Why is 6Z not a prime ideal of Z?

Conditions

• The ambient ring is the ring of integers Z.
• The ideal under consideration is 6Z.
• 6 is a composite number.

Reasoning, step by step

1. State the definition of a prime ideal: I is a proper ideal, and for all a, $b \in R$, if ab ∈ I, then $a \in I$ or $b \in I$.
2. Select a counterexample with $a = 2$ and $b = 3$.
3. Calculate the product ab = $2 * 3 = 6$.
4. Verify that $6 \in 6Z$ because 6 is a multiple of 6.
5. Check the membership of the factors: 2 ∉ 6Z and 3 ∉ 6Z because neither is a multiple of 6.
6. Conclude that since ab ∈ 6Z but a ∉ 6Z and b ∉ 6Z, 6Z fails the prime ideal condition.

Example

The video provides this exact counterexample: 'Example: when m is composite, mZ is not a prime ideal: $a = 2$ and $b = 3$, ab ∈ 6Z, but a ∉ 6Z and b ∉ 6Z.' The voiceover explains: 'This is because m is composite and can be factored.'

Common misconceptions

• Assuming that any principal ideal mZ is a prime ideal, forgetting that m must be prime or zero.
• Confusing the failure of the product condition with the failure of the proper ideal condition; 6Z is a proper ideal, but it fails the product condition.

Watch the explanation

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