§10. Complements on flat modules

For the proofs of the properties stated without proof in nos. (10.1) and (10.2), we refer the reader to Bourbaki, Alg. comm., chap. II and III.

10.1. Relations between flat modules and free modules

10.1.1.

Let $A$ be a ring, $\mathfrak{J}$ an ideal of $A$, $M$ an $A$-module; for every integer $p\ge 0$, one has a canonical homomorphism of $(A/\mathfrak{J})$-modules

  φ_p : (M/𝔍 M) ⊗_{A/𝔍} (𝔍^p/𝔍^{p+1}) → 𝔍^p M/𝔍^{p+1} M               (10.1.1.1)

which is obviously surjective. We shall denote by $gr(A)={\oplus }_{p\ge 0}{\mathfrak{J}}^p/{\mathfrak{J}}^{p+1}$ the graded ring associated to $A$ filtered by the ${\mathfrak{J}}^p$, and by $gr(M)={\oplus }_{p\ge 0}{\mathfrak{J}}^{p}M/{\mathfrak{J}}^{p+1}M$ the graded $gr(A)$-module associated to $M$ filtered by the ${\mathfrak{J}}^{p}M$; we therefore have $g{r}_p(A)={\mathfrak{J}}^p/{\mathfrak{J}}^{p+1}$, $g{r}_p(M)={\mathfrak{J}}^{p}M/{\mathfrak{J}}^{p+1}M$; the ${\varphi }_p$ define a surjective homomorphism of graded $gr(A)$-modules

  φ : gr_0(M) ⊗_{gr_0(A)} gr(A) → gr(M).                                 (10.1.1.2)

10.1.2.

Suppose that one of the following hypotheses holds:

(i) $\mathfrak{J}$ is nilpotent;

(ii) $A$ is Noetherian, $\mathfrak{J}$ is contained in the radical of $A$, and $M$ is of finite type.

Then the following properties are equivalent:

a) $M$ is a free $A$-module.

b) $M/\mathfrak{J}M=M{\otimes }_A(A/\mathfrak{J})$ is a free $(A/\mathfrak{J})$-module, and $To{r}_1^A(M,A/\mathfrak{J})=0$.

c) $M/\mathfrak{J}M$ is a free $(A/\mathfrak{J})$-module and the canonical homomorphism (10.1.1.2) is injective (hence bijective).

10.1.3.

Suppose that $A/\mathfrak{J}$ is a field (in other words, that $\mathfrak{J}$ is maximal), and that one of the hypotheses (i), (ii) of (10.1.2) holds. Then the following properties are equivalent:

a) $M$ is a free $A$-module.

b) $M$ is a projective $A$-module.

c) $M$ is a flat $A$-module.

d) $To{r}_1^A(M,A/\mathfrak{J})=0$.

e) The canonical homomorphism (10.1.1.2) is bijective.

This result will apply in particular in the following two cases:

(i) $M$ is an arbitrary module over a local ring $A$ whose maximal ideal $\mathfrak{J}$ is nilpotent (for example an Artinian local ring).

(ii) $M$ is a module of finite type over a Noetherian local ring.

10.2. Local criteria of flatness

10.2.1.

The hypotheses and notation being those of (10.1.1), consider the following conditions:

a) $M$ is a flat $A$-module.

b) $M/\mathfrak{J}M$ is a flat $(A/\mathfrak{J})$-module and $To{r}_1^A(M,A/\mathfrak{J})=0$.

c) $M/\mathfrak{J}M$ is a flat $(A/\mathfrak{J})$-module and the canonical homomorphism (10.1.1.2) is bijective.

d) For every $n\ge 1$, $M/{\mathfrak{J}}^{n}M$ is a flat $(A/{\mathfrak{J}}^n)$-module.

One then has the implications

  a) ⟹ b) ⟹ c) ⟹ d)

and if $\mathfrak{J}$ is nilpotent, the four conditions a), b), c), d) are equivalent. The same holds if $A$ is Noetherian and if moreover $M$ is ideally separated, that is, if for every ideal $\mathfrak{a}$ of $A$, the $A$-module $\mathfrak{a}{\otimes }_{A}M$ is separated for the $\mathfrak{J}$-preadic topology.

