Mathematics.

ring theory

Ring Homomorphisms

Abstract Algebra II30 minDifficulty6 out of 10

You should know: rings

Overview

A ring homomorphism is a function between two rings that preserves both addition and multiplication (and typically the multiplicative identity), the ring-theoretic analogue of a group homomorphism. Its kernel — the elements mapping to zero — is always an ideal of the domain, and by the First Isomorphism Theorem for rings, the quotient by the kernel is isomorphic to the image. Ring homomorphisms are the structure-preserving maps used to compare rings, define quotient constructions, and evaluate polynomials at specific points.

Intuition

A ring homomorphism is a translator between two arithmetic systems that respects both operations at once — it doesn't just care that sums map to sums, but that products map to products too, so the entire multiplicative structure carries over consistently. The kernel behaves like the 'blind spot' of the map, the elements that get erased to zero, and — crucially — this blind spot is always an ideal, not just any subring, because absorbing a kernel element into any ring element via multiplication still lands you in the kernel (φ(ra)=φ(r)φ(a)=φ(r)·0=0).

Formal Definition

Definition

A function φ: R → S between rings is a ring homomorphism if for all a, b ∈ R:

φ(a+b)=φ(a)+φ(b)\varphi(a+b) = \varphi(a) + \varphi(b)
Preserves addition
φ(ab)=φ(a)φ(b)\varphi(ab) = \varphi(a)\varphi(b)
Preserves multiplication
φ(1R)=1S(if unital homomorphism)\varphi(1_R) = 1_S \quad \text{(if unital homomorphism)}
Preserves multiplicative identity
ker⁡(φ)={a∈R:φ(a)=0S} is an ideal of R,R/ker⁡(φ)≅im⁡(φ)\ker(\varphi) = \{ a \in R : \varphi(a) = 0_S \} \text{ is an ideal of } R,\quad R/\ker(\varphi) \cong \operatorname{im}(\varphi)
Kernel is an ideal; First Isomorphism Theorem for rings

Worked Examples

  1. Addition is preserved: (a+b) mod n = (a mod n) + (b mod n) mod n, by definition of modular arithmetic.

    φ(a+b)=φ(a)+φ(b)\varphi(a+b) = \varphi(a) + \varphi(b)

  2. Multiplication is preserved similarly: (ab) mod n = (a mod n)(b mod n) mod n.

    φ(ab)=φ(a)φ(b)\varphi(ab) = \varphi(a)\varphi(b)

  3. The kernel is all integers congruent to 0 mod n, i.e. multiples of n.

    ker⁡(φ)=nZ\ker(\varphi) = n\mathbb{Z}

Answer: φ is a ring homomorphism with ker(φ) = nℤ, and by the First Isomorphism Theorem, ℤ/nℤ ≅ ℤₙ.

Practice Problems

Difficulty 4/10

Prove that φ(0_R) = 0_S for any ring homomorphism φ: R → S.

Difficulty 5/10

Show that the kernel of a ring homomorphism φ: R → S is always an ideal of R, not merely a subring.

Difficulty 6/10

Is the map φ: ℤ → ℤ, φ(a) = 2a, a ring homomorphism? Check both additivity and multiplicativity.

Quiz

A ring homomorphism φ: R → S must preserve:
The kernel of a ring homomorphism is always:
For φ: ℤ → ℤₙ the reduction mod n map, ker(φ) equals:

Summary

  • A ring homomorphism φ: R → S preserves both addition and multiplication: φ(a+b)=φ(a)+φ(b), φ(ab)=φ(a)φ(b).
  • ker(φ) is always an ideal of R (absorbs multiplication by arbitrary ring elements, not just its own).
  • First Isomorphism Theorem for rings: R/ker(φ) ≅ im(φ), e.g. ℤ/nℤ ≅ ℤₙ via the reduction map.

References

  1. WebsiteWikipedia — Ring homomorphism