Proposition 5

*Let A be an integral domain, with field of fractions K, and let L be an algebraic extension of finite degree of K. We denote by B the integral closure of A in L and by p a prime ideal of A.

Suppose that the local ring $A_p$ is noetherian and regular; let t be an element of L such that $L = K(t)$ and suppose that there exists in $A[T]$ an element P, an Eisenstein polynomial for $pA_p$, having t as a root.

a) There exists in B a unique prime ideal q above p. b) The local ring $B_q$ is noetherian and regular, of the same dimension as $A_p$. c) One has $B_q = A_p[t]$. d) The canonical homomorphism of $A/p$ into $B/q$ induces an isomorphism of the fields of fractions of these rings.*

Put $C = A_p[t]$ and let d denote the degree of P. By prop. 3 applied to the ring $A_p$, the Eisenstein polynomial P is irreducible in $K[T]$ and $(1, t, ..., t^{d-1})$ is a basis of L over K, hence of C over $A_p$. Since P is monic, the kernel of the canonical homomorphism of $A_p[T]$ onto C is equal to (P). By the corollary to prop. 4 above, C is therefore a regular noetherian local ring of the same dimension as $A_p$, the maximal ideal $m_C$ of C is generated by $p \cup \{ t \}$ and the field $\kappa_C$ is a trivial extension of the field of fractions of $A/p$. In order to prove prop. 5, it is therefore enough to show that there exists a unique prime ideal q of B above p, and that one has $C = B_q$.

Put S = A − p. One knows (V, § 1, n° 5, prop. 16) that the integral closure of