Why is mZ a principal ideal of the integer ring Z?
mZ is a principal ideal of Z because it is generated by a single element m. By definition, the principal ideal generated by m in a commutative ring is the set of all multiples rm where r is in the ring. For the integer ring Z, mZ is exactly the set {rm | }, which perfectly matches the definition of the principal ideal (m).
Conditions
- The ambient ring is the ring of integers Z.
- m is an integer.
- The set under consideration is mZ.
Reasoning, step by step
- Recall the definition of a principal ideal in a commutative ring: (a) = {ra | }.
- Apply this definition to the ring Z with generator m: (m) = {rm | }.
- Observe that the set mZ is defined as the set of all integer multiples of m, i.e., mZ = {rm | }.
- Conclude that mZ is identical to the principal ideal (m), and therefore mZ is a principal ideal of Z.
Example
The video states: 'Example: mZ is a principal ideal of Z (m is an integer).' The voiceover explains: 'Take an integer m from Z; multiplying this integer by every element of Z yields mZ, which exactly matches the definition of a principal ideal.'
Common misconceptions
- Assuming that mZ is just a subset and not an ideal, forgetting that it satisfies the additive closure and absorption properties required of ideals.
- Confusing the generator m with the elements of the ideal; m is the generator, while the elements are the multiples rm.
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