    # Ring Homomorphisms

“You are not even an insignificant and useless human being, but just a low-level, primitive, bug-like, and oxygen-thief life-form. You really are an impossible to underestimate loose collection of personal flaws, unresolved personal traumas, a fully-fledged package of dark areas and unconscious biases, combined into an unholy, grotesque, and humiliating succession of frightening mistakes and unfortunate circumstances,” the healthcare worker took a few seconds to breath, Apocalypse, Anawim, #justtothepoint.

# Definition

A ring homomorphism Φ from a ring R to a ring S is a structure-preserving function or mapping between two rings. More explicitly, it is a function Φ: R → S such that Φ is addition, multiplication, and unit (multiplicative identity) preserving, Φ(a + b) = Φ(a) + Φ(b), Φ(ab) = Φ(a)Φ(b), Φ(1R) = 1S, ∀a, b ∈ R. Figure 1.a.

💣In our original definition, a ring does no need to have a multiplicative identity, other authors define a ring to have a multiplicative identity and, without the requirement for a multiplicative identity, such a construct is instead called a rng, a non-unital ring or pseudoring. Φ(1R) = 1S only obviously applies when a ring do have a multiplicative identity. 1. A ring isomorphism is a ring homomorphism that is both one-to-one and onto. It is used to show that two rings are algebraically identical.
2. The kernel of a ring homomorphism is the set of elements of R which are mapped to zero, Ker(Φ) = {r ∈ R| Φ(r) = 0}, where 0 is the additive identity.
3. The image of a ring homomorphism is the set of elements in the codomain S that are imaged of some elements in the domain R, im(Φ) = Φ(R) = {s ∈ S | s = Φ(r) for some r ∈ R}.

# Examples

• ∀n ∈ ℤ, n > 0, the function Φ: ℤ → ℤn defined by Φ(m) = [m] = m mod n is a surjective (it is very obvious that we can hit every element on ℤn) ring homomorphism.
1. Addition. This has already being taken care of because we have checked it previously regarding group homomorphisms.
2. ∀m1, m2 ∈ ℤ, Φ(m1 · m2) = [m1m2] = [m1][m2] = Φ(m1)Φ(m2).
3. Ker(Φ) = nℤ.
• Φ: ℝ[x]→ ℂ, Φ(f(x)) = f(i) is a ring homomorphism. If f(x) ∈ Ker(Φ) ⇒ Φ(f(x)) = f(i) = 0 ⇒ f(x) = (x-i)g(x) where g(x) ∈ ℂ[x] ⇒[By assumption, f(x) need to “live” in ℝ[x], Φ domain] f(x) = (x-i)(x+i)f’(x) where f’(x) ∈ ℝ[x] and we are not talking about derivate ⇒ f(x) = (x2+1)f’(x) where f’(x) ∈ ℝ[x]. Therefore, Ker(Φ) = (x2+1)ℝ[x].

• The complex conjugation Φ: ℂ → ℂ, Φ(a + bi) = a - bi is a ring homomorphism.

• Let ℚ[x] be the ring of all polynomials with rational coefficients, the function ℚ[x] → ℝ, Φ(f(x)) = f($\sqrt{2}$) is a ring homomorphism. Im(Φ) = ℚ[$\sqrt{2}$] = {a + b$\sqrt{2}$ | a, b ∈ ℚ}

Proof:

1. ∀ p(x), q(x) ∈ ℚ[x], Φ(p(x) + q(x)) = $p(\sqrt{2}) + q(\sqrt{2})$ = Φ(p(x)) + Φ(q(x)).

2. Φ(p(x)·q(x)) = $p(\sqrt{2})q(\sqrt{2})$ = Φ(p(x))Φ(q(x)).

3. Ker(Φ) = {p(x) ∈ ℚ[x] | $p(\sqrt{2})=0$)}. Notice that p(x) ∈ Ker(Φ) ⇒ $p(\sqrt{2})=0 ⇒ [\sqrt{2}~ is~ a~ root~ of~ this~ polynomial] p(x)=(x-\sqrt{2})q(x)$ this is going on over ℝ[x] ⇒ $p(x)= (x-\sqrt{2})(x+\sqrt{2})q’(x)$, q’(x) ∈ ℚ[x] ⇒ p(x) = (x2 -2)q’(x). Therefore, Ker(Φ) = {(x2 -2)p(x) | p(x) ∈ ℚ[x]} = (x2 -2)ℚ[x].

