RELATIVE ALGEBRAIC STRUCTURES

Amir M. Rahimi

Abstract. The concept and some of the algebraic properties of the rejective and non–absorptive sets of a subgroup, subring, and subgroup of a module over a ring are investigated. It is shown that the set theoretic complement of a non– absorptive set in the above mentioned algebraic substructures is a normal subgroup (respectively, (left, right) ideal, submodule) of its underlying algebraic structure. The invariant property of the non–absorptive sets under the operation of inversion in the related underlying algebraic structure is proved. G \ R(H), the set theoretic complement of the rejective set of a subgroup H in a group G, is closed under the product in G and whenever |G| the order of the group G is finite, |R(H)| = (k−s)|H| where each of the k and s is the index of H in G and in G \ R(H), respectively. For the case of rings and modules, the set theoretic complement of the rejective set of a substructure in the underlying ring is a subring of the underlying ring. For any subring S of a ring R, examples and some of the properties of S–relative (left) ideals and S–relative submodules are given and also it is shown that S is contained in the set theoretic complement of the rejective set of that S–relative (left) ideal (respectively, submodule). Finally, some of the properties of the relative homomorphisms of R–modules, and the rejective (respectively, non–absorptive) sets of the group homomorphisms of R–modules are investigated.

1. Rejective and Non–Absorptive Sets of Some Algebraic Substructures.

\underline{Definition 1.1}. For a subgroup H of a group G, the set of all elements h in H such that for each h there exists an element g in G with ghg^{−1} ∉ H is called the non–absorptive set of H in G and is denoted by N(H, G) or N(H) whenever there is no confusion in the context.

\underline{Remark}. In the above definition, it is clear that H is a normal subgroup of G if and only if N(H) is the empty set.

\underline{Theorem 1.1}. Let N(H) be the non–absorptive set of a subgroup H in a group G. Then H \ N(H) is a normal subgroup of G.

\underline{Proof}. Suppose to the contrary that a and b are in H \ N(H) and ab is in N(H). Thus, for some g ∈ G, gabg^{−1} = gag^{−1}gbg^{−1} ∉ H which is a contradiction since both gag^{−1} and gbg^{−1} are in H. Next, b ∈ H \N(H) implies gbg^{−1} ∈ H for all g ∈ G and this makes (gbg^{−1})^{−1} = gb^{−1}g^{−1} to be in H for all g ∈ G which implies b^{−1} ∈ H \N(H). Finally, it remains to show that H \N(H) is normal in G. Suppose for some a ∈ H \ N(H) there exists g ∈ G such that gag^{−1} ∉ H \ N(H). Hence, by definition, gag^{−1} ∈ N(H) and this forces k(gag^{−1})k^{−1} = (kg)a(kg)^{−1} ∉ H for some k in G and this is a contradiction to the choice of a in H \ N(H).

\underline{Theorem 1.2}. Let N(H) be the non–absorptive set of a subgroup H in a group G. Then for each x ∈ G \ N(H), x^{−1} is also in G \ N(H).

\underline{Proof}. Suppose to the contrary that x ∈ G \ N(H) and x^{−1} is in N(H). Consequently, x is in H \ N(H) which implies (gxg^{−1})^{−1} = gx^{−1}g^{−1} ∈ H for all g ∈ G and this is a contradiction to the choice of x^{−1} in N(H).

\underline{Remark}. From the above theorem, it is clear that x^{−1} ∈ N(H) whenever x ∈ N(H). In other words, N(H) is invariant under the group operation of inversion. Furthermore, N(H) can never be closed under the product in the group since e the group identity element is not in N(H).

\underline{Corollary 1.1}. Let N(H) be the non–absorptive set of a subgroup H in a group G. Then G \ N(H) is a subgroup of G if and only if G \ N(H) is closed under the product in G.

