A study of modules over rings and their extensions
We study the transfer of properties of some types of modules under identity preserving ring homomorphisms. These studies seem to be overlooked in literature. We have picked a few types of modules and provided proofs for these properties that are transferable and suitable counterexamples for the prop...
Збережено в:
| Дата: | 2018 |
|---|---|
| Автори: | , , , |
| Формат: | Стаття |
| Мова: | Англійська |
| Опубліковано: |
Institute of Mathematics, NAS of Ukraine
2018
|
| Онлайн доступ: | https://umj.imath.kiev.ua/index.php/umj/article/view/1636 |
| Теги: |
Додати тег
Немає тегів, Будьте першим, хто поставить тег для цього запису!
|
| Назва журналу: | Ukrains’kyi Matematychnyi Zhurnal |
| Завантажити файл: |  |
Репозитарії
Ukrains’kyi Matematychnyi Zhurnal| _version_ | 1860507459867639808 |
|---|---|
| author | Al-Hashmi, S. A. Nauman, S. K. Аль, - Хашмі С. А. Науман, С. К. |
| author_facet | Al-Hashmi, S. A. Nauman, S. K. Аль, - Хашмі С. А. Науман, С. К. |
| author_sort | Al-Hashmi, S. A. |
| baseUrl_str | https://umj.imath.kiev.ua/index.php/umj/oai |
| collection | OJS |
| datestamp_date | 2019-12-05T09:21:55Z |
| description | We study the transfer of properties of some types of modules under identity preserving ring homomorphisms. These studies
seem to be overlooked in literature. We have picked a few types of modules and provided proofs for these properties that
are transferable and suitable counterexamples for the properties that are not transferable. |
| first_indexed | 2026-03-24T02:09:39Z |
| format | Article |
| fulltext |
UDC 512.5
S. A. Al-Hashmi, S. K. Nauman (King Abdulaziz Univ., Jeddah, KSA)
A STUDY OF MODULES OVER RINGS AND THEIR EXTENSIONS
ДОСЛIДЖЕННЯ МОДУЛIВ НАД КIЛЬЦЯМИ ТА ЇХ РОЗШИРЕННЯ
We study the transfer of properties of some types of modules under identity preserving ring homomorphisms. These studies
seem to be overlooked in literature. We have picked a few types of modules and provided proofs for these properties that
are transferable and suitable counterexamples for the properties that are not transferable.
Вивчається проблема передачi властивостей деяких типiв модулiв пiд дiєю гомоморфiзмiв, що зберiгають одиницю.
Подiбнi дослiдження, здається, вiдсутнi в лiтературi. Вибрано кiлька типiв модулiв та наведено доведення для тих
властивостей, що передаються, та придатнi контрприклади для тих властивостей, що не передаються.
Introduction. In this note we consider the ordinary ring extension under identity preserving ring
homomorphisms between two rings with identity and use it to study the transfer of properties of dif-
ferent types of modules and their submodules over these rings. These studies seem to be overlooked
in literature and in several cases may not be trivial to prove. Moreover, several properties do not
transfer under such extensions. In this paper we have picked some very commonly studied classes of
modules and planned to cover more in future. Almost all statements and counter statements are de-
fended and followed by examples and counterexamples. As a summary, the lists of types of modules
that we have studied here are given at the end.
1. Preliminaries. Throughout this note, unless otherwise specified, the term ring means an
associative ring with identity 1 \not = 0, homomorphisms are identity preserving, and all modules are
unital. For a ring R, we shall denote the Jacobson radical of R by J(R). We will denote the
annihilator of a module M by ann M. The symbols \BbbN \ast , \BbbZ , \BbbQ , \BbbR (\BbbZ n, n \in \BbbN \ast ) will denote the set
of all positive integers, integers, rational numbers, real numbers and integers modulo n, respectively.
Suppose that R and S are rings and \Phi : R \rightarrow S is a ring homomorphism, then S is called an
extension of R by \Phi . Let VS be a right S -module. Then VS can be considered as a right R-module
by the action
x \cdot r = x\Phi (r) \forall x \in VS \forall r \in R (1.1)
and is denoted by V| R. As usual, we will write x \cdot r = xr.
Note that, if \Phi (1) = 1, and VS is a unital S -module, then V| R is a unital R-module too.
However, in general this statement is not true. For instance, suppose that the ring R contains a
nontrivial idempotent e and consider the ring homomorphism \Phi : \BbbZ - \rightarrow R, which is defined by
\Phi (n) = ne for all n \in \BbbZ . Then R is a unital R-module, but not a unital \BbbZ -module.
For any right R-module M, a pair (E, i) is an injective envelope of M in case E is an injective
right R-module and i : M - \rightarrow E is an essential monomorphism. A projective cover of M means an
epimorphism \theta : P - \rightarrow M, where PR is a projective module and \mathrm{k}\mathrm{e}\mathrm{r} \theta is a small (= superfluous)
submodule of P. A projective module M is called semiperfect if every factor module of M has a
projective cover.
The following propositions list some useful and known results.
c\bigcirc S. A. AL-HASHMI, S. K. NAUMAN, 2018
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10 1299
1300 S. A. AL-HASHMI, S. K. NAUMAN
Proposition 1.1 ([2, p. 439], Theorem 3.7). (a) The following conditions on a nonzero ring R
are equivalent:
(i) R is semisimple;
(ii) every right R-module is projective;
(iii) every right R-module is injective;
(iv) every right R-module is semisimple.
(b) Let M be a semisimple right R-module and N \subseteq M be a submodule. Then N is simple iff
N is indecomposable.
Proposition 1.2. (a) Let I be an ideal in a ring R and R = R\diagup I. Let M be a right R-module,
which is, therefore a right R- module. If MR has a projective cover over R, then MR also has a
projective cover over R .
(b) Every module over an Artinian ring has a projective cover; hence, every projective module
over an Artinian ring is semiperfect.
Proof. See [6] (Lemma 24.15) and [7] (Corollary 5.4).
2. Main results. Some elementary observations are entered in the following theorem.
Theorem 2.1. Suppose that R and S are rings and \Phi : R \rightarrow S is a ring homomorphism. Then
the following statements hold. The converse of each statement also holds provided \Phi is surjective.
(a) If V| R is generated by X, then VS is generated by X.
(b) If V| R is a cyclic R-module, then VS is a cyclic S -module.
(c) If V \prime is an S -submodule of VS , then V \prime is an R-submodule of V| R.
(d) If V| R is a simple R-module, then VS is a simple S -module.
(e) If V| R is an indecomposable R-module, then VS is an indecomposable S -module.
(f) If V| R is Artinian (Noetherian), then VS is Artinian (Noetherian).
Proof. (a) Since V| R is generated by X , by (1.1),
v =
n\sum
i=1
xiri =
n\sum
i=1
xi\Phi (ri),
where v \in VS , n \in \BbbN \ast , xi \in X, ri \in R. Thus, VS is also generated by X .
Conversely, suppose that VS is generated by X and \Phi is surjective, then si = \Phi (ri), si \in S,
ri \in R. Hence, by (1.1),
v =
n\sum
i=1
xisi =
n\sum
i=1
xi\Phi (ri) =
n\sum
i=1
xiri,
where v \in VS , n \in \BbbN \ast , xi \in X. Thus, V| R is also generated by X .
