216
Appendix: Review of Basic Algebra
Remark A.2 Note that we assume in Definition A.6 that 1 = 0, but some authors
do not put this constraint (we adopt the notation of [137]).
Another characterization of integral domains is given in the next proposition.
Proposition A.5 A nonzero commutative ring (R, +, ·) with unit 1, where 1 = 0,
is a integral domain if and only if for all x, y, z ∈ R, if x · y = x · z, with x = 0,
implies that y = z (cancelation law for multiplication).
Proof See Proposition 3.5 in [137].
In the following we define the concept of field.
Definition A.7 Let (F, +, ·) be a commutative ring with unit 1, where 1 = 0. We say
that (F, +, ·) is a field if every nonzero element x ∈ F has a multiplicative inverse,
that is, for all x ∈ F, x = 0, there exists x
−1
∈ F such that x · x
−1
= x
−1
· x = 1,
where 1 is the unit of the field.
Definition A.8 Let (R, +, ·) be a commutative ring. An ideal in R is a nonempty
subset I ⊆ R such that:
(I1) For all x, y ∈ I , it follows that x − y ∈ I .
(I2) For all x ∈ I and for all a ∈ R, one has a ∗ x ∈ I .
In words, an ideal I is a subgroup of the additive (abelian) group (R, +), closed
under the multiplication of elements of R by the elements of I .
Ideals have suitable properties, as shown in the following proposition.
Proposition A.6 Let I be an ideal in a commutative ring (R, +, ·). Then the following hold:
(i) 0 ∈ I .
(ii) If a ∈ I , then −a ∈ I .
(iii) If a, b ∈ I , then a + b ∈ I .
(iv) If the ring has identity 1 and if an invertible element of R belongs to I , then
I = R.
Proof The proofs are immediate and left to the reader.
Let (R, +, ·) be a commutative ring with identity and I ⊆ R an ideal of R. We
know that I is a normal subgroup of the additive group R, since (R, +) is abelian.
We then consider the additive abelian group R/I and define a multiplication on it as
follows: for all a, b ∈ R, (a + I ) · (b + I ) = (a · b) + I . Considering what we have
said, we can show that R/I with these two operations is a commutative ring with
identity (the identity of R/I is the class 1 + I , where 1 is the identity of R).
Theorem A.2 If (R, +, ·) is a commutative ring and I an ideal of R, then the
additive abelian group R/I is a commutative ring with identity, where the addition
if defined by (a + I ) + (b + I ) = (a + b) + I and the multiplication is given by
(a + I ) · (b + I ) = (a · b) + I , for all a, b ∈ R.
Proof See the proof of Theorem 3.110 in [137].
Appendix: Review of Basic Algebra
Remark A.2 Note that we assume in Definition A.6 that 1 = 0, but some authors
do not put this constraint (we adopt the notation of [137]).
Another characterization of integral domains is given in the next proposition.
Proposition A.5 A nonzero commutative ring (R, +, ·) with unit 1, where 1 = 0,
is a integral domain if and only if for all x, y, z ∈ R, if x · y = x · z, with x = 0,
implies that y = z (cancelation law for multiplication).
Proof See Proposition 3.5 in [137].
In the following we define the concept of field.
Definition A.7 Let (F, +, ·) be a commutative ring with unit 1, where 1 = 0. We say
that (F, +, ·) is a field if every nonzero element x ∈ F has a multiplicative inverse,
that is, for all x ∈ F, x = 0, there exists x
−1
∈ F such that x · x
−1
= x
−1
· x = 1,
where 1 is the unit of the field.
Definition A.8 Let (R, +, ·) be a commutative ring. An ideal in R is a nonempty
subset I ⊆ R such that:
(I1) For all x, y ∈ I , it follows that x − y ∈ I .
(I2) For all x ∈ I and for all a ∈ R, one has a ∗ x ∈ I .
In words, an ideal I is a subgroup of the additive (abelian) group (R, +), closed
under the multiplication of elements of R by the elements of I .
Ideals have suitable properties, as shown in the following proposition.
Proposition A.6 Let I be an ideal in a commutative ring (R, +, ·). Then the following hold:
(i) 0 ∈ I .
(ii) If a ∈ I , then −a ∈ I .
(iii) If a, b ∈ I , then a + b ∈ I .
(iv) If the ring has identity 1 and if an invertible element of R belongs to I , then
I = R.
Proof The proofs are immediate and left to the reader.
Let (R, +, ·) be a commutative ring with identity and I ⊆ R an ideal of R. We
know that I is a normal subgroup of the additive group R, since (R, +) is abelian.
We then consider the additive abelian group R/I and define a multiplication on it as
follows: for all a, b ∈ R, (a + I ) · (b + I ) = (a · b) + I . Considering what we have
said, we can show that R/I with these two operations is a commutative ring with
identity (the identity of R/I is the class 1 + I , where 1 is the identity of R).
Theorem A.2 If (R, +, ·) is a commutative ring and I an ideal of R, then the
additive abelian group R/I is a commutative ring with identity, where the addition
if defined by (a + I ) + (b + I ) = (a + b) + I and the multiplication is given by
(a + I ) · (b + I ) = (a · b) + I , for all a, b ∈ R.
Proof See the proof of Theorem 3.110 in [137].