\underline{Remark}. For a family \{G_{i} | i ∈ I\} of groups with H_{i} a subgroup of G_{i} for each i ∈ I, it is not difficult to show that ∏ N(H_{i}, G_{i}) ⊆ N(∏ H_{i}, ∏ G_{i}) and for |I| = 2, (N(H_{1}) × H_{2}) ∪ (H_{1} × N(H_{2})) = N(H_{1} × H_{2}). In general, ∪_{i∈I} ∏^{j} H_{i} = N(∏ H_{i}) where ∏^{j} H_{i} is the Cartesian product of all subgroups H_{i} (i ≠ j) and N(H_{j}) for i = j.

\underline{Definition 1.2}. For a subgroup H of a group G, the set of all g ∈ G such that for each g there exists an element h in H with ghg^{−1} ∉ H is called the rejective set of H in G and is denoted by R(H, G) or R(H) whenever there is no confusion in the context.

\underline{Remark}. It is clear that R(H, G) = ∅ if and only if H is a normal subgroup of G. From the definition, it is obvious that R(H, G) ⊆ G \ H.

\underline{Theorem 1.3}. For any subgroup H of a group G, G \ R(H) the set theoretic complement of the rejective set of H in G is closed under the group operation of product.

\underline{Proof}. Suppose to the contrary that a, b ∈ G \ R(H) and ab ∈ R(H). Thus, for some h in H, abhb^{−1}a^{−1} = a(bhb^{−1})a^{−1} ∉ H and this is a contradiction.

\underline{Corollary 1.2}. Let R(H) be the rejective set of a subgroup H in a group G. Then G \ R(H) is a subgroup of G if and only if R(H) is invariant under the operation of inversion in G.

\underline{Corollary 1.3}. Let R(H) be the rejective set of a subgroup H in a finite group G. Then G\R(H) is a subgroup of G and |R(H)| = (k −s)|H| where k is the index of H in G and s is the index of H in G \ R(H).

\underline{Proof}. Since every non–empty finite subset of a group G is a subgroup of G if and only if it is closed under the product in G, consequently, by applying the above theorem together with Lagrange’s Theorem, G \ R(H) is a subgroup of G and each of |G \ R(H)| and |G| is divisible by |H|. Thus, s|H| = |G \ R(H)| = |G| − |R(H)| = k|H| − |R(H)| which implies |R(H)| = (k − s)|H| where s is the index of H in G \ R(H) and k is the index of H in G.

\underline{Example}. As an application of the above corollary, it is easy to conclude that any subgroup H of a group G with a finite order 2n is normal in G whenever the order of H is n and there exists an element g in G \ H such that ghg^{−1} ∈ H for all h ∈ H.

\underline{Definition 1.3}. For a subring A of a ring R, the set of all elements a ∈ A such that for each a there exists an element r in R with ra ∉ A (respectively, ar ∉ A) is called the left (respectively, right) non–absorptive set of A in R and is denoted by N_{l}(A, R) or N_{l}(A) (respectively, N_{r}(A, R) or N_{r}(A)) whenever there is no confusion in the context.

\underline{Remark}. From the above definition, it is clear that A is a left (respectively, right) ideal in R if and only if N_{l}(A) (respectively, N_{r}(A)) is the empty set.

\underline{Theorem 1.4}. Let A be a subring of a ring R. Then A \ N_{l}(A) (respectively, A \ N_{r}(A)) is a left (respectively, right) ideal of R.

\underline{Proof}. Let each of a and b be an element in A \ N_{l}(A) and suppose to the contrary that (a − b) ∉ A \ N_{l}(A). Then for some r in R, r(a − b) = ra − rb ∉ A which is a contradiction since both ra and rb are in A. Now, for any a ∈ A \ N_{l}(A) and r ∈ R if ra ∉ A \ N_{l}(A), then ra must be in N_{l}(A). Hence, for some s in R, s(ra) = (sr)a ∉ A which is a contradiction to the choice of a in A \ N_{l}(A). A proof for the case of N_{r}(A) can be followed analogously.

\underline{Theorem 1.5}. Let N_{l}(A) be the left non–absorptive set of a subring A in a ring R. Then for each element a ∈ R \ N_{l}(A), −a is also in R \ N_{l}(A). In other words, a ∈ N_{l}(A) implies −a ∈ N_{l}(A).