(b) This is a special case of (a).
(c) This holds by following the definition (1.1).
(d) Follows by applying (c).
(e) By (c), VS is a decomposable S -module iff VS = MS \oplus NS , where MS and NS are S -
submodules of VS iff V| R = MR \oplus NR, where MR and NR are R-submodules of V| R iff V| R is a
decomposable R-module.
(f) Again by (c), any descending (ascending) chain of S -submodules of VS is a descending
(ascending) chain of R-submodules of V| R and vice versa.
Theorem 2.1 is proved.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
A STUDY OF MODULES OVER RINGS AND THEIR EXTENSIONS 1301
Corollary 2.1. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. Then V| R is a semisimple R-module iff VS is a semisimple S -module.
Proof. By Theorem 2.1(c) and Theorem 2.1(d), V| R is a semisimple R-module iff V| R =
=
\sum
i\in I
\oplus Vi, where V \prime
iS are simple R-submodules of V| R iff VS =
\sum
i\in I
\oplus Vi, where V \prime
iS are
simple S -submodules of VS iff VS is a semisimple S -module.
Some examples and counterexamples relevant to the above observations are as under.
Example 2.1. Let R =
\biggl\{ \biggl(
a b
0 a
\biggr)
: a, b \in \BbbZ 2
\biggr\}
. Define \Phi : R \rightarrow \BbbZ 2 by \Phi
\biggl( \biggl(
a b
0 a
\biggr) \biggr)
= a.
Then \Phi is a surjective ring homomorphism and \BbbZ 2 is a cyclic \BbbZ 2-module generated by 1. Consider
\BbbZ 2 as an R-module by the action
x
\Biggl(
a b
0 a
\Biggr)
= x\Phi
\Biggl( \Biggl(
a b
0 a
\Biggr) \Biggr)
= xa, x, a, b \in \BbbZ 2.
Since (1)
\biggl(
1 0
0 1
\biggr)
= 1 and (1)
\biggl(
0 1
0 0
\biggr)
= 0, \BbbZ 2 is also a cyclic R-module generated by 1.
Example 2.2. Moreover, in the case of simple modules, obviously in Example 2.1, \BbbZ 2 is a
simple \BbbZ 2-module as well as an R-module.
Example 2.3. Let R = UTM2(\BbbZ 6) be the ring of all 2 \times 2 upper triangular matrices over \BbbZ 6 .
Define \Phi : R \rightarrow \BbbZ 6 by \Phi
\biggl( \biggl(
a b
0 c
\biggr) \biggr)
= a. Then \Phi is a surjective ring homomorphism. Since 3\BbbZ 6
and 4\BbbZ 6 are simple \BbbZ 6-submodules of \BbbZ 6 and \BbbZ 6 = 3\BbbZ 6 \oplus 4\BbbZ 6, by Theorem 2.1(d), 3\BbbZ 6| R and
4\BbbZ 6| R are simple R-modules. Therefore, \BbbZ 6 is a semisimple R-module.
The following is a counterexample of Corollary 2.1 for \Phi not to be surjective.
Example 2.4. Let R = UTM2(\BbbZ 6) and \Phi : \BbbZ 6 - \rightarrow R be defined by \Phi (a) =
\biggl(
a 0
0 a
\biggr)
. Then
\Phi is a ring homomorphism. Since J(R) =
\biggl(
0 \BbbZ 6
0 0
\biggr)
, R is not a semisimple R-module. Consider
R as a \BbbZ 6-module by the action\Biggl(
a b
0 c
\Biggr)
x =
\Biggl(
a b
0 c
\Biggr)
\Phi (x) =
\Biggl(
ax bx
0 cx
\Biggr)
, x, a, b, c \in \BbbZ 6.
Since \BbbZ 6 is semisimple, by Proposition 1.1(a), R is a semisimple \BbbZ 6-module.
The following is a counterexample of Theorem 2.1(a, c, d, f), and Corollary 2.1 for \Phi not to be
surjective.
Example 2.5. Let \Phi : \BbbZ - \rightarrow \BbbQ be defined by \Phi (a) = a. Then \Phi is a ring homomorphism.
Consider \BbbQ as a \BbbQ -module. Since it is a field, it is a simple \BbbQ -module. Therefore, it is a semisimple
and a cyclic \BbbQ -module. Consider \BbbQ as a \BbbZ -module. Then it is not a cyclic \BbbZ -module. Since it
contains \BbbZ as a \BbbZ -submodule, which is neither semisimple nor Artinian, it is not a semisimple or an
Artinian \BbbZ -module.
Example 2.6. Let R = UTM2(\BbbZ 6). Define \Phi : R - \rightarrow \BbbZ 6 by \Phi
\biggl( \biggl(
a b
0 c
\biggr) \biggr)
= a. Since 3\BbbZ 6 is
a simple \BbbZ 6-module and R-module, it is an indecomposable \BbbZ 6-module and R-module.
The following is a counterexample of Theorem 2.1(e, f) for \Phi not to be surjective.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
1302 S. A. AL-HASHMI, S. K. NAUMAN
Example 2.7. Let \Phi : \BbbQ - \rightarrow \BbbR be defined by \Phi (a) = a. Then \Phi is a ring homomorphism.
Since \BbbR \BbbR is simple, it is indecomposable. Consider \BbbR as a \BbbQ -module. Since \BbbQ is semisimple,
by Proposition 1.1 (a), \BbbR is a semisimple \BbbQ -module, which is not simple, since it contains \BbbQ as a
\BbbQ -submodule. Therefore, by Proposition 1.1(b), it is decomposable. Let si =
\bigl\{ \surd
pi : pi is a prime
number
\bigr\}
, i \in \BbbN \ast . Then the chain of \BbbQ -submodules of \BbbR , \langle s1\rangle \subseteq
\biggl\langle
2
\cup
i=1
si
\biggr\rangle
\subseteq
\biggl\langle
3
\cup
i=1
si
\biggr\rangle
\subseteq . . . does
not terminate. Thus, \BbbR is not a Noetherian \BbbQ -module, but since it is a field, it is a Noetherian
\BbbR -module.
Theorem 2.2. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism, then:
(a) V \prime
R is an essential R-submodule of V| R iff V \prime
S is an essential S -submodule of VS .
(b) \mathrm{S}\mathrm{o}\mathrm{c}(V| R) = \mathrm{S}\mathrm{o}\mathrm{c}(VS), where \mathrm{S}\mathrm{o}\mathrm{c}(V| R) and \mathrm{S}\mathrm{o}\mathrm{c}(VS) denote, respectively, the socle of V| R
and VS .
Proof. (a) Suppose that V \prime
R is an essential R-submodule of V| R. Since \Phi is surjective, by
Theorem 2.1(c), V \prime
S is an S -submodule of VS . Suppose that MS is an S -submodule of VS such that
V \prime
S \cap MS = 0. By Theorem 2.1(c), MR is an R-submodule of V| R. Therefore, V \prime
R \cap MR = 0. Since
V \prime
R is an essential R-submodule of V| R, MR = 0. Therefore, MS = 0. Thus, V \prime
S is an essential
S -submodule of VS . Similarly, we can verify the only if part.