• Let ℝ[x] be the ring of all polynomials with real coefficients, the function Φ: ℝ[x] → ℝ, Φ(f(x)) = f(1) is a ring homomorphism, too.
• Φ: ℤ2 → ℤ2, Φ(x) = x2.
1. Φ(x + y) = (x + y)2 = x2 + 2xy + y2 = [2xy=0 because 2 times anything is 0 in ℤ2] x2 + y2 = Φ(x) + Φ(y).
2. Φ(xy) = (xy)2 = x2y2 = Φ(x)Φ(y)
• Φ: ℤ4 → ℤ10, Φ(x) = 5x.
1. Φ(x + y) = [x + y = 4q1 + r1, 0 ≤ r1 < 4] 5·4·q1 + 5·r1 = 5·r1. Φ(x) + Φ(y) = 5x + 5y = 5(x + y) = 5·r1.
2. Φ(xy) = [xy = 4q2 + r2, 0 ≤ r2 < 4] 5·4·q2 + 5·r2 = 5·r2. Φ(x)Φ(y) = 5x · 5y = 5·5(xy) = 5·(5·r2) = (5·5)·r2 [5·5 = 5 in ℤ10] 5·r2.
• For a ring R of prime characteristic p, Φ: R → R defined by Φ(r) = rp is a ring endomorphism called the Frobenius endomorphism.

Φ(a + b) = (a + b)p =[(a + b)p can be expanded using the binomial theorem] $\sum_{i=0}^{i=p} {{p}\choose{i}} a^ib^{p-i}$ where ${{p}\choose{i}}=\frac{p!}{i!(n-i)!}$ 0 ≤ i ≤ p. If 1 ≤ i ≤ p-1, p divides ${{p}\choose{i}}=\frac{p!}{i!(n-i)!}$ so the coefficients of all the terms except ap and bb vanish ⇒ Φ(a + b) = (a + b)p = ap + bp = Φ(a) + Φ(b). Φ(a·b) = (a·b)p = ap·bp = Φ(a)·Φ(b).

• Non-examples: There is no non-trivial ring homomorphism ℤ → ℤ/nℤ for any n ≥ 1. If m ≠ n, then mℤ is not isomorphic to nℤ.

Consider a ring homomorphism Φ: ℤ → ℤ/nℤ, 12 = 1 ⇒ Φ(12)= Φ(1) ⇒ Φ(12) = Φ(1·1) = Φ(1)·Φ(1) = Φ(1)2 = Φ(1). The only element in ℤ/nℤ that is identical to its square is zero, so Φ(1) = 0. However, Φ(k) = Φ(1+ ··· (k times) ··· + 1) = kΦ(1) = k·0 = 0, Φ = 0. The only ring homomorphism is the trivial one.

Consider a ring isomorphism Φ: nℤ → mℤ, a ring isomorphism must take generators to generators, n would have to be mapped to ±m. Consider the case of Φ(n) = m (without losing generality). Φ(n·n)= Φ(n + n + ··n times ·· + n) = Φ(n) + Φ(n) + ··n times ·· + Φ(n) = m + m + ··n times ·· + m = nm. However, Φ(n·n) = Φ(n)Φ(n) = mm. Therefore nm = mm, but n ≠ m ⊥

A similar reasoning shows that ℤ and 2ℤ are not isomorphic, since 1 is a generator of ℤ, Φ(1) is a generator of 2ℤ, Φ(1) = ±2. Then Φ(1) = Φ(1·1)= Φ(1)Φ(1) = 4 in both cases, but the same element in order to be one-to-one cannot be mapped to two different elements, namely ±2 and 4⊥.

• There is no ring isomorphism between 2ℤ and 3ℤ. Suppose Φ: 2ℤ → 3ℤ, Φ(2) ∈ 3ℤ, Φ(2) = 3n, n ∈ ℤ. Φ(4)= Φ(2+2) = Φ(2) + Φ(2) = 3n + 3n = 6n. Besides, Φ(4)= Φ(2·2) = Φ(2) · Φ(2) = 3n · 3n = 9n2 ⇒ 6n = 9n2 ⇒ [Φ(2) = 3n, n ≠ 0. If n = 0, Φ(2) = Φ(0) = 0 which contradicts Φ is one-to-one] 2 = 3n ⇒ n = 2/3 ∉ ℤ ⊥
• There is no ring isomorphism between ℚ[$\sqrt{2}$] and ℚ[$\sqrt{3}$]. Suppose Φ: ℚ[$\sqrt{2}$] → ℚ[$\sqrt{3}$] is a ring isomorphism, Φ($\sqrt{2}$) = a + b$\sqrt{3}$ for some a, b ∈ ℚ, not both zero.