10.2.2.

Let $A$ be a Noetherian ring, $B$ a commutative Noetherian $A$-algebra, $\mathfrak{J}$ an ideal of $A$ such that $\mathfrak{J}B$ is contained in the radical of $B$, $M$ a $B$-module of finite type. Then, when $M$ is considered as an $A$-module, the conditions a), b), c), d) of (10.2.1) are equivalent.

This result applies above all when $A$ and $B$ are Noetherian local rings, the homomorphism $A\to B$ a local homomorphism. More particularly, if $\mathfrak{J}$ is then the maximal ideal of $A$, one may, in conditions b) and c), drop the hypothesis that $M/\mathfrak{J}M$ is flat, which is automatically satisfied, and condition d) means that the modules $M/{\mathfrak{J}}^{n}M$ are free over the $A/{\mathfrak{J}}^n$.

10.2.3.

The hypotheses on $A$, $B$, $\mathfrak{J}$, $M$ being those formulated at the beginning of (10.2.2), let Â be the Hausdorff completion of $A$ for the $\mathfrak{J}$-preadic topology, $\hat{M}$ the Hausdorff completion of $M$ for the $\mathfrak{J}B$-preadic topology. Then, for $M$ to be a flat $A$-module, it is necessary and sufficient that $\hat{M}$ be a flat Â-module.

10.2.4.

Let $\rho :A\to B$ be a local homomorphism of Noetherian local rings, $k$ the residue field of $A$, $M$, $N$ two $B$-modules of finite type, $N$ being assumed to be $A$-flat. Let $u:M\to N$ be a $B$-homomorphism. Then the following conditions are equivalent:

a) $u$ is injective and $Coker(u)$ is a flat $A$-module.

b) $u\otimes 1:M{\otimes }_{A}k\to N{\otimes }_{A}k$ is injective.

10.2.5.

Let $\rho :A\to B$, $\sigma :B\to C$ be local homomorphisms of Noetherian local rings, $k$ the residue field of $A$, $M$ a $C$-module of finite type. Suppose that $B$ is a flat $A$-module. Then the following conditions are equivalent:

a) $M$ is a flat $B$-module.

b) $M$ is a flat $A$-module, and $M{\otimes }_{A}k$ is a flat $(B{\otimes }_{A}k)$-module.

Proposition (10.2.6).

Let $A$, $B$ be two Noetherian local rings, $\rho :A\to B$ a local homomorphism, $\mathfrak{J}$ an ideal of $B$ contained in the maximal ideal, $M$ a $B$-module of finite type. Suppose that for every $n\ge 0$, $M_n=M/{\mathfrak{J}}^{n+1}M$ is a flat $A$-module. Then $M$ is a flat $A$-module.

Proof. We must prove that for every injective homomorphism $u:N^{\prime }\to N$ of $A$-modules of finite type, $v=1\otimes u:M{\otimes }_A{N}^{\prime }\to M{\otimes }_{A}N$ is injective. Now, $M{\otimes }_A{N}^{\prime }$ and $M{\otimes }_{A}N$ are $B$-modules of finite type, hence separated for the $\mathfrak{J}$-preadic topology $(0_I,\mathrm{7.3.5})$; it therefore suffices to prove that the homomorphism $\hat{v}:(M{\otimes }_A{N}^{\prime }{)}^{\wedge }\to (M{\otimes }_{A}N{)}^{\wedge }$ between Hausdorff completions is injective. Now, one has $\hat{v}=\lim v_n$, where $v_n$ is the homomorphism $1\otimes u:M_n{\otimes }_A{N}^{\prime }\to M_n{\otimes }_{A}N$; since by hypothesis $M_n$ is $A$-flat, $v_n$ is injective for every $n$, and so the same holds for $\hat{v}$, the functor lim being left exact.

Corollary (10.2.7).

