Then, she continued, “You’re a living proof against both evolutionary biologists and believers, that is, in itself, quite remarkable. You are physically repulsive, morally reprehensible and bankrupt, intellectually impaired, socially awkward, embarrassing, and inept, religiously condemned, psychologically handicapped, diagnosed as an incurable narcissist, subtle as a brick to the head, sensitive as an ATM machine, rude, gross, pathetic, and you even smell,” Apocalypse, Anawim, #justtothepoint.
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:
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).
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) = a_{0}+a_{1}x + ··· + a_{n}x^{n}| Φ(f(x)) = f(0) = a_{0} = 0} = {f(x) ∈ ℤ[x], f(x) = a_{1}x + a_{2}x^{2} + ··· + a_{n}x^{n}} = {f(x) ∈ ℤ[x], f(x) = x(a_{1} + a_{2}x + ··· + a_{n}x^{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
Claim: Ker(Φ) = ⟨x^{2} + 1⟩
By the first isomorphism theorem for rings, ℝ[x]/⟨x^{2}+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:
Proof.
∀ i_{1} + j_{1}, i_{2} + j_{2} ∈ I + J, (i_{1} + j_{1}) - (i_{2} + j_{2}) = (i_{1} -i_{2}) + (j_{1} -j_{2}) ∈ I + J. (i_{1} + j_{1})·(i_{2} + j_{2}) = i_{1}i_{2} + (i_{1}j_{2} + j_{1}i_{2} + j_{1}j_{2}) ∈ I + J because i_{1}i_{2} ∈ I and i_{1}j_{2} + j_{1}i_{2} + j_{1}j_{2} ∈ J (J is an ideal of R and it absorbs these products) ⇒ I + J is a subring of S.
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.
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).
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.
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.
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, $\frac{I}{I ∩ J} ≋ \frac{I + J}{J}$