Φ(2) = $Φ(\sqrt{2})Φ(\sqrt{2})= (a + b\sqrt{3})^{2}=a^{2}+3b^{2}+2ab\sqrt{3}$

Φ(2) = Φ(1+1) = Φ(1) + Φ(1) = [By assumption, then the identity is carried] 1 + 1 = 2 ⇒[2 = $a^{2}+3b^{2}+2ab\sqrt{3}$] 2$+0\sqrt{3} =a^{2}+3b^{2}+2ab\sqrt{3}$ ⇒ $2=a^{2}+3b^{2}, 2ab = 0$ that is a = 0 or b = 0. If a = 0 ⇒ 3b2= 2 ⇒ b = $\sqrt{\frac{2}{3}}$ ∈ ℚ ⊥. If b = 0 ⇒ a2 = 2 ⇒ a = $\sqrt{2}$ ∈ ℚ ⊥

• Test for Divisibility by 9. Let n ∈ ℤ with decimal representation akak-1···a0, n = ak10k + ak-110k-1 + ··· + a0. Let Φ be the natural homomorphism from ℤ to ℤ9, Φ(n) = n mod 9.

n is divisible by 9 ↭ 0 = Φ(n) ↭ 0 = Φ(ak10k + ak-110k-1 + ··· + a0) = Φ(ak)(Φ(10))k + Φ(ak-1)(Φ(10))k-1 + ··· + Φ(a0) = [Φ(n) = n mod 9, in particular Φ(10) = 1] Φ(ak) + Φ(ak-1) + ··· + Φ(a0) = Φ(ak + ak-1 + ··· + a0) ↭ ak + ak-1 + ··· + a0 is divisible by 9

• Φ:ℂ → M2x2(ℝ) is a homomorphism defined by Φ(a + bi) = $(\begin{smallmatrix}a & -b\\ b & a\end{smallmatrix})$
1. Φ((a + bi) + (c + di)) = Φ((a+c) + (b+d)i) = $(\begin{smallmatrix}a + c & -(b + d)\\ (b + d) & a + c\end{smallmatrix}) = (\begin{smallmatrix}a & -b\\ b & a\end{smallmatrix}) + (\begin{smallmatrix}c & -d\\ d & c\end{smallmatrix})$ = Φ(a + bi) + Φ(c + di).
2. Φ((a + bi)(c + di)) = Φ((ac-bd) + (ad+bc)i) = $(\begin{smallmatrix}ac-bd & -(ad+bc)\\ ad+bc & ac-bd\end{smallmatrix}) = (\begin{smallmatrix}a & -b\\ b & a\end{smallmatrix}) (\begin{smallmatrix}c & -d\\ d & c\end{smallmatrix})$ = Φ(a + bi)Φ(c + di).
3. a + bi ∈ Ker(Φ) ⇒ Φ(a + bi) = $(\begin{smallmatrix}a & -b\\ b & a\end{smallmatrix}) = (\begin{smallmatrix}0 & 0\\ 0 & 0\end{smallmatrix})$ then a = b = 0 ⇒ Ker(Φ) = {0} ⇒ Φ is injective.

Im(Φ) = {$(\begin{smallmatrix}a & -b\\ b & a\end{smallmatrix})$ | a, b ∈ ℝ} ⊆ M2x2(ℝ) ⇒ ℂ ≋ Im(Φ).

• Φ, Φ’, Φ’’:M2x2(ℝ) → ℝ defined by Φ$(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})$ = a, Φ’$(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}) = det(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix})$, and Φ’’$(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}) = trace(\begin{smallmatrix}a & b\\ c & d\end{smallmatrix}) = a + d$ are not homomorphisms, because Φ(ab)≠Φ(a)Φ(b).

• Exercise. Find all ring homomorphisms, Φ: ℤxℤ → ℤ, Φ(1, 0) = m, Φ(0, 1) = n ⇒ Φ(0, 0) = Φ((0, 1)·(1, 0)) = m·n. Φ(0, 0) = [Φ ring homomorphism, Φ transports the addition identity] 0 ⇒ mn = 0 ⇒ m = 0 or n = 0. Without any loss of generality, let’s assume m = 0.

