Corollary

Let R be a noetherian integral domain and n an integer. If R is Japanese, the ring $R[[T_1, ..., T_n]]$ is Japanese.

Reasoning by induction, we may suppose that $n = 1$. Let S denote the integral closure of R; if R is Japanese, S is a finite algebra over R, hence a Japanese ring (No. 1, prop. 2). The ring S[[T]] is noetherian and integrally closed (V, § 1, No. 4, prop. 14); applying Theorem 1 to $A = S[[T]]$ and $a = T$, we deduce that S[[T]] is Japanese. Consequently R[[T]] is Japanese (No. 1, prop. 2).