MATH 401, ASSIGNMENT 8

DANIEL R. GRAYSON

15.1: 3. The field k(x) couldn’t be generated by f_{1}/g_{1}, …, f_{n}/g_{n}, since 1/(1 + xg_{1} … g_{n}) isn’t a polynomial P (f_{1}/g_{1}, …, f_{n}/g_{n}), because its denominator is a non-unit relatively prime to the other denominators.

15.2: 21. (a) It’s the complement in V of the closed subset V ∩ Z(f). (Lift f to the polynomial ring to make sense of that.)

(b) Let R = k[V]. The ring k[x_{1}, …, x_{n+1}]/J is isomorphic to R[x_{n+1}]/(x_{n+1}f − 1), which in turn is isomorphic to R_{f}, as we prove in problem 10 below. The ring map R → R_{f} gets us a continuous function Q : Z(J) → V. Its image is contained in V_{f}, because when (a_{1}, …, a_{n+1}) ∈ Z(J), the equation a_{n+1}f(a_{1}, …, a_{n}) − 1 = 0 tells us that f(a_{1}, …, a_{n}) ≠ 0. Its image contains V_{f} because if we have a point (a_{1}, …, a_{n}) ∈ V_{f}, we can set a_{n+1} = 1/f(a_{1}, …, a_{n}) to obtain a point (a_{1}, …, a_{n+1}) ∈ Z(J) that maps onto it. In fact, the number a_{n+1} is uniquely determined by the equation a_{n+1} = 1/f(a_{1}, …, a_{n}), so the map Q is one-to-one. To show it’s a homeomorphism, it’s enough to show that Q maps closed sets to closed sets. So suppose L is an ideal in R_{f}, and let’s apply Q to the closed subset Z(L) ⊆ Z(J). Pick generators for L which are fractions in R_{f}, but clear the denominators, so they are of the form g_{1}/1, …, g_{m}/1 ∈ R_{f}. Then Q(Z(L)) = Z(g_{1}, …, g_{m}) ∩ V_{f}, hence is closed. (Whew!)

(c) Suppose U = V − Z(I). Since the ring R := k[V] is noetherian, I can be generated by a finite number of elements f_{1}, …, f_{n} ∈ R. Thus U = V − Z(f_{1}, …, f_{n}) = V − Z(f_{1}) ∩ ⋯ ∩ Z(f_{n}) = (V − Z(f_{1})) ∪ ⋯ ∪ (V − Z(f_{m})) = V_{f_{1}} ∪ ⋯ ∪ V_{f_{m}}.

15.4: 10. (We don’t need to know that f is not nilpotent for this problem.) There is a map ψ : R_{f} → R[x]/(xf − 1) because x̄ ∈ R[x]/(xf − 1) provides a reciprocal for f̄. There is a map ϕ : R[x]/(xf − 1) → R_{f} because we can set ϕ(x) = 1/f. The composite ϕ ∘ ψ is the identity because it’s the identity on fractions of the form r/1; the fractions with denominators have to go along for the ride. The composite ψ ∘ ϕ is the identity because it’s the identity on elements of R, and sends x̄ to x̄.

15.5: 1. The continuous map mentioned in the hint is a bijection because every prime contains N. It’s closed because the image of Z(I/N) is Z(I). Thus it’s a homeomorphism.

15.5: 14. Since (1, 0) · (0, 1) = (0, 0) = 0 and (1, 0) + (0, 1) = (1, 1) = 1, every prime ideal contains (0, 1) or (1, 0) but not both. This partitions Spec(R) into two closed sets. Let’s identify one of them. Let I_{1} be the ideal 0 × R_{2} ⊆ R; it’s also the ideal generated by (0, 1). The continuous injective map Spec(R/I_{1}) → Spec(R) sends the closed set Z(J/I_{1}) to the closed set Z(J), so it’s a homeomorphism onto its image, and Spec(R/I_{1}) is homeomorphic to Spec(R_{1}) because the rings are isomorphic. Similarly on the other side.

Date: November 17, 1999.

15.5: 17. Students who came to the problem session should recognize this group ring as the ring ℤ[T]/(T^{2} − 1) that we studied.

(a) The ideal 0 is not prime because (1 − σ)(1 + σ) = 1 − σ^{2} = 0 but 1 − σ ≠ 0 and 1 + σ ≠ 0. Indeed, any prime will contain (1 − σ)(1 + σ), so must contain one of its factors.

(b) This is done just as in the previous problems. (Since ℤ[G] → ℤ is surjective, it’s essentially projection onto a quotient ring.)

(c) Compute ℤ[G]/(1 − σ, 1 + σ) ≅ ℤ[T]/(1 − T^{2}, 1 − T, 1 + T) ≅ ℤ[T]/(1 − T, 1 + T) ≅ ℤ[T]/(2, T − 1) ≅ (ℤ/2)[T]/(T − 1) ≅ ℤ/2 and observe that the result is a field, hence the ideal I = (1 − σ, 1 + σ) is maximal. The primes of X^{+} ∩ X^{−} must contain I.

(d) Every point of X must be contained in X^{+} or X^{−}, and is closed there iff it’s closed in X. But X^{+} and X^{−} are both homeomorphic with Spec(ℤ), which has the prime ideal 0 as the unique non-closed point, whose closure is all of Spec(ℤ), as we saw in class. The closed points are the maximal ideals of the form (p, 1 − σ) or (p, 1 + σ), where p is a prime integer, and those two ideals are distinct unless p = 2. That leads to the picture in the book, except you may wish to make the two nonclosed points fuzzier, as I did in class.

University of Illinois at Urbana-Champaign

   E-mail address: dan@math.uiuc.edu
   URL: http://www.math.uiuc.edu/~dan