Ideals and Factor Rings

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💍 A ring R is a non-empty set with two binary operations, addition (a + b) and multiplication (ab), such that ∀ a, b, c ∈ R:

1. Both operations are closed: a + b ∈ R, a·b ∈ R.
2. Commutative under addition: a + b = b + a.
3. Associative under addition: (a + b) + c = a + (b + c).
4. There is an additive identity 0 ∈ R such that a + 0 = a, ∀ a ∈ R.
5. There are inverse elements for addition, ∃-a ∈ R: a + (-a) = (-a) + a = 0.
6. Associative under product: a(bc) = (ab)c.
7. Multiplication is distributive over addition: a(bc) = ab + ac, (b + c)a = ba + ca.

Motivation for the definition of an ideal

Let R be a ring and let A ⊆ R be a subring. In a ring, cosets are always respect to addition, r ∈ R, r + A = {r + a | a ∈ A}.

Notice that since (R, +) is an Abelian group, if A is a subgroup under addition ⇒ A is an Abelian subgroup under addition, and therefore, A is a normal subgroup of R, i.e., A ◁ R and we can form a quotient group (R/A, +). The quotient group is an Abelian group with respect to addition.

What condition do we need to satisfy on the subring A so that R/A = {r + A | r ∈ R} is a ring?

Basically, given a, b ∈ R, we need (a + A)(b + A) = ab + A to be well defined, that is, this operation does not depend on the chosen representatives.

Let a + S = c + S and b + S = d + S, (a + S)(b + S) = ab + S = cd + S or equivalently ab -cd ∈ S?

a + S = c + S ⇒ a -c ∈ S ⇒ ∃s₁ ∈ S, a - c = s₁ ⇒ a = c + s₁. Similarly, ∃s₂ ∈ S, b = d + s₂.

ab = (c + s₁)(d + s₂) = cd + cs₂ +s₁d + s₁s₂ ⇒ ab -cd = cs₂ +s₁d + s₁s₂. So we need ab -cd ∈ S ↭ cs₂ +s₁d + s₁s₂ ∈ S ↭[s₁s₂ ∈ S, S is a subring] cs₂, s₁d ∈ S. This is called the absorption property.

Definition

A subring A of a ring R is a left ideal of R if it absorbs multiplication from the left from R, that is, ∀r ∈ R, ∀a ∈ A, ra ∈ A.

A subring A of a ring R is a right ideal of R if it absorbs multiplication from the right from R, that is, ∀r ∈ R, ∀a ∈ A, ar ∈ A.

A subring A of a ring R is called a (two-sided) ideal of R if it absorbs multiplication from the left and right from R, that is, ∀r ∈ R, a ∈ A, ra ∈ A and ar ∈ A. In other words, ∀r ∈ R, rA = {ra | a ∈ A} ⊆ A and Ar = {ar | a ∈ A} ⊆ A.

Let A ⊂ R, A ≠ R, R be a ring, A is an ideal of R if the following conditions are satisfied:

1. a - b ∈ A, ∀a, b ∈ A.
2. ∀r ∈ R, ∀a ∈ A, ra ∈ A and ar ∈ A.

Let R be a commutative ring with unity. A principal ideal is an ideal generated by a single element a of R through multiplication by every element of R, ⟨a⟩ = Ra = {ra | r ∈ R} = {ar | r ∈ R}., e.g., ⟨x⟩ = ⟨x, x²⟩ is a principal ideal of ℤ[x], but ⟨x, 3⟩ is not.

Proof. ⟨a⟩ is an ideal.

∀i ∈ ⟨a⟩ ⇒ ∃r’ ∈ R such that i = r’a. ∀r ∈ R, ir = (r’a)r = [R is a commutative ring] (ar’)r = a(r’r) ∈ ⟨a⟩.

Remark. If R is non-commutative or does not contain 1, the situation is most complicated, the closest concept would be $\bigcap_{I~ ideal, a ∈ I} I$.