\underline{Proof}. Suppose a ∈ R \ N_{l}(A) and −a ∈ N_{l}(A). Then a must be in A \ N_{l}(A) since A is an additive subgroup of R. Thus, r(−a) = −r(a) ∈ A for all r in R and this is a contradiction to the choice of −a ∈ N_{l}(A).

\underline{Remark}. Let \{R_{i} | i ∈ I\} be a family of rings with A_{i} a subring of R_{i} for each i ∈ I. Then ∏ N_{l}(A_{i}, R_{i}) ⊆ N_{l}(∏ A_{i}, ∏ R_{i}), and for |I| = 2, we have (N_{l}(A_{1})×A_{2})∪(A_{1}×N_{l}(A_{2})) = N_{l}(A_{1}×A_{2}). In general, ∪_{i∈I} ∏^{j} A_{i} = N_{l}(∏_{i∈I} A_{i}) where ∏^{j} A_{i} is the Cartesian product of all subrings A_{i} (i ≠ j) and N_{l}(A_{j}) for i = j.

\underline{Remark}. In any commutative ring, it is obvious that the left and right non– absorptive sets of a subring coincide with each other.

\underline{Definition 1.4}. For a subring A of a ring R, the set of all r ∈ R such that for each r there exists an element a in A with ra ∉ A (respectively, ar ∉ A) is called the left (respectively, right) rejective set of A in R and is denoted by R_{l}(A, R) or R_{l}(A) (respectively, R_{r}(A, R) or R_{r}(A)) whenever there is no confusion in the context.

\underline{Remark}. It is clear that R_{l}(A) (respectively, R_{r}(A)) is a subset of R \ A and A is a left (respectively, right) ideal of R if and only if R_{l}(A) (respectively, R_{r}(A)) is the empty set. Note that in a commutative ring R, both the left and right rejective sets of any subring A of R coincide with each other.

\underline{Theorem 1.6}. Let R_{l}(A) (respectively, R_{r}(A)) be the left (respectively, right) rejective set of a subring A in a ring R. Then R \ R_{l}(A) (respectively, R \ R_{r}(A)) is a subring of R.

\underline{Proof}. Suppose r, s ∈ R \ R_{l}(A) and r − s ∉ R \ R_{l}(A). Then for some a in A, (r − s)a = ra − sa ∉ A and this is a contradiction since both ra and sa are in A. Now, suppose for some r, s ∈ R \ R_{l}(A), rs is not in R \ R_{l}(A). Thus, for some a in A, (rs)a = r(sa) ∉ A and this is a contradiction to the choice of r and s.

\underline{Defintion 1.5}. Let A be a subgroup of an R–module M over a ring R. The set of all elements a ∈ A such that for each a there exists an element r ∈ R with ra ∉ A is called the non–absorptive set of A in M and is denoted by N(A, M) or N(A) whenever there is no confusion in the context.

\underline{Remark}. From the above definition, it is clear that A is a submodule of M if and only if N(A) is the empty set.

\underline{Theorem 1.7}. Let N(A) be the non–absorptive set of a subgroup A in an R–module M over a ring R. Then A \ N(A) is a submodule of M.

\underline{Proof}. Similar to the proof of Theorem 1.4.

\underline{Theorem 1.8}. Let N(A) be the non–absorptive set of a subgroup A in an R– module M over a ring R. Then a ∈ M \ N(A) implies −a ∈ M \ N(A). In other words, a ∈ N(A) implies −a ∈ N(A).

\underline{Proof}. Similar to the proof of Theorem 1.5.

\underline{Remark}. Let \{M_{i} | i ∈ I\} be a family of R–module over a ring R and A_{i} a subgroup of M_{i} for each i ∈ I. Then ∏ N(A_{i}) ⊆ N(∏ A_{i}) and for |I| = 2, (N(A_{1}) × A_{2}) ∪ (A_{1} × N(A_{2})) = N(A_{1} × A_{2}). In general, we have ∪_{i∈I} ∏^{j} A_{i} = N(∏_{i∈I} A_{i}) where ∏^{j} A_{i} is the Cartesian product of all subgroups A_{i} (i ≠ j) and N(A_{j}) for i = j.