Φ(a, b) = Φ((1, 0) + ··· (a times) ··· (1, 0) + (0, 1) + ··· (b times) ··· + (0, 1)) = Φ(1, 0) + ··· (a times) ··· Φ(1, 0) + Φ(0, 1) + ··· (b times) ··· + Φ(0, 1) = am + bn = [By assumption, m = 0] bn

1. Case 1. m = 0. Φ(a, b) = bn. Ker(Φ) = ℤ x {0}. Im(Φ) = nℤ.
2. Case 2. n = 0. Φ(a, b) = am. Ker(Φ) = {0} x ℤ. Im(Φ) = mℤ.

# Properties

Let Φ be a ring homomorphism from a ring R to another S, Φ: R → S. Let A be a subring of R and let B be an ideal of S.

• Φ(0R) = 0S.

0R = 0R + 0R ⇒ Φ(0R) = Φ(0R + 0R) =[Φ is addition preserving, Φ(a+b) = Φ(a) + Φ(b)] Φ(0R) + Φ(0R) ⇒ Φ(0R) = Φ(0R) + Φ(0R) ⇒[S is a ring, Φ(0R) has an additive inverse] 0S = Φ(0R) -Φ(0R) = (Φ(0R) + Φ(0R)) -Φ(0R) =[Associative] Φ(0R) + (Φ(0R) -Φ(0R)) ⇒ 0S = Φ(0R)∎

• Let R and S have multiplicative identities 1R and 1S respectively, then Φ(-1R) = -1S.

0S =[Previous property] Φ(0R) = Φ(1R -1R) =[Φ is addition preserving, Φ(a+b) = Φ(a) + Φ(b)] Φ(1R) + Φ(-1R)⇒ 0S = Φ(1R) + Φ(-1R) =[Φ is unit or multiplicative identity preserving] 1S + Φ(-1R) ⇒[0S = 1S + Φ(-1R)] Φ(-1R) = -1S.

• ∀r∈R, n∈ℤ+, Φ(nr) = nΦ(r) and Φ(rn) = (Φ(r))n.

∀r∈R, n∈ℤ+, Φ(nr) = Φ(r + r + ··n times·· + r) = Φ(r) + Φ(r) + ··n times·· + Φ(r) = nΦ(r).

∀r∈R, n∈ℤ+, Φ(rn) = Φ(r · r · ··n times·· · r) = Φ(r) · Φ(r) · ··n times·· · Φ(r) = Φ(r)n.

• ∀r ∈ R, Φ(-r) = -Φ(r).

∀r ∈ R, 0S = Φ(0R) = Φ(r + (-r)) = Φ(r) + Φ(-r) ⇒ Φ(r) + Φ(-r) = 0S ⇒ Φ(-r) = -Φ(r).

• If r ∈ Rx then Φ(r) ∈ Sx and Φ(r-1) = Φ(r)-1.

∀ r ∈ Rx, ∃r-1 ∈ R: r·r-1 = r-1·r = 1R ⇒ Φ(r)·Φ(r-1) = Φ(r-1)·Φ(r) = Φ(1R) =[Φ is unit (multiplicative identity) preserving] 1S ⇒ Φ(r) has a multiplicative inverse and it is Φ(r-1).