Examples

• Let R be a ring, the set R itself and {0_R} are ideals. They are called the unit and the trivial ideal respectively. An (left, right or two-sided) ideal that is not the unit ideal is called a proper ideal of R, that is, A is a proper subset of R.
• The even integers 2ℤ form an ideal in the ring ℤ of all integers because the sum of any even integers is even, and the product of any integer with an even integer is also even.
• Counterexample. ℚ ⊆ ℝ, ℚ is not an ideal because 1/2 ∈ ℚ, π ∈ ℝ and 1/2·π ∉ ℚ. Suppose π/2 = a/b for some a ∈ ℤ, b ∈ ℤ \ {0} ⇒ π = 2a/b ∈ ℚ ⊥
• Counterexample. ℤ is a subring of ℝ. It contains 0, is closed under addition (under taking additive inverses, too) and multiplication. With regard to multiplication, note that the product of two integers is an integer. However, ℤ is not an ideal in ℝ, $\sqrt{2}$ ∈ ℝ, 3 ∈ ℤ, but $3·\sqrt{2}$ ∉ ℤ.
• ∀n ∈ ℤ, n > 0, nℤ is an ideal of ℤ.

Suppose m ∈ ℤ, a ∈ nℤ ⇒ ∃a’ ∈ ℤ: a = na’ ⇒ ma = m(na’) = n(ma’) ∈ nℤ ∎

Therefore, the principal ideals in ℤ are of the form ⟨n⟩ = nℤ.

• In ℤ[x], I = {a₂x² + a₃x³+···+a_nxⁿ| n ≥ 2, a_i ∈ ℤ} is an ideal of ℤ[x], e.g., 3·x²+4x³, x⁵+4x³+3·x²+4x³ ∈ I.

Take f(x) ∈ ℤ[x], g(x) ∈ I, deg(f(x)·g(x)) = deg(f(x)) + deg(g(x)) ≥ deg(g(x)) ≥ 2 ⇒ deg(f(x) + g(x)) ≥ 2 ⇒ f(x) + g(x) ∈ I ∎

• Let ℝ[x] be the set of all polynomials with coefficients in ℝ. The subset of all polynomials with constant term 0, i.e., ⟨x⟩ is an ideal of R[x].
• In general, let F be a field, p(x) ∈ F[x], ⟨p(x)⟩ is the set of all multiples of p(x), the principal ideal generated by p(x). F[x]/⟨p(x)⟩ is the quotient ring. If a(x) ∈ F[x], then a(x) + ⟨p(x)⟩ is the coset of ⟨p(x)⟩ represented by a(x). a(x) = b(x) (mod (p(x))) ↭[Recall: aH = bH ↭ a - b ∈ H ⇒ a(x) - b(x) ∈ ⟨p(x)⟩] p(x) | a(x) - b(x).
• Let R be any ring. The ideal generated by a is defined as Ra = ⟨a⟩ = {ra | r ∈ R}. Let R be a commutative ring with unity, let a₁, a₂, ···, a_n be a family of elements of R. Then, the ideal generated by the family a₁, a₂, ···, a_n is the set of elements of R of all finite linear combinations of a₁, a₂, ···, a_n, i.e, r₁a₁ + r₂a₂ + ··· + r_na_n. Hence, I = ⟨a₁, a₂, ···, a_n⟩ = {r₁a₁ + r₂a₂ + ··· + r_na_n | r_i ∈ R}.

For R = ℤ, the ideal ⟨4, 6⟩ = {4a + 6b | a, b ∈ ℤ} = {2a | a ∈ ℤ} = ⟨2⟩

• Let R be the ring of all real-valued functions of a real variable. The subset I of all differentiable functions is a subring of R but not an ideal of R.

It’s not an ideal because it’s not true that ri ∈ I for every r ∈ R and i ∈ S. Consider $r(x) = \begin{cases} 1, &x ∈ ℚ \\\ -1, &x ∉ ℚ \end{cases}$ and i = e^x, $ri(x) = \begin{cases} e^x, &x ∈ ℚ \\\ -e^x, &x ∉ ℚ \end{cases}$ is not differentiable. In fact it’s not even continuous ⇒ ri ∉ I ⇒ I is not an ideal of R. However, it’s a subring since it’s nonempty and the product and difference of differentiable functions are differentiable functions, too.

