Theorem 1 (Tate)

Let A be an integrally closed noetherian ring, a an element of A. It is assumed that the ideal $aA$ is prime, that the ring $A/aA$ is Japanese, and that A is complete for the $aA$-adic topology. Then the ring A is Japanese.

a) The field of fractions of A being denoted by K, the assertion being trivial when K is of characteristic 0 (No. 1, corollary of prop. 1), one may suppose K of characteristic $p > 0$. One may also suppose $a \neq 0$.

Let L be a finite radicial extension of K and q a power of p such that $L \subset K^{1/q}$. Put $x = a^{1/q}$ and $M = L(x)$. By prop. 1 of No. 1, it is enough to prove that the integral closure B of A in M is a finite A-algebra.

b) Let us first prove that the ideal $xB$ is the unique prime ideal of B over $aA$. In fact there exists at least one prime ideal of B over $aA$ (V, § 2, No. 1, th. 1). Let q be one of these ideals. We have $x^q = a \in q$, whence $xB \subset q$ since q is prime. Conversely, let y be an element of q; the element $y^q$ of K is integral over A, and therefore belongs to A since A is integrally closed. Since $q \cap A = aA$, there exists an element $\alpha$ of A such that $y^q = a\alpha = x^q\alpha$. Consequently the element $y/x$ of M is integral over A, and therefore belongs to B; thus one has $y \in xB$, whence $q = xB$, which proves our assertion.

c) It follows that the ring $B_{xB}$ is the integral closure in M of the ring $A_{aA}$ (V, § 1, No. 5, prop. 16 and § 2, No. 1, prop. 2). By VI, § 3, No. 6, prop. 9, $A_{aA}$ is a discrete valuation ring; one then deduces from the Krull-Akizuki theorem (VII, § 2, No. 5, prop. 5) that the field $\kappa(xB)$ is a finite extension of $\kappa(aA)$ and that $B_{xB}$ is noetherian.

d) The ring $B/xB$ is integral over the Japanese ring $A/aA$ and its field of fractions is a finite extension of the field of fractions of the latter. Consequently, $B/xB$ is a finitely generated $(A/aA)$-module. For every integer $i \geq 0$, the same is true of the module $x^iB/x^{i+1}B$; it follows that the $(A/aA)$-module $B/aB$ possesses a composition series of length q whose quotients are finitely generated $(A/aA)$-modules, and is therefore itself a finitely generated $(A/aA)$-module.

e) Let us endow the ring A with the $(aA)$-adic filtration and the ring B with the $(aB)$-adic filtration. Then A is complete by hypothesis; since $B_{xB}$ is an integral domain and noetherian, the $aB_{xB}$-adic filtration of $B_{xB}$ is separated (III, § 3, No. 2, corollary to prop. 5); consequently one has $\bigcap a^nB \subset \bigcap a^nB_{xB} = \{0\}$, and the $aB$-adic filtration of B is separated; the gr(A)-module gr(B) is generated by gr_0(B), hence is finitely generated by d). It then follows from III, § 2, No. 9, cor. 1 to prop. 12, that B is a finitely generated A-module, which completes the proof.