\underline{Definition 1.6}. For a group A of an R–module M over a ring R, the set of all elements r ∈ R such that for each r there exists an element a in A with ra ∉ A is called the rejective set of A in M and is denoted by R(A, M) or R(A) whenever there is no confusion in the context.

\underline{Remark}. For the above definition, it is clear that A is a submodule of M if and only if R(A) is the empty set.

\underline{Theorem 1.9}. If R(A) is the rejective set of a subgroup A in an R–module M over a ring R, then R \ R(A) is a subring of R.

\underline{Proof}. Similar to the proof of Theorem 1.6.

\underline{Theorem 1.10}. For any two subgroups (respectively, subrings, subgroups) A and B of a group (respectively, a ring, an R–module), A − N(A) ⊆ B − N(B) (respectively, A − N_{l}(A) ⊆ B − N_{l}(B), A − N(A) ⊆ B − N(B)) whenever A ⊆ B.

\underline{Proof}. We just give a proof for the subgroups A and B of a group G and leave the other cases to the reader. Suppose to the contrary that there exists an element a ∈ A − N(A) with a ∉ B − N(B). Thus, a ∈ N(B) and this implies gag^{−1} ∉ B for some g ∈ G. Consequently, gag^{−1} is not in A which implies a ∈ N(A) and this is a contradiction to the choice of a in A − N(A).

2. Relative Ideals.

\underline{Definition 2.1}. Let S be a subring of a ring R. A subring A of R is an S– relative left (respectively, right) ideal of R provided s ∈ S and a ∈ A imply sa ∈ A (respectively, as ∈ A). A is an S–relative ideal of R if it is both an S–relative left and an S–relative right ideal of R. A subring A of R is said to be a strictly S–relative left (respectively, right) ideal of R whenever A is an S–relative left (respectively, right) ideal of R and it is not an R–relative left (respectively, right) ideal of R

\underline{Remark}. Whenever a statement is made about the S–relative left ideals, it is to be understood that the analogous statement holds for the S–relative right ideals. It is clear that any left ideal A of a ring R is an S–relative left ideal of R for any subring S of R. Also, A contains S whenever 1_{R} the identity element of R is in A.

\underline{Example}. Let M be the ring of all 2 × 2 matrices over a ring R. Then A the subring of all matrices of the form

( \begin{matrix}x & 0 \\ 0 & 0 \end{matrix} )

is neither a left nor a right ideal of M. Let S be the set of all 2 × 2 matrices with zero (2, 1) entries and T the set of all 2 × 2 matrices with zero (1, 2) entries. Now, it is not difficult to show that A is an S–relative left and a T –relative right ideal of M. Furthermore, A is neither an S–relative right ideal nor a T –relative left ideal of M.

\underline{Theorem 2.1}. If S and A are two subrings of a ring R with A an S–relative left ideal of R, then R \ R_{l}(A) contains S and A is also an R \ R_{l}(A)–relative left ideal of R where R_{l}(A) is the left rejective set of A in R.

\underline{Proof}. See Theorem 1.6.

\underline{Corollary 2.1}. For a subring S of a ring R if \{A_{i} | i ∈ I\} is a family of S– relative left ideals of R, then A_{l} is a ∩_{i∈I}R \ R_{l}(A_{i})–relative left ideal of R for each i in I.

\underline{Theorem 2.2}. The following results can be proved directly from the definition.

a) If A is an S–relative left ideal of a ring R, then S ∩ R_{l}(A) = ∅.

b) If f: R → T is a homomorphism of rings and A is an S–relative left ideal of R, then f(A) is an f(S)–relative left ideal of T.

c) If A is both an S–relative left and a T –relative right ideal of a ring R, then A is an S ∩ T –relative ideal of R.

d) Let {S_{i} | i ∈ I} be a family of subrings of a ring R. Then A is a ∩_{i∈I}S_{i}–relative left ideal of R if A is an S_{i}–relative left ideal of R for each i ∈ I.

e) If S_{1} ⊆ S_{2} are two subrings of a ring R and A is an S_{2}–relative left ideal of R, then A is an S_{1}–relative left ideal of R.