Let $A$ be a Noetherian ring, $B$ a Noetherian local ring, $\rho :A\to B$ a homomorphism, $f$ an element of the maximal ideal of $B$, $M$ a $B$-module of finite type. Suppose that the homothety $f_M:x\mapsto fx$ of $M$ is injective and that $M/fM$ is a flat $A$-module. Then $M$ is a flat $A$-module.

Proof. Set $M_i=f^{i}M$ for $i\ge 0$; since $f_M$ is injective, $M_i/M_{i+1}$ is isomorphic to $M/fM$, hence $A$-flat for every $i\ge 0$; from the exact sequence

$$0\to M_i/M_{i+1}\to M/M_{i+1}\to M/M_i\to 0$$

one deduces by induction on $i$ that $M/M_i$ is $A$-flat for every $i\ge 0$ $(0_I,\mathrm{6.1.2})$; one can therefore apply (10.2.6). One may also argue directly as follows: for every $A$-module $N$ of finite type, $M{\otimes }_{A}N$ is a $B$-module of finite type; since $f$ belongs to the radical $\mathfrak{n}$ of $B$, the (f)-adic topology on $M{\otimes }_{A}N$ is finer than the $\mathfrak{n}$-adic topology, and the latter is known to be separated $(0_I,\mathrm{7.3.5})$.

Moreover, since $M/M_i$ is $A$-flat, one has $f^i(M{\otimes }_{A}N)=Im(M_i{\otimes }_{A}N\to M{\otimes }_{A}N)=Ker(M{\otimes }_{A}N\to (M/M_i){\otimes }_{A}N)$ $(0_I,\mathrm{6.1.2})$. Let then $N$ be an $A$-module of finite type, $N^{\prime }$ a submodule of $N$, $j:N^{\prime }\to N$ the canonical injection; in the commutative diagram

  M ⊗_A N'    →    (M/M_i) ⊗_A N'
     │                  │
     │ 1_M ⊗ j           │ 1_{M/M_i} ⊗ j
     ↓                  ↓
  M ⊗_A N     →    (M/M_i) ⊗_A N

$1_{M/M_i}\otimes j$ is injective since $M/M_i$ is $A$-flat; one concludes that

  Ker(M ⊗_A N' → M ⊗_A N) ⊂ Ker(M ⊗_A N' → (M/M_i) ⊗_A N')

whatever the value of $i$; since the intersection of the right-hand sides is reduced to 0 as seen above, the same holds for the left-hand side, and consequently $M$ is $A$-flat.

Proposition (10.2.8).

Let $A$ be a reduced Noetherian ring, $M$ an $A$-module of finite type. Suppose that for every $A$-algebra $B$ which is a discrete valuation ring, $M{\otimes }_{A}B$ is a flat $B$-module (hence free (10.1.3)). Then $M$ is a flat $A$-module.

Proof. It is known that for $M$ to be flat, it is necessary and sufficient that for every maximal ideal $\mathfrak{m}$ of $A$, $M_{\mathfrak{m}}$ be a flat $A_{\mathfrak{m}}$-module $(0_I,\mathrm{6.3.3})$; one may therefore restrict to the case where $A$ is local $(0_I,\mathrm{1.2.8})$. Let then $\mathfrak{m}$ be the maximal ideal of $A$, ${\mathfrak{p}}_i$ $(1\le i\le r)$ its minimal prime ideals, $k=A/\mathfrak{m}$ the residue field. It is known (II, 7.1.7) that for each $i$ there exists a discrete valuation ring $B_i$ having the same field of fractions $K_i$ as the integral ring $A/{\mathfrak{p}}_i$, and dominating the latter. Set $M_i=M{\otimes }_A{B}_i$. By hypothesis, $M_i$ is free over $B_i$, so one has, denoting by $k_i$ the residue field of $B_i$,

  rg_{k_i}(M_i ⊗_{B_i} k_i) = rg_{K_i}(M_i ⊗_{B_i} K_i).                 (10.2.8.1)