• The homomorphic image of a subring is also a subring, that is, ∀A subring of R, Φ(A) = {Φ(a)| a ∈ A} is a subring of S.
1. ∀a’1, a’2 ∈ Φ(A), ∃a1, a2 ∈ A: Φ(a1) = a’1, Φ(a2) = a’2. a’1 - a’2 = Φ(a1) - Φ(a2) =[∀r ∈ R, Φ(-r) = -Φ(r).] Φ(a1) + Φ(-a2) = Φ(a1 - a2) ∈ Φ(A) because a1 -a2 ∈ A subring of R.
2. ∀a’1, a’2 ∈ Φ(A), ∃a1, a2 ∈ A: Φ(a1) = a’1, Φ(a2) = a’2. a’1 · a’2 = Φ(a1) · Φ(a2) = Φ(a1 · a2) ∈ Φ(A) because a1·a2 ∈ A subring of R.
3. 1R ∈ A ⇒ Φ(1R) = 1S ∈ Φ(A).
• An onto ring homomorphism maps ideals on ideals, i.e., if A is an ideal of R and Φ is onto, then Φ(A) is an ideal of S.
Generally speaking, the homomorphism image of an ideal is not an ideal, e.g., let i: ℤ → ℚ, be the natural injection given by i(n) = n. ℚ is a field ⇒ [The only ideals of a field are {0} and the field itself] its only ideals are {0} and ℚ. Take any ideal ⟨n⟩ = nℤ ⊆ ℤ with n ≠ 0, i(⟨n⟩) = ⟨n⟩ = nℤ is not an ideal of ℚ.
1. ∀a’1, a’2 ∈ Φ(A), ∃a1, a2 ∈ A: Φ(a1) = a’1, Φ(a2) = a’2. a’1 - a’2 = Φ(a1) - Φ(a2) =[∀r ∈ R, Φ(-r) = -Φ(r).] Φ(a1) + Φ(-a2) = Φ(a1 - a2) ∈ Φ(A) because a1 -a2 ∈ A ideal of R.
2. ∀a’ ∈ Φ(A), ∃a ∈ A: Φ(a) = a’, ∀s ∈ S ⇒[Φ is onto] ∃r ∈ R: Φ(r) = s. Therefore, sa’ = Φ(r)Φ(a) = Φ(ra) ∈ Φ(A) because ra ∈ A (A is an ideal), hence sa’∈ Φ(A)∎
• The preimage of an ideal by a ring homomorphism is an ideal, i.e., Φ-1(B) = {r ∈ R | Φ(r) ∈ B} is an ideal of R.

I = Φ-1(B) = {r ∈ R | Φ(r) ∈ B}

1. ∀a, b ∈ Φ-1(B), a - b ∈ Φ-1(B)? Φ(a), Φ(b) ∈ B, B is an ideal ⇒ Φ(a) - Φ(b) ∈ B ⇒ [Φ is a ring homomorphism] Φ(a) - Φ(b) = Φ(a) + Φ(-b) = Φ(a - b) Λ Φ(a) - Φ(b) ∈ B ⇒ a - b ∈ Φ-1(B).
2. ∀r ∈ R, a ∈ Φ-1(B), then Φ(a) ∈ B, Φ(r) ∈ S, B is an ideal of S ⇒ Φ(r)Φ(a) ∈ B ⇒ [Φ is a ring homomorphism] Φ(r)Φ(a) = Φ(ra) ∈ B ⇒ ra ∈ Φ-1(B). Therefore, Φ-1(B) is an ideal of R.
• The homomorphic image of a commutative ring is commutative. In other words, if R is commutative, then Φ(R) is commutative.

Φ(R) = {Φ(r): r ∈ R}. ∀s1, s2 ∈ Φ(R), ∃r1, r2 ∈ R: Φ(r1) = s1, Φ(r2) = s2. s1·s2 =[By definition of Φ(R)] Φ(r1)Φ(r2) =[Φ is a ring homomorphism] Φ(r1r2) =[By assumption, R is commutative] Φ(r2r1) = [Φ is a ring homomorphism] Φ(r2)Φ(r1) = s2s1

• Let Φ: R → S be an onto homomorphism from a ring R with unit element. Then, Φ(1) is the unit element of S.

By assumption, Φ is onto, ∀s ∈ S, ∃r: Φ(r) = s. Φ(r)Φ(1) =[Φ is a ring homomorphism] Φ(r·1) =[R is a ring with unit element 1] Φ(r). Analogously, Φ(1)Φ(r) =[Φ is a ring homomorphism] Φ(1·r) =[R is a ring with unit element 1] Φ(r). Therefore, Φ(r)Φ(1) = Φ(1)Φ(r) = Φ(r) or ∀s ∈ S, ∃Φ(1)∈ S such that s·Φ(1) = Φ(1)·s = s ∎

• A ring homomorphism is injective iff its kernel is trivial. In other words, Φ is injective ↭ Ker(Φ) = {r ∈ R | Φ(r) = 0} = {0}.

Corollary. Φ is a ring isomorphism ↭ Φ is onto and its kernel is trivial.

⇒) Suppose Φ is injective. ∀r ∈ Ker(Φ), Φ(r) = 0S =[Φ is a ring homomorphism ⇒ Φ(0R) = 0S] Φ(0R) ⇒[Φ(r) = Φ(0S), Φ injective] r = 0R.

