By H. Tachikawa, Claus M. Ringel

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Suppose that I[α] = R. Then there exists p ∈ Dp1 (R) such that I[α] ⊆ p because I[α] is a divisorial ideal of R. Since R[α] p is integral over R p by assumption, R[α] p is a flat extension of R p . As R[α] p is flat over R p , R[α] p is a free R p -module of rank d. Here we want to show that R[α] p = R p + R p α + · · · + R p α d−1 . To this end, we have only to show that 1 , α , . , α d−1 ∈ R[α] p / p R[α] p are linearly independent over k( p), where α denotes the residue class of α in R[α] p / p R[α] p .

Let ϕα (X ) = 32 Simple Extensions of High Degree X d + η1 X d−1 + · · · + ηd be the minimal polynomial of α. Put η0 = 1. Then there exists j such that v(η j ) ≤ v(ηi ) for all i. Thus ηi /η j = ai /b ∈ R p , where b ∈ R \ p, ai ∈ R. In particular, a j = b ∈ p. Hence ϕα (X ) = η j (a0 /η j )X d + · · · + η j (ad /η j )ηd Note that I[α] R p = Iη j R p by the choice of η j . Hence f (X ) := (b/η j )ϕα (X ) = a0 X d + · · · + ad ∈ Iη j ϕα (X )R p [X ] = I[α] ϕα (X )R p [X ]. Since a j = b ∈ p, we have c( f (X )) ⊆ p R p .

3 Let A = R[α] with α ∈ K . Suppose that R is strictly closed in A. Then α is integral if and only if Iα−1 ⊆ Iα . Proof (⇒): By the determinant trick, this implication is clear. n (⇐): Let C be the conductor of A over R. Then it is easy to see that C = i=1 Iα i , for some n. 2. Hence it holds that Iα ⊆ Iαi for all i, and Iα ⊆ C. Thus C = Iα . Since the conductor C is an ideal of R, Iα is an ideal in A, and α Iα ⊆ Iα . 5 Upper-Prime, Upper-Primary, or Upper-Quasi-Primary Ideals Let R be a Noetherian domain and K its quotient field.