Let E be a finite extension of K. Let N be a finite quasi-Galois extension of K containing E (A, V, p. 54, cor. 1), and L the field of invariants of the group of K-automorphisms of N. Then (A, V, p. 73, prop. 13), L is a radicial extension of K and N is a separable extension of L. The integral closure B of A in L is therefore by hypothesis a finite A-algebra; the integral closure C of B in N is a finite B-algebra (V, § 1, No. 6, cor. 1 to prop. 18), hence a finite A-algebra. The integral closure of A in E is contained in C, and is therefore a finite A-algebra since A is noetherian.

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.

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.

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).

By Theorem 3 of § 2, No. 5 and Theorem 2 of § 3, No. 3, there exist an integer $n \geq 0$, a ring R which is a field or a discrete valuation ring with field of fractions of characteristic 0, and a subring B of A, isomorphic to R[[T₁, ..., Tₙ]] and such that A is a finite B-algebra. Then R is Japanese (No. 1, example and corollary to prop. 1), hence B is Japanese (corollary to Theorem 1), and A is Japanese (No. 1, prop. 2).

Suppose first that A is local and complete, and let p₁, ..., pₙ be the minimal (distinct) prime ideals of A; for i = 1, ..., n, let Kᵢ denote the field of fractions of A/pᵢ and Aᵢ' the integral closure of A/pᵢ. Since A is reduced, R is the product of the rings Kᵢ and A' the product of the rings Aᵢ' (V, § 1, No. 2, cor. 1 to prop. 9). Since the local rings A/pᵢ are integral domains and complete, they are Japanese (Theorem 2), so that each Aᵢ' is a finite A-algebra, and A' is a finite A-algebra.

If A is semi-local and complete, it is isomorphic to a finite product of complete local rings (III, § 2, No. 13, corollary to prop. 19), and we conclude at once from the preceding.

Let us pass to the general case, and note that the completion  of A is a semi-local, complete, noetherian ring faithfully flat over A (III, loc. cit., § 3, No. 4, corollary to prop. 8 and § 3, No. 5, prop. 9). Let S be the set of non-zero-divisors of A; one has R = S⁻¹A. Since  is flat over A, the elements of S are non-zero-divisors in Â, and S⁻¹Â is identified with a subring of the total ring of fractions T of Â. Again because  is flat over A, the ring A' ⊗_A  is identified with a subring of R ⊗_A  = S⁻¹Â, hence also with a subring of T integral over Â. By the first part of the proof, A' ⊗_A  is therefore a finitely generated Â-module; hence A' is a finitely generated A-module (I, § 3, No. 6, prop. 11).

Recall (A, V, p. 114, def. 1) that an algebra E over a field K is said to be separable if the ring L ⊗_K E is reduced for every extension L of K; it is enough that this be so for every finite extension of K. The following proposition generalizes Theorem 2:

Let L be a finite extension of K and B the integral closure of A in L. Let F be a finite subset of B such that L = K[F] (V, § 1, No. 5, cor. 2 to prop. 16); let C denote the (finite) A-algebra generated by F. Since L is the field of fractions of C, the ring B is the integral closure of C (V, § 1, No. 1, prop. 6) and it is enough to prove that B is a finite C-algebra. Now, C is a semi-local noetherian ring (IV, § 2, No. 5, cor. 3 to prop. 9 ); its completion is identified with C ⊗_A Â (III, § 3, No. 4, th. 3 (ii)), hence also with a subring of the reduced ring L ⊗_A Â = L ⊗_K (K ⊗_A Â) and therefore is reduced. Proposition 3 therefore follows from the corollary to Theorem 2.

The canonical homomorphism of B into $\prod_{i=1}^n \hat{B}_{m_i}$ extends to an isomorphism of $\hat{A} \otimes_A B$ onto $\prod_{i=1}^n \hat{B}_{m_i}$.

By IV, § 2, No. 5, cor. 3 to prop. 9, the ring B is semi-local and $m_A B$ is an ideal of definition of it. By III, § 3, No. 4, th. 3, (ii), the ring $\hat{A} \otimes_A B$ is the completion of B for the topology defined by its radical; one then applies III, § 2, No. 13, corollary to prop. 19.

Let indeed m be a nonzero element of M; then Ann(m) is contained in a prime ideal p of A associated with M (IV, § 1, No. 1, prop. 2), and the image of m in $M_p$ is nonzero (II, § 2, No. 2, prop. 4).

Let $N$ denote the submodule of $M$ formed by the elements $m$ such that $pm = 0$. We have $N \cap xM = xN$; in fact, if an element $m$ of $M$ is such that $pxm = 0$, then $pm = 0$ since $x_M$ is injective, hence $m \in N$. Consequently, the $A$-module $N/xN$ is isomorphic to the submodule $(N + xM)/xM$ of $M/xM$, and it is enough to prove that $q$ is associated with $N/xN$. Since $p$ is associated with $M$, there exists an element $m$ of $M$ such that $p = \mathrm{Ann}(m)$; we have $m \in N$, whence $p = \mathrm{Ann}(N)$ and therefore $\mathrm{Supp}(N/xN) = V(p + xA)$ by II, § 4, No. 4, corollary to prop. 18; consequently, $q$ is associated with $N/xN$ (IV, § 1, No. 4, th. 2).

Suppose there exists a nonzero nilpotent element $x$ of $B$, and let $\pi$ be a uniformizer of $A$. Since one has $\pi B \subset m_B$, the ring $B$ is separated for the $\pi B$-adic topology. There therefore exist $n \in \mathbf{N}$ and $y \in B$ with $x = \pi^n y$ and $y \notin \pi B$. Since $B$ is flat over $A$, multiplication by $\pi$ is injective in $B$. The class of $y$ in $B/\pi B$ is therefore a nonzero nilpotent element, which contradicts the assumption.

(R$_n$) $\left\{ \begin{array}{l} \text{for every local noetherian Nagata ring } C \text{ of dimension } < n \text{ and every prime ideal } r \text{ of } C, \text{ the ring } \kappa(r) \otimes_C \hat{C} \text{ is reduced.} \end{array} \right.$

Let $A$ be a semi-local noetherian Nagata ring of dimension $n$, let $p$ be a prime ideal of $A$ and $L$ a finite extension of the field $\kappa(p)$; it is enough to prove,