This has H = 1


 f\in A(Y-P)

  • Consider  V(f) \subset Y-p. If p is in the clsoure (taken in Y), then extend by zero.
  • Otherwise p is not in the closure, argue  1/f \in A(U - p) for a sufficiently small U \ni p disjoint from V(f)
  • Argue in fact 1/f \in \mathcal{O}_p; for all  q \in U-p,  1/f = g/h near q
    • If h(p) = 0 use  \dim Y \ge 2 to show  V(f) \cap U \ne \emptyset; contradiction.
  • Consider J \subset \{g \in \mathcal{O}_p| fg \in \mathcal{O}_p\}
    • By the above, conclude f^nJ \subset J, \forall n.
  • Use the determinant trick to show f is integral over \mathcal{O}_p.

Commutative Algebra

Determinant Trick: Let M be a finitely generated (say by n elements)  A module and  \phi \colon M \to M be endomorphism such that there is an ideal  J \subset A such that  \phi(M) \subset JM, then  \phi satifies a monic polynomial relation of degree n.

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