(b) Since \mathrm{S}\mathrm{o}\mathrm{c}(V| R) = \cap
\bigl\{
L \leq V| R : L is essential inV| R
\bigr\}
, the result is satisfied immediately
from (a).
Theorem 2.2 is proved.
Example 2.8. Let R = UTM2(\BbbZ 8) and \Phi : R \rightarrow \BbbZ 8 be defined by \Phi
\biggl( \biggl(
a b
0 c
\biggr) \biggr)
= a. Then
\Phi is a surjective ring homomorphism. Consider \BbbZ 8 as an R-module by the action
x
\biggl( \biggl(
a b
0 c
\biggr) \biggr)
= x\Phi
\biggl( \biggl(
a b
0 c
\biggr) \biggr)
= xa, x, a, b, c \in \BbbZ 8.
Since \BbbZ 8 and R are Artinian, where J(\BbbZ 8) = 2\BbbZ 8 and J(R) =
\biggl(
2\BbbZ 8 \BbbZ 8
0 2\BbbZ 8
\biggr)
, by [1] (Corol-
lary 15.21),
\mathrm{S}\mathrm{o}\mathrm{c}(\BbbZ 8\BbbZ 8
) = \mathrm{S}\mathrm{o}\mathrm{c}(\BbbZ 8| R) = \{ x \in \BbbZ 8 : xa = 0, a \in 2\BbbZ 8\} = 4\BbbZ 8.
Theorem 2.3. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism, then:
(a) V \prime
R is a maximal R-submodule of V| R iff V \prime
S is a maximal S -submodule of VS ;
(b) V \prime
R is a small R-submodule of V| R iff V \prime
S is a small S -submodule of VS ;
(c) J(V| R) = J(VS), where J(V| R) and J(VS) denote, respectively, the Jacobson radicals of
V| R and VS .
Proof. (a) Suppose that V \prime
R is a maximal submodule of V| R. Then V \prime
R \not = V| R. Hence, V \prime
S \not = VS .
Since \Phi is surjective, by Theorem 2.1(c), V \prime
S is a submodule of VS . Now suppose that there exists a
submodule MS of VS such that V \prime
S \subseteq MS \subseteq VS . By Theorem 2.1(c), MR is a submodule of V| R.
Since, V \prime
R \subseteq MR \subseteq V| R and V \prime
R is a maximal submodule of V| R, MR = V \prime
R or MR = V| R . Thus,
MS = V \prime
S or MS = VS . Therefore, V \prime
S is is a maximal submodule of VS . Similarly, we can verify
the only if part.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
A STUDY OF MODULES OVER RINGS AND THEIR EXTENSIONS 1303
(b) Suppose that V \prime
R is a small submodule of V| R. Since \Phi is surjective, by Theorem 2.1(c),
V \prime
S is a submodule of VS . Suppose that MS is a submodule of VS such that V \prime
S + MS = VS . By
Theorem 2.1(c), MR is a submodule of V| R. Therefore, V \prime
R + MR = V| R. Since V \prime
R is a small
submodule of V| R, MR = V| R. Therefore, MS = VS . Thus, V \prime
S is a small submodule of VS .
Similarly, we can verify the only if part.
(c) By definition of the Jacobson radical of a module and (a) or (b), we have J(V| R) = J(VS).
Theorem 2.3 is proved.
Example 2.9. Let R =
\biggl(
\BbbZ 6 0
0 \BbbZ 6
\biggr)
and \Phi : R - \rightarrow \BbbZ 6 be defined by \Phi
\biggl( \biggl(
a 0
0 b
\biggr) \biggr)
= a.
Then \Phi is a surjective ring homomorphism. Since J(\BbbZ 6) = 0, J(R) = 0. Consider \BbbZ 6 as an
R-module. Since R is semisimple, by Proposition 1.1(a), \BbbZ 6 is a projective R-module. Thus,
J(\BbbZ 6| R) = \BbbZ 6J(R) = 0 = J(\BbbZ 6) [6] (Theorem 24.7).
The following example shows that if \Phi is not surjective, then
\mathrm{S}\mathrm{o}\mathrm{c}(V| R) \not = \mathrm{S}\mathrm{o}\mathrm{c}(VS) and J(V| R) \not = J(VS).
Example 2.10. Let \Phi : \BbbZ - \rightarrow \BbbQ be defined by \Phi (a) = a. Then \mathrm{S}\mathrm{o}\mathrm{c}(\BbbQ \BbbQ ) = \BbbQ and J(\BbbQ \BbbQ ) = 0.
Since \BbbQ \BbbZ has no maximal and no minimal \BbbZ -submodules, \mathrm{S}\mathrm{o}\mathrm{c}(\BbbQ \BbbZ ) = 0 and J(\BbbQ \BbbZ ) = \BbbQ .
Theorem 2.4. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If V| R is a faithful R-module, then VS is a faithful S -module.
Proof. Suppose s \in S such that vs = 0 for all v \in VS . Since \Phi is surjective, there exists r \in R
such that s = \Phi (r). Hence, v\Phi (r) = 0, which implies by (1.1), vr = 0 for all v \in V| R. Since V| R
is faithful, r = 0. Therefore, s = \Phi (0) = 0. Thus, VS is a faithful S -module.
Theorem 2.4 is proved.
Example 2.11. Let R = UTM2(\BbbZ 6) and \Phi : R - \rightarrow \BbbZ 6 be defined by \Phi
\biggl( \biggl(
a b
0 c
\biggr) \biggr)
= a.
Consider 3\BbbZ 6 = \{ 0, 3\} as a \BbbZ 6-module. Then ann 3\BbbZ 6 = \{ 0, 2, 4\} . Thus, 3\BbbZ 6 is not a faithful
\BbbZ 6-module. Now consider 3\BbbZ 6 as an R-module by the action
x
\Biggl(
a b
0 c
\Biggr)
= x\Phi
\Biggl( \Biggl(
a b
0 c
\Biggr) \Biggr)
= xa, x \in 3\BbbZ 6, a, b, c \in \BbbZ 6.
Then
ann3\BbbZ 6| R =
\Biggl\{ \Biggl(
a b
0 c
\Biggr)
\in R : a \in \{ 0, 2, 4\}
\Biggr\}
.
Thus, 3\BbbZ 6| R is not a faithful R-module.
The following example shows that the converse of Theorem 2.4 is not true in general.
Example 2.12. Let R =
\biggl\{ \biggl(
a b
b a
\biggr)
: a, b \in \BbbR
\biggr\}
. Then R is a subring of M2(\BbbR ). Define \Phi :
R - \rightarrow \BbbR by \Phi
\biggl( \biggl(
a b
b a
\biggr) \biggr)
= a - b. Then \Phi is a surjective ring homomorphism. Let A =
\biggl(
\BbbR
\BbbR
\biggr)
.
