b) If $A_\alpha$ is noetherian for every $\alpha \in A$, then A is noetherian.

a) Put $m = \lim \overrightarrow{m_\alpha}$; this is an ideal of A. The quotient ring $A/m$ is the inductive limit of the fields $A_\alpha/m_\alpha$, hence is a field (A, I, p. 116, Prop. 3). On the other hand, every element of $A - m$ is invertible in A: in fact, let $x \in A - m$; there exist $\alpha \in I$ and $\xi \in A_\alpha$ such that $x = \varphi_\alpha(\xi)$; one has $\xi \notin m_\alpha$, hence $\xi$ is invertible in $A_\alpha$ and x is invertible in A. Consequently, A is a local ring, with maximal ideal m. Let $\alpha \in I$. From the relations $\varphi_{\beta\alpha}(m_\alpha)$ $A_\beta = m_\beta$ for $\beta \geq \alpha$, one deduces, by passing to the inductive limit, $\varphi_\alpha(m_\alpha)$ $A = m$; finally, the homomorphism $\varphi_\alpha$ is flat by I, § 2, No. 7, Prop. 9.

(b) Let $\hat{A}$ be the separated completion of A for the $m$-adic topology and $\pi$ the canonical mapping of A into $\hat{A}$. Suppose that the rings $A_\alpha$ are noetherian. Fix $\alpha \in I$ and prove that the ring $\hat{A}$ is noetherian and flat over $A_\alpha$. By hypothesis, $m_\alpha$ is a finitely generated ideal of $A_\alpha$, hence $m = \varphi_\alpha(m_\alpha)$. A is a finitely generated ideal of A. It follows that the maximal ideal $\hat{m}$ of $\hat{A}$ is equal to $mA$ (III, § 2, No. 12, Cor. 2 to Prop. 16 and No. 13, Prop. 19), hence is finitely generated. Consequently, the ring $\hat{A}$ is noetherian (loc. cit., No. 10, Cor. 5 to Th. 2). On the other hand, for every $n \in \mathbf{N}$, the quotient $\hat{A}/\hat{m}^n$ is isomorphic to $A/m^n$ (loc. cit., No. 12, Cor. 2 to Prop. 16 and formula (21)), which means that $\hat{A}/\pi \circ \varphi_\alpha(m_\alpha^n)$ $\hat{A}$ is isomorphic to $A \otimes_A (A_\alpha/m_\alpha^n)$; since A is a flat $A_\alpha$-module, the $(A_\alpha/m_\alpha^n)$-module $\hat{A}/\pi \circ \varphi_\alpha(m_\alpha^n)$ $\hat{A}$ is flat for every $n \in \mathbf{N}$. By III, § 5, No. 4, Prop. 2, the $A_\alpha$-module $\hat{A}$ is ideally separated for $m_\alpha$; by loc. cit., No. 2, Th. 1, the $A_\alpha$-module $\hat{A}$ is therefore flat. It follows by passing to the inductive limit that $\hat{A}$ is (faithfully) flat over A (I, § 2, No. 7, Prop. 9), hence that A is noetherian (I, § 3, No. 5, corollary to Prop. 8).

Since the polynomial $P$ is monic, the $A$-module $B$ is free of finite type (A, IV, p. 10). The ring $B/m_A B$ is identified with $\kappa_A[X]/(\overline{P})$, and therefore is a field; hence the ideal $m_A B$ is maximal. Let $q$ be a maximal ideal of $B$; then the ideal $\rho^{-1}(q)$ is maximal (V, § 2, no 1, prop. 1); hence we have $\rho^{-1}(q) = m_A$, whence $q \supset m_A B$ and finally $q = m_A B$. Thus the ring $B$ is local. Lemma 1 follows at once.

a) The ring B is local and one has $m_A B = m_B$.

b) The A-algebra B is faithfully flat.

c) The canonical homomorphism

is bijective.

d) If A is noetherian, the same is true of B and the Hilbert-Samuel series (VIII, § 4, No. 3) of A and B are equal.

Let $(B_\lambda)_{\lambda \in \Lambda}$ be a family of subalgebras of B satisfying conditions a), b) and c) of Def. 2.

Let $\Lambda'$ be the set of indices $\lambda \in \Lambda$ such that, for every $\mu \leq \lambda$ in $\Lambda$, the A-algebra $B_\mu$ is local and faithfully flat, and such that one has $m_{B_\mu} = m_A B_\mu$. Suppose that $\Lambda' \neq \Lambda$ and let $v$ be the smallest element of $\Lambda - \Lambda'$. One has $\alpha \in \Lambda'$, hence $v \neq \alpha$. Now $S_v$ is contained in $\Lambda'$. If $S_v$ has no greatest element, one has $B_v = \bigcup_{\lambda \in S_v} B_\lambda$ and $v$ belongs to $\Lambda'$ by Prop. 1 of No. 1. If $S_v$ has a greatest element $\mu$, one has $\mu \in \Lambda'$ and $B_v$ is an elementary inflation of $B_\mu$: one again has $v \in \Lambda'$ by the remarks following Def. 1, whence a contradiction.

When A is noetherian, one proves analogously that the set $\Lambda''$ of indices $\lambda \in \Lambda$ such that the ring $B_\lambda$ is noetherian is equal to $\Lambda$.

Thus one has $\omega \in \Lambda'$, whence assertions a) and b). When A is noetherian, one has $\omega \in \Lambda''$, hence $B = B_\omega$ is noetherian.

Assertion c) follows from a), b), and Theorem 1 of III, § 5, No. 2. Suppose that A (hence B) is noetherian; since one has

for every $n \in \mathbf{N}$, the Hilbert-Samuel series of A and B are equal.

Let us then put $B = \tilde{B}_{\omega}$ and, for every $\lambda \in \Lambda$, denote by $B_{\lambda}$ the image of $\tilde{B}_{\lambda}$ under the canonical injection $\tilde{B}_{\lambda} \to B$. The family $(B_{\lambda})_{\lambda \in \Lambda}$ satisfies conditions a), b), c) of Definition 2, and B is an inflation of A. On the other hand, the homomorphism $u_{\omega}$ is an A'-isomorphism of $B \otimes_{A} A'$ onto $B'$.

In fact, the homomorphism $\kappa_{A} \to K$ is an inflation (Example 1). Applying Theorem 1 with $A' = \kappa_{A}$ and $B' = K$, one obtains the existence of an inflation B of A and of an A-isomorphism of $B/m_{A}B$ onto K, whence the corollary.

We may regard k as an extension of the residue field $\mathbf{Z}/p\mathbf{Z}$ of the local ring $\mathbf{Z}_{(p)}$. By the corollary to Theorem 1, there exists a local ring B, an inflation of $\mathbf{Z}_{(p)}$, such that $\kappa_{B}$ is isomorphic to k. The local ring $\mathbf{Z}_{(p)}$ is regular and $\{ p \}$ is a system of coordinates of $\mathbf{Z}_{(p)}$. By the corollary to Proposition 2 of No. 2, the ring B is regular and $\{ pl_{B} \}$ is a system of coordinates of B. In other words, B is a discrete valuation ring, with maximal ideal $pB$. The completion C of B is then a $p$-ring of infinite length and the residue field $\kappa_{C}$ is isomorphic to $\kappa_{B}$, hence to k. Moreover, for every integer $n \geq 1$, $C/p^{n}C$ is a $p$-ring of length n, with residue field isomorphic to $\kappa_{C}$, hence to k.

Exercises