Hereditary Properties between a Ring and its Maximal Subrings
We study the existence of maximal subrings and hereditary properties between a ring and its maximal subrings. Some new techniques for establishing the existence of maximal subrings are presented. It is shown that if R is an integral domain and S is a maximal subring of R, then the relation dim(R) = ...
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| author | Azarang, A. Karamzadeh, O. A. S. Namazi, A. Азаранг, А. Карамзадех, О. А. С. Намазі, А. |
| author_facet | Azarang, A. Karamzadeh, O. A. S. Namazi, A. Азаранг, А. Карамзадех, О. А. С. Намазі, А. |
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| description | We study the existence of maximal subrings and hereditary properties between a ring and its maximal subrings. Some new techniques for establishing the existence of maximal subrings are presented. It is shown that if R is an integral domain and S is a maximal subring of R, then the relation dim(R) = 1 implies that dim(S) = 1 and vice versa if and only if (S : R) = 0. Thus, it is shown that if S is a maximal subring of a Dedekind domain R integrally closed in R; then S is a Dedekind domain if and only if S is Noetherian and (S : R) = 0. We also give some properties of maximal subrings of one-dimensional valuation domains and zero-dimensional rings. Some other hereditary properties, such as semiprimarity, semisimplicity, and regularity are also studied. |
| first_indexed | 2026-03-24T02:24:06Z |
| format | Article |
| fulltext |
UDC 512.5
A. Azarang, O. A. S. Karamzadeh, A. Namazi (Chamran Univ., Ahvaz, Iran)
HEREDITARY PROPERTIES BETWEEN A RING
AND ITS MAXIMAL SUBRINGS
СПАДКОВI ВЛАСТИВОСТI МIЖ КIЛЬЦЕМ
ТА ЙОГО МАКСИМАЛЬНИМИ ПIДКIЛЬЦЯМИ
We study the existence of maximal subrings and hereditary properties between a ring and its maximal subrings. Some new
techniques for establishing the existence of maximal subrings are given. It is shown that if R is an integral domain and S is
a maximal subring of R, then the relation dim(R) = 1 implies that dim(S) = 1 and vice versa if and only if (S : R) = 0.
Consequently, we show that if S is a maximal subring of a Dedekind domain R that is integrally closed in R, then S is a
Dedekind domain if and only if S is Noetherian and (S : R) = 0. We also give some properties of maximal subrings of
one-dimensional valuation domains and zero-dimensional rings. Some other hereditary properties such as semiprimarity,
semisimplicity, and regularity are also studied.
Вивчається iснування максимальних пiдкiлець та спадковi властивостi мiж кiльцем та його максимальними пiд-
кiльцями. Наведено деякi новi методи встановлення iснування максимальних пiдкiлець. Показано, що якщо R —
iнтегральна область, а S — її максимальне пiдкiльце, то iз спiввiдношення dim(R) = 1 випливає, що dim(S) = 1,
i навпаки тодi i тiльки тодi, коли (S : R) = 0. Як наслiдок показано, що, якщо S є максимальним пiдкiльцем
дедекiндової областi R, яка є iнтегрально замкненою в R, то S є дедекiндовим пiдкiльцем тодi i тiльки тодi, коли
S є нетеровим та (S : R) = 0. Наведено також деякi властивостi максимальних пiдкiлець одновимiрних областей
нормування та нульвимiрних кiлець. Також вивчено деякi iншi спадковi властивостi, такi як напiвпримарнiсть,
напiвпростота та регулярнiсть.
1. Introduction. All rings in this article are commutative with 1 6= 0; all modules are unital. If S is a
subring of a ring R, then 1R ∈ S. In this paper the characteristic of a ring R is denoted by Char(R).
Let us call a ring with a maximal subring, a submaximal ring. Recently the existence of maximal
subrings and also some hereditary properties between a ring and any of its maximal subrings are
studied, see [1 – 5]. Let us, for the sake of the reader, first cite some old and new hereditary properties
which are shared between a ring and each of its maximal subrings. M. L. Modica proved that, if D is
a maximal subring of a ring R such that D is an integral domain and is integrally closed in R, then R
is an integral domain too, see [12] (Theorem 10). The latter result is also an immediate consequence
of [7] (Theorem 2.7). We remind the reader that all finite rings except Zn, where n is a natural
number, are submaximal. It is also interesting to note that whenever S is a finite maximal subring of
a ring R, then R must be finite, see [6] (Theorem 8), hence finiteness is a nice hereditary property
shared between a ring R and any maximal subring of R. The latter interesting fact is also an easy
consequence of [3] (the proof of Theorem 2.9) or [4] (Theorem 3.8). In [4] (Theorem 3.8) it is shown
that if S is a maximal subring of a commutative ring R, then S is Artinian if and only if R is Artinian
and is integral over S. Some other hereditary properties such as Noetherianity, zero-dimensionality
and von Neumann regularity are also studied in [4]. In [2], it is also shown that if S is a maximal
subring of an integral domain R, then S is a G-domain if and only if R is a G-domain. It is also
proved that if S is a maximal subring of R, then S is never a field and, similar to R, the only ring
endomorphism of S is the identity, see [2].