But it is clear that the composite homomorphism $A\to A/{\mathfrak{p}}_i\to B_i$ is local, so $k_i$ is an extension of $k$, and one has $M_i{\otimes }_{B_i}{k}_i=M{\otimes }_A{k}_i=(M{\otimes }_{A}k){\otimes }_k{k}_i$, and on the other hand $M_i{\otimes }_{B_i}{K}_i=M{\otimes }_A{K}_i$. The equality (10.2.8.1) therefore yields

  rg_k(M ⊗_A k) = rg_{K_i}(M ⊗_A K_i)                  for 1 ≤ i ≤ r

and since $A$ is reduced, this condition is known to imply that $M$ is a free $A$-module (Bourbaki, Alg. comm., chap. II, § 3, no. 2, prop. 7).

10.3. Existence of flat extensions of local rings

Proposition (10.3.1).

Let $A$ be a Noetherian local ring, $\mathfrak{J}$ its maximal ideal, $k=A/\mathfrak{J}$ its residue field. Let $K$ be an extension of the field $k$. There exists a local homomorphism from $A$ into a Noetherian local ring $B$, such that $B/\mathfrak{J}B$ is $k$-isomorphic to $K$, and such that $B$ is a flat $A$-module.

Proof. We shall prove this proposition in several steps.

10.3.1.1.

Suppose first that $K=k(T)$, where $T$ is an indeterminate. In the polynomial ring $A^{\prime }=A[T]$, consider the prime ideal ${\mathfrak{J}}^{\prime }=\mathfrak{J}{A}^{\prime }$ formed by the

polynomials with coefficients in the ideal $\mathfrak{J}$; it is clear that $A^{\prime }/{\mathfrak{J}}^{\prime }$ is canonically isomorphic to k[T]. Let us show that the ring of fractions $B=A_{{\mathfrak{J}}^{\prime }}^{\prime }$ answers the question; it is evidently a Noetherian local ring whose maximal ideal is $\mathfrak{L}=\mathfrak{J}B$. Furthermore, $B/\mathfrak{L}=(A^{\prime }/{\mathfrak{J}}^{\prime }{)}_{{\mathfrak{J}}^{\prime }}=(k[T]{)}_{(0)}$ is nothing other than the field of fractions $K$ of k[T]. Finally, $B$ is a flat $A^{\prime }$-module and $A^{\prime }$ a free $A$-module, hence $B$ is a flat $A$-module $(0_I,\mathrm{6.2.1})$.

10.3.1.2.

Suppose next that $K=k(t)=k[t]$, where $t$ is algebraic over $k$; let $f\in k[T]$ be the minimal polynomial of $t$; there exists a unitary polynomial $F\in A[T]$ whose canonical image in k[T] is $f$. Set again $A^{\prime }=A[T]$, and let ${\mathfrak{J}}^{\prime }$ be the ideal $\mathfrak{J}{A}^{\prime }+(F)$ in $A^{\prime }$. We are going to see that the quotient ring $B=A^{\prime }/(F)$ answers the question this time. First of all, it is clear that $B$ is a free $A$-module, hence flat. The ring $A^{\prime }/{\mathfrak{J}}^{\prime }$ is isomorphic to $(A^{\prime }/\mathfrak{J}{A}^{\prime })/((\mathfrak{J}{A}^{\prime }+(F))/\mathfrak{J}{A}^{\prime })=k[T]/(f)=K$; the image $\mathfrak{L}$ of ${\mathfrak{J}}^{\prime }$ in $B$ is therefore maximal and one obviously has $\mathfrak{L}=\mathfrak{J}B$. Finally, $B$ is a semi-local ring, since it is an $A$-module of finite type (Bourbaki, Alg. comm., chap. IV, § 2, no. 5, cor. 3 of prop. 9), and its maximal ideals are in bijective correspondence with those of $B/\mathfrak{J}B$ ([13], vol. I, p. 259); what precedes therefore proves that $B$ is a local ring.

Lemma (10.3.1.3).

