Math 2.0

Associative rings and algebras Proof

Proof that a ring homomorphism is injective iff its kernel is zero

Builds on Ring homomorphismKernel of a ring homomorphismInjective function

Proof

  1. Let f ⁣:R→Sf \colon R \to S be a ring homomorphism with kernel ker⁡f={a∈R:f(a)=0}\ker f = \set{a \in R : f(a) = 0}.
  2. We have f(0)=f(0+0)=f(0)+f(0)f(0) = f(0 + 0) = f(0) + f(0). Subtracting f(0)f(0) gives f(0)=0f(0) = 0. So 0∈ker⁡f0 \in \ker f.
  3. For a∈Ra \in R, f(a)+f(−a)=f(a+(−a))=f(0)=0f(a) + f(-a) = f(a + (-a)) = f(0) = 0. So f(−a)=−f(a)f(-a) = -f(a).
  4. Hence f(a−b)=f(a)+f(−b)=f(a)−f(b)f(a - b) = f(a) + f(-b) = f(a) - f(b) for all a,b∈Ra, b \in R.
  5. Suppose that ff is injective, and let a∈ker⁡fa \in \ker f. Then f(a)=0=f(0)f(a) = 0 = f(0), so a=0a = 0. With step 2, ker⁡f={0}\ker f = \set{0}.
  6. Conversely, suppose that ker⁡f={0}\ker f = \set{0}. Let a,b∈Ra, b \in R with f(a)=f(b)f(a) = f(b).
  7. By step 4, f(a−b)=f(a)−f(b)=0f(a - b) = f(a) - f(b) = 0. So a−b∈ker⁡fa - b \in \ker f, and a−b=0a - b = 0.
  8. Thus a=ba = b, and ff is injective.

Idea

Two elements have the same image exactly when their difference lies in the kernel. If only 00 lies in the kernel, the difference is 00 and the elements are equal. The argument uses only that ff preserves addition.

Reading path

To understand this page from the ground up, read these 19 pages in order. The foundations come first.

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Element, definition, no evidence levelElementdefinitionGroup, concept, no evidence levelGroupconceptProposition, concept, no evidence levelPropositionconceptSet, definition, no evidence levelSetdefinitionRing, definition, no evidence levelRingdefinitionImplication, definition, no evidence levelImplicationdefinitionPredicate, definition, no evidence levelPredicatedefinitionOrdered pair, definition, no evidence levelOrdered pairdefinitionRing homomorphism, definition, no evidence levelRing homomorphismdefinitionExistential quantifier, definition, no evidence levelExistential quantifierdefinitionUniversal quantifier, definition, no evidence levelUniversal quantifierdefinitionCartesian product, definition, no evidence levelCartesian productdefinitionKernel of a ring homomorphism, definition, no evidence levelKernel of a ring homo…definitionSubset, definition, no evidence levelSubsetdefinitionUniqueness quantifier, definition, no evidence levelUniqueness quantifierdefinitionBinary relation, definition, no evidence levelBinary relationdefinitionFunction, definition, no evidence levelFunctiondefinitionInjective function, definition, no evidence levelInjective functiondefinitionProof that a ring homomorphism is injective iff its kernel is zero, proof, no evidence levelProof that a ring hom…proof
Each page uses the pages above it. The colour of the bar on a box shows how strong the evidence is, and the box also gives it as text. The list has the same pages.
  • no evidence level
  1. Element definition
  2. Group concept
  3. Proposition concept
  4. Set definition
  5. Ring definition
  6. Implication definition
  7. Predicate definition
  8. Ordered pair definition
  9. Ring homomorphism definition
  10. Existential quantifier definition
  11. Universal quantifier definition
  12. Cartesian product definition
  13. Kernel of a ring homomorphism definition
  14. Subset definition
  15. Uniqueness quantifier definition
  16. Binary relation definition
  17. Function definition
  18. Injective function definition
  19. Proof that a ring homomorphism is injective iff its kernel is zero proof

Where this leads

73 pages build on this one.

Further reading

  1. Web https://proofwiki.org/wiki/Kernel_is_Trivial_iff_Monomorphism/Ring
About this page
TagAT76ZBC9ZM
URIm2://AT76ZBC9ZM
Kindproof
WritingW0 stub
MSC16B99
ProvesA ring homomorphism is injective iff its kernel is zero
UsesRing homomorphism, Kernel of a ring homomorphism, Injective function
Changed4 November 2026

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