Now, let us sketch a brief outline of this paper. In Section 2, the existence of maximal subrings in
fields (via G-domain), Artinian rings, infinite direct product (via ideals) and finally the existence of
c© A. AZARANG, O. A. S. KARAMZADEH, A. NAMAZI, 2013
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 7 883
884 A. AZARANG, O. A. S. KARAMZADEH, A. NAMAZI
maximal subrings via R-modules are studied. In Section 3, we study hereditary properties between a
ring and its maximal subrings and properties of maximal subrings in some special rings such as zero-
dimensional rings and one-dimensional valuation domains. We first prove that under some conditions
the Hilbert property can be shared between a ring and its maximal subrings. Next we prove that if R
is an integral domain and S is a maximal subring of it, then dim(R) = 1 implies dim(S) = 1 and
vice versa if and only if (S : R) = 0. Consequently, we show that if S is a maximal subring of a
Dedekind domain R, which is integrally closed in R, then S is a Dedekind domain if and only if S
is Noetherian and (S : R) = 0. We also show that if S is a maximal subring of a zero-dimensional
ring R and S is integrally closed in R, then dim(S) = 1 and we give the structures of Max(S)
and Spec(S). The properties of maximal subrings of one-dimensional valuation domains are also
studied. Some other hereditary properties such as semiprimarity, semisimplicity and regularity are
also studied in Section 3.
Next, let us recall some standard definitions and notation for commutative ring theory which will
be used throughout the paper, see [9]. An integral domain D is called G-domain if the quotient field
of D is finitely generated as a ring over D. A prime ideal P of a ring R is called G-ideal if
R
P
is
a G-domain. A ring R is called Hilbert if every G-ideal of R is maximal. We also call a ring R,
not necessarily Noetherian, semi-local (resp. local) if Max(R) is finite (resp. |Max(R)| = 1). The
set of minimal prime ideals and the prime ideals, of a ring R are denoted by Min(R) and Spec(R),
respectively. As usual, let U(R) denote the set of all units of a ring R. The Jacobson and the nil
radical of a ring R are also denoted by J(R) and N(R), respectively. If D is an integral domain,
then the quotient field and the integral closure of D are denoted by Q(D) and D′, respectively. More
generally, if S is a subring of a ring R, then the integral closure of S in R is denoted by S′R. Finally,
let S ⊆ R be two rings, then the conductor ideal of this extension is denoted by (S : R), that is the
largest ideal of R which is contained in S, i.e., (S : R) = {x ∈ R | Rx ⊆ S}.
The following two theorems, whose proofs could be found in either [8] or [12], are needed.
Before presenting them, let us recall that whenever S is a maximal subring of a ring R, then one can
easily see that either R is integral over S or S is integrally closed in R.
Theorem 1.1. Let S be a maximal subring of a ring R. Then the following statements are true:
(1) (S : R) ∈ Spec(S).
(2) (S : R) ∈ Max(S) if and only if R is integral over S.
(3) If S is integrally closed in R, then (S : R) ∈ Spec(R).
Let us before presenting the second theorem, recall that if S is a maximal subring of a ring R and
X is a multiplicatively closed set in S, then one can easily see that either SX , the ring of fractions
of S with respect to X, is a maximal subring of RX or SX = RX .
Theorem 1.2. Let S be a maximal subring of a ring R. Then there exists a unique maximal
ideal M of S such that SM is a maximal subring of RS\M and for any other prime ideal P of S we
have SP = RS\P . Moreover, (S : R) ⊆M.
The unique maximal ideal in the previous theorem is called the crucial maximal ideal of the
extension S ⊆ R and we denote it by cruR(S).
2. Existence of maximal subrings and generalizations. In [3], it is proved that uncountable
fields, nonabsolutely algebraic fields (i.e., fields which are not algebraic over their prime subfields),
and fields with zero characteristic are submaximal. In what follows, for the sake of completeness, we
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HEREDITARY PROPERTIES BETWEEN A RING AND ITS MAXIMAL SUBRINGS 885
give new proofs to some slight extensions of these results, which are basic in our study. First, we give
the following theorem about the extension of G-domains. Let us recall that a ring R is submaximal
if and only if there exists a proper subring S of R and α ∈ R\S with S[α] = R. In particular, R has
a maximal subring containing S but not α, see [1] (Theorem 2.5).
Theorem 2.1. Let R be a G-domain with the quotient field F 6= R. Then any algebraic field
extension K of F is the quotient field of a nontrivial G-domain.
Proof. First, note that by the preceding comment, we may assume that R is a maximal subring of
F. Let R̄ be the integral closure of R in K. By [1] (Theorem 3.4) R is a G-domain and is integrally
closed in its quotient field F. We may assume that K 6= F. Since R is integrally closed in F, we
infer that R̄ 6= K. Now we claim that there exists an element u ∈ R with K = R̄[u−1] and this
completes the proof. To this end, we note that since R is a G-domain we have F = R[u−1], where u
is a nonunit element of R. We show that we also have K = R̄[u−1]. Clearly, R̄[u−1] ⊆ K. Hence let
x ∈ K be any element and we are to show that x ∈ R̄[u−1]. Since K is algebraic over F we infer that
xn + an−1x
n−1 + . . .+ a1x+ a0 = 0 for some a0, a1, . . . , an−1 ∈ F. But F = R[u−1] implies that
for any ai, i = 1, 2, . . . , n− 1, there exists a nonnegative integer mi such that ai = u−mibi for some
bi ∈ R. Now let m be a natural number such that m ≥ mi for all i and multiply the above equation by
umn. Hence we have the equation (umx)n+uman−1(u
mx)n−1+. . .+um(n−1)a1(u
mx)+umna0 = 0.
This implies that umx is integral over R (note, since ai = u−mibi, bi ∈ R, the coefficient of the
above equation are in R) and therefore umx ∈ R̄ which implies that x ∈ R̄[u−1], and we are done.
Theorem 2.1 is proved.
Corollary 2.1. Let F be a field with a nonfield maximal subring. Then any field extension of
F is submaximal. In particular, every field with zero characteristic and any nonabsolutely algebraic
field (such as uncountable fields) are submaximal.
Proof. Let K/F be a field extension. If K is algebraic over F, then we are done by the previous
theorem and the comment preceding it. Hence assume that K is not algebraic over F. Suppose that
S is a transcendence basis for K over F. Then K is algebraic over F (S) and hence we are done by
the previous theorem and the fact that F (S) has a maximal subring which is not a field. The final
part is evident.
