Questions: Associated Primes

Question 1

What are the associated primes of Z/12Z as a Z-module?

• A. {(2), (3)}
• B. {(2), (3), (12)}
• C. {(2), (4), (3)}
• D. {(0), (2), (3)}

Answer: (A)

The primary decomposition of (12) in Z is (12) = (4) ∩ (3). The associated primes are the radicals of the primary components: √(4) = (2) and √(3) = (3). So Ass(Z/12Z) = {(2), (3)}. These are both minimal primes of (12); there are no embedded primes in this case.

Question 2

In a Noetherian ring R, the set of zero divisors equals the union of the associated primes of R (as a module over itself).

• T. True
• F. False

Answer: True

An element r is a zero divisor on R iff r annihilates some nonzero element, iff r lies in ann(x) for some x ≠ 0. In a Noetherian ring, every such annihilator is contained in an associated prime (an annihilator that is itself prime). So the zero divisors are exactly the union of Ass(R). This characterization is much more precise than just 'the set of zero divisors.'

Question 3

What is the difference between minimal and embedded associated primes?

Model answer: Minimal associated primes are the minimal elements of Ass(M) under inclusion -- they correspond to irreducible components of Supp(M). Embedded associated primes are non-minimal elements of Ass(M) -- they are 'hidden' inside larger components and represent non-reduced or non-generic structure.

For example, in k[x,y] the ideal I = (x^2, xy) has Ass(R/I) = {(x), (x,y)}. The prime (x) is minimal (it defines the line x = 0), while (x,y) is embedded (the origin is 'inside' the line). Embedded primes depend on the particular decomposition and are the source of much subtlety in algebraic geometry.

Question 4

For a Noetherian module M over a Noetherian ring R, Ass(M) is always a finite set.

• T. True
• F. False

Answer: True

This follows from the existence of primary decomposition for submodules of Noetherian modules. The zero submodule 0 ⊂ M has a primary decomposition 0 = Q_1 ∩ ... ∩ Q_n in M, and Ass(M) equals the set of primes {√ann(M/Q_i)}. Since primary decomposition is finite, so is Ass(M). The minimal primes are independent of the decomposition; the embedded ones may vary.

Question 5

Explain the relationship between Ass(M), Supp(M), and the minimal primes of ann(M).

Model answer: Ass(M) ⊆ Supp(M), and the minimal elements of both sets coincide (they equal the minimal primes of ann(M)). Supp(M) is the closure of Ass(M) in Spec R. Every prime in Supp(M) contains an associated prime.

Supp(M) = {P in Spec R : M_P ≠ 0} = V(ann(M)), the set of primes containing ann(M). The minimal primes of Supp(M) are exactly the minimal associated primes. Supp(M) can be much larger than Ass(M) -- it is the closure, while Ass(M) picks out the 'generic points' of the irreducible components plus any embedded structure.