Let $(A_{\lambda },f_{\mu \lambda })$ be a filtered inductive system of local rings, such that the $f_{\mu \lambda }$ are local homomorphisms; let ${\mathfrak{m}}_{\lambda }$ be the maximal ideal of $A_{\lambda }$, and let $K_{\lambda }=A_{\lambda }/{\mathfrak{m}}_{\lambda }$. Then $A^{\prime }=\lim A_{\lambda }$ is a local ring whose maximal ideal is ${\mathfrak{m}}^{\prime }=\lim {\mathfrak{m}}_{\lambda }$, and whose residue field is $K=\lim K_{\lambda }$. Furthermore, if ${\mathfrak{m}}_{\mu }={\mathfrak{m}}_{\lambda }{A}_{\mu }$ for $\lambda <\mu$, then ${\mathfrak{m}}^{\prime }={\mathfrak{m}}_{\lambda }{A}^{\prime }$ for every $\lambda$. If, in addition, $A_{\mu }$ is a flat $A_{\lambda }$-module for $\lambda <\mu$, and if all the $A_{\lambda }$ are Noetherian, then $A^{\prime }$ is Noetherian and is a flat $A_{\lambda }$-module for every $\lambda$.

Proof. Since $f_{\mu \lambda }({\mathfrak{m}}_{\lambda })\subset {\mathfrak{m}}_{\mu }$ for $\lambda <\mu$ by hypothesis, the ${\mathfrak{m}}_{\lambda }$ form an inductive system, and its limit ${\mathfrak{m}}^{\prime }$ is obviously an ideal of $A^{\prime }$. Furthermore, if $x^{\prime }\notin {\mathfrak{m}}^{\prime }$, there exists $\lambda$ such that $x^{\prime }=f_{\lambda }(x_{\lambda })$ for some $x_{\lambda }\in A_{\lambda }$ ($f_{\lambda }:A_{\lambda }\to A^{\prime }$ denoting the canonical homomorphism); since $x^{\prime }\notin {\mathfrak{m}}^{\prime }$, we necessarily have $x_{\lambda }\notin {\mathfrak{m}}_{\lambda }$, so $x_{\lambda }$ admits an inverse $y_{\lambda }$ in $A_{\lambda }$, and $y^{\prime }=f_{\lambda }(y_{\lambda })$ is the inverse of $x^{\prime }$ in $A^{\prime }$, which proves that $A^{\prime }$ is a local ring with maximal ideal ${\mathfrak{m}}^{\prime }$; the assertion concerning $K$ follows immediately from the fact that lim is an exact functor. The hypothesis ${\mathfrak{m}}_{\mu }={\mathfrak{m}}_{\lambda }{A}_{\mu }$ means that the canonical map ${\mathfrak{m}}_{\lambda }{\otimes }_{A_{\lambda }}{A}_{\mu }\to {\mathfrak{m}}_{\mu }$ is surjective; the relation ${\mathfrak{m}}^{\prime }={\mathfrak{m}}_{\lambda }{A}^{\prime }$ thus follows again from the exactness of the functor lim and the fact that it commutes with the tensor product.

Suppose now that for $\lambda <\mu$ one has ${\mathfrak{m}}_{\mu }={\mathfrak{m}}_{\lambda }{A}_{\mu }$ and that $A_{\mu }$ is a flat $A_{\lambda }$-module. Then $A^{\prime }$ is a flat $A_{\lambda }$-module for every $\lambda$, by virtue of $(0_I,\mathrm{6.2.3})$; since $A^{\prime }$ and $A_{\lambda }$ are local rings and ${\mathfrak{m}}^{\prime }={\mathfrak{m}}_{\lambda }{A}^{\prime }$, $A^{\prime }$ is even a faithfully flat $A_{\lambda }$-module $(0_I,\mathrm{6.6.2})$. Suppose finally, in addition, that the $A_{\lambda }$ are Noetherian; the ${\mathfrak{m}}_{\lambda }$-preadic topologies are then separated $(0_I,\mathrm{7.3.5})$; let us show that it follows first that on $A^{\prime }$ the ${\mathfrak{m}}^{\prime }$-adic topology is separated. Indeed, if $x^{\prime }\in A^{\prime }$ belongs to every ${\mathfrak{m}}^{\prime n}$ $(n>0)$, it is the image of some $x_{\mu }\in A_{\mu }$ for a certain index $\mu$, and since the inverse image in $A_{\mu }$ of ${\mathfrak{m}}^{\prime n}={\mathfrak{m}}_{\mu }^n{A}^{\prime }$ is ${\mathfrak{m}}_{\mu }^n$ $(0_I,\mathrm{6.6.1})$, $x_{\mu }$ belongs to every ${\mathfrak{m}}_{\mu }^n$, hence $x_{\mu }=0$ by hypothesis, and consequently $x^{\prime }=0$. Denote by Â' the completion of $A^{\prime }$ for the ${\mathfrak{m}}^{\prime }$-adic topology; what precedes shows that $A^{\prime }\subset {\hat{A}}^{\prime }$. We shall show that Â' is Noetherian and $A_{\lambda }$-flat for every $\lambda$; it will