Corollary 2.2. Let F be a field with a nonfield maximal subring. Then any F -algebra is sub-
maximal.
Proof. Let R be an F -algebra and M be a maximal ideal of R, then R/M contains a copy of
F. Thus R/M and therefore R are submaximal by Corollary 2.1.
In [4] (Theorem 3.8), it is proved that whenever S is a maximal subring of a ring R, then S is
Artinian if and only if R is Artinian and integral over S. The following is now in order.
Corollary 2.3. Let R be an Artinian ring with a non-Artinian maximal subring. Then any R-
algebra is submaximal.
Proof. Let S be a non-Artinian maximal subring of R. Then we infer that R is not integral over
S, by [4] (Theorem 3.8). Therefore P = (S : R) ∈ Max(R) but P /∈ Max(S), by Theorem 1.1.
Hence S/P is a nonfield maximal subring of the field R/P. Now, let T be any R-algebra, since P
is a prime ideal of R, we infer that there exists a prime ideal Q of T, such that Q ∩ R ⊆ P. But
dim(R) = 0 and Q ∩ R ∈ Spec(R), hence Q ∩ R = P. Thus T/Q contains a copy of R/P and
therefore we are done by Corollary 2.2.
The next proposition is the converse of Theorem 1.2. Before presenting it, we need a well-known
fact which follows. Let R be a ring and M be an R-module. If N is an R-submodule of M and for
any prime ideal (or maximal ideal) P of R we have NP = MP , then M = N, see [9, p. 24] (Ex. 5).
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 7
886 A. AZARANG, O. A. S. KARAMZADEH, A. NAMAZI
Proposition 2.1. Let S be a proper subring of a ring R. If there exists a unique maximal ideal
M of S such that SM is a maximal subring of RS\M ; and for any maximal ideal N 6= M of S, we
have SN = RS\N , then S is a maximal subring of R.
Proof. Let L be a subring of R with S ⊆ L ⊆ R and we are to show that L = S or L = R. For
each maximal ideal N 6= M of S, Since SN = RS\N , we infer that SN = LS\N = RS\N . We also
have SM ⊆ LS\M ⊆ RS\M . Hence either SM = LS\M , in which case S = L (note, SP = LP for
all maximal ideals P of S, where S ⊆ L are considered as S-modules), or LS\M = RS\M which
also implies that SP = LP for all maximal ideals P of S, i.e., L = R and we are done.
Proposition 2.1 is proved.
In [3] (Corollary 2.3), it is shown that whenever R1, . . . , Rn are rings and n ≥ 2, then the ring
R1 × . . .×Rn is submaximal if and only if either at least one of the Ri is submaximal or Ri/Mi
∼=
∼= Rj/Mj for two distinct i and j and maximal idealsMk of Rk, k = i, j. Also, in [4] (Theorem 3.17),
it is proved that if {Ri}i∈I is an infinite family of rings, then
∏
i∈I
Ri is submaximal. The following
is in order.
Proposition 2.2. Let R be a ring which is not submaximal and {Mi}i∈I be a family of maximal
ideals in R. Then T =
∏
i∈I
R/Mi is submaximal if and only if |I| =∞.
Proof. If I is infinite then we are done by the preceding comments . Hence assume that T is
submaximal but |I| < ∞. Thus T = R/M1 × R/M2 × . . . × R/Mn is submaximal. Therefore, by
the first part of the above comments, either there exists i, 1 ≤ i ≤ n, such that R/Mi is submaximal,
or there exist i 6= j, 1 ≤ i, j ≤ n, such that R/Mi
∼= R/Mj . The first case is not possible since R is
not submaximal. We show that the second case is not possible either. To this end, if the second case
holds, then we infer that R/Mi×R/Mj is submaximal, see [1]. Hence R/(Mi ∩Mj) is submaximal
which is absurd.
Proposition 2.2 is proved.
Next, we prove a generalization of [3] (Corollary 2.3). Before doing this generalization, we need
some observations. In the proof of Theorem 3.17 in [4], it is shown that if {Ei}∞i=1 is a family of
infinite fields which are not submaximal and Char(Ei) = p, where p is a fixed prime number, then
the ring R =
∏∞
i=1
Ei is not integral over Zp. Moreover, it is proved that there exists a maximal
ideal M of R such that R/M is not algebraic over Zp. See also [5] for a more general result. The
following is now in order.
Theorem 2.2. Let {Ri}i∈I be a family of rings. Then R =
∏
i∈I
Ri is submaximal if and only
if at least one of the following conditions holds:
(1) There exists i ∈ I, such that Ri is submaximal.
(2) There exist i 6= j in I, and maximal ideals Mk of Rk, k = i, j, such that
Ri
Mi
∼=
Rj
Mj
(
hence
in this case R has distinct maximal ideals M and N such that
R
M
∼=
R
N
)
.
(3) There exists a maximal ideal M of R, such that
R
M
is submaximal.
Proof. If one of the above conditions holds then it is clear that R is submaximal (note, if item (2)
holds, then Ri × Rj is submaximal, by the first part of the comment preceding Proposition 2.2,
therefore R is submaximal too). Conversely, if R is submaximal, but conditions (1) and (2) do not
hold. We show that item (3) holds. First note that by [3] (Corollary 2.3), I must be an infinite set.
Now, For any i ∈ I, let Mi be a maximal ideal of Ri. Since for each i ∈ I, Ri is not submaximal,
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HEREDITARY PROPERTIES BETWEEN A RING AND ITS MAXIMAL SUBRINGS 887
we infer that
Ri
Mi
is not submaximal for each i ∈ I. Therefore by Corollary 2.1, Char
(
Ri
Mi
)
6= 0
for all i ∈ I. Also, note that for each i, j ∈ I, i 6= j, we have
Ri
Mi
�
Rj
Mj
. Now we have two cases.