Then A is a left \BbbR -module by the action r
\biggl(
a
b
\biggr)
=
\biggl(
ra
rb
\biggr)
, r \in \BbbR . If 0 \not =
\biggl(
a
b
\biggr)
\in A and r
\biggl(
a
b
\biggr)
= 0,
then ra = 0 and rb = 0. Since \BbbR is a field, r = 0. Thus, A is a faithful \BbbR -module. Consider A as
a left R-module by the action
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
1304 S. A. AL-HASHMI, S. K. NAUMAN\Biggl(
a b
b a
\Biggr) \Biggl(
a\prime
b\prime
\Biggr)
= \Phi
\Biggl( \Biggl(
a b
b a
\Biggr) \Biggr) \Biggl(
a\prime
b\prime
\Biggr)
=
\Biggl(
(a - b)a\prime
(a - b)b\prime
\Biggr)
.
If 0 \not =
\biggl(
a\prime
b\prime
\biggr)
\in A and
\biggl(
a b
b a
\biggr) \biggl(
a\prime
b\prime
\biggr)
= 0, then (a - b)a\prime = 0 and (a - b)b\prime = 0. Since \BbbR is a field,
a = b. Hence, annA =
\biggl\{ \biggl(
a a
a a
\biggr)
: a \in \BbbR
\biggr\}
. Thus, A is not a faithful R-module.
The following is a counterexample for \Phi not to be surjective.
Example 2.13. Let R = UTM2(\BbbZ 6) and \Phi : 3\BbbZ 6 - \rightarrow R be defined by \Phi (a) =
\biggl(
a 0
0 a
\biggr)
,
a \in 3\BbbZ 6 = \{ 0, 3\} . Consider A =
\biggl(
3\BbbZ 6
3\BbbZ 6
\biggr)
. Then A is a left R-module by the action
\Biggl(
a b
0 c
\Biggr) \Biggl(
a\prime
b\prime
\Biggr)
=
\Biggl(
aa\prime + bb\prime
cb\prime
\Biggr)
\in A, a, b, c \in \BbbZ 6, a\prime , b\prime \in 3\BbbZ 6.
Also A is a left 3\BbbZ 6-module by the action
a
\Biggl(
b
c
\Biggr)
= \Phi (a)
\Biggl(
b
c
\Biggr)
=
\Biggl(
ab
ac
\Biggr)
, a, b, c \in 3\BbbZ 6.
If 0 \not =
\biggl(
b
c
\biggr)
and a
\biggl(
b
c
\biggr)
= 0, then ab = 0 and ac = 0. Since a, b \in 3\BbbZ 6, a = 0. Thus, A is a faithful
3\BbbZ 6-module. But A is not a faithful R-module, since for all
\biggl(
a
b
\biggr)
\in A,
\biggl(
4 4
0 4
\biggr) \biggl(
a
b
\biggr)
= 0.
Lemma 2.1. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If s \in S is not a zero divisor, then there exists r \in R, which is not a zero divisor such that
s = \Phi (r).
Proof. Suppose s \in S is not a zero divisor. Then for all 0 \not = s\prime \in S, we have ss\prime \not = 0 and
s\prime s \not = 0. Since \Phi is surjective, there exist r, r\prime \in R such that s = \Phi (r), s\prime = \Phi (r\prime ). Since \Phi (0) = 0
and s\prime \not = 0, r\prime \not = 0. Therefore, ss\prime \not = 0 implies \Phi (rr\prime ) \not = 0 and so rr\prime \not = 0. Similarly, if s\prime s \not = 0,
then r\prime r \not = 0. Hence, r is not a zero divisor.
Lemma 2.1 is proved.
Theorem 2.5. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If V| R is a divisible R-module, then VS is a divisible S -module.
Proof. Suppose s \in S is not a zero divisor and v \in VS . Then by Lemma 2.1, there exists
r \in R, which is not a zero divisor such that s = \Phi (r). Since V| R is divisible, there exists v\prime \in V| R
such that v = v\prime r, so by (1.1), v = v\prime r = v\prime \Phi (r) = v\prime s. Thus, VS is divisible.
Theorem 2.5 is proved.
Example 2.14. Let R =
\biggl(
\BbbZ 0
0 \BbbZ
\biggr)
and \Phi : R \rightarrow \BbbZ be defined by \Phi
\biggl( \biggl(
a 0
0 b
\biggr) \biggr)
= a. Then \Phi
is a surjective ring homomorphism. Consider \BbbZ as a \BbbZ -module. Then \BbbZ is not divisible, since there
does not exist x \in \BbbZ such that 3 = 2x. Now consider \BbbZ as an R-module by the action
z
\Biggl(
a 0
0 b
\Biggr)
= z\Phi
\Biggl( \Biggl(
a 0
0 b
\Biggr) \Biggr)
= za, z, a, b \in \BbbZ .
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
A STUDY OF MODULES OVER RINGS AND THEIR EXTENSIONS 1305
Then \BbbZ is not a divisible R-module, since for r =
\biggl(
2 0
0 3
\biggr)
\in R, there does not exist x \in \BbbZ such
that 5 = x
\biggl(
2 0
0 3
\biggr)
.
The converse of Theorem 2.5 is not true in general and the following example illustrates this.
Example 2.15. Let \Phi : \BbbZ \rightarrow \BbbZ 3 be defined by \Phi (a) = a. Then \Phi is a surjective ring homomor-
phism. For 1, 2 \in \BbbZ 3, we have x = 1(x) for all x \in \BbbZ 3, 0 = 2(0), 1 = 2(2) and 2 = 2(1). Thus,
\BbbZ 3 is a divisible \BbbZ 3-module. Consider \BbbZ 3 as a \BbbZ -module by the action az = a\Phi (z) = az. Then \BbbZ 3
is not a divisible \BbbZ -module, since there does not exist a \in \BbbZ 3 such that 2 = 3a.
Remark 2.1. Since every Abelian group is a unital \BbbZ -module in a unique way [1, p. 27], the
action of \BbbZ on \BbbZ 3 in the previous example is the usual action by simple calculations.
The following is a counterexample for \Phi not to be surjective.
Example 2.16. Let \Phi : \BbbR \rightarrow \BbbR [x] be defined by \Phi (a) = a. Then \Phi is a ring homomorphism.
Consider the ideal V = \langle x2\rangle of \BbbR [x]. Let f(x) = x \in \BbbR [x]. Then there does not exist g(x) \in V
such that x2 = f(x)g(x), since \mathrm{d}\mathrm{e}\mathrm{g}(g(x)) \geq 2. Thus, V is not a divisible \BbbR [x]-module. Consider V
as a left \BbbR -module by the action af(x) = \Phi (a)f(x) \in V, a \in \BbbR , f(x) \in V. Then for all 0 \not = r \in \BbbR
and f(x) \in V, there exists r - 1f(x) \in V such that f(x) = r[r - 1f(x)]. Thus, V is a divisible
\BbbR -module.
Theorem 2.6. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism.
(a) If VS is a torsion S -module, then V| R is a torsion R-module.
(b) If V| R is a torsion free R-module, then VS is a torsion free S -module.
Proof. (a) Suppose v \in VS . Since VS is a torsion S -module, there exists s \in S, which is
not a zero divisor such that vs = 0. Therefore, by Lemma 2.1, there exists r \in R, which is not a
zero divisor such that s = \Phi (r). Hence, by (1.1), 0 = vs = v\Phi (r) = vr. Thus, V| R is a torsion
R-module.
(b) Suppose s \in S, which is not a zero divisor such that vs = 0. Then by Lemma 2.1, there exists
r \in R, which is not a zero divisor such that s = \Phi (r). Therefore, by (1.1), 0 = vs = v\Phi (r) = vr.
