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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.

The unit ideal (1) = Z is excluded from being a prime or maximal ideal because both definitions require the ideal to be a proper ideal. A proper ideal must be a strict subset of the ring ($I \ne R$).

Conditions: The ambient ring is the ring of integers Z.; The ideal under consideration is the unit ideal (1) = Z.

The unit ideal (1) = Z is excluded from being a prime or maximal ideal because both definitions require the ideal to be a proper ideal. A proper ideal must be a strict subset of the ring ($I \ne R$).

Conditions: The ambient ring is the ring of integers Z.; The ideal under consideration is the unit ideal (1) = Z.

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.