherian, local domain with integral closure D0 .
Then the following assertions are equivalent:
(1)
(2)
(3)
(4)
D is analytically irreducible.
D is unibranched and analytically unramified.
D is unibranched and D0 is finitely generated as a D-module.
D is unibranched and if m0 denotes the maximal ideal of the integral closure, then the
m-adic topology on D coincides with subspace topology induced by m0 .
b is a domain and therefore is a reduced ring.
Proof. (1) ⇒ (2): By assumption, D
Assume that D is not unibranched and let M1 6= M2 be two different maximal ideals of the
integral closure D0 of D. Let a1 ∈ M1 \ M2 and a2 ∈ M2 \ M1 . Then R = D[a1 , a2 ] ⊆ D0 is an integral extension and therefore finitely generated as D-module. Hence, according to Proposition 3.8,
R is a semilocal domain. However, the extension R ⊆ D0 is integral and therefore Ni = Mi ∩ R are
maximal ideals of R for i = 1, 2 according to Fact 2.4. Due to the choice of a1 and a2 , N1 6= N2
b is not a domain.
and R is semilocal but not local. It follows from Proposition 3.6 that D
0
(2) ⇒ (3): If D is not finitely generated as D-module then there exists an infinite strictly
ascending chain of subrings Di of D0 which are finitely generated as D-modules.
A NOTE ON ANALYTICALLY IRREDUCIBLE DOMAINS
13
ci denote the m-adic completion of Di .
Let K be the quotient field of D and Di for all i and D
c
c
ci ∩ K = bDi and hence a ∈ Di .
If ∈ Di ∩ K, then a ∈ bDi ∩ K. However, by Corollary 3.3, bD
b
ci ∩ K = Di ( Di+1 = D
c
[
[
Hence D
i+1 ∩ K which implies that Di ( Di+1 for all i.
ci ' Di ⊗D D
b is a finitely generated D-module.
b
ci is
Moreover, according to Fact 3.4, D
Hence D
0
b
b
contained in the integral closure D of D in its total ring of quotients. Consequently, the extension
b ⊆D
b 0 contains the infinite strictly ascending chain of intermediate rings D
ci . Hence D
b 0 is not
D
b
b is not reduced by Corollary 5.4.
finitely generated as D-module
which implies that D
(3) ⇒ (4): The assertion immediately follows from Proposition 3.8.
c0 be the m0 -adic completion of D. Since the m0 -adic topology induces the
(4) ⇒ (1): Let D
c0 . Hence D
b is the topological
m-adic topology on D, it follows that D is a topological subspace of D
c0 .
closure of D in D
By assumption D0 is local, so D0 is a discrete valuation domain according to Corollary 2.6. As
c0 of D0 is also a discrete valuation domain.
shown in Example 3.1, the m0 -adic completion D
0
b
c
Hence D is a subring of the domain D and hence it is a domain itself.
a
b
Remark. There are examples of one-dimensional, Noetherian, local domains which are unibranched but not analytically irreducible, cf. [8].
References
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Reading, Mass.-London-Don Mills, Ont., 1969.
[2] Paul-Jean Cahen and Jean-Luc Chabert. Integer-Valued Polynomials. Mathematical Surveys and Monographs
48. American Mathematical Society, 1997.
[3] David Eisenbud. Commutative algebra, volume 150 of Graduate Texts in Mathematics. Springer-Verlag, New
York, 1995. With a view toward algebraic geometry.
[4] Serge Lang. Algebra, volume 211 of Graduate Texts in Mathematics. Springer-Verlag New York, 2002.
[5] Hideyuki Matsumura. Commutative Ring Theory. Cambridge University Press, 1986.
[6] Masayoshi Nagata. Local rings, volume 13 of Interscience Tracts in Pure and Applied Mathematics. Interscience
Publishers a division of John Wiley & Sons, 1962.
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[8] Bruce Olberding. One-dimensional bad Noetherian domains. Trans. Amer. Math. Soc., 366(8):4067–4095, 2014.
Institut für Analysis und Zahlentheorie, TU Graz, Kopernikusgasse 24, 8010 Graz, Austria
E-mail address: [email protected]