Consider the ideal (6) in ℤ. Is it prime? Is it maximal?
APrime and maximal, because 6 is an integer and all nonzero ideals in ℤ are prime and maximal
BPrime but not maximal, because ℤ/(6) has no zero divisors but is not a field
CNeither prime nor maximal, because 2·3 = 6 ∈ (6) while 2 ∉ (6) and 3 ∉ (6), and (6) ⊊ (2) ⊊ ℤ
DMaximal but not prime, because (6) cannot be extended to a larger proper ideal
For (6) to be prime, whenever ab ∈ (6) we would need a ∈ (6) or b ∈ (6). But 2·3 = 6 ∈ (6), yet neither 2 nor 3 is divisible by 6 — so (6) is not prime. For (6) to be maximal, there would need to be no proper ideal strictly between (6) and ℤ. But (6) ⊊ (2) ⊊ ℤ is a chain of proper inclusions, so (6) is not maximal either. In ℤ, an ideal (n) is prime iff n is prime, and maximal iff n is prime — both require n to be a prime number.
Question 2 Multiple Choice
A student argues: 'Since every field is an integral domain, and R/P is an integral domain when P is prime, it follows that R/P is always a field.' What is wrong?
ANothing is wrong — prime ideals always give field quotients
BThe argument confuses sufficient and necessary conditions: R/P being an integral domain requires P prime; R/P being a field requires the stronger condition that P is maximal
CR/P is never an integral domain unless R is itself an integral domain
DThe argument is correct for commutative rings but fails for noncommutative rings
Every field is an integral domain, but not every integral domain is a field. The correspondence mirrors this: maximal ↔ field (stronger), prime ↔ integral domain (weaker). In ℤ, the ideal (2) is both prime and maximal — ℤ/(2) ≅ F₂ is a field. But (0) is prime (ℤ/(0) ≅ ℤ is an integral domain) yet not maximal. The student's argument illegitimately reverses the implication 'field ⟹ integral domain' into 'integral domain ⟹ field.'
Question 3 True / False
In any commutative ring with unity, every maximal ideal is also a prime ideal.
TTrue
FFalse
Answer: True
If M is maximal, then R/M is a field. Every field is an integral domain (it has no zero divisors). Since R/M is an integral domain, M is prime. The chain is: M maximal ⟹ R/M is a field ⟹ R/M is an integral domain ⟹ M is prime. The reverse implication fails: prime does not imply maximal.
Question 4 True / False
In a commutative ring with unity, nearly every prime ideal is also maximal.
TTrue
FFalse
Answer: False
In ℤ, the ideal (0) is prime because ℤ has no zero divisors (if ab = 0 in ℤ, then a = 0 or b = 0). But (0) is not maximal because (0) ⊊ (2) ⊊ ℤ — the ideal (2) sits strictly between (0) and all of ℤ. However, in a principal ideal domain, every nonzero prime ideal is maximal. The claim holds in special settings but fails in general.
Question 5 Short Answer
Why does the correspondence 'R/M is a field ⟺ M is maximal' hold, and what does it reveal about the relationship between algebraic structure and ideal size?
Think about your answer, then reveal below.
Model answer: A field has no proper nonzero ideals. Under the correspondence between ideals of R/I and ideals of R containing I, the ideals of R/M correspond exactly to ideals J with M ⊆ J ⊆ R. If M is maximal, no such J exists strictly between M and R, so R/M has no nontrivial ideals — it is a field. Conversely, if R/M is a field, it has no nontrivial ideals, so no ideal of R sits strictly between M and R, making M maximal. The 'size' of the ideal (how close it is to all of R) directly controls the 'simplicity' of the quotient ring.
This is a deep structural insight: collapsing a maximal portion of a ring produces a maximally simple structure (a field). The hierarchy maximal → prime → general ideal mirrors the hierarchy field → integral domain → general ring, and studying which quotients have which properties is the essential technique of commutative algebra.