⇐) Suppose Ker(Φ) = {0R}. ∀r1, r2 ∈ R such that Φ(r1) = Φ(r2) ⇒ Φ(r1)-Φ(r2) = Φ(r1)+Φ(-r2) = Φ(r1-r2) = 0S ⇒ r1 - r2∈ Ker(Φ) = {0R} ⇒ r1 = r2 ⇒ Φ is injective ∎

• The inverse of a ring isomorphism is also an isomorphism or, in other words, Φ is a ring isomorphism ⇒ Φ-1 is isomorphism.
1. ∀s1, s2 ∈ S ⇒[Φ is isomorphism ⇒ bijective] ∃r1, r2 ∈ R: Φ(r1) = s1, Φ(r2) = s2. Φ-1(s1s2) = Φ-1(Φ(r1)Φ(r2)) = Φ-1(Φ(r1r2)) =[Definition of an inverse mapping] r1r2 = Φ-1(s1-1(s2)
2. ∀s1, s2 ∈ S ⇒[Φ is isomorphism ⇒ bijective] ∃r1, r2 ∈ R: Φ(r1) = s1, Φ(r2) = s2. Φ-1(s1+s2) = Φ-1(Φ(r1)+Φ(r2)) = Φ-1(Φ(r1+r2)) =[Definition of an inverse mapping] r1+r2 = Φ-1(s1)+Φ-1(s2)
3. Φ-1(1S) =[Φ(1R) = 1S] 1R.
4. By 1, 2 and 3, Φ-1 is a ring homomorphism, and since inverse of a bijective function Φ is bijective, Φ-1 is also bijective, hence Φ is a ring isomorphism.
• The kernel of a ring homomorphism Φ: R → S is an ideal in R.

Proof.

1. The Kernel of a group homomorphism (a ring homomorphism is a group homomorphism) is a subgroup, (Ker(Φ), +) ≤ (R, +).
2. Ker(Φ) is a subring of R. ∀x, y ∈ Ker(Φ) -Ker(Φ) ≤ R-, x -y ∈ Ker(Φ), x · y ∈ Ker(Φ)? Φ(x·y) = [Φ is a ring homomorphism] Φ(x)·Φ(y) = [x, y ∈ Ker(Φ)] 0·0 = 0 ⇒ x · y ∈ Ker(Φ)
3. Ker(Φ) is an ideal of R. Let x ∈ Ker(Φ), r ∈ R, Φ(x·r) = [Φ is ring homomorphism] Φ(x)·Φ(r) = [x ∈ Ker(Φ)] 0·Φ(r) = 0 ⇒ x·r ∈ Ker(Φ) and similarly Φ(r.x) = 0, so r.x ∈ Ker(Φ).
• Let Φ: R → S be a ring homomorphism and let R'⊆ R be a subring of R ⇒ Φ(R') is a subring of S. In particular, Φ(R) is a subring of S.
1. 1R ∈ R’ ⇒ 1S = Φ(1R) ∈ Φ(R’). This requirement depends on the definition of a ring.
2. ∀s1, s2 ∈ Φ(R’) ⇒ ∃r1, r2 ∈ R’ such that Φ(r1) = s1 and Φ(r2) = s2. s1 - s2 = Φ(r1) - Φ(r2) =[Φ is a ring homomorphism] Φ(r1 - r2) ∈ Φ(R’) because R’ is a subring, r1 - r2 ∈ R'.
3. ∀s1, s2 ∈ Φ(R’) ⇒ ∃r1, r2 ∈ R’ such that Φ(r1) = s1 and Φ(r2) = s2. s1 · s2 = Φ(r1)·Φ(r2) =[Φ is a ring homomorphism] Φ(r1·r2) ∈ Φ(R’) because R’ is a subring, r1 ·r2 ∈ R'.

# Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License. This post relies heavily on the following resources, specially on NPTEL-NOC IITM, Introduction to Galois Theory, Michael Penn, and Contemporary Abstract Algebra, Joseph, A. Gallian.
1. NPTEL-NOC IITM, Introduction to Galois Theory.
2. Algebra, Second Edition, by Michael Artin.
3. LibreTexts, Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
4. Field and Galois Theory, by Patrick Morandi. Springer.
5. Michael Penn (Abstract Algebra), and MathMajor.
6. Contemporary Abstract Algebra, Joseph, A. Gallian.
7. Andrew Misseldine: College Algebra and Abstract Algebra.
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