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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.

The zero ideal {0} is not a maximal ideal of Z because every ideal of Z contains the integer 0, and therefore contains {0}. This means {0} can be properly contained in a larger ideal, violating the definition of a maximal ideal.

Conditions: The ambient ring is the ring of integers Z.; The ideal under consideration is the zero ideal {0}.

The zero ideal {0} is not a maximal ideal of Z because every ideal of Z contains the integer 0, and therefore contains {0}. This means {0} can be properly contained in a larger ideal, violating the definition of a maximal ideal.

Conditions: The ambient ring is the ring of integers Z.; The ideal under consideration is the zero ideal {0}.