follow that Â' is $A^{\prime }$-flat $(0_I,\mathrm{6.2.3})$, and since ${\mathfrak{m}}^{\prime }{\hat{A}}^{\prime }\ne {\hat{A}}^{\prime }$, Â' is a faithfully flat $A^{\prime }$-module $(0_I,\mathrm{6.6.2})$, whence one will finally conclude that $A^{\prime }$ is Noetherian $(0_I,\mathrm{6.5.2})$, which will complete the proof of the lemma.

One has ${\hat{A}}^{\prime }={\lim }_n{A}^{\prime }/{\mathfrak{m}}^{\prime n}$; on account of the fact that $A^{\prime }$ is $A_{\lambda }$-flat, one has

  𝔪'^n/𝔪'^{n+1} = (𝔪_λ^n/𝔪_λ^{n+1}) ⊗_{A_λ} A'
                = (𝔪_λ^n/𝔪_λ^{n+1}) ⊗_{K_λ} (K_λ ⊗_{A_λ} A')
                = (𝔪_λ^n/𝔪_λ^{n+1}) ⊗_{K_λ} K;

since ${\mathfrak{m}}_{\lambda }^n/{\mathfrak{m}}_{\lambda }^{n+1}$ is a $K_{\lambda }$-vector space of finite dimension, ${\mathfrak{m}}^{\prime n}/{\mathfrak{m}}^{\prime n+1}$ is a $K$-vector space of finite dimension for every $n\ge 0$. It therefore follows from $(0_I,\mathrm{7.2.12})$ and $(0_I,\mathrm{7.2.8})$ that Â' is Noetherian. We further know that the maximal ideal of Â' is ${\mathfrak{m}}^{\prime }{\hat{A}}^{\prime }$ and that ${\hat{A}}^{\prime }/{\mathfrak{m}}^{\prime n}{\hat{A}}^{\prime }$ is isomorphic to $A^{\prime }/{\mathfrak{m}}^{\prime n}$; since $A^{\prime }/{\mathfrak{m}}^{\prime n}=(A_{\lambda }/{\mathfrak{m}}_{\lambda }^n){\otimes }_{A_{\lambda }}{A}^{\prime }$, $A^{\prime }/{\mathfrak{m}}^{\prime n}$ is a flat $(A_{\lambda }/{\mathfrak{m}}_{\lambda }^n)$-module $(0_I,\mathrm{6.2.1})$; criterion (10.2.2) is therefore applicable to the Noetherian $A_{\lambda }$-algebra Â', and shows that Â' is $A_{\lambda }$-flat.

10.3.1.4.