1. The set
{
Char
(
Ri
Mi
)
: i ∈ I
}
is infinite. In this case let Char
(
Rin
Min
)
= qn, where qn,
n ∈ N, are distinct prime numbers and in ∈ I. Put T =
∏∞
n=1
Rin
Min
. Thus T is a zero-dimensional
ring with Char(T ) = 0. Hence there exists a prime (and therefore maximal) ideal N of T such that
T∩Z = 0 (note Z\{0} is a multiplicatively closed set in T ). Thus
T
N
is submaximal by Corollary 2.1.
2. The set
{
Char
(
Ri
Mi
)
: i ∈ I
}
is finite. Thus, in this case, there exists an infinite subset
{in : n ∈ N} of I, such that Char
(
Rin
Min
)
= p 6= 0 for each n ≥ 1. Also, we may assume that
each
Rin
Min
is infinite field. (Note, the only finite fields with characteristic p which is not submaximal
is Zp, up to isomorphism. Also note that item (2) is not hold.) Thus by the comments preceding
Theorem 2.2, we infer that there exists a maximal ideal N in the ring T =
∏∞
n=1
Rin
Min
such that
T/N is not algebraic over Zp and therefore T/N is submaximal by Corollary 2.1.
Finally, let us put J = I \ {in : n ∈ N} and M = N ×
∏
j∈J
Rj
Mj
. Now it is clear that M is
a maximal ideal of S =
∏
i∈I
Ri∏
i∈I
Mi
and
S
M
is submaximal. This immediately implies that (3) holds
and we are done.
Theorem 2.2 is proved.
We conclude this section with some observations about the existence of maximal subrings in
a ring R, via R-modules. For a ring R let Simp(R) be the set of all simple R-modules, up to
isomorphism. Now, the following is in order.
Proposition 2.3. Let R be a ring which satisfies at least one of the following conditions:
(1) There exists a simple R-module M, such that either |M | ≥ 2ℵ0 or M is a torsion free
Z-module.
(2) | Simp(R)| > 2ℵ0 .
Then R is submaximal.
Proof. If (1) holds, then since M is a simple R-module, there exists a maximal ideal m in R,
such that M ' R/m as an R-module. Hence either |R/m| ≥ 2ℵ0 or Char(R/m) = 0 and therefore
by Corollary 2.1, R/m and R are both submaximal. If (2) holds, then as we see in (1), for any simple
R-module M, there exists a (unique) maximal ideal m of R such that M ' R/m as R-module. This
implies that | Simp(R)| = |Max(R)|, hence |Max(R)| > 2ℵ0 and therefore R is submaximal, by
[2] (Proposition 2.6).
Proposition 2.3 is proved.
The next results are generalization of [3] (Proposition 2.4) and [4] (Theorem 2.9), respectively.
Proposition 2.4. Let R be a ring. If there exists an uncountable Artinian R-module M, then
R is submaximal.
ISSN 1027-3190. Укр. мат. журн., 2013, т. 65, № 7
888 A. AZARANG, O. A. S. KARAMZADEH, A. NAMAZI
Proof. We first recall that Artinian modules over commutative rings are countably generated,
see [10] (Corollary 4.2) for a generalization of this fact. Consequently, we may assume that M =
= Rm1+Rm2+. . .+Rmk+. . . . Hence we infer that there exists i ∈ N, such that |Rmi| ≥ 2ℵ0 . Now
note that Rmi ' R/ ann(mi), which in turn implies that R/ ann(mi) is an uncountable Artinian
ring. Thus, by [3] (Proposition 2.4), R/ ann(mi) and therefore R are both submaximal.
Proposition 2.4 is proved.
Proposition 2.5. Let R be a ring. If there exists an R-module M, such that every cyclic R-
submodule of M is Noetherian and M =
∑
i∈I
Rmi, where |I| ≤ 2ℵ0 < |M | (in particular, if M is
a Noetherian R-module with |M | > 2ℵ0), then R is submaximal.
Proof. Since |M | > 2ℵ0 , we infer that there exists i ∈ I such that |Rmi| > 2ℵ0 . Now, we have
Rmi ' R/ ann(mi), as R-modules. Hence R/ ann(mi) is a Noetherian ring with |R/ ann(mi)| >
> 2ℵ0 . Thus we are done by [4] (Theorem 2.9).
Proposition 2.5 is proved.
Let R be a ring. By Soc(R) we mean the sum of all minimal ideals of R. Now the following
corollary is immediate.
Corollary 2.4. Let R be a ring which is not submaximal and Soc(R) 6= (0). Then either
| Soc(R)| ≤ 2ℵ0 or Soc(R) is not λ-generated where λ ≤ 2ℵ0 .
3. Hereditary properties between a ring and its maximal subrings. We begin this section with
the following interesting result about sharing the Hilbert property between a ring and its maximal
subrings. We remind the reader that recently in [5] it is proved that every ring is either submaximal
or it is a Hilbert ring. In particular, every Hilbert ring R with | Spec(R)| > 22
ℵ0 is submaximal. Now
the following is in order.
Theorem 3.1. Let S be a maximal subring of a ring R. Then the following are valid:
(1) If S is Hilbert, then R is Hilbert too.
(2) If R is Hilbert and is integral over S, then S is Hilbert too.
Proof. (1) Assume that S is a Hilbert ring, and take α ∈ R \ S. Thus R = S[α] and therefore
R is Hilbert (note, if S is Hilbert ring, then the polynomial ring S[x] and the epimorphic image of S
are Hilbert, see [9]).
(2) We must show that every prime ideal P of S is an intersection of maximal ideals. Since R is
integral over S, there exists a prime ideal Q of R such that P = Q∩R. Since R is Hilbert, we infer
that there exists a family {Mi}i∈I of maximal ideals of R such that Q =
⋂
i∈I Mi. Inasmuch as R is
integral over S, we infer that Mi ∩S ∈ Max(S), for each i ∈ I. Hence P = Q∩S =
⋂
i∈I(Mi ∩S)
and we are done.