• The principal ideal ⟨2⟩ in ℤ₁₅. ⟨2⟩ = {2n | n ∈ ℤ₁₅} = {0, 2, 4, 6, 8, 10, 12, 14, (2·8_ℤ15) 1, 3, 5, 7, 9, 11, 13} = ℤ₁₅. ⟨5⟩ = {5n | n ∈ ℤ₁₅} = {0, 5, 10}. Notice that 2 is relatively prime to 15, so 2 is a unit in ℤ₁₅ (2·8 = 1).

Proposition. Suppose R is a commutative ring with unity. ⟨a⟩ = R ↭ a ∈ R is a unit.

Proof:

⇒) Suppose that ⟨a⟩ = R ⇒[By assumption, R is a commutative ring with unity] 1 ∈ ⟨a⟩ ⇒ ∃r∈R: ar = 1 ⇒ a is a unit (it has a multiplicative inverse, namely r).

⇐) Suppose a ∈ R is a unit ⇒ ∃r∈R: ar = 1 ⇒ 1 ∈ ⟨a⟩ ⇒ ∀b ∈ R, b = b·1 ∈ ⟨a⟩ ⇒ R = ⟨a⟩.

Corollary. The only ideals of a field F are the zero ideal and the unit ideal.

Proof.

Let I be an ideal of F. If I is the zero ideal, we are done.

Suppose I is a non-zero ideal of F ⇒ ∃a ∈ I, a ≠ 0 (the zero identity -addition- of F) ⇒[F is field and in a field, all non-zero elements are units] a is a unit ⇒[Proposition. Suppose R is a commutative ring with unity. ⟨a⟩ = R ↭ a ∈ R is a unit.] I = F, i.e., I is the unit ideal.

Proposition. Every ideal of ℤ is a principal ideal.

Proof: Suppose that I ⊆ ℤ is an ideal of ℤ.

1. If I = {0}, the trivial ideal, then it is certainly principal.
2. Otherwise, we can assume that it contains a non-zero element. Let n ∈ I to be its smallest non-negative (n > 0) element (it does contain non-negative elements, ∃m < 0, m ∈ I ⇒ -1·m ∈ I). We claim that I = ⟨n⟩. Obviously, ⟨n⟩ ⊆ I.
3. Let’s prove that I ⊆ ⟨n⟩. Let m ∈ I ⇒ [By the division algorithm] m = nq + r where 0 ≤ r < n ⇒ r = [m ∈ I, nq ∈ I (n ∈ I, it is its smallest non-negative number)] m - nq ∈ I ⇒ [r < n, but by n’s minimality] r = 0 ⇒ m = nq ⇒ m ∈ ⟨n⟩ ⇒ I ⊆ ⟨n⟩, therefore, I = ⟨n⟩∎

Factor Rings

Let R be a ring and let I be an ideal of R. In particular, R is an Abelian group under addition, and I is a normal subgroup of R, I ◁ R, therefore we could form the factor group R/I = {r + I | r ∈ R}.

As in the groups case the cosets partition the ring into disjoint subsets. The element r is called the coset representative of the coset. Besides, r₁ and r₂ represent the same coset ↭ r₁ - r₂ ∈ I. The ideal I itself is a coset, 0 + I.

Theorem. Let R be a ring and let I be a subring of R. The set of cosets R/I = {r + I | r ∈ R} is a ring under the operations (s + I) + (t + I) = (s + t) + I and (s + I)(t + I) = st + I iff I is an ideal of R.

Proof.

⇐) We only need to check that these operations are well-defined. Suppose that I is an ideal.

Let’s suppose s + I = s’ + I, t + I = t’ + I ⇒[s - s’ ∈ I, t - t’ ∈ I] s = s’ + a, t = t’ + a’, where a, a’ ∈ I.

(s - s’) + (t - t’) = (s + t) - (s’ + t’) ∈ I (I is closed under addition and s - s’ ∈ I, t - t’ ∈ I) ⇒[aH = bH ↭ a^-1b ∈H ↭_{Additive notation} a-b ∈ H] (s + t) + I = (s’ + t’) + I.

st + I = (s’+a)(t’ + a’) + I = s’t’ + at’ + s’a’ + aa’ + I =[I is an ideal, a, a’ ∈ I, and therefore it absorbs at’ + s’a’ + aa’] s’t’ + I , hence st + I = s’t’ + I. We already know that the set of cosets forms a group under addition and it is trivial to prove that the multiplication is associative and distributive over addition.

