Isomorphism Theorems for Rings

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The reader may want to recall that in the case of a homomorphism Φ of groups, Φ: G → G’, the cosets of ker(Φ) have the structure of a group (the quotient group, G/Ker(Φ)≋ Φ(G)). In this case, ker(Φ) is the identity of the associated quotient group. Futhermore, every kernel is a normal subgroup of the domain (Ker(Φ) ◁ G) and every normal subgroup can be realized as the kernel of some group homomorphism. Is there anything similar for rings?

Let (R, +, ·) be a ring, let A be an ideal of R. Let R/A be the coset space of R modulo A with respect to addition +. Let’s define two operations:

1. ∀r,s ∈ R: (r + A) + (s + A) = (r + s) + A
2. ∀r,s ∈ R: (r + A) · (s + A) = (r · s) + A The algebraic structure (R/A, +, ·) is called the quotient ring of R by A. In other words, the additive quotient group R/A is a ring under the previously defined operations.

Theorem. Every ideal of a ring R is the kernel of a ring homomorphism of R.

Proof.

Let A be an ideal of a ring R. The function η: R → R/A defined by η(r) = r + A is a ring homomorphism (it is called the canonical o natural homomorphism for A) and its kernel is A.

∀r, s ∈ A, η(r) + η(s) = (r + A) + (s + A) = (r + s) + A = η(r + s)

η(r)η(s) = (r + A)(s + A) = rs + A = η(rs), and therefore Φ is a ring homomorphism.

Futhermore, r ∈ ker(η) ↭ η(r) = 0_R + A = A ↭ η(r) = r + A = A ↭ r ∈ A ↭ Ker(μ) = A.

First Isomorphism Theorem for Rings. Let R and S be rings, and let Φ be a ring homomorphism from R to S. Then, R/Ker(Φ) ≋ Φ(R), i.e., the mapping ψ: R/Ker(Φ) → Φ(R), given by r + Ker(Φ) → Φ(r) is an isomorphism, (Figure 1.c).  

Proof: Let’s define the mapping ψ: R/Ker(Φ) → Φ(R) given by ψ(r + Ker(Φ)) = Φ(r).

1. Well-defined. Let’s take two representatives of the same cosets, so suppose that r + Ker(Φ) = r’ + Ker(Φ) ⇒ r’- r ∈ Ker(Φ) ⇒ Φ(r’ -r) = 0. ψ(r + Ker(Φ)) = Φ(r) = Φ(r) + 0 = [Φ(r’ -r) = 0] Φ(r) + Φ(r’ -r) = Φ(r + r’ -r) =[(R, +) is an Abelian group] Φ(r’) = ψ(r’ + Ker(Φ)).
2. Ring homomorphism. ψ((a + Ker(Φ) + (b + Ker(Φ)) = ψ((a + b) + Ker(Φ)) = Φ(a + b) = Φ(a) + Φ(b) = ψ(a + Ker(Φ)) + ψ(b + Ker(Φ)). Besides, ψ((a + Ker(Φ)) · (b + Ker(Φ)) = ψ((a · b) + Ker(Φ)) = Φ(a · b) = Φ(a) · Φ(b) = ψ(a + Ker(Φ)) · ψ(b + Ker(Φ)).
3. Injective. Suppose r + Ker(Φ) ∈ Ker(ψ) ⇒ ψ(r + Ker(Φ)) = 0, we also know that ψ(r + Ker(Φ)) = Φ(r), therefore Φ(r) = 0, hence r ∈ Ker(Φ) ⇒ r + Ker(Φ) = 0 + Ker(Φ). Ker(ψ) = 0 + Ker(Φ) = Ker(Φ), that is, the kernel is the trivial identity in the coset group R/Ker(Φ), so ψ is injective.
4. Surjective. Suppose an element from the image, Φ(r) ∈ Φ(R)=Im(Φ), obviously ψ(r + Ker(Φ)) = Φ(r), in other words, it was built-in to be surjective.

Examples

• Let Φ: ℤ → ℤ_n, Φ(m) = [m]_ℤn is clearly an onto (im(Φ) = ℤ_n) ring homomorphism. Ker(Φ) = {m ∈ ℤ: Φ(m) = [0]_ℤn} = {m ∈ ℤ: m ≡ 0 (mod n)} = nℤ. Therefore, by the First Isomorphism Theorem for Rings, ℤ/nℤ ≋ ℤ_n.

• Let Φ: ℤ[x] → ℤ given by Φ(f(x)) = f(0). It is a ring homomorphism with Ker(Φ) = ⟨x⟩, Φ is obviously onto (∀n ∈ ℤ, n ∈ ℤ[x] is a polynomial of degree zero and Φ(n) = n). Therefore, by the First Isomorphism Theorem for Rings, ℤ[x]/⟨x⟩ ≋ ℤ. ℤ is an integral domain but not a field, then the ideal ⟨x⟩ is prime (A is prime ↭ R/A integral domain) but not maximal (R/A is a field ↭ A is maximal) in ℤ[x].

  Ker(Φ) = {f(x) ∈ ℤ[x], f(x) = a₀+a₁x + ··· + a_nxⁿ| Φ(f(x)) = f(0) = a₀ = 0} = {f(x) ∈ ℤ[x], f(x) = a₁x + a₂x² + ··· + a_nxⁿ} = {f(x) ∈ ℤ[x], f(x) = x(a₁ + a₂x + ··· + a_nx^n-1)} = {xq(x)| q(x)∈ ℤ[x]} = ⟨x⟩.

• Fix an arbitrary real number a. Consider the evaluation map μ: ℝ[x] → ℝ, μ(f) = f(a). It is a surjective ring homomorphism, e.g., it is surjective, ∀b ∈ ℝ, let f(x) be the constant polynomial, f(x) = b, then μ(f) = f(a) = b.

