When m is a prime number, mZ is a maximal ideal of Z because any ideal I containing mZ must be of the form kZ (since all ideals of Z are principal). The inclusion mZ ⊆ kZ implies that k divides m.
Conditions: The ambient ring is the ring of integers Z.; m is a prime number.; The ideal under consideration is mZ.
When m is a prime number, mZ is a maximal ideal of Z because any ideal I containing mZ must be of the form kZ (since all ideals of Z are principal). The inclusion mZ ⊆ kZ implies that k divides m.
Conditions: The ambient ring is the ring of integers Z.; m is a prime number.; The ideal under consideration is mZ.
6Z is not a prime ideal of Z because 6 is a composite number. The definition of a prime ideal requires that if a product ab lies in the ideal, then at least one factor (a or b) must lie in the ideal.
Conditions: The ambient ring is the ring of integers Z.; The ideal under consideration is 6Z.; 6 is a composite number.
6Z is not a prime ideal of Z because 6 is a composite number. The definition of a prime ideal requires that if a product ab lies in the ideal, then at least one factor (a or b) must lie in the ideal.
Conditions: The ambient ring is the ring of integers Z.; The ideal under consideration is 6Z.; 6 is a composite number.
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.
Conditions: The ambient ring is the ring of integers Z.; m is an integer.; The set under consideration is mZ.
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.
Conditions: The ambient ring is the ring of integers Z.; m is an integer.; The set under consideration is mZ.