Proposition 4

Let $A$ be a Nagata ring. a) Every finite $A$-algebra is a Nagata ring. b) For every multiplicative subset $S$ of $A$, the ring $S^{-1}A$ is a Nagata ring. a) Let $B$ be a finite $A$-algebra, $\rho : A \to B$ the canonical homomorphism. For every prime ideal $p$ of $B$, the ring $B/p$, which is a finite algebra over the Japanese ring $A/\rho^{-1}(p)$, is Japanese (No. 1, prop. 2). b) Let $q$ be a prime ideal of $S^{-1}A$; then there exists a prime ideal $p$ of $A$ such that $q = S^{-1}p$. The ring $(S^{-1}A)/q$ is a ring of fractions of the Japanese ring $A/p$, and is therefore Japanese (No. 1, remark 2).