f) For any ascending chain {S_{i} | i ∈ I} of subrings S_{i} of a ring R, A is a ∪_{i∈I}S_{i}– relative left ideal of R if and only if A is an S_{i}–relative left ideal of R for each i ∈ I.

g) For a family {S_{i} | i ∈ I} of subrings S_{i} of a ring R, ∩_{i∈I}A_{i} is a ∩_{i∈I}S_{i}–relative left ideal of R whenever A_{i} is an S_{i}–relative left ideal of R for each i ∈ I.

h) Let {R_{i} | i ∈ I} be a family of rings and S_{i} a subring of R_{i} for each i ∈ I. If A_{i} is an S_{i}–relative left ideal of R_{i} for each i ∈ I, then ∏_{i∈I} A_{i} is a ∏_{i∈I} S_{i}– relative left ideal of ∏_{i∈I} R_{i} the direct product of the rings.

\underline{Example}. In the ring R of n × n matrices over a division ring D, let I_{k} be the set of all matrices that have nonzero entries only in column k and J′_{k} the set of all matrices with zero kth rows. Then I_{k} is a left ideal and a J′_{k}–relative right ideal but not a right ideal of R. If J_{k} consists of those matrices with nonzero entries only in row k and I′_{k} the set of all matrices with zero kth columns, then J_{k} is a right ideal and an I′_{k}–relative left ideal but not a left ideal in R.

\underline{Theorem 2.3}. For any subring A of a ring R, the left rejective set of A in R is R \ A whenever A contains 1_{R} the identity element of R.

\underline{Proof}. A = R \ R_{l}(A) since A is always contained in R \ R_{l}(A) and 1_{R} ∈ A implies R \ R_{l}(A) ⊆ A.

\underline{Example}. As an application of the above theorem let R[X] be the ring of all polynomials over an integral domain R and A the ring of all polynomials with zero X–coefficients, then R(A, R[X]) the rejective set of A in R[X] is R[X] \ A which is exactly the set of all polynomials with nonzero X–coefficients.

\underline{Definition 2.2}. Let X be a subset of a ring R and S a subring of R. If {A_{i} | i ∈ I} is the family of all S–relative left ideals of R containing X, then ∩_{i∈I}A_{i} is called the S–relative left ideal generated by X in R and is denoted by (X)_{S}. The elements of X are called S–relative generators of (X)_{S}. If X = {x_{1}, x_{2}, …, x_{n}}, then the S–relative left ideal (X)_{S} is denoted by (x_{1}, x_{2}, …, x_{n})_{S} and is said to be an S–relative finitely generated left ideal. An S–relative left ideal (x)_{S} generated by a single element x is called an S–relative principal left ideal of R.

\underline{Theorem 2.4}. Let S be a subring of a ring R, a an element in R, and K the set of all elements of the form ra + as + na + ∑_{i=1}^{m} r_{i}as_{i} where r, s, r_{i}, s_{i} ∈ S, n an integer, and m runs over the set of non–negative integers. Then we have the following results:

1) K ⊆ (a)_{S} the S–relative principal ideal generated by a in R. Moreover, a ∈ S implies (a)_{S} = K = (a)^{S} the principal ideal generated by a in S.

2) x ∈ (a)_{S} \ K implies −x ∈ (a)_{S} \ K.

3) If R is a commutative ring and a ∈ S, then (a)_{S} consists of all elements of the form sa + na where s ∈ S and n ∈ Z the ring of rational integers.

\underline{Proof}. The proof is an immediate consequence of the definition and we leave it to the reader as an exercise.

\underline{Theorem 2.5}. For a subring S of a ring R if A is an S–relative ideal of R, then 1) S + A = {s + a | s ∈ S, a ∈ A} is also an S–relative ideal in R.

2) S ∪ A is a multiplicative system in R.

3) S ∩ A is an S–relative ideal of R, and also it is an S–relative left ideal of R whenever A is an S–relative left ideal of R.

\underline{Proof}. The proof is a direct consequence of the definition and we leave it to the reader.