Since V| R is torsion free, v = 0. Thus, VS is a torsion free S -module.
Theorem 2.6 is proved.
The converse of Theorem 2.6(a) or Theorem 2.6(b) is not true in general and the following
example illustrates this.
Example 2.17. Let \Phi : \BbbZ \rightarrow \BbbZ 3 be defined by \Phi (a) = a. Since \BbbZ 3 is a field, it is a torsion free
\BbbZ 3-module, whence it is not a torsion \BbbZ 3-module. But \BbbZ 3 is clearly a torsion \BbbZ -module, whence, it
is not a torsion free \BbbZ -module.
The following is a counterexample of both cases of Theorem 2.6 for \Phi not to be surjective.
Example 2.18. Let \Phi : \BbbR \rightarrow \BbbR [x] be defined by \Phi (a) = a. Consider the \BbbR [x]-module \BbbR 2. Then
\BbbR 2 is a torsion \BbbR [x]-module [3, p. 185]. Therefore, it is not a torsion free \BbbR [x]-module. Consider \BbbR 2
as an \BbbR -module by the action r(a, b) = \Phi (r)(a, b) = (ra, rb). If 0 \not = r \in \BbbR and r(a, b) = 0, then
ra = 0 and rb = 0 implies a = 0 and b = 0. Hence, \BbbR 2 is a torsion free \BbbR -module. Therefore, it is
not a torsion \BbbR -module.
Lemma 2.2. Suppose that R and S are rings and \Phi : R \rightarrow S is a ring homomorphism, then:
(a) If MS and VS are S -modules and \theta : MS \rightarrow VS is an S -homomorphism, then \theta : M| R \rightarrow
\rightarrow V| R is an R-homomorphism.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
1306 S. A. AL-HASHMI, S. K. NAUMAN
(b) If \Phi is surjective and \theta : M| R \rightarrow V| R is an R-homomorphism, then \theta : MS \rightarrow VS is an
S -homomorphism.
(c) \mathrm{E}\mathrm{n}\mathrm{d}S(VS) \subseteq \mathrm{E}\mathrm{n}\mathrm{d}R(V| R). Equality holds if \Phi is surjective.
Proof. (a) Suppose m \in M| R, r \in R. Then by (1.1),
\theta (mr) = \theta (m\Phi (r)) = \theta (m)\Phi (r) = \theta (m)r.
Therefore, \theta is an R-homomorphism.
(b) Suppose m \in MS ; s \in S. Since \Phi is surjective, there exists r \in R such that s = \Phi (r).
Hence, by (1.1),
\theta (ms) = \theta (m\Phi (r)) = \theta (mr) = \theta (m)r = \theta (m)\Phi (r) = \theta (m)s.
Therefore, \theta is an S -homomorphism.
(c) Put MS = VS and M| R = V| R , then by (a) \mathrm{E}\mathrm{n}\mathrm{d}S(VS) \subseteq \mathrm{E}\mathrm{n}\mathrm{d}R(V| R) and if \Phi is surjective,
then by (b) \mathrm{E}\mathrm{n}\mathrm{d}S(VS) = \mathrm{E}\mathrm{n}\mathrm{d}R(V| R).
Lemma 2.2 is proved.
The following is a counterexample for \Phi not to be surjective.
Example 2.19. Let R = UTM2(\BbbZ 6). Define \Phi : \BbbZ 6 \rightarrow R by \Phi (a) =
\biggl(
a 0
0 a
\biggr)
. Consider \BbbZ 6
as an R-module by the action
x
\Biggl(
a b
0 c
\Biggr)
= xc \in \BbbZ 6. (2.1)
Then \BbbZ 6| \BbbZ 6
is a \BbbZ 6-module by the action
xz = x\Phi (z) = x
\Biggl(
z 0
0 z
\Biggr)
, x, z \in \BbbZ 6. (2.2)
Consider \BbbZ 6 \oplus \BbbZ 6 as an R-module by the action
(x, y)
\Biggl(
a b
0 c
\Biggr)
= (xa, xb+ yc) \in \BbbZ 6 \oplus \BbbZ 6. (2.3)
Then \BbbZ 6 \oplus \BbbZ 6 is a \BbbZ 6-module by the action
(x, y)z = (x, y)\Phi (z) = (xz, yz), x, y, z \in \BbbZ 6. (2.4)
Now define \theta : \BbbZ 6| \BbbZ 6
- \rightarrow \BbbZ 6 \oplus \BbbZ 6| \BbbZ 6
by \theta (x) = (x, 0). It is clear that \theta is a group homomor-
phism. By (2.4), \theta (xz) = (xz, 0) = (x, 0)z = \theta (x)z. Hence, \theta is a \BbbZ 6-homomorphism. But \theta is
not an R-homomorphism, since by (2.1) and (2.3),
\theta
\Biggl(
x
\Biggl(
a b
0 c
\Biggr) \Biggr)
= \theta (xc) = (xc, 0),
while
\theta (x)
\Biggl(
a b
0 c
\Biggr)
= (x, 0)
\Biggl(
a b
0 c
\Biggr)
= (xa, xb),
\Biggl(
a b
0 c
\Biggr)
\in R, x, z \in \BbbZ 6.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
A STUDY OF MODULES OVER RINGS AND THEIR EXTENSIONS 1307
Theorem 2.7. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homo-
morphism, which is an R-homomorphism also. If \theta \in \mathrm{H}\mathrm{o}\mathrm{m}R(V| R, R) = V \ast
| R, then \Phi \circ \theta \in
\in \mathrm{H}\mathrm{o}\mathrm{m}S(VS , S) = V \ast
S .
Proof. Since S is an R-module by the action sr = s\Phi (r), s \in S; r \in R, we can consider \Phi as
an R-homomorphism. Suppose v \in VS . Since \Phi is surjective, s = \Phi (r) for s \in S, r \in R. Hence,
by (1.1),
(\Phi \circ \theta )(vs) = \Phi (\theta (vs)) = \Phi (\theta (v\Phi (r)) = \Phi (\theta (vr)) = \Phi [(\theta (v)r] =
= \Phi [(\theta (v)]r = [(\Phi \circ \theta )(v)]r = [(\Phi \circ \theta )(v)]\Phi (r) = [(\Phi \circ \theta )(v)]s.
Thus, \Phi \circ \theta \in \mathrm{H}\mathrm{o}\mathrm{m}S(VS , S) = V \ast
S .
Theorem 2.7 is proved.
Theorem 2.8. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If V| R is an injective R-module, then VS is an injective S -module.
Proof. Let LS and MS be S -modules, f : LS - \rightarrow MS be an S -monomorphism and g : LS - \rightarrow
- \rightarrow VS be an S -homomorphism. Then by Lemma 2.2, f : L| R - \rightarrow M| R is an R-monomorphism and
g : L| R - \rightarrow V| R is an R-homomorphism. Since V| R is injective, there exists an R-homomorphism h :
M| R - \rightarrow V| R such that h \circ f = g. Since \Phi is surjective, by Lemma 2.2, h is an S -homomorphism.
Thus, VS is an injective S -module.
Theorem 2.8 is proved.
