Introduction to Algebraic Geometry by Serge Lang

Author:Serge Lang
Language: eng
Format: epub
Publisher: Courier Publishing
Published: 2019-03-12T16:00:00+00:00

PROPOSITION 2. Let T : U → V be a birational correspondence defined over a field k. Let (x) be a generic point of U over k, and T(x) = (y). Assume that k[x] = R is contained in k[y] = S and that k[y] is integral over k[x]. Then the set of points on U or V where T is not biholomorphic is the k-closed set of zeros of the conductor a of S over R.

Proof: Our statement depends on the fact that is an ideal of S and R, and hence determines an algebraic set both on U and V.

Let us begin with U. Let (x’) be a point of U where T is not defined. If (x’) is not a zero of the conductor, then there exists f(x) in such that f(x’) ≠ 0. By definition, for each coordinate yj of (y) we have f(x)yj = gj(x). This shows that yj = gj(x)/f(x), and that T is defined at (x’), a contradiction.

Conversely, suppose that T is defined at (x’). By the lemma, we can write S = Rw1 + . . . + Rwm, and by hypothesis Wi = fi(x)/gi(x), where gix’) ≠ 0. Hence gi(x)wi = fi(x), and by definition, we see that the product g of the gi(x) must be in the conductor. But g(x’) ≠ 0, and hence (x’) cannot be a zero of the conductor.

Let us now go over to V. Let (y’) be a point of V where T–1 is notbiholomorphic. If we put T–1(y’) = (x’), then Pcannot be defined at (x’). If (y’) is not a zero of the conductor (viewed as an ideal of k[y]) then (x’) cannot be a zero of the conductor (viewed as an ideal of k[x] and we have a contradiction as before. The converse is proved the same way.

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