Theorem 3.1 is proved.
Definition 3.1. Let S be a subring of a ring R. We say that S is essential in R and denote
it by S ⊆e R, if S ∩ I 6= 0, for every nonzero ideal I of R. Also, we say that S is an essential
S-submodule of R, if it intersects every nonzero S-submodule of R nontrivially and we denote it by
S ≤e R.
It is manifest that in the above definition S ≤e R implies that S ⊆e R. The following is now in
order.
Lemma 3.1. Let S be a maximal subring of a ring R. Then either S ⊆e R or Soc(R) 6= 0
(i.e., R has a nonzero minimal ideal).
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Proof. Assume that S is not essential in R. So there exists a nonzero ideal I of R such that
S ∩ I = 0 and therefore R = S ⊕ I. We show that I is a minimal ideal of R and we are done. Let
A be a nonzero ideal of R such that A ⊆ I. Hence we have S ∩ A = 0. Thus R = S ⊕ A. But
I = A+ (I ∩ S) = A+ 0, i.e., I = A and we are done.
Corollary 3.1. Let R be a reduced ring without nontrivial idempotents. Then any maximal
subring S of R is essential in R. In particular, any maximal subring of an integral domain is
essential.
Proof. Let S be a maximal subring of R which is not essential. So we infer that Soc(R) 6= 0 by
the previous lemma. Hence assume that I is a nonzero minimal ideal in R. Since R is reduced, we
infer that I2 = I. Now, let 0 6= x ∈ I, thus we have I = Rx = Rx2 and therefore x = rx2 for some
r ∈ R. Now by putting e = rx we infer that e is a nontrivial idempotent, which is a contradiction.
The last part is now evident.
Remark 3.1. We recall that whenever S is a maximal subring of a ring R, then R is algebraic
over S (note, for any x ∈ R, either x2 ∈ S or x ∈ S[x2]). Now assume that R is an integral domain
and S is a maximal subring of R which is integrally closed in R. Then one can easily see that R must
be an overring of S, i.e., S ⊂ R ⊆ Q(S). In particular, in the latter case Q(R) = Q(S). Moreover,
if R is integrally closed, and S is a maximal subring of R, which is integrally closed in R, then
S is integrally closed too. To see this, let E = Q(S) = Q(R), then we have S ⊆ R ⊆ E. Hence
S ⊆ S′ ⊆ R′ = R, where S′ and R′ are the integral closures of S and R in E, respectively. Thus
S′ = S, since S is a maximal subring of R which is integrally closed in it.
It is clear that whenever S is a subring of a ring R, then for any prime ideal P of S, there exists
a prime ideal Q of R such that S ∩Q ⊆ P (note, we might have 0 = S ∩Q ⊆ P ). In what follows
we show that if R is an integral domain and S is a maximal subring of R which is integrally closed
in R, then S ∩Q 6= 0 whenever P 6= 0, cruR(S).
Lemma 3.2. Let R be an integral domain and S be a maximal subring of R which is integrally
closed in R. Then for any 0 6= P ∈ Spec(S) \ {cruR(S)} there exists a nonzero prime ideal Q of R
such that 0 6= Q ∩ S ⊆ P.
Proof. First note that, we have SP = RS\P , by Theorem 1.2. Since SP is a local integral domain
which is not a field, we infer that RS\P is also a local ring with a nonzero unique maximal ideal
QS\P . Hence Q 6= 0 and therefore 0 6= Q ∩ S ⊆ P, by Corollary 3.1.
Lemma 3.2 is proved.
The following interesting result is now in order.
Proposition 3.1. Let R be an integral domain and S be a maximal subring of R which is
integrally closed in R. Then we have the following statements:
(1) If dim(R) = 1, then dim(S) = 1 if and only if (S : R) = 0.
(2) If dim(S) = 1, then dim(R) = 1 if and only if (S : R) = 0.
Proof. (1) Assume that dim(S) = dim(R) = 1. We show that (S : R) = 0. We have (S :
R) ∈ Spec(S) \ Max(S), by Theorem 1.1, therefore (S : R) = 0. Conversely, let 0 6= P ∈
∈ Spec(S) \ {cruR(S)}, thus by the above lemma, there exists a nonzero prime ideal Q of R, such
that 0 6= Q ∩ S ⊆ P. But Q * S, since (S : R) = 0. Thus we infer that Q ∩ S ∈ Max(S), (note,
S+Q = R, R/Q ∼= S/(S ∩Q) and Q ∈ Max(R), since dim(R) = 1). Thus Q∩S = P ∈ Max(S)
and we are done. The proof of part (2) is similar to the proof of part (1).
Proposition 3.1 is proved.
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890 A. AZARANG, O. A. S. KARAMZADEH, A. NAMAZI
Now we are ready to present the following fact about sharing the Dedekind property between a
ring and its maximal subrings, which is an immediate corollary of Remark 3.1, Proposition 3.1 and
[9] (Theorem 96).
Corollary 3.2. Let R be an integral domain and S be a maximal subring of R which is integrally
closed in R. Then the following statements hold:
(1) If S is a Dedekind domain, then R is a Dedekind domain and (S : R) = 0.
(2) If R is a Dedekind domain, then S is a Dedekind domain if and only if S is Noetherian and
(S : R) = 0.
Remark 3.2. Let us record the simple needed fact which follows. If R is an integral domain and
S is a maximal subring of R. Then the field Q(R) is a simple (finite) algebraic extension of the field
Q(S) (note, we may have Q(R) = Q(S), e.g., when S is integrally closed in R, see Remark 3.1),
and when R is integral over S, then in fact for any x ∈ R \ S, we have Q(R) = Q(S)[x], for
R = S[x] implies that R ⊆ Q(S)[x], hence Q(R) = Q(S)[x].