Ker(μ) = {f ∈ ℝ[x] : μ(f) = f(a) = 0}. By the factor theorem, f(a) = 0 ↭ (x -a) is a factor of f, hence Ker(μ) = {(x -a)g(x) | g ∈ ℝ[x]} = ⟨x-a⟩. By the first isomorphism theorem, ℝ[x]/⟨x-a⟩ ≋ ℝ

• Φ: ℝ[x] → ℂ, given by f → f(i). Φ is a surjective ring homomorphism. ∀x ∈ ℂ, x = a + bi, consider the polynomial a + bx, Φ(a + bx) = a + bi = x, so Φ is surjective.

Claim: Ker(Φ) = ⟨x² + 1⟩

1. ⟨x² + 1⟩ ⊆ Ker(Φ). x² + 1 ∈ Ker(Φ) because Φ(x² +1) = i² + 1 = -1 + 1 = 0. Besides, ∀Φ ring homomorphism, Ker(Φ) is an ideal of R, in our particular case, ℝ[x], therefore every polynomial multiple of x² +1 is in the Kernel, so ⟨x² + 1⟩ ⊆ Ker(Φ)
2. f ∈ Ker(Φ) ⇒ By the division algorithm, f = (x² +1)q(x) + r(x) where r = 0 or deg(r) < 2. Let’s apply Φ, Φ(f) = Φ(x² +1)Φ(q(x)) + Φ(r(x)) ⇒[f and x²+1 are in the kernel] 0 = 0·Φ(q(x)) + Φ(r(x)) ⇒ Φ(r(x)) = 0 ⇒[deg(r)<2, r = a +bx, Φ(r) = a + bi = 0 ⇒ a = b = 0] r = 0 ⇒ f = (x² +1)q(x) ⇒ f ∈ ⟨x² +1⟩.

By the first isomorphism theorem for rings, ℝ[x]/⟨x²+1⟩ ≋ ℂ

Second Isomorphism Theorem for Rings. Let R be a ring, I be a subring of R, and J be an ideal of R. Then:

1. I + J = {i + j: i ∈ I, j∈ J} is a subring of R
2. I ∩ J is an ideal of I.
3. $\frac{I}{I ∩ J} ≋ \frac{I + J}{J}$

Proof.

• I + J is a subring of R. I is a subring (1 ∈ I), J is an ideal of R (an ideal is an Abelian subgroup of R, 0 ∈ J) ⇒ 1 = 1 + 0 ∈ I + J.

∀ i₁ + j₁, i₂ + j₂ ∈ I + J, (i₁ + j₁) - (i₂ + j₂) = (i₁ -i₂) + (j₁ -j₂) ∈ I + J. (i₁ + j₁)·(i₂ + j₂) = i₁i₂ + (i₁j₂ + j₁i₂ + j₁j₂) ∈ I + J because i₁i₂ ∈ I and i₁j₂ + j₁i₂ + j₁j₂ ∈ J (J is an ideal of R and it absorbs these products) ⇒ I + J is a subring of S.

• I ∩ J is an ideal of I. I ∩ J is not empty since 0 is contained in I and J. Besides, ∀a₁, a₂ ∈ I ∩ J, a₁ + a₂ ∈ I ∩ J since I and J are both closed under addition.

Suppose a ∈ I ∩ J, b ∈ I, ab ∈ I ∩ J and ba ∈ I ∩ J?

a ∈ I ∩ J ⇒ a ∈ I, a ∈ J, [J is an ideal of R] ab ∈ J, [I is subring, so there’s closure under multiplication, a ∈ I, b ∈ I] ab ∈ I ⇒ ab ∈ I ∩ J.

Mutatis mutandis, the same reasoning applies to ba ∈ I ∩ J.

• Let’s define the following map, Φ: I → I+J/J, Φ(a) = a + J.

1. Φ is homomorphism. Φ(a + b) = (a + b) + J = [By definition of addition in quotient ring] (a + J) + (b + J) = Φ(a) + Φ(b). Similarly, Φ(ab) = Φ(a)Φ(b).

2. Φ is onto. Suppose an arbitrary coset of the codomain a + J ∈ I + J/J, where a ∈ I + J, that is, a = i + j, i ∈ I, j ∈ J ⇒ a + J = (i + j) + J = [j ∈ J, aH = H ↭ a ∈ H] i + J. Therefore, Φ(i) = i + J = a + J.

3. Ker(Φ) = I ∩ J

   Suppose a ∈ I ∩ J ⇒ a ∈ I, a ∈ J. Because a ∈ I, our domain, it makes sense apply to it our homomorphism, Φ(a) = a + J = [a ∈ J] 0 + J = J, therefore a ∈ Ker(Φ) ⇒ I ∩ J ⊆ Ker(Φ)

   Suppose a ∈ Ker(Φ), Ker(Φ) is an ideal of the domain of Φ, that is, I, so a ∈ I. Besides Φ(a) = [Definition of the map Φ] a + J = [a ∈ Ker(Φ)] 0 + J = J ⇒ a ∈ J ⇒ a ∈ I ∩ J ⇒ Ker(Φ) ⊆ I ∩ J ⇒[We have previously demonstrated that I ∩ J ⊆ Ker(Φ)] Ker(Φ) = I ∩ J.

4. By the first Isomorphic Theorem of Ring, $\frac{I}{Ker(Φ)}=\frac{I}{I ∩ J} ≋ Im(Φ)$ = [Φ is onto] $\frac{I + J}{J}$ Therefore, $\frac{I}{I ∩ J} ≋ \frac{I + J}{J}$

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