A corrigendum to “Hereditary properties between a ring and its maximal subrings”
Let $R$ be a commutative ring with identity. In [2] (Proposition 3.1), Azarang proved that if $R$ is an integral domain and $S$ is a maximal subring of $R$, and is integrally closed in $R$, then $\mathrm{d}\mathrm{i}\mathrm{m}(S) = 1$ implies that $\mathrm{d}\mathrm{i}\mathrm{m}(R) = 1$ if and only...
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Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507383193665536 |
|---|---|
| author | Gaur, A. Kumar, R. Гаур, А. Кумар, Р. |
| author_facet | Gaur, A. Kumar, R. Гаур, А. Кумар, Р. |
| author_sort | Gaur, A. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T09:19:04Z |
| description | Let $R$ be a commutative ring with identity. In [2] (Proposition 3.1), Azarang proved that if $R$ is an integral domain and $S$ is a maximal subring of $R$, and is integrally closed in $R$, then $\mathrm{d}\mathrm{i}\mathrm{m}(S) = 1$ implies that $\mathrm{d}\mathrm{i}\mathrm{m}(R) = 1$ if and only if
$(S : R) = 0$. An example is given which shows the above mentioned proposition is not correct. |
| first_indexed | 2026-03-24T02:08:26Z |
| format | Article |
| fulltext |
К О Р О Т К I П О В I Д О М Л Е Н Н Я
UDC 512.5
R. Kumar*, A. Gaur** (Univ. Delhi, India)
A CORRIGENDUM TO ”HEREDITARY PROPERTIES BETWEEN A RING
AND ITS MAXIMAL SUBRINGS”
ПОПРАВКА ДО РОБОТИ „СПАДКОВI ВЛАСТИВОСТI МIЖ КIЛЬЦЕМ
ТА ЙОГО МАКСИМАЛЬНИМИ ПIДКIЛЬЦЯМИ”
Let R be a commutative ring with identity. In [2] (Proposition 3.1), Azarang proved that if R is an integral domain and
S is a maximal subring of R, and is integrally closed in R, then \mathrm{d}\mathrm{i}\mathrm{m}(S) = 1 implies that \mathrm{d}\mathrm{i}\mathrm{m}(R) = 1 if and only if
(S : R) = 0. An example is given which shows the above mentioned proposition is not correct.
Нехай R — комутативне кiльце з одиницею. В роботi [2] (твердження 3.1) Азаранг довiв, що у випадку, коли
R — iнтегральна множина, а S — максимальне пiдкiльце R, iнтегрально замкнене в R, iз рiвностi \mathrm{d}\mathrm{i}\mathrm{m}(S) = 1
випливає, що \mathrm{d}\mathrm{i}\mathrm{m}(R) = 1 тодi i тiльки тодi, коли (S : R) = 0. Наведено приклад, який показує, що це твердження
є неправильним.
1. Introduction. All rings considered throughout are commutative with nonzero identity; all ring
extensions, ring homomorphisms, and algebra homomorphisms are unital. Given rings S \subseteq R, the
conductor (S : R) = \{ r \in R : rR \subseteq S\} . Also, dimension(al) refers to Krull dimension. If S is
a proper subring of a ring R, then S is a maximal subring of R if there is no ring T such that
S \subset T \subset R where \subset denotes proper inclusion.
In [2], Azarang proved in Proposition 3.1 that if R is an integral domain and S is a maximal
subring of R, and is integrally closed in R, then \mathrm{d}\mathrm{i}\mathrm{m}(S) = 1 implies \mathrm{d}\mathrm{i}\mathrm{m}(R) = 1 if and only if (S :
R) = 0. The importance of Proposition 3.1 in [2] is witnessed by the abstract of [2]. We have given
an example which shows that the reverse implication of above proposition is not correct.
2. Corrigendum. The following result was proved in [2].
Theorem 2.1 ([2], Proposition 3.1). Let R be an integral domain and S be a maximal subring
of R, and is integrally closed in R. Then the following statements are true:
(1) If \mathrm{d}\mathrm{i}\mathrm{m}(R) = 1, then \mathrm{d}\mathrm{i}\mathrm{m}(S) = 1 if and only if (S : R) = 0.
(2) If \mathrm{d}\mathrm{i}\mathrm{m}(S) = 1, then \mathrm{d}\mathrm{i}\mathrm{m}(R) = 1 if and only if (S : R) = 0.
We now present the counter example to show that (2) is not correct. However, (1) is correct.