It is manifest that if R is an integral domain then Soc(R) = 0. In what follows we give a kind
of the converse of this fact. Assume that S is an integral domain which is a maximal subring of a
ring R (note, one can easily show the interesting fact, essentially due to D. E. Dobbs, that any ring
S is a maximal subring of a bigger ring R, see the introduction of [4]), then R is an integral domain
if and only if Soc(R) = 0. For the converse, note that by [7] (Theorem 2.7), since Soc(R) = 0 we
infer that R must be an integral domain.
Theorem 3.2. Let S be a nonfield maximal subring of a ring R, such that S is a Dedekind
domain. Then the following statements hold:
(1) If R is integral over S, then R is a Dedekind domain if and only if R = S′E where E = Q(R).
(2) If R is an integral domain, then R/I is an Artinian ring for any nonzero ideal I of R.
Proof. For item (1), if R is a Dedekind domain, then since S ⊂ R ⊆ E, we infer that R ⊆ S′E ⊆
⊆ R′E = R. Thus S′E = R. Conversely, if R = S′E , then by [9] (Theorem 98) and the previous
remark R is a Dedekind domain too. For item (2), assume that I is a nonzero ideal of R, if I ⊆ S,
then S/I is an Artinian maximal subring of R/I (note, S is a Dedekind domain). Hence by [4]
(Theorem 3.8) we conclude that R/I is an Artinian ring too. If I * S, then S+ I = R and therefore
R/I ∼= S/(S∩I), and we note that S∩I 6= 0 by Lemma 3.1. Hence R/I is Artinian and we are done.
Theorem 3.2 is proved.
In the next two results we give some properties of maximal subrings in zero-dimensional rings
and one-dimensional valuation domains. We recall that, if S is a maximal subring of a ring R, then
S is zero-dimensional if and only if R is zero-dimensional and integral over S, see [4] and [5] for
more similar properties. The following result, which concerns the integrally closed maximal subrings
of a zero-dimensional ring, is now in order.
Proposition 3.2. Let R be a zero-dimensional ring and S be a maximal subring of it which
is integrally closed in R. Then the following statements hold: (1) S/(S : R) is one-dimensional
valuation domain with a unique nonzero prime ideal cruR(S)/P. (2) Max(S) = {cruR(S)}∪{N∩S |
N ∈ Max(R), N * S}. (3) Spec(S) = {(S : R)} ∪ Max(S). (4) dim(S) = 1 and in fact
(S : R) ⊂ cruR(S) is the only chain of length 1 in S.
Proof. Assume that P = (S : R) and M = cruR(S). Hence by Theorem 1.1, we infer that P ∈
∈ Spec(R)∩ Spec(S) but P /∈ Max(S); also by Theorem 1.2, we have P ⊂M. Since dim(R) = 0,
we infer that P is a maximal ideal of R. Thus S/P is a maximal subring of the field R/P. Hence
S/P is one-dimensional valuation domain with a unique nonzero prime ideal M/P. This proves (1).
Now by Theorem 1.1, for any prime ideal Q ∈ Spec(S) \ {M}, we have SQ = RS\Q. But SQ is
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HEREDITARY PROPERTIES BETWEEN A RING AND ITS MAXIMAL SUBRINGS 891
a local ring, hence RS\Q is local too. Thus let Q′S\Q be the unique prime ideal of RS\Q. Hence
Q′ ∈ Max(R). Now we have the following two cases.
1. If Q = P, then Q′ = P.
2. If Q 6= P, then we first show that Q′ * S. To see this, let Q′ ⊆ S, hence Q′ = P. But
P = Q′∩S = Q′ ⊆ Q, hence P ⊆ Q and therefore by (1) we have Q = M, which is a contradiction
(note, SM = SQ is a maximal subring of RS\M , but SQ = RS\Q). Thus we have Q′ * S and
therefore Q′ ∩ S ∈ Max(S) (note S +Q′ = R). Since Q′ ∩ S ⊆ Q, we infer that Q′ ∩ S = Q.
Hence, every prime ideal Q 6= P of S is maximal. Thus (2), (3) and therefore (4) hold.
Proposition 3.2 is proved.
Proposition 3.3. Let (V,Q) be a one-dimensional valuation domain. If R is a maximal subring
of V, then either R is one dimensional or R is a local integral domain with dim(R) = 2.
Proof. If V is integral over R or R is integrally closed in V and (R : V ) = 0, then dim(R) = 1,
by Proposition 3.1. Hence assume that R is integrally closed in V, (R : V ) 6= 0 and M = cruR(S).
Hence by Theorems 1.1 and 1.2, we infer that (R : V ) = Q ⊂ M and therefore R/Q is a one-
dimensional valuation domain with a unique nonzero prime ideal M/Q (note, R/Q is a nonfield
maximal subring of the field V/Q). Now assume that P 6= 0,M be any prime ideal in R. Hence
by Theorem 1.2, we have RP = VR\P . But RP is not a field, hence we infer that VR\P is also not
a field. Since V is a maximal subring of Q(V ), see [9, p. 43] (Ex. 29), we infer that V = VR\P .
Thus RP = V and therefore ht(P ) = 1. This shows that the only chain of prime ideals of maximum
length in R is 0 ⊂ Q ⊂ M and therefore dim(R) = 2. Now it remains to be shown that R is local.
Note that RM is a maximal subring of VM and also R ⊆ RM ⊂ VM and V ⊆ VM ⊆ Q(V ). Since V
is a maximal subring of Q(V ) we infer that either VM = Q(V ) which implies that RM is a maximal
subring of Q(V ) or V = VM . Since dim(RM ) = 2, we infer that RM cannot be a maximal subring
of Q(V ), see [9, p. 43] (Ex. 29). Therefore V = VM and RM is a maximal subring of V which
contains R, hence R = RM and we are done.
