In ℤ, the ideal (12) decomposes as (4) ∩ (3). Which statement correctly describes this decomposition?
A(4) is primary with radical (2) and (3) is primary with radical (3), giving an irredundant primary decomposition of (12)
B(4) and (3) are both prime ideals, so this is a prime decomposition
C(12) = (4) · (3) is a product decomposition, not an intersection
DThis decomposition is redundant because (4) ⊇ (3)
The ideal (4) = (2²) is primary: if ab ∈ (4), then 4 | ab; if 4 ∤ a, then 2 ∤ a (since we need a²... actually) — more precisely, (4) is (2)-primary because √(4) = (2) is prime, and (4) satisfies the primary condition. The ideal (3) is prime (hence primary with radical (3)). The intersection (4) ∩ (3) = (12) because lcm(4,3) = 12. This is irredundant: neither component contains the other. In ℤ, primary decomposition corresponds to the prime-power factorization of integers.
Question 2 Multiple Choice
Which of the following is the correct analog of primary decomposition in ℤ?
AFactoring n into a product of primes: n = p₁^{a₁} ··· pₖ^{aₖ}
BWriting (n) as an intersection of prime-power ideals: (n) = (p₁^{a₁}) ∩ ··· ∩ (pₖ^{aₖ})
CWriting n as a sum of prime numbers (Goldbach's conjecture)
DFactoring the ideal (n) into a product of prime ideals
For n = p₁^{a₁} ··· pₖ^{aₖ}, the ideal (n) decomposes as (p₁^{a₁}) ∩ ··· ∩ (pₖ^{aₖ}). Each (pᵢ^{aᵢ}) is (pᵢ)-primary. This is an irredundant primary decomposition. Note the distinction between intersection and product: in ℤ, (p₁^{a₁}) ∩ ··· ∩ (pₖ^{aₖ}) = (p₁^{a₁}) ··· (pₖ^{aₖ}) when the primes are distinct, but in general rings, the intersection and product of ideals differ.
Question 3 True / False
In a Noetherian ring, every ideal has a primary decomposition.
TTrue
FFalse
Answer: True
This is the Lasker-Noether theorem, the fundamental existence result for primary decomposition. Lasker proved it for polynomial rings in 1905, and Noether generalized it to all Noetherian rings in 1921. The proof uses the Noetherian property to show that every ideal is a finite intersection of irreducible ideals, and every irreducible ideal in a Noetherian ring is primary. The decomposition may not be unique, but the associated primes and their primary components (for minimal primes) are unique.
Question 4 True / False
The primary decomposition of an ideal in a Noetherian ring is always unique.
TTrue
FFalse
Answer: False
While the associated primes (the radicals of the primary components) are uniquely determined, and the primary components corresponding to minimal associated primes are unique, the primary components corresponding to embedded (non-minimal) primes may not be unique. For example, in k[x,y], the ideal (x², xy) has associated primes (x) and (x,y), and the (x,y)-primary component can vary. The first uniqueness theorem says the set of associated primes is unique; the second says minimal primary components are unique.
Question 5 Short Answer
Explain the difference between a prime ideal and a primary ideal, and why primary ideals are the 'right' building blocks for ideal decomposition.
Think about your answer, then reveal below.
Model answer: A prime ideal P satisfies: ab ∈ P implies a ∈ P or b ∈ P. A primary ideal Q satisfies the weaker condition: ab ∈ Q implies a ∈ Q or bⁿ ∈ Q for some n. Equivalently, Q is primary if every zero-divisor in R/Q is nilpotent. The radical √Q is always prime, and Q is 'concentrated at' this prime. Primary ideals are the right building blocks because prime ideals alone are too restrictive — the ideal (4) in ℤ is not prime but is (2)-primary, and (12) = (4) ∩ (3) requires the non-prime component (4).
Primary ideals relate to prime ideals as prime powers relate to primes in ℤ. The 'defect' of Q from being prime — that b might not be in Q when ab ∈ Q — is controlled: some power of b must be in Q. This means the zero-divisors in R/Q are all nilpotent (they become zero when raised to a high enough power), which is a manageable kind of 'non-primeness.' Without primary ideals, you cannot decompose ideals like (x², xy) in k[x,y], which require embedded components.