Questions: Spec and the Zariski Topology

Question 1

What are the points of Spec Z?

• A. The positive integers
• B. The prime numbers 2, 3, 5, 7, ... together with the generic point (0)
• C. The maximal ideals (2), (3), (5), ... only
• D. The elements of Z

Answer: (B)

Spec Z consists of all prime ideals of Z: the zero ideal (0) and the ideals (p) for each prime number p. The point (0) is the 'generic point' -- its closure is all of Spec Z. The points (p) are the closed points (maximal ideals). The topology has closed sets V(n) = {(p) : p divides n} for each integer n.

Question 2

In the Zariski topology on Spec R, the closed sets are exactly the sets of the form V(I) for some ideal I.

• T. True
• F. False

Answer: True

The Zariski topology is defined by declaring V(I) = {P in Spec R : P ⊇ I} to be closed for each ideal I. One verifies: V(R) = ∅, V(0) = Spec R, V(∩I_α) = ∪V(I_α), and V(I + J) = V(I) ∩ V(J). Moreover, V(I) = V(√I), so closed sets correspond bijectively to radical ideals.

Question 3

Describe the Zariski topology on Spec k[x] where k is an algebraically closed field.

Model answer: Spec k[x] has one generic point (0) and closed points (x - a) for each a in k. The closed sets are: all of Spec k[x], finite sets of closed points, and the empty set. The generic point (0) is dense -- it lies in every nonempty open set.

Since k[x] is a PID over an algebraically closed field, the nonzero primes are (x - a) for a in k (the maximal ideals). The closed sets V(f) for nonzero f consist of the finitely many (x - a) where a is a root of f. So the topology on closed points is the cofinite topology. The generic point (0) has closure V(0) = Spec k[x], making it dense.

Question 4

The stalk of the structure sheaf O_X at a prime P in Spec R is the local ring R_P.

• T. True
• F. False

Answer: True

The structure sheaf is defined so that O_X(D(f)) = R_f (the localization inverting f), and the stalk at P is the direct limit over all open neighborhoods of P, which computes to R_P (the localization at the complement of P). This is the key property making (Spec R, O_X) a locally ringed space.

Question 5

Explain what it means for Spec R to be irreducible and relate this to a property of R.

Model answer: Spec R is irreducible (not the union of two proper closed subsets) if and only if the nilradical √(0) is a prime ideal. For a reduced ring (no nilpotents), this is equivalent to R being a domain. An irreducible Spec R has a unique generic point, namely the nilradical.

If Spec R = V(I) ∪ V(J) with neither equal to Spec R, then IJ ⊆ √(0) but neither I nor J is in √(0), contradicting primality. Conversely, if √(0) is prime, any decomposition Spec R = V(I) ∪ V(J) = V(IJ) forces IJ ⊆ √(0), so I or J lies in √(0), making one factor all of Spec R. The generic point η = √(0) is the unique point whose closure is Spec R.