Proposition 3.3 is proved.
Corollary 3.3. Let R be a ring with dim(R) = n <∞ and S be a maximal subring of R. Then
dim(S) ≤ n+ 1.
Proof. If R is integral over S then we are done. Hence assume that R is not integral over S
and M = cruR(S). Thus for any prime ideal P 6= M of S we have SP = RS\P , by Theorem 1.2.
Therefore ht(P ) = dim(RS\P ) ≤ dim(R) = n. This also shows that ht(M) ≤ n+ 1 and therefore
dim(S) ≤ n+ 1 and we are done.
Proposition 3.4. Let S be a maximal subring of a ring R. Then the following statements hold:
(1) S is reduced if and only if either R is reduced or R/N(R) ∼= S.
(2) If S is reduced and S is integrally closed in R, then R is reduced.
Proof. (1) It is clear that if R is reduced or R/N(R) ∼= S, then S is reduced too. Conversely,
if S is reduced, then either R is reduced and we are done or N(R) 6= 0. Now since S is reduced
we infer that S ∩ N(R) = 0. Also, since S is maximal subring of R we have S + N(R) = R and
therefore S ∼= R/N(R).
(2) If S is integrally closed in R and x ∈ N(R), then x is integral over S and therefore x ∈ S,
i.e., x = 0 and we are done.
Proposition 3.4 is proved.
It is clear that if S ⊆ R is an integral extension of rings, then J(R) ∩ S = J(S). Also we
recall that if S is a maximal subring of a ring R, then S is Artinian if and only R is Artinian and
it is integral over S, see [4] (Theorem 3.8). By a semisimple ring we mean an Artinian ring whose
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892 A. AZARANG, O. A. S. KARAMZADEH, A. NAMAZI
Jacobson radical is zero. It is clear that a commutative ring is semisimple if and only if it is a finite
direct product of fields.
Theorem 3.3. Let S be a maximal subring of a ring R. Then the following statements hold:
(1) If S is a semisimple ring, then either R is semisimple or R is Artinian and R/J(R) ∼= S as
rings.
(2) If R is a semisimple ring, then S is a semisimple ring if and only if S is zero-dimensional.
Proof. (1) Since S is semisimple we infer that R is Artinian and integral over S by [4] (The-
orem 3.8). Hence by the preceding comments, we conclude that J(R) ∩ S = J(S) = 0. Now we
have two cases, either J(R) = 0 and therefore R is semisimple, or J(R) 6= 0, thus J(R) * S and
therefore S + J(R) = R, hence we infer that R/J(R) ∼= S and we are done. (2) If S is semisimple
then S is zero-dimensional and we are done. Conversely, assume that S is zero-dimensional. Since R
is reduced we infer that S is reduced and therefore S is von Neumann regular (note, it is well-known
that a zero-dimensional reduced ring is von Neumann regular). Also since S is zero dimensional we
infer that (S : R) ∈ Max(S) and therefore R is integral over S, by Theorem 1.1. Hence again by
the preceding comments, we conclude that S is Artinian too and since it is von Neumann regular, we
infer that S is semisimple.
Theorem 3.3 is proved.
The ring R is called semiprimary if J(R) is nilpotent and R/J(R) is Artinian (semisimple).
One can easily see that if R is a semiprimary ring, then R must be zero-dimensional. Next, we are
interested in maximal subrings which are semiprimary.
Theorem 3.4. Let S be a maximal subring of a ring R. Then the following statements hold:
(1) If R is a semiprimary ring, then S is a semiprimary ring if and only if R is integral over S.
(2) If S is a semiprimary ring, then R is semiprimary if and only if J(R) is nilpotent.
Proof. (1) If S is semiprimary, then S is zero-dimensional. Hence (S : R) ∈ Max(S) and there-
fore R is integral over S, by Theorem 1.1. Conversely, if R is integral over S, then J(S) = J(R)∩S,
by the comment preceding Theorem 3.3. Hence J(S) is nilpotent. Now we have two cases.
(1) J(R) ⊆ S and therefore J(R) = J(S) = J. Hence S/J is a maximal subring of R/J and
R/J is integral over S/J. Thus S/J is Artinian by [4] (Theorem 3.8), and therefore S is semiprimary.
(2) J(R) * S, hence S + J(R) = R, since S is a maximal subring of R. The latter equality
implies that S/J(S) ∼= R/J(R) and therefore S/J(S) is Artinian. Consequently, S is semiprimary
and we are done. Finally, the proof of (2) is similar to the part (1).
Theorem 3.4 is proved.
For the next result we need to recall some definitions, see [9]. A Noetherian local ring (R,M)
is called a regular ring, if dim(R) = v dimK(M/M2), where K = R/M. Now assume that R is a
Noetherian ring, not necessary local, R is called regular if Rm is regular for every maximal ideal m of
R. Now let R and S be rings, f : S −→ R be a ring homomorphism and pdS(R) = n (the projective
dimension of R as an S-module). Then for any right R-module M, we have pdS(M) ≤ n+pdR(M),
see [11, p. 204] (Ex. 25).
Theorem 3.5. Let S be a maximal subring of a regular ring R. If R is integral over S and
pdS(R) <∞, then S is regular too.
Proof. It is manifest that S is Noetherian, see [4]. Hence to show that S is regular it suffices to
prove that for any maximal ideal M of S we have pdS(M) <∞, see [11] (Theorem 5.94). Consider-
ing the exact sequence 0→M → S → S/M → 0, we may show that pdS(S/M) <∞, for any max-
imal ideal M of S, see [11] (Theorem 5.20). Hence assume that M is a maximal ideal of S, since R is
integral over S there exists a maximal ideal N of R such that M = S ∩N. Now we have two cases:
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HEREDITARY PROPERTIES BETWEEN A RING AND ITS MAXIMAL SUBRINGS 893
(1) If N ⊆ S, then N = M, hence by the previous comment we have pdS(N) ≤ pdS(R) +
pdR(N), thus pdS(N) <∞ (note, pdR(N) <∞, for R is regular).
