Why is mZ a maximal ideal of Z when m is a prime number?

Conditions

• The ambient ring is the ring of integers Z.
• m is a prime number.
• The ideal under consideration is mZ.

Reasoning, step by step

1. Assume there exists an ideal I of Z such that mZ ⊆ I.
2. Recall that all ideals of Z are principal, so write I = kZ for some integer k.
3. Translate the ideal inclusion mZ ⊆ kZ into the divisibility relation k | m.
4. Apply the definition of a prime number: since m is prime, k must equal 1 or m.
5. Substitute $k=1$ to get $I = 1Z = Z$.
6. Substitute $k=m$ to get I = mZ.
7. Conclude that any ideal containing mZ is either mZ itself or the entire ring Z, satisfying the equivalent definition of a maximal ideal.

Example

The video proves this by contradiction: 'Suppose there exists an ideal I := kZ of Z such that mZ ⊆ I. Then mZ ⊆ kZ, so k divides m. Since m is prime, $k = 1$ or m. If $k = 1$, then $I = Z$; if $k = m$, then I = mZ. Q.E.D.!"

Common misconceptions

• Assuming that mZ is maximal for any integer m, forgetting that the primality of m is crucial for restricting the divisors k to 1 and m.
• Confusing maximal ideals with prime ideals; while mZ is both when m is prime, the proofs rely on different properties (divisibility vs. product membership).

Watch the explanation

Explore next

• Why is 6Z not a prime ideal of Z?
• What is the equivalent definition of a maximal ideal used in this proof?
• Why does mZ ⊆ kZ imply k | m?