⇒) Suppose for the sake of contradiction, R/I is a ring and yet I is not an ideal of R ⇒ ∃a ∈ I, r ∈ R: ar ∉ I or ra ∉ I. Without loss of generality, let’s suppose that ar ∉ I, a ∈ I ⇒[a + H = H = 0 + H ↭ a ∈ H] a + I = 0 + I.

(a + I)(r + I) = ar + I, (0 + I)(r + I) = 0r + I = 0 + I = I. Then, ar + I = I ⊥ ar ∉ I, so the multiplication is not well-defined.

Examples

• ℤ/4ℤ = {0 + 4ℤ, 1 + 4ℤ, 2 + 4ℤ, 3 +4ℤ}

(2 + 4ℤ) + (3 + 4ℤ) = 5 + 4ℤ = 1 + 4 + 4ℤ =[aH = H ↭ a ∈ H. In addition notation, a + H = H ↭ a ∈ H] 1 + 4ℤ. (2 + 4ℤ)(3 + 4ℤ) = 6 + 4ℤ = 2 + 4 + 4ℤ = 2 + 4ℤ.

• ℤ/6ℤ = {0 + 6ℤ, 1 + 6ℤ, 2 + 6ℤ, 3 +6ℤ, 4 +6ℤ, 5 +6ℤ}. (5 +6ℤ)(5 +6ℤ) = 25 + 6ℤ = 1 +24 + 6ℤ =[24 ∈ 6ℤ ⇒ 24 + 6ℤ = 6ℤ] 1 + 6ℤ. (3 + 6ℤ)(4 + 6ℤ) = 12 + 6ℤ = 0 + 6ℤ. Notice that 3 + 6ℤ and 4 + 6ℤ are zero divisors, and 5 + 6ℤ is a unit (it is its own multiplicative inverse).
• ℤ₆/⟨3⟩ = {[0], [1], [2]}

0 + ⟨3⟩ = {0, 3} = [0]
1 + ⟨3⟩ = {1, 4} = [1]
2 + ⟨3⟩ = {2, 5} = [2]
(1 + ⟨3⟩) + (1 + ⟨3⟩) = (1 + 1) + ⟨3⟩ = 2 + ⟨3⟩
(1 + ⟨3⟩)(1 + ⟨3⟩) = (1 · 1) + ⟨3⟩ = 1 + ⟨3⟩  

• ℤ[x]/I where I = {a₂x² + a₃x³+···+a_nxⁿ| n ≥ 2, a_i ∈ ℤ}, e.g., (x⁴+2x²+4x+2) + I =[x⁴+2x² ∈ I] (4x+2) + I. In general, all cosets have the form (ax + b) + I = [Some authors will use another notation] $\overline {ax + b}$. ((ax +b) + I)((cx +d) + I) = (acx² + (ad + bc)x + bd) + I = [acx² ∈ I] ((ad + bc)x + bd) + I.

• Let R = M₂(ℤ) = {$(\begin{smallmatrix}a_1 & a_2\\\ a_3 & a_4\end{smallmatrix})$ | a_i ∈ ℤ}, and let I be the subset of R with even entries. I is an ideal of R (it is a similar reasoning behind 2ℤ being an ideal of the ring ℤ).

1. Subring. Let x, y ∈ I : x = $(\begin{smallmatrix}2a_1 & 2a_2\\\ 2a_3 & 2a_4\end{smallmatrix})$, y = $(\begin{smallmatrix}2b_1 & 2b_2\\\ 2b_3 & 2b_4\end{smallmatrix})$. Then, x - y = $(\begin{smallmatrix}2a_1 & 2a_2\\\ 2a_3 & 2a_4\end{smallmatrix}) - (\begin{smallmatrix}2b_1 & 2b_2\\\ 2b_3 & 2b_4\end{smallmatrix}) = (\begin{smallmatrix}2(a_1-b_1) & 2(a_2-b_2)\\\ 2(a_3-b_3) & 2(a_4-b_4)\end{smallmatrix})$∈ I, x · y = $(\begin{smallmatrix}2a_1 & 2a_2\\\ 2a_3 & 2a_4\end{smallmatrix}) · (\begin{smallmatrix}2b_1 & 2b_2\\\ 2b_3 & 2b_4\end{smallmatrix}) = (\begin{smallmatrix}4(a_1b_1+a_2b_3) & 4(a_1b_2+a_2b_4)\\\ 4(a_3b_1+a_4b_3) & 4(a_3b_2+a_4b_4)\end{smallmatrix})$∈ I.