\underline{Theorem 2.6}. In a commutative ring R, let each of S_{1}, S_{2}, …, S_{n} be a subring of R and A_{i} an S_{i}–relative ideal of R for each i = 1, 2, …, n, respectively. Then A_{1}A_{2}A_{3} ⋯ A_{n} is an S_{i_{1}}S_{i_{2}} ⋯ S_{i_{k}}–relative ideal of R where {i_{1}, i_{2}, …, i_{k}} is a subset of the set {1, 2, …, n}.

\underline{Proof}. The proof follows directly from the definition and we leave it to the reader.

\underline{Definition 2.3}. Let S be a subring of a ring R. An S–relative left ideal P of R is said to be an S–relative prime left ideal of R if P ≠ R and for any S–relative left ideals A and B of R, AB ⊆ P implies A ⊆ P or B ⊆ P.

\underline{Theorem 2.7}. For a subring S of a ring R if P is an S–relative left ideal of R such that P ≠ R and for all elements a, b ∈ R, ab ∈ P implies a ∈ P or b ∈ P, then P is an S–relative prime left ideal of R. Conversely if R is commutative and P is an S–relative prime ideal of R, then for any a, b ∈ S, ab ∈ P implies either a ∈ P or b ∈ P.

\underline{Proof}. Let A and B be two S–relative left ideals of R such that AB ⊆ P. Suppose A ⊈ P, then there exists an element a ∈ A with a ∋ P. Now, for each b ∈ B, ab ∈ AB ⊆ P implies a ∈ P or b ∈ P which implies b ∈ P and consequently B ⊆ P. Conversely, ab ∈ P implies (ab)_{S} ⊆ P. Now since R is commutative and a, b ∈ S, then (a)_{S}(b)_{S} ⊆ (ab)_{S} ⊆ P which implies the desired conclusion.

\underline{Remark}. From the above result it is clear that if P is an S–relative prime ideal of a commutative ring R, then S \ P is a multiplicative system in R.

\underline{Remark}. Any S–relative (left) ideal A of a ring R is a (left) S–module.

3. Relative Submodules.

\underline{Definition 3.1}. For a ring R, let M be an R–module and S a subring of R. A non–empty subset A of M is an S–relative submodule of M provided that A is an additive subgroup of M and sa ∈ A for all s ∈ S and a ∈ A.

\underline{Example}. A subring S of a ring R is an S–relative submodule of R. In general, any S–relative (left) ideal of a ring R is an S–relative submodule of R whenever R is assumed to be an R–module over itself.

\underline{Remark}. An S–relative submodule B of an R–module A over a ring R need not be a subring of R whenever A = R.

\underline{Example}. Let A be an S–relative left ideal of a ring R and M an R–module. If X is a non–empty subset of M, then AX = {∑_{i=1}^{n} a_{i}x_{i} | a_{i} ∈ A, x_{i} ∈ X, and n a positive integer} forms an S–relative submodule of M. Similarly, for any x ∈ M, Ax = {ax | a ∈ A} is an S–relative submodule of M.

\underline{Theorem 3.1}. For a subring S of a ring R if A is an S–relative submodule of an R–module M, then A is an R \ R(A)–relative submodule of M and S is contained in R \ R(A) where R(A) is the rejective set of A in M.

\underline{Proof}. See Theorem 1.9.

\underline{Corollary 3.1}. For a subring S of a ring R if \{A_{i} | i ∈ I\} is a family of S–relative submodules of an R–module M, then A_{i} is a ∩_{i∈I}(R \ R(A_{i}))–relative submodule of M for each i in I.

