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Ring Homomorphisms

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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.


Image 

  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

  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.

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].

  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)
  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.

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).

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⊥.

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 ⊥

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

  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(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.

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)∎

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

∀ 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).

  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).
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)∎

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.

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

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 ∎

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 ∎

  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.

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).
  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, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, Andrew Misseldine, blackpenredpen, and MathMajor, YouTube’s channels.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. MIT OpenCourseWare, 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007, YouTube.
  8. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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