We now take up the general case. There exist an ordinal $\gamma$ and, for every ordinal $\lambda \le \gamma$, a subfield $k_{\lambda }$ of $K$ containing $k$, such that: 1° For every $\lambda <\gamma$, $k_{\lambda +1}$ is an extension of $k_{\lambda }$ generated by a single element; 2° For every ordinal $\mu$ without predecessor, $k_{\mu }={\bigcup }_{\lambda <\mu }{k}_{\lambda }$; 3° $K=k_{\gamma }$. Indeed, it suffices to consider a bijection $\xi \mapsto t_{\xi }$ of the set of ordinals $\xi \le \beta$ (for a suitable $\beta$) onto $K$, to define $k_{\lambda }$ by transfinite induction (for $\lambda \le \beta$) as the union of the $k_{\mu }$ for $\mu <\lambda$ if $\lambda$ has no predecessor, and, if $\lambda =\nu +1$, as $k_{\nu }(t_{\xi })$, where $\xi$ is the smallest ordinal such that $t_{\xi }\notin k_{\nu }$; $\gamma$ is then by definition the smallest ordinal $\le \beta$ such that $k_{\gamma }=K$.

This being so, we shall define, by transfinite recursion, a family of Noetherian local rings $A_{\lambda }$ for $\lambda \le \gamma$, and local homomorphisms $f_{\mu \lambda }:A_{\lambda }\to A_{\mu }$ for $\lambda \le \mu$, satisfying the following conditions:

(i) $(A_{\lambda },f_{\mu \lambda })$ is an inductive system and $A_0=A$.

(ii) For every $\lambda$, one has a $k$-isomorphism $A_{\lambda }/\mathfrak{J}{A}_{\lambda }\stackrel{\sim }{\to }{k}_{\lambda }$.

(iii) For $\lambda \le \mu$, $A_{\mu }$ is a flat $A_{\lambda }$-module.

Suppose then that the $A_{\lambda }$ and the $f_{\mu \lambda }$ are defined for $\lambda <\mu <\xi$, and suppose first that $\xi =\zeta +1$, so that $k_{\xi }=k_{\zeta }(t)$. If $t$ is transcendental over $k_{\zeta }$, one defines $A_{\xi }$ following the procedure of (10.3.1.1) as equal to $(A_{\zeta }[T]{)}_{\mathfrak{J}{A}_{\zeta }[T]}$; $f_{\xi \zeta }$ is the canonical map, and for $\lambda <\zeta$ one takes $f_{\xi \lambda }=f_{\xi \zeta }\circ f_{\zeta \lambda }$; the verification of conditions (i) to (iii) is then immediate, in view of what was proved in (10.3.1.1). Suppose next that $t$ is algebraic, and let $h$ be its minimal polynomial in $k_{\zeta }[T]$, $H$ a unitary polynomial of $A_{\zeta }[T]$ whose image in $k_{\zeta }[T]$ is $h$; one then takes $A_{\xi }$ equal to $A_{\zeta }[T]/(H)$, the $f_{\xi \lambda }$ being defined as before; the verification of conditions (i) to (iii) then follows from what was seen in (10.3.1.2).

Suppose now that $\xi$ has no predecessor; one then takes for $A_{\xi }$ the inductive limit of the inductive system of local rings $(A_{\lambda },f_{\mu \lambda })$ for $\lambda <\xi$; $f_{\xi \lambda }$ is defined as the canonical map for $\lambda <\xi$. The fact that $A_{\xi }$ is local Noetherian, that the $f_{\xi \lambda }$ are local homomorphisms, and conditions (i) to (iii) for $\lambda \le \xi$

then follow from the inductive hypothesis and from Lemma (10.3.1.3). This construction being done, it is clear that the ring $B=A_{\gamma }$ satisfies the statement of (10.3.1).

One should note that by virtue of (10.2.1, c)), one has a canonical isomorphism

  gr(A) ⊗_k K ⥲ gr(B).                                                  (10.3.1.5)

On the other hand, one may replace $B$ by its $\mathfrak{J}B$-adic completion $\hat{B}$ without changing the conclusions of (10.3.1), since $\hat{B}$ is a flat $B$-module $(0_I,\mathrm{7.3.3})$, hence a flat $A$-module $(0_I,\mathrm{6.2.1})$.

We have moreover proved the

Corollary (10.3.2).

If $K$ is an extension of finite degree, one may suppose that $B$ is a finite $A$-algebra.