\underline{Theorem 3.2}. The following results can be proved directly from the definition.

a) If A is an S–relative submodule of a module M over a ring R, then S∩R(A) = ∅ where R(A) is the rejective set of A in M.

b) If R is a ring and f: M → N an R–module homomorphism, then the homomorphic image (respectively, inverse image) of any S–relative submodule of M (respectively, N) is again an S–relative submodule of N (respectively, M).

c) For a ring R if {S_{i} | i ∈ I} is a family of subrings of R, A an R–module, and B_{i} an S_{i}–relative submodule of A for each i ∈ I, then ∩_{i∈I}B_{i} is a ∩_{i∈I}S_{i}–relative submodule of A.

d) If S_{1} ⊆ S_{2} are two subrings of a ring R and A is an S_{2}–relative submodule of an R–module M, then A is an S_{1}–relative submodule of M.

e) For any ascending chain {S_{i} | i ∈ I} of subrings S_{i} of a ring R, A is a ∪_{i∈I}S_{i}– relative submodule of an R–module M if and only if A is an S_{i}–relative submodule of M for each i ∈ I.

f) For a family {S_{i} | i ∈ I} of subrings S_{i} of a ring R, ∩_{i∈I}A_{i} is a ∩_{i∈I}S_{i}–relative submodule of an R–module whenever A_{i} is an S_{i}–relative submodule of M for each i ∈ I.

g) For a family of rings {R_{i} | i ∈ I}, assume S_{i} is a subring of R_{i}, M_{i} an R_{i}– module, and A_{i} an S_{i}–relative submodule of M_{i} for each i ∈ I, then ∏_{i∈I} A_{i} is a ∏_{i∈I} S_{i}–relative submodule of ∏_{i∈I} M_{i}.

h) Let S be a subring of a ring R and {A_{i} | i ∈ I} an ascending chain of subgroups of an R–module M. Then ∪_{i∈I}A_{i} is an S–relative submodule of M whenever A_{i} is an S–relative submodule of M for each i ∈ I.

\underline{Definition 3.2}. If X is a subset of a module M over a ring R and S is a subring of R, then the intersection of all S–relative submodules of M containing X is called the S–relative submodule generated by X or spanned by X and is denoted by ⟨X⟩_{S}. If X is finite and X generates the S–relative submodule A in M, then A is said to be S–relative finitely generated. If X = {a}, then ⟨a⟩_{S} is called the S–relative cyclic submodule generated by a. Finally if {B_{i} | i ∈ I} is a family of S–relative submodules of M, then the S–relative submodule generated by X = ∪_{i∈I}B_{i} is called the sum of the S–relative submodules B_{i}. If the index set I is finite, then the sum of B_{1}, B_{2}, …, B_{n} is denoted by B_{1} + B_{2} + ⋯ + B_{n}.

\underline{Theorem 3.3}. Let S be a subring of a ring R, A an R–module, X a subset of A, {B_{i} | i ∈ I} a family of S–relative submodules of A, a an element in A, and Sa = {sa | s ∈ S}.

1) Sa is an S–relative submodule of A and the map S → Sa given by s ⊢ sa is an S–module epimorphism.

2) The S–relative cyclic submodule C generated by a is {sa + na | s ∈ S and n ∈ ℤ the ring of integers}. If S has an identity 1_{S} and 1_{S}a = a, then C = Sa.

3) The S–relative submodule D generated by X is the set of all elements of the form ∑_{i=1}^{n} s_{i}a_{i} + ∑_{j=1}^{m} n_{j}b_{j} where n, m are non–negative integers, n_{j} ∈ ℤ, s_{i} ∈ S and a_{i}, b_{j} ∈ X. If S has an identity 1_{S} and for each x ∈ X, 1_{S}x = x, then D = SX = {∑_{i=1}^{n} s_{i}a_{i} | s_{i} ∈ S, a_{i} ∈ X, and n a non–negative integer}.

4) The sum of the family {B_{i} | i ∈ I} consists of all finite sums b_{i_{1}} +b_{i_{2}} +⋯+b_{i_{n}} where b_{i_{k}} is an element of B_{i_{k}}.

\underline{Proof}. The proof follows directly from the definition.

\underline{Definition 3.3}. Let S and T with S ⊆ T be two subrings of a ring R, A an R–module and B a T –module. A group homomorphism f: A → B is said to be an S–relative homomorphism of modules if for all s ∈ S and a ∈ A, f(sa) = sf(a).

\underline{Theorem 3.4}. Let S be a subring of a ring R and B an S–relative submodule of a module A over R. Then the quotient group A/B is an S-module with the action of S on A/B given by s(a + B) = sa + B for all s ∈ S and a ∈ A. The map π_{S}: A → A/B given by a ⊢ a + B is an S–relative epimorphism of modules with the kernel B. The map π_{S} is called the S–relative canonical epimorphism or projection.