2. Absorption. x ∈ I, y ∈ R, xy = $(\begin{smallmatrix}2b_1 & 2b_2\\\ 2b_3 & 2b_4\end{smallmatrix})(\begin{smallmatrix}a_1 & a_2\\\ a_3 & a_4\end{smallmatrix}) = (\begin{smallmatrix}2(b_1a_1 + b_2a_3) & 2(b_1a_2 + b_2a_4)\\\ 2(b_3a_1 + b_4a_3) & 2(b_3a_2 + b_4a_4)\end{smallmatrix})$ ∈ I. Analogously, yx ∈ I (this step is absolutely necessarily given that R is a non-commutative ring), therefore I is indeed an ideal of R.

3. Futhermore, every member of R can be written in the form $(\begin{smallmatrix}2q_1 +r_1 & 2q_2 + r_2\\\ 2q_3 + r_3 & 2q_4 + r_4\end{smallmatrix})$. Therefore, $(\begin{smallmatrix}2q_1 +r_1 & 2q_2 + r_2\\\ 2q_3 + r_3 & 2q_4 + r_4\end{smallmatrix})+I = (\begin{smallmatrix}r_1 & r_2\\\ r_3 & r_4\end{smallmatrix})+(\begin{smallmatrix}2q_1 & 2q_2\\\ 2q_3 & 2q_4\end{smallmatrix})+I = (\begin{smallmatrix}r_1 & r_2\\\ r_3 & r_4\end{smallmatrix})+ I$ since an ideal absorbs its own elements.

   In other words, R/I = {$(\begin{smallmatrix}r_1 & r_2\\\ r_3 & r_4\end{smallmatrix}) + I|r_i∈${0, 1}}, |R/I| = 2⁴ = 16, e.g., $(\begin{smallmatrix}3 & 6\\\ 5 & -3\end{smallmatrix})+I = (\begin{smallmatrix}1 & 0\\\ 1 & 1\end{smallmatrix}) + (\begin{smallmatrix}2 & 6\\\ 4 & -4\end{smallmatrix}) + I = (\begin{smallmatrix}1 & 0\\\ 1 & 1\end{smallmatrix}) + I$ since an ideal absorbs its own elements as it was stated previously.

Proposition. Given a ring R with ideals I, J ⊆ R, I + J = {i + j | i ∈ I, j ∈ J}, IJ = {i₁j₁ + ··· + i_nj_n | i_k ∈ I, j_k ∈ J, ∀k: 1 ≤ k ≤ n}, I ∩ J, I + J, IJ are all ideals.

Proof.

It is left as an exercise to prove that I ∩ J, I + J, IJ are all subrings.

∀r ∈ R, a ∈ I ∩ J ⇒[a∈ I, a∈ J, I and J ideals] ra ∈ I, ar ∈ I, ra ∈ J, ar ∈ J ⇒ ra ∈ I ∩ J and ar ∈ I ∩ J ⇒ I ∩ J is an ideal of R.

∀r ∈ R, a ∈ I + J ⇒ a = i + j, i ∈ I, j ∈ J ⇒ ra = r(i + j) =[Distributivity] ri + rj ∈ I + J, ri ∈ I, rj ∈ J. Mutatis mutadis, ar ∈ I + J, and hence I + J is an ideal.

∀r ∈ R, a ∈ IJ ⇒ a = i₁j₁ + ··· + i_nj_n, i_k ∈ I, j_k ∈ J, ∀k: 1 ≤ k ≤ n ⇒ ra = r(i₁j₁ + ··· + i_nj_n) =[Distributivity] r(i₁j₁) + ··· + r(i_nj_n) =[Associativity] (ri₁)j₁ + ··· + (ri_n)j_n ∈ IJ because ri_k ∈ I (By assumption, I is a ring), j_k ∈ J, ∀k: 1 ≤ k ≤ n. Mutatis mutandis, ar ∈ IJ.