Example 2.1. Let R = \BbbQ and S = \BbbZ 2\BbbZ . We assert that S is a maximal subring of R. Suppose
there is a ring T such that S \subset T \subseteq R. Choose
p
2nq
\in T \setminus S, where p and 2nq are coprime, n \in \BbbN .
Thus, (2n - 1q)
\biggl(
p
2nq
\biggr)
\in T, which gives 1/2 \in T. Therefore, T = R. Hence, S is a maximal
subring of R. Since S is a one dimensional valuation domain with quotient field R, S is integrally
* Was supported by a grant from UGC, India.
** Was supported by R & D grant, University of Delhi, India.
c\bigcirc R. KUMAR, A. GAUR, 2018
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4 583
584 R. KUMAR, A. GAUR
closed in R. Now, suppose p/q \in (S : R), where p and q are coprime. Clearly, q must be odd.
Also,
p
2nq
\in S for all n \in \BbbN , implies that p = 0. Therefore, (S : R) = 0. Clearly, \mathrm{d}\mathrm{i}\mathrm{m}(S) = 1 but
\mathrm{d}\mathrm{i}\mathrm{m}(R) = 0. This counters (2) of above mentioned theorem.
Remark 2.1. Note that under the stated conditions of Theorem 2.1, if \mathrm{d}\mathrm{i}\mathrm{m}(S) = 1, then
\mathrm{d}\mathrm{i}\mathrm{m}(R) \leq 1 by [1] (Proposition 4.1), and (S : R) = 0 by [3] (Theorem 7).
References
1. Ayache A. Minimal overrings of an integrally closed domain // Communs Algebra. – 2003. – 31, № 12. – P. 5693 –
5714.
2. Azarang A., Karamzadeh O. A. S., Namazi A. Hereditary properties between a ring and its maximal subrings // Ukr.
Math. J. – 2013. – 65, № 7. – P. 883 – 893.
3. Modica M. L. Maximal subrings: Ph. D. Dissertation. – Chicago, 1975.
Received 16.01.17
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 4
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| spelling | umjimathkievua-article-15782019-12-05T09:19:04Z A corrigendum to “Hereditary properties between a ring and its maximal subrings” Поправка до роботи „спадковi властивостi мiж кiльцем та його максимальними пiдкiльцями” Gaur, A. Kumar, R. Гаур, А. Кумар, Р. Let $R$ be a commutative ring with identity. In [2] (Proposition 3.1), Azarang proved that if $R$ is an integral domain and $S$ is a maximal subring of $R$, and is integrally closed in $R$, then $\mathrm{d}\mathrm{i}\mathrm{m}(S) = 1$ implies that $\mathrm{d}\mathrm{i}\mathrm{m}(R) = 1$ if and only if $(S : R) = 0$. An example is given which shows the above mentioned proposition is not correct. Нехай $R$ — комутативне кiльце з одиницею. В роботi [2] (твердження 3.1) Азаранг довiв, що у випадку, коли $R$ — iнтегральна множина, а $S$ — максимальне пiдкiльце $R$, iнтегрально замкнене в $R$, iз рiвностi $\mathrm{d}\mathrm{i}\mathrm{m}(S) = 1$ випливає, що $\mathrm{d}\mathrm{i}\mathrm{m}(R) = 1$ тодi i тiльки тодi, коли $(S : R) = 0$. Наведено приклад, який показує, що це твердження є неправильним. Institute of Mathematics, NAS of Ukraine 2018-04-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1578 Ukrains’kyi Matematychnyi Zhurnal; Vol. 70 No. 4 (2018); 583-584 Український математичний журнал; Том 70 № 4 (2018); 583-584 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1578/560 Copyright (c) 2018 Gaur A.; Kumar R. |
| spellingShingle | Gaur, A. Kumar, R. Гаур, А. Кумар, Р. A corrigendum to “Hereditary properties between a ring and its maximal subrings” |
| title | A corrigendum to “Hereditary properties between a ring and its maximal subrings” |
| title_alt | Поправка до роботи „спадковi властивостi мiж кiльцем
та його максимальними пiдкiльцями” |
| title_full | A corrigendum to “Hereditary properties between a ring and its maximal subrings” |
| title_fullStr | A corrigendum to “Hereditary properties between a ring and its maximal subrings” |
| title_full_unstemmed | A corrigendum to “Hereditary properties between a ring and its maximal subrings” |
| title_short | A corrigendum to “Hereditary properties between a ring and its maximal subrings” |
| title_sort | corrigendum to “hereditary properties between a ring and its maximal subrings” |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1578 |
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