Bourbaki

Exercise 21, § 2

Commutative Algebra, V, 369

Let AA be an integrally closed domain containing the field Q\mathbf{Q}, KK its field of fractions, LL an algebraic extension of KK and BB the integral closure of AA in LL. Show that, for every ideal aa of AA, A∩aB=aA \cap_a B = a.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.