Remark 2.2. (a) The converse of Theorem 2.8 is not true in general.
(b) The injective envelopes of V| R and VS , which are denoted, respectively, by E(V| R) and
E(VS), may not be the same. These facts are illustrated in the following example.
Example 2.20. Let \Phi : \BbbZ \rightarrow \BbbZ 3 be defined by \Phi (a) = a. Then \BbbZ 3 is an injective \BbbZ 3-module
and E(\BbbZ 3) = \BbbZ 3. Consider \BbbZ 3 as a \BbbZ -module. By Example 2.15, \BbbZ 3 is not a divisible \BbbZ -module.
Thus, \BbbZ 3 is not an injective \BbbZ -module and E(\BbbZ 3) = \BbbZ 3\infty (the Prüfer 3-group) [5] (Proposition 3.19,
Example 3.36).
The following is a counterexample for \Phi not to be surjective.
Example 2.21. Let \Phi : \BbbR \rightarrow \BbbR [x] be defined by \Phi (a) = a and V = \langle x2\rangle . By Example 2.16, V
is not a divisible \BbbR [x]-module. Therefore, V is not an injective \BbbR [x]-module [3] (Proposition 5.2.11).
Since \BbbR is semisimple, by Proposition 1.1(a), V is an injective \BbbR -module.
Lemma 2.3. Suppose that R and S are rings and \Phi : R \rightarrow S is a ring homomorphism. If M
is a right S -module, N is a left S -module and B is an Abelian group, then:
(a) If \tau : MS \times S N \rightarrow B is an S -balanced map, then \tau : M| R \times R| N \rightarrow B is an R-balanced
map.
(b) If \Phi is surjective and \tau : M| R\times R| N \rightarrow B is an R-balanced map, then \tau : MS \times S N \rightarrow B
is an S -balanced map.
Proof. (a) Since \tau : MS \times S N \rightarrow B is an S -balanced map,
\tau (m+m\prime , n) = \tau (m,n) + \tau (m\prime , n),
\tau (m,n+ n\prime ) = \tau (m,n) + \tau (m,n\prime ),
\tau (ms, n) = \tau (m, sn) \forall m,m\prime \in M \forall n, n\prime \in N \forall s \in S.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
1308 S. A. AL-HASHMI, S. K. NAUMAN
So, to make \tau an R-balanced map, we need only to verify the third condition. Suppose r \in R.
Then by (1.1),
\tau (mr, n) = \tau (m\Phi (r), n) = \tau (m,\Phi (r)n) = \tau (m, rn).
Thus, \tau is an R-balanced map.
(b) Since \tau is an R-balanced map and \Phi is surjective, by (1.1), we have
\tau (ms, n) = \tau (m\Phi (r), n) = \tau (m,\Phi (r)n) = \tau (m, sn), s \in S, r \in R, with s = \Phi (r).
Hence, \tau is an S -balanced map.
Lemma 2.3 is proved.
Lemma 2.4. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If M = MS = M| R and N = SN = R| N, then
M \otimes R N \sim = M \otimes S N as groups.
Proof. The pair (\tau , M \otimes R N) is a tensor product of M and N over R, where \tau : M \times N \rightarrow
\rightarrow M \otimes R N is an R-balanced map. Since \Phi is surjective, by Lemma 2.3, \tau is an S -balanced
map. Now suppose that A is an Abelian group and \beta : M \times N \rightarrow A is an S -balanced map. Then,
by Lemma 2.3, \beta is an R-balanced map. By definition of a tensor product, there exists a unique
\BbbZ -homomorphism f : M \otimes R N - \rightarrow A such that the diagram
M \times N
\tau \swarrow \searrow \beta
M \otimes R N
f - \rightarrow A
commutes. Since \tau and \beta are S -balanced maps, the pair (\tau , M \otimes R N) is a tensor product of M
and N over S. Since the tensor product is unique to within isomorphism, M \otimes R N \sim = M \otimes S N.
Lemma 2.4 is proved.
Example 2.22. Let R be a ring and \Phi : R - \rightarrow R\diagup I be the natural epimorphism, where I is
an ideal of R. Let B = U\diagup I, where U is an ideal of R such that I \subseteq U. Then B is an ideal
of R\diagup I. Since IB = \{ I\} , R\diagup I \otimes R B \sim = B\diagup IB \sim = B\diagup \{ I\} \sim = B [2, p. 217] (Execise 9). Also,
R\diagup I \otimes R\diagup I B \sim = B.
The following is a counterexample for \Phi not to be surjective.
Example 2.23. Let \Phi : \BbbR - \rightarrow \BbbC be defined by \Phi (a) = a. Then \Phi is a ring homomorphism. By
considering \BbbC as a vector space over \BbbC , we obviously have
\BbbC \otimes \BbbC \BbbC \sim = \BbbC .
and as vector spaces over \BbbR ,\BbbC \sim = \BbbR 2. Hence,
\BbbC \otimes \BbbR \BbbC \sim = \BbbC \otimes \BbbR \BbbR 2 \sim = \BbbC 2.
Theorem 2.9. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If V| R is a flat R-module, then VS is a flat S -module.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
A STUDY OF MODULES OVER RINGS AND THEIR EXTENSIONS 1309
Proof. Let
0 - \rightarrow A - \rightarrow B - \rightarrow C - \rightarrow 0
be a short exact sequence of left S -modules. Then by Lemma 2.2,
0 - \rightarrow R| A - \rightarrow R| B - \rightarrow R| C - \rightarrow 0
is a short exact sequence of left R-modules. Since V| R is a flat R-module,
0 - \rightarrow V| R \otimes R A - \rightarrow V| R \otimes R B - \rightarrow V| R \otimes R C - \rightarrow 0
is a short exact sequence of Abelian groups. Hence, by Lemma 2.4,
0 - \rightarrow VS \otimes S A - \rightarrow VS \otimes S B - \rightarrow VS \otimes S C - \rightarrow 0
is a short exact sequence of Abelian groups. Thus, VS is a flat S -module.
Theorem 2.9 is proved.
Example 2.24. Let R =
\biggl(
\BbbZ 2 0
0 \BbbZ 2
\biggr)
. Define \Phi : R \rightarrow \BbbZ 2 by \Phi
\biggl( \biggl(
a 0
0 b
\biggr) \biggr)
= a. Then \Phi is
a surjective ring homomorphism. Since R and \BbbZ 2 are semisimple, by Proposition 1.1(a) and [1]
(Proposition 19.16), \BbbZ 2 is a flat \BbbZ 2-module and R-module.
The following example shows that the converse of Theorem 2.9 is not true in general.
Example 2.25. Let \Phi : \BbbZ \rightarrow \BbbZ 3 be defined by \Phi (a) = a. Then \BbbZ 3 is a flat \BbbZ 3-module. Since
\BbbZ 3 is not a torsion free \BbbZ -module, it is not a flat \BbbZ -module [8] (Corollary 3.51).
Theorem 2.10. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If V| R is a free R-module, then VS is a free S -module.