\underline{Proof}. If a+B = a′+B, then a−a′ ∈ B. Since B is an S–relative submodule of A, then sa−sa′ = s(a−a′) is an element in B for all s in S. Thus, sa+B = sa′ +B which implies that the action of S on A/B is well defined. The remainder of the proof is left to the reader.

\underline{Definition 3.4}. Let A and B be two R–modules over a ring R and f: A → B a group homomorphism. The set of all r in R such that for each r there exists an element a in A with f(ra) ≠ rf(a) is called the rejective set of f in R and is denoted by R(f, R) or R(f) whenever there is no confusion in the context.

\underline{Remark}. From the above definition, it is clear that f: A → B is an R–module homomorphism if and only if R(f, R) is the empty set. Note that zero is always in R \ R(f) since f(0a) = f(0) = 0 = 0f(a). In addition if f: A → B is a group homomorphism of two unitary R–modules A and B, then f(1_{R}a) = f(a) = 1_{R}f(a) which implies 1_{R} ∈ R \ R(f).

\underline{Theorem 3.5}. Assume each of A and B is an R–module over a ring R. If f: A → B is a group homomorphism, then R \ R(f) the set theoretic complement of the rejective set of f in R is a subring of R and f is an R \ R(f)–relative homomorphism of A and B. In addition, R \ R(f) is a subfield of R whenever R is a field and A and B are unitary R–modules.

\underline{Proof}. For any r, s ∈ R \ R(f) and a ∈ A, f((r − s)a) = f(ra − sa) = f(ra) + f(−sa) = rf(a) − sf(a) = (r − s)f(a) which implies (r − s) is in R \ R(f). Similarly, f((rs)a) = f(r(sa)) = rf(sa) = (rs)f(a) implies rs is in R \ R(f). Now suppose R is a field and r is an arbitrary nonzero element of R \ R(f). Thus, for any a in A, f(a) = f(rr^{−1}a) = rf(r^{−1}a) which implies r^{−1}f(a) = f(r^{−1}a).

\underline{Corollary 3.2}. Let S be a subring of a ring R and f: A → B a group homomorphism of the R–modules A and B. Then f is an R\R(f)–relative homomorphism of the R–modules A and B and S is contained in R\R(f) whenever f is an S–relative homomorphism of A and B.

\underline{Definition 3.5}. Let A and B be two R–modules over a ring R and f: A → B a homomorphism of the groups. The set of all a in A such that for each a there exists an element r in R with f(ra) ≠ rf(a) is called the non–absorptive set of f in R and is denoted by N(f, R) or N(f) whenever there is no confusion in the context.

\underline{Remark}. In the above definition, it is clear that f is an R–module homomorphism of A and B if and only if N(f) is the empty set.

\underline{Theorem 3.6}. Let A and B be two R–modules over a ring R and f: A → B a group homomorphism. Then A \ N(f) the set theoretic complement of the non– absorptive set of f in A is a submodule of A.

\underline{Proof}. Note that A\N(f) is a non–empty set since it contains the zero element f(r0) = f(0) = 0 = rf(0) for any r in R. For any r ∈ R and a, b ∈ A \ N(f), f(r(a−b)) = f(ra−rb) = f(ra)+f(−rb) = rf(a)+rf(−b) = rf(a−b) which implies a−b is in A\N(f). Now suppose for some a in A\N(f) there exists an r in R such that ra ∋ A\N(f). Then there exists s ∈ R such that f(s(ra)) ≠ sf(ra) = (sr)f(a) which is a contradicition to the choice of a in A \ N(f).

In conclusion, it should be noted that the above ideas are new to the author and a search of the literature found no mention of such a concept as presented here. It is entirely possible, however, that a reader might know of a source of similar ideas.

Amir Rahimi Inst. for Studies in Theoretical Phys. & Math. Department of Mathematics Niavaran Bldg. P. O. Box 19395–5746 Tehran, IRAN