Bourbaki

Exercise 29, § 1

Commutative Algebra, VII, 554

Let A be an integrally closed domain and K its field of fractions. (a) Let L be an algebraic extension of K and B the integral closure of A in L. Show that for every fractional ideal a of A, if we write b=aB,b∩A=a′′b = aB, b \cap A = a''. (Reduce it to the case where A⊂a~A \subset \tilde{a}, in other words A: a⊂Aa \subset A and prove that B: b⊂Bb \subset B; for this, let x∈Bx \in B: bb and let ci(1≤i≤n)c_i (1 \leq i \leq n) be the coefficients of its minimal polynomial over K; note that for ally ∈a\in a the elements ciyi(1≤i≤n)c_i y^i (1 \leq i \leq n) belong to A and deduce that the cic_i belong to A.) (b) Let C be a polynomial ring A[Xλ]λ∈LA[X_\lambda]_{\lambda \in L} in any family of indeterminates. Show that for every fractional ideal a of A, if we write c=aC,c∩A=a′′c = aC, c \cap A = a'' (same method).

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Transcribed from the printed volume. Not translated.