Equivalence relations in a ring; Quotient rings
Let (R, +, ⋅) a ring. Analogously to the definition of congruences on ℤ we give the following definition:
6.3.0 Definition A relation ρ defined in a ring R is said to be compatible with the operations of R if for each a1, a2, b1, b2 ∈ R.
| a1 ρ a2, b1 ρ b2 ⇒ | (a1 + b1) ρ (a2 + b2) |
| (a1 ⋅ b1) ρ (a2 ⋅ b2) |
The equivalence relation defined on ℤ as we have seen is compatible with both operations of ℤ.
We observe now, that assigning in a ring R an equivalence relation combpatible with both operation is equivalent to assigning to it a two-sided ideal.
6.3.1 Definition Let ρ an equivalence relation definef on R compatible with its operations. Then the subset
I := {x ∈ R | x ρ 0}
is a two-sided ideal.
Proof. We have to prove that for each x,y ∈ I, and r ∈ R, we have x − y ∈ I and xr ∈ I, rx ∈ I.
x,y ∈ I ⇒ x ρ 0, y ρ 0 ⇒ x ρ y, −y ρ −y.
since ρ is compatible with regards to addition, it follows that (x − y) ρ 0, hence xr ∈ I. Thus, if x ∈ I and r ∈ R,
x ρ 0
r ρ r
imply owning to the compatibility of ρ with regards to multiplication, that xr ρ 0, hence xr ∈ I. Analogously, we prove that rx ∈ I.□
6.3.2 Proposition. Let I a two-sided ideal of a ring R. The relation ρ defined on R as
x ρ y ⇐⇒ x − y ∈ I
is an equivalence relation compatible with the operation of R called the congruence modulo I.
Proof. That ρ is an equivalence relation is easy to verify. Let's prove the compatibility with the operations. Let x1 ρ x2 and y1 ρ y2. Then
x1 ρ x2 ⇐⇒ x1 − x2 ∈ I
y1 ρ y2 ⇐⇒ i1 − y2 ∈ I
yields since an ideal is an additive subgroup to
(x1 + y1) − (x2 + y2) = (x1 − x2) + (y1 − y2) ∈ I.
thus x1 + y1 ρ x2 + y2 and ρ is compatible with addition. Moreover
x1y1 = x1y1 − x1y2 + x1y2 = x1 (y1 − y2) + (x1 − x2) y2 ∈ I.
hence ρ is compatible with multiplication too.□
We define the following sets
𝓡 := {equivalence relations defined on R, compatible with the operation on R}
𝓘 := {two-ideals of R}
The map
Ψ: 𝓡 ⟶ 𝓘
ρ ⟼ I = {x ∈ R | x ρ 0}
is a biunique map between 𝓡 and 𝓘 which inverse is.
Ψ*: 𝓘 ⟶ 𝓡
I ⟼ ρ where x ρ y ⇐⇒ x − y ∈ I
To sum up, we've the following proposition
6.3.3 Proposition. All and the only relations compatible defined on a ring R are the congruences modulo an ideal.
6.3.4 Proposition. The relations in ℤ compatible with addition and multiplicatoin are exactly congrunces modulo n. The ideals are all and only the subset In of the form nℤ
x ρ y ⇐⇒ x − y ∈ In = nℤ
hence x − y = kn for some k ∈ ℤ.
If ρ is an equivalence relation defined on the ring R and compatible with the operations of R, in the quotient set
R/ρ = {a̅ | a ∈R}, with a̅ = {x ∈ R | x ρ a}
we introduce the following operations
a̅ + b̅ := a + b, a̅ ⋅ b̅ := a⋅b
compatibility of equivalence relations guarantee that these two operations are well-defined: indeed if they are well-defined it is implied that given a2 related to a1 and given b2 related to b1, then a1 + b1 is related to a2 + b2 and a1 ⋅ b1 is related to a2 ⋅ b2 which is exaclty the meaning of compatible equivalence relation.
