Math 610, 2nd Assignment

Proof. We will calculate the integral closure of A inside K. We begin with a more general discussion. Suppose C is an irreducible, affine curve, that is, C is some irreducible affine variety of dimension 1. Let A be the ring of regular functions on C, so that A is a noetherian domain of Krull dimension 1. Localization preserves the noetherian condition, as well as the domain condition. Thus, the local ring Ox at any point x ∈ C is a noetherian domain. It also has Krull dimension 1 as the dimension of any local ring of an irreducible variety is equal to the dimension of the variety. If C has a singular point x, then Ox is not regular. We know that for noetherian local domains of dimension 1, regularity is equivalent to integrally closed, so that Ox is not integrally closed. Since the localization of an integrally closed domain is integrally closed, if C has singular points, we can conclude that A is not integrally closed. On the other hand, if C is smooth, then A must be integrally closed, and thus a Dedekind domain. Now suppose we resolve the singularities of C to obtain some smooth curve C ′ and a map C ′ C of varieties. Suppose in addition that C ′ happens to be affine, with coordinate ring A′, which by the above discussion is a Dedekind domain. This induces an injection A ↪→ A′. I will show that in the case of this problem, the above injection makes A′ a finitely generated A-module, which is enough to show A′ is integral over A. To see this, suppose A′ is finitely generated as an A-module. Let f ∈ A′. Then if {b1, . . . , bn} are generators, let b = (b1, . . . , bn) ∈ A′n, and notice that fb = Mb for some M ∈ Mn(A). It follows that (fIn −M)b = 0 ∈ A′n. Multiplying on the left by the adjoint matrix gives det(fIn −M)b = 0. By definition, of b, this implies that det(fIn −M) is in the annihilator of A′, but A′ is a faithful A module, so det(fIn −M) = 0. Expanding the determinant gives the desired monic polynomial with coefficients in A that f satisfies. Now we specialize to the case of the plane curve Z(y − x). Let C = Z(y − x). By problem 2 (below), we can resolve the singularity of C to obtain the affine line, A = C ′. Specifically, we have the blowup