Proof. Suppose that X = \{ xi\} i\in I is an R-basis for V| R. Then by Theorem 2.1(a), X generates
VS , so it remains to prove that X is linearly independent over S. Suppose si \in S and
\sum
i\in I
xisi =
= 0, which implies
\sum
i\in I
xi\Phi (ri) = 0, ri \in R, since \Phi is surjective. By (1.1), we get
\sum
i\in I
xiri =
= 0. Since X is linearly independent over R, ri = 0 \forall i \in I. Therefore, si = \Phi (ri) = \Phi (0) = 0
\forall i \in I. Thus, VS is a free S -module.
Theorem 2.10 is proved.
Example 2.26. Let \Phi : \BbbZ [x] \rightarrow \BbbZ be defined by \Phi (f(x)) = a, where a is the constant term
of f(x). Then \Phi is a surjective ring homomorphism. Consider \BbbZ [x] as a \BbbZ -module. Then \BbbZ [x]
is a free \BbbZ -module with basis \{ xn : n \in \BbbZ +\} . Consider \BbbZ [x] as a \BbbZ [x]-module by the action
f(x)g(x) = f(x)\Phi (g(x)) = f(x)a, f(x), g(x) \in \BbbZ [x], a is the constant term of g(x). Then \BbbZ [x]
is a free \BbbZ [x]-module with the same basis.
The following is a counterexample for \Phi not to be surjective.
Example 2.27. Let R = UTM2(\BbbZ 2). Define \Phi : \BbbZ 2 \rightarrow R by \Phi (a) =
\biggl(
a 0
0 a
\biggr)
. Consider
\BbbZ 2 \oplus \BbbZ 2 as an R-module by the action
(x, y)
\Biggl(
a b
0 c
\Biggr)
= (xa, xb+ yc) \in \BbbZ 2 \oplus \BbbZ 2.
Since \BbbZ 2 \oplus \BbbZ 2 and Rn, for any n \in \BbbN \ast , are of different orders, they can not be isomorphic.
Hence, \BbbZ 2 \oplus \BbbZ 2 is not a free R-module [2, p. 181] (Theorem 2.1). On the other hand, \BbbZ 2 \oplus \BbbZ 2 is a
free \BbbZ 2-module.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
1310 S. A. AL-HASHMI, S. K. NAUMAN
Theorem 2.11. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If V| R is a projective R-module, then VS is a projective S -module.
Proof. Suppose that V| R is a projective R-module and let
0 - \rightarrow LS
f - \rightarrow MS
g - \rightarrow VS - \rightarrow 0
be a short exact sequence of S -modules. Since f and g are S -homomorphism, by Lemma 2.2, f
and g are R-homomorphism. Thus,
0 - \rightarrow L| R
f - \rightarrow M| R
g - \rightarrow V| R - \rightarrow 0
is a short exact sequence of R-modules. Since V| R is a projective R-module, this sequence splits.
Therefore, there exists an R-homomorphism g\prime : V| R - \rightarrow M| R such that gg\prime = 1V| R . Since \Phi is
surjective, by Lemma 2.2, g\prime is an S -homomorphism and 1V| R = 1VS
. Thus, VS is a projective
S -module [2, p. 192] (Theorem 3.4).
Example 2.28. Let \Phi : \BbbZ [x] \rightarrow \BbbZ be defined by \Phi (f(x)) = a, where a is the constant term of
f(x). By Example 2.26, \BbbZ [x] is a free \BbbZ -module and \BbbZ [x]-module. Therefore, \BbbZ [x] is a projective
\BbbZ -module and \BbbZ [x]-module [2, p. 191] (Theorem 3.2).
The following is a counterexample of Theorems 2.9 and 2.11 for \Phi not to be surjective.
Example 2.29. Let \Phi : \BbbR \rightarrow \BbbR [x] be defined by \Phi (a) = a. Since \BbbR is semisimple, by Propo-
sition 1.1(a), \BbbR 2 is a projective \BbbR -module, therefore, it is a flat \BbbR -module [1] (Proposition 9.16).
Since \BbbR 2 is not a torsion free \BbbR [x]-module by Example 2.18, it is not a flat \BbbR [x]-module [8] (Corol-
lary 3.51), so it is not a projective \BbbR [x]-module [1] (Proposition 19.16).
The converse of Theorem 2.10 and the converse of Theorem 2.11 are not true in general. The
following example illustrates these facts.
Example 2.30. Let R =
\biggl\{ \biggl(
a b
0 a
\biggr)
: a \in \BbbZ 2
\biggr\}
. Define \Phi : R \rightarrow \BbbZ 2 by \Phi
\biggl( \biggl(
a b
0 a
\biggr) \biggr)
= a.
Then \BbbZ 2 is a free (projective) \BbbZ 2-module. Consider \BbbZ 2 as an R-module. Since \BbbZ 2 and Rn, for
any n \in \BbbN \ast , are of different orders, they can not be isomorphic. Hence, \BbbZ 2 is not a free R-
module [2, p. 181] (Theorem 2.1). Since J(R) =
\biggl\{ \biggl(
0 0
0 0
\biggr)
,
\biggl(
0 1
0 0
\biggr) \biggr\}
, R is a local ring [1]
(Proposition 15.15). Therefore, \BbbZ 2 is not a projective R-module [4] (Theorem 2).
Theorem 2.12. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If V| R has a projective cover over R, then VS has a projective cover over S also.
Proof. Since \Phi is surjective, by the first isomorphism theorem
R\diagup \mathrm{k}\mathrm{e}\mathrm{r}\Phi \sim = S.
Therefore, by Proposition 1.2(a), we get the result.
Theorem 2.12 is proved.
Example 2.31. Let R =
\biggl(
\BbbZ 3 0
0 \BbbZ 3
\biggr)
and \Phi : R \rightarrow \BbbZ 3 be defined by \Phi
\biggl( \biggl(
a 0
0 b
\biggr) \biggr)
= a. Since
R and \BbbZ 3 are semisimple rings, by Proposition 1.1(a), \BbbZ 3 is a projective \BbbZ 3-module and R-module.
Therefore, it has a projective cover over \BbbZ 3 and R [6] (Remark 24.11(5)).
The converse of Theorem 2.12 is not true in general and the following example illustrates this.
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
A STUDY OF MODULES OVER RINGS AND THEIR EXTENSIONS 1311
Example 2.32. Let \Phi : \BbbZ \rightarrow \BbbZ 3 be defined by \Phi (a) = a. Since \BbbZ 3 is a projective \BbbZ 3-module
and J -semisimple, it has a projective cover over \BbbZ 3. On the other hand, \BbbZ 3 is not a projective
\BbbZ -module, it also does not have a projective cover over \BbbZ [6] (Remark 24.11(5)).
Theorem 2.13. Suppose that R and S are rings and \Phi : R \rightarrow S is a surjective ring homomor-
phism. If V| R is a semiperfect R-module, then VS is a semiperfect S -module.
Proof. Since V| R is projective, by Theorem 2.11, VS is projective. Let V = VS\diagup M be a factor
module of VS , where M is an S -submodule of VS . By Theorem 2.1(c), M is an R-submodule
of V| R. Hence, V| R\diagup M is an R-module by the action (v + M)r = vr + M = v\Phi (r) + M =
= (v + M)\Phi (r). Therefore, V| R\diagup M = (V )| R. Since V| R is semiperfect, (V )| R has a projective
cover over R. Thus, by Theorem 2.12, V has a projective cover over S. Hence, VS is a semiperfect
S -module.