Since an equivalence relation on R, corresponds to an ideal I of R, in such a way that ρ is defined by
x ρ y ⇐⇒ x − y ∈ I
the quotient R/ρ can be indicated as R/I, where
R/ρ := {a + I | a ∈ R}
with
[a] = a + I := {a + i | i ∈ I} = {x ∈ R | x ρ a}
Endowed with the operations defined above, R/I has the structure of ring called quotient ring modulo an ideal I or factor ring. The zero of this ring is the class I, the opposite of the class a + I is the class −a + I, etc. Note that
r + I = s + I iff r − s ∈ I.
Example 6.3.5. For R = ℤ, I = 4ℤ, we have that
I = {..., −4, 0, 4, 8,...}
1 + I = {...,,−3, 1, 5, 9 ,...}
2 + I = {..., −2, 2, 6, 10,...}
3 + I = {...,,−1, 3, 7, 11, ...}
There are 4 different cosets, the equivalence classes for
[r] ∼ [s] ⇐⇒ r ≡ s mod 4.
So 2 + I = 6 + I, 3 + I = −1 + I, etc.
Moreover owning to the definition of the operations in R/I, the application π from R to R/I which maps every element a of R to the class a + I to which it belongs and called quotient homomorphism or canonical projection. This is a surjective ring homomorphism hence an epimorphism.
To sum up what presented in this paragraph we state the following proposition
6.3.6 Proposition. Let I an ideal of a ring R.
The relation
x ρ y ⇐⇒ x − y ∈ I
is an equivalence relation compatible with the operation in R.
Indicating the equivalence classes as
a̅ = a + I = {a + i | i ∈ I}
and defining the operations
a̅ + b̅ := a + b, a̅ ⋅ b̅ := a ⋅ b
or equivantely as
(a + I) + (b + I) = (a + b) + I, (a + I) (b + I) = ab + I
The quotient set R/I = {a̅ | a ∈ R} becomes a ring, known as the quotient ring of R modulo an ideal I.
The canonical projection
π: a ∈ R ⟼ a̅ ∈ R/I
is easily seen to be an epimorphism of rings, with kernel I.
Lemma 6.3.7 Lemma R a ring with unit with quotient ring R/I. Let a be a unit of R. Then a + I is a unit of R/I.
Proof. (a + I)(b + I) = ab + I if a is the unit we get b + I.□
Example 6.3.8 (ℤ/nℤ). The most familiar quotient ring is ℤ/nℤ, for which we will continue to use the notation ℤn (they are the same) whose elements can be written as cosets of the form a + nℤ.
a + nℤ = [a] = {..., a −2n, a − n, a, a + n, a + 2n, ...}
and we recognize [a] as a residue class of ℤn.
Addition and multiplication of cosets is then just addition and multiplication modulo n, as we can see, that this is just a formalization of the ring ℤn, that we have already looked at. Recall that ℤn is an integral domain if and only if n is prime and ℤn is a field for precisely the same n. If n = 0 then ℤ/nℤ is the same as ℤ.
Operations with ideals
If A and I are two ideals of the ring R, then their sum A + I is the set
A + I = {a + i | a ∈ A, i ∈ I}
It is easy to see that A + I is an ideal of R, the minimal ideal of A which contains both I and A. Equivalentely, A + I is the ideal generated by A ∪ I.
The product AI is the ideal generated by all products ai, the set of all possible sums
a1i1 + a2i2 + ... + anin
that is
IJ := {∑ni=0 aibi | n ∈ ℕ, ai ∈ I, bi ∈ J}
The product AI is contained in the ideal intersection A ∩ I.
6.3.9 Proposition. Let R a ring, I one of its ideal and A a subring of R such that A ⊇ I. Then A = (A + I).
Proof. We prove by showing double inclusion. For a ∈ A, and i ∈ I, a + i in A, so we have A + I ⊂ A; this is because A is an additive subgroup of R so if you sum two elements of it, it remains in A (closusre under +). Conversely, we have for a ∈ A that a = a + 0, and 0 is an element of I, so a ∈ A + I, which gives us A ⊂ A + I. □