Math 2.0

statement proposition

A ring homomorphism is injective iff its kernel is zero

Statement

Let f ⁣:R→Sf \colon R \to S be a ring homomorphism. Then ff is injective if and only if its kernel is {0}\{0\}.

Remarks

The proof uses only the addition: f(a)=f(b)f(a) = f(b) holds if and only if f(a−b)=0f(a - b) = 0, that is, a−b∈ker⁡fa - b \in \ker f. The same criterion holds for group homomorphisms and for linear maps.

Explanations

L1 human

If two elements have the same image, then their difference goes to 00. So when only 00 goes to 00, no two different elements have the same image.

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Group, concept, no evidence levelGroupconceptRing, definition, no evidence levelRingdefinitionRing homomorphism, definition, no evidence levelRing homomorphismdefinitionKernel of a ring homomorphism, definition, no evidence levelKernel of a ring homo…definitionA ring homomorphism is injective iff its kernel is zero, statement, no evidence levelA ring homomorphism i…statement
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  1. Group concept
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  4. Kernel of a ring homomorphism definition
  5. A ring homomorphism is injective iff its kernel is zero statement

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Sources

  1. Web https://en.wikipedia.org/wiki/Ring_homomorphism, Properties

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