A student claims that the ideal I = {2f(x) + xg(x) : f, g ∈ ℤ[x]} in ℤ[x] must be principal because both 2 and x are irreducible elements. Which response is correct?
AThe student is right — gcd(2, x) = 1 so the ideal is the whole ring ℤ[x]
BThe student is right — the ideal is generated by 2, since x = x·1 is a multiple of x
CThe student is wrong — the ideal (2, x) requires two generators and cannot be generated by any single element, showing ℤ[x] is not a PID
DThe student is wrong — 2 and x are not irreducible in ℤ[x], so the claim is based on a false premise
The ideal (2, x) contains all polynomials with even constant term. No single polynomial d(x) generates this ideal: if d were the generator, d would have to divide 2 (so d = ±1 or ±2) and divide x (so d = ±1). But (1) = ℤ[x], and the ideal (2, x) is a proper ideal (it doesn't contain 1, since 1 has an odd constant term). So no single element generates it, and ℤ[x] is not a PID. This is the canonical example showing the PID property is strictly stronger than being a UFD.
Question 2 Multiple Choice
Which of the following correctly describes the relationship between the three classes: Euclidean domains (ED), principal ideal domains (PID), and unique factorization domains (UFD)?
AED ⊇ PID ⊇ UFD: every UFD is a PID and every PID is a Euclidean domain
BED ⊆ PID ⊆ UFD: every Euclidean domain is a PID and every PID is a UFD
CThey are all equivalent — the three conditions always hold together or not at all
DUFD ⊆ ED ⊆ PID: unique factorization is the strongest condition
The containments go ED ⊆ PID ⊆ UFD. Every Euclidean domain has a division algorithm, which implies every ideal is principal (PID). Every PID has sufficient structure to guarantee unique factorization into irreducibles (UFD). But the containments are strict: there exist PIDs that are not Euclidean domains, and there exist UFDs that are not PIDs. The canonical example for the latter is ℤ[x]: it is a UFD (polynomials factor uniquely into irreducibles) but not a PID, because (2, x) is not principal.
Question 3 True / False
In any principal ideal domain, every irreducible element is also a prime element.
TTrue
FFalse
Answer: True
This is one of the key theorems about PIDs and is what makes their divisibility theory so clean. In a general integral domain, prime implies irreducible but not conversely. In a PID, both directions hold: an element p is irreducible if and only if it is prime. The proof uses the fact that in a PID, the ideal (p) is maximal if and only if p is irreducible, and maximal ideals are prime ideals, so (p) is a prime ideal, making p a prime element. This equivalence fails in ℤ[√-5], for example, where 3 is irreducible but not prime.
Question 4 True / False
Most unique factorization domain is a principal ideal domain.
TTrue
FFalse
Answer: False
ℤ[x] is the standard counterexample. It is a UFD — every polynomial over ℤ factors uniquely into irreducibles — but it is not a PID because the ideal (2, x) cannot be generated by a single polynomial. Being a UFD guarantees unique factorization of elements, but it says nothing about whether every *ideal* (which is a set closed under ring operations, not just a single factored element) is principal. PIDs have the stronger property that every ideal is singly generated.
Question 5 Short Answer
Why does the ideal (2, x) in ℤ[x] fail to be principal, and what does this demonstrate about where ℤ[x] falls in the hierarchy of integral domains?
Think about your answer, then reveal below.
Model answer: The ideal (2, x) consists of all polynomials whose constant term is even. Any single generator d would have to generate both 2 and x, so d must divide 2 and divide x in ℤ[x]. The only divisors of both are ±1, but (1) = ℤ[x] — the whole ring — not this proper ideal. So no single generator works. This demonstrates that ℤ[x] is a UFD (polynomials factor uniquely into irreducibles via Gauss's lemma and the fact that ℤ is a UFD) but not a PID, confirming that the UFD → PID implication is strictly one-directional: every PID is a UFD, but not every UFD is a PID.
The key insight is that 'unique factorization of elements' and 'every ideal is principal' are different structural properties. PIDs have both; UFDs only guarantee the first. The example ℤ[x] shows why polynomial rings over ℤ are harder to work with than polynomial rings over fields: F[x] is a PID (division algorithm works), but ℤ[x] is merely a UFD.