Theorem 2.13 is proved.
Example 2.33. Let R =
\biggl(
\BbbZ 3 0
0 \BbbZ 3
\biggr)
. Define \Phi : R \rightarrow \BbbZ 3 by \Phi
\biggl( \biggl(
a 0
0 b
\biggr) \biggr)
= a. Since R and
\BbbZ 3 are semisimple, by Proposition 1.1(a), \BbbZ 3 is a projective \BbbZ 3-module and R-module. Thus, by
Proposition 1.2(b), it is a semiperfect \BbbZ 3-module and R-module.
The following is a counterexample for \Phi not to be surjective.
Example 2.34. Let \Phi : \BbbR \rightarrow \BbbR [x] be defined by \Phi (a) = a. By Example 2.29, \BbbR 2 is not a
projective \BbbR [x]-module. Therefore, it is not a semiperfect \BbbR [x]-module. Since \BbbR is Artinian and \BbbR 2
is a projective \BbbR -module, by Proposition 1.2(b) it is a semiperfect \BbbR -module.
Remark 2.3. The converse of Theorem 2.13 is not true in general, since if VS is a projective
S -module, V| R may not be a projective R-module (see Example 2.30).
3. Conclusion. A summary of above studies is entered in the following tables.
3.1. First column on left-hand side of the following table represents the type of the module V| R,
and the remaining two columns illustrate whether VS is of the same type or not.
Type of module General case Surjective case
1-generated yes yes
2-simple yes yes
3-semisimple no (Example 2.4) yes
4-indecomposable yes yes
5-Artinian yes yes
6-Noetherian yes yes
7-faithful no (Example 2.13) yes
8-divisible no (Example 2.16) yes
9-torsion no no (Example 2.17)
10-torsion free no (Example 2.18) yes
11-injective no (Example 2.21) yes
12-flat no (Example 2.29) yes
13-free no (Example 2.27) yes
14-projective no (Example 2.29) yes
15-semiperfect no (Example 2.34) yes
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
1312 S. A. AL-HASHMI, S. K. NAUMAN
3.2. First column on left-hand side of the following table represents the type of the module VS ,
and the remaining two columns illustrate whether V| R is of the same type or not.
Type of module General case Surjective case
1-generated no (Example 2.5) yes
2-simple no (Example 2.5) yes
3-semisimple no (Example 2.5) yes
4-indecomposable no (Example 2.7) yes
5-Artinian no (Example 2.5) yes
6-Noetherian no (Example 2.7) yes
7-faithful no no (Example 2.12)
8-divisible no no (Example 2.15)
9-torsion no (Example 2.18) yes
10-torsion free no no (Example 2.17)
11-injective no no (Example 2.20)
12-flat no no (Example 2.25)
13-free no no (Example 2.30)
14-projective no no (Example 2.30)
15-semiperfect no no (Example 2.30)
References
1. Anderson F. W., Fuller K. R. Rings and categories of modules // Grad. Texts Math. – 1992. – 13.
2. Hungerford T. W. Algebra // Grad. Texts Math. – 1989. – 73.
3. Hazewinkel M., Gubareni N., Kirichenko V. V. Algebra, rings and modules // Math. and Appl. – 2005. – Vol. 1.
4. Kaplansy I. Projective modules // Ann. Math. – 1958. – 68. – P. 372 – 377.
5. Lam T. Y. Lectures on modules and rings // Grad. Texts Math. – 1999. – 189.
6. Lam T. Y. A first course in noncommutative rings. – New York: Springer-Verlag, 2001.
7. Mares E. A. Semi-perfect modules // Math. Z. – 1963. – 82. – S. 347 – 360.
8. Rotman J. An introduction to homological algebra. – Springer Science & Business Media, 2008.
Received 02.02.16
ISSN 1027-3190. Укр. мат. журн., 2018, т. 70, № 10
|
| id | umjimathkievua-article-1636 |
| institution | Ukrains’kyi Matematychnyi Zhurnal |
| keywords_txt_mv | keywords |
| language | English |
| last_indexed | 2026-03-24T02:09:39Z |
| publishDate | 2018 |
| publisher | Institute of Mathematics, NAS of Ukraine |
| record_format | ojs |
| resource_txt_mv | umjimathkievua/ee/fdd6162ae0c6f7750494cb5ddb6f3dee.pdf |
| spelling | umjimathkievua-article-16362019-12-05T09:21:55Z A study of modules over rings and their extensions Дослiдження модулiв над кiльцями та їх розширення Al-Hashmi, S. A. Nauman, S. K. Аль, - Хашмі С. А. Науман, С. К. We study the transfer of properties of some types of modules under identity preserving ring homomorphisms. These studies seem to be overlooked in literature. We have picked a few types of modules and provided proofs for these properties that are transferable and suitable counterexamples for the properties that are not transferable. Вивчається проблема передачi властивостей деяких типiв модулiв пiд дiєю гомоморфiзмiв, що зберiгають одиницю. Подiбнi дослiдження, здається, вiдсутнi в лiтературi. Вибрано кiлька типiв модулiв та наведено доведення для тих властивостей, що передаються, та придатнi контрприклади для тих властивостей, що не передаються. Institute of Mathematics, NAS of Ukraine 2018-10-25 Article Article application/pdf https://umj.imath.kiev.ua/index.php/umj/article/view/1636 Ukrains’kyi Matematychnyi Zhurnal; Vol. 70 No. 10 (2018); 1299-1312 Український математичний журнал; Том 70 № 10 (2018); 1299-1312 1027-3190 en https://umj.imath.kiev.ua/index.php/umj/article/view/1636/618 Copyright (c) 2018 Al-Hashmi S. A.; Nauman S. K. |
| spellingShingle | Al-Hashmi, S. A. Nauman, S. K. Аль, - Хашмі С. А. Науман, С. К. A study of modules over rings and their extensions |
| title | A study of modules over rings and their extensions |
| title_alt | Дослiдження модулiв над кiльцями та їх розширення |
| title_full | A study of modules over rings and their extensions |
| title_fullStr | A study of modules over rings and their extensions |
| title_full_unstemmed | A study of modules over rings and their extensions |
| title_short | A study of modules over rings and their extensions |
| title_sort | study of modules over rings and their extensions |
| url | https://umj.imath.kiev.ua/index.php/umj/article/view/1636 |
| work_keys_str_mv | AT alhashmisa astudyofmodulesoverringsandtheirextensions AT naumansk astudyofmodulesoverringsandtheirextensions AT alʹhašmísa astudyofmodulesoverringsandtheirextensions AT naumansk astudyofmodulesoverringsandtheirextensions AT alhashmisa doslidžennâmodulivnadkilʹcâmitaíhrozširennâ AT naumansk doslidžennâmodulivnadkilʹcâmitaíhrozširennâ AT alʹhašmísa doslidžennâmodulivnadkilʹcâmitaíhrozširennâ AT naumansk doslidžennâmodulivnadkilʹcâmitaíhrozširennâ AT alhashmisa studyofmodulesoverringsandtheirextensions AT naumansk studyofmodulesoverringsandtheirextensions AT alʹhašmísa studyofmodulesoverringsandtheirextensions AT naumansk studyofmodulesoverringsandtheirextensions |