Proposition 2

Let B be a noetherian integral ring and A a noetherian subring of B, such that B is a finite A-algebra. In order that A be Japanese, it is necessary and sufficient that B be Japanese.

Let K (resp. L) denote the field of fractions of A (resp. B). Suppose first that A is Japanese, and let M be a finite extension of L. Let C denote the integral closure of B in M. By V, § 1, No. 1, prop. 6, C is the integral closure of A in M, and therefore is a finite A-algebra since M is a finite extension of K and A is Japanese. A fortiori, C is a finite B-algebra. This proves that B is Japanese.

Conversely, suppose B Japanese and let N be a finite extension of K. Let D denote the integral closure of A in N. Let E be an extension of K compounded of L and N; since B is Japanese, the integral closure D’ of B in E is a finite B-algebra, therefore a finite A-algebra; the A-module D, which is a submodule of D’, is therefore finitely generated, whence it follows that A is Japanese.