Proposition. Let m, n ∈ ℕ, l = lcm(m, n), d = gcd(m, n). Then, mℤ ∩ nℤ = lℤ, mℤ + nℤ = dℤ ↭ mℤ ∩ nℤ = ⟨m⟩ ∩ ⟨n⟩ = ⟨l⟩, ⟨m⟩ + ⟨n⟩ = ⟨d⟩.

Proof.

Suppose, a ∈ mℤ ∩ nℤ ⇒ a ∈ mℤ and a ∈ nℤ ⇒ ∃x, y ∈ ℤ such that a = mx = ny ⇒ m | a and n | a ⇒[l = lcm(m, n)] l | a ⇒ ∃z ∈ ℤ, a = lz ⇒ a ∈ ⟨l⟩ = lℤ ⇒ mℤ ∩ nℤ ⊆ lℤ.

Suppose, a ∈ lℤ ⇒ ∃x ∈ ℤ, a = lx ⇒[Since l = lcm(m, n), ∃b, c ∈ ℤ: l = mb = nc] a = (mb)x = (nc)x ⇒[Associativity] a = m(bx) = n(cx) ∈ mℤ ∩ nℤ ⇒[We have already proved that mℤ ∩ nℤ ⊆ lℤ] mℤ ∩ nℤ = lℤ.

Suppose a ∈ mℤ + nℤ ⇒ ∃x, y ∈ ℤ such that a = mx + ny ⇒[d = gcd(m, n) ⇒∃z, w ∈ ℤ: m = dz, n = dw] a = (dz)x + (dw)y =[Associativity] d(zx) + d(wy) =[Distributivity] d(zx + wy) ∈ ⟨d⟩ = dℤ ⇒ mℤ + nℤ ⊆ dℤ.

Suppose a ∈ dℤ ⇒ ∃t ∈ ℤ: a = dt ⇒[d = gcd(m, n) ⇒∃x, y ∈ ℤ: d = mx + ny] a = (mx + ny)t =[Distributivity] (mx)t + (ny)t =[Associativity] m(xt) + n(yt) ∈ mℤ + nℤ ⇒[We have already proved that mℤ + nℤ ⊆ dℤ] mℤ + nℤ = dℤ ∎

Proposition. Let R be an integral domain. ⟨a⟩ = ⟨b⟩ ↭ ∃u ∈ R, u unit such that a = bu.

Proof.

⇒) Suppose ⟨a⟩ = ⟨b⟩ ⇒[R is an integral domain, i.e., a commutativity ring with unity ⇒ 1 ∈ R] a = 1·a ∈ ⟨a⟩ ⇒[By assumption, ⟨a⟩ = ⟨b⟩, hence a ∈ ⟨b⟩] ∃r ∈ R: a = br.

Similarly b ∈ ⟨b⟩ = ⟨a⟩, ∃s ∈ S: b = as ⇒ a = br = (as)r =[Associativity] a(sr) ⇒ a = a(sr) ⇒[Cancellation laws apply in integral domains] sr = 1 ⇒ s and r are units and a = br.

⇐) Suppose ∃u ∈ R, u unit such that a = bu.

∀r ∈ ⟨a⟩ ⇒ ∃x ∈ R: r = ax ⇒[a = bu] r = (bu)x =[Associativity] b(ux) ∈ ⟨b⟩ ⇒ ⟨a⟩ ⊆ ⟨b⟩.

a = bu ⇒[u is a unit ⇒ ∃u^-1 ∈ R such that uu^-1 = u^-1u = 1] b = au^-1. ∀s ∈ ⟨b⟩ ⇒ ∃y ∈ R: s = by ⇒[b = au^-1] s = (au^-1)b =[Associativity] a(u^-1b) ∈ ⟨a⟩ ⇒ ⟨b⟩ ⊆ ⟨a⟩ ⇒ ⟨a⟩ = ⟨b⟩ ∎

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