(2) If N * S, then we consider two cases, either S ∩ N = 0, in which case R/N ∼= S (note,
S + N = R), hence S is field and we are done, or S ∩ N 6= 0. In the latter case, we note that
R/N = (S +N)/N ' S/(S ∩N) = S/M as S-modules. But pdS(R/N) is finite, for by the above
comment pdS(R/N) ≤ pdS(R) + pdR(R/N) <∞. Hence pdS(S/M) <∞ and we are done.
Theorem 3.5 is proved.
We conclude this article with the next result.
Proposition 3.5. Let S be a maximal subring of a ring R such that S is a von Neumann
regular and R is self-injective. Then RS is injective. Moreover at least one of the following holds:
(1) If S *e R, then S is a self-injective ring.
(2) If S ⊆e R, then R is a von Neumann regular ring.
Proof. Since S is von Neumann regular, we infer that RS is flat, by [11] (Theorem 4.21). Hence
RS is injective by [11] (Corollary 3.6A). Now consider two cases.
(1) If S *e R, then there exists a nonzero ideal I of R, such that I ∩ S = 0. Hence R = I ⊕ S,
therefore S is injective as an S-module and we are done.
(2) If S ⊆e R, then I∩S 6= 0, for each nonzero ideal I of R. But we have J(R)∩S = 0 (note, S
is a von Neumann regular subring of R), hence J(R) = 0. Thus R is reduced. Therefore Z(R) = 0
(the singular ideal of R), by [11] (Lemma 7.8). To complete the proof, let us recall a general fact
(it is even true in the noncommutative case and it is attributed to many people), namely, if R is
self-injective then the Jacobson radical of R coincides with its singular ideal and R/J(R) is a self-
injective von Neumann regular ring, see the comment preceding [11] (Theorem 13.1), for some of the
names of the above people and a generalization of this fact. Hence R is a von Neumann regular ring.
Proposition 3.5 is proved.
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Received 24.05.12
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| institution | Ukrains’kyi Matematychnyi Zhurnal |
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| language | English |
| last_indexed | 2026-03-24T02:24:06Z |
| publishDate | 2013 |
| publisher | Institute of Mathematics, NAS of Ukraine |
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| resource_txt_mv | umjimathkievua/85/9f5af79f2dc8dbbff474e210aecd7a85.pdf |
| spelling | umjimathkievua-article-24752020-03-18T19:16:28Z Hereditary Properties between a Ring and its Maximal Subrings Спадкові властивоcті мiж кільцем та його максимальними підкільцями Azarang, A. Karamzadeh, O. A. S. Namazi, A. Азаранг, А. Карамзадех, О. А. С. Намазі, А. We study the existence of maximal subrings and hereditary properties between a ring and its maximal subrings. Some new techniques for establishing the existence of maximal subrings are presented. It is shown that if R is an integral domain and S is a maximal subring of R, then the relation dim(R) = 1 implies that dim(S) = 1 and vice versa if and only if (S : R) = 0. Thus, it is shown that if S is a maximal subring of a Dedekind domain R integrally closed in R; then S is a Dedekind domain if and only if S is Noetherian and (S : R) = 0. We also give some properties of maximal subrings of one-dimensional valuation domains and zero-dimensional rings. Some other hereditary properties, such as semiprimarity, semisimplicity, and regularity are also studied. Вивчається існування максимальних підкілєць та спадкові властивості між кільцєм та його максимальними підкільцями. Наведено деякі нові методи встановлення існування максимальних підкілець. Показано, що якщо $R$ — інтегральна область, а $S$ — її максимальне підкільце, то із співвідношення $\dim(R) = 1$ випливає, що $\dim(S) = 1$, і навпаки тоді i тільки тоді, коли $(S : R) = 0$. Як наслідок показано, що, якщо $S$ є максимальним підкільцем дедекіндової області $R$, яка є інтегрально замкненою в $R$, то $S$ є дедекіндовим підкільцем тоді i тільки тоді, коли $S$ є нетеровим та $(S : R) = 0$. Наведено також деякі властивості максимальних підкілець одновимірних областей нормування та нульвимірних кілець. Також вивчено деякі інші спадкові властивості, такі як напівпримарність, напівпростота та регулярність. Institute of Mathematics, NAS of Ukraine 2013-07-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/2475 Ukrains’kyi Matematychnyi Zhurnal; Vol. 65 No. 7 (2013); 883–893 Український математичний журнал; Том 65 № 7 (2013); 883–893 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/2475/1716 https://umj.imath.kiev.ua/index.php/umj/article/view/2475/1717 Copyright (c) 2013 Azarang A.; Karamzadeh O. A. S.; Namazi A. |
| spellingShingle | Azarang, A. Karamzadeh, O. A. S. Namazi, A. Азаранг, А. Карамзадех, О. А. С. Намазі, А. Hereditary Properties between a Ring and its Maximal Subrings |
| title | Hereditary Properties between a Ring and its Maximal Subrings |
| title_alt | Спадкові властивоcті мiж кільцем та його максимальними підкільцями |
| title_full | Hereditary Properties between a Ring and its Maximal Subrings |
| title_fullStr | Hereditary Properties between a Ring and its Maximal Subrings |
| title_full_unstemmed | Hereditary Properties between a Ring and its Maximal Subrings |
| title_short | Hereditary Properties between a Ring and its Maximal Subrings |
| title_sort | hereditary properties between a ring and its maximal subrings |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/2475 |
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