Éléments de Géométrie Algébrique — English

§19. Formally smooth algebras and Cohen rings

19.0. Introduction

(19.0.1) In Chapter IV we shall introduce and study, among other things, an important class of morphisms of preschemes, the smooth morphisms.¹ One of their fundamental properties (which, together with a finiteness condition, characterizes them) is a property of lifting morphisms: if $f:X\to Y$ is a smooth morphism, $g:Y^{\prime }\to Y$ a morphism, then for every morphism $h:Y^{\prime \prime }\to Y^{\prime }$ making Y'' a prescheme "little different" from $Y^{\prime }$, every $Y$-morphism $Y^{\prime \prime }\to X$ factors as $Y^{\prime \prime }\stackrel{h}{\to }{Y}^{\prime }\to X$. More precisely, restricting ourselves to the case where $Y=\mathrm{Spec}(A)$, $X=\mathrm{Spec}(B)$ are affine, $B$ is then called a formally smooth $A$-algebra if, for every $A$-algebra $C$ and every nilpotent ideal $\mathfrak{J}$ of $C$, every $A$-homomorphism $B\to C/\mathfrak{J}$ lifts to $B\to C\to C/\mathfrak{J}$. In other words, the map

                              Hom_A(B, C) → Hom_A(B, C/𝔍)

is surjective. In many applications $X$ will appear as an object representing a representable contravariant functor $Y^{\prime }\mapsto F(Y^{\prime })$ from the category of $Y$-preschemes into that of sets, so that one has $(0_{III},\mathrm{8.1.8})$ $F(Y^{\prime })\cong {\mathrm{Hom}}_Y(Y^{\prime },X)$. In the affine case, if one sets $F^{\circ }(C)=F(\mathrm{Spec}(C))$, the verification that, under the conditions above, $F^{\circ }(C)\to F^{\circ }(C/\mathfrak{J})$ is surjective (which can be done even without knowing that $F$ is representable) will establish that $f$ is smooth, provided the additional finiteness condition is satisfied.

(19.0.2) For topological algebras over a topological ring $A$, there is an analogous notion of formally smooth algebra which we shall not make precise here (cf. (19.3.1)). The study of these notions is first carried out from an elementary point of view in §19, then, by means of the properties of differentials which will be developed in §§20 and 21, more delicate theorems will be proved in §22. We summarize here the principal results on formally smooth algebras:

I. — Let $A$, $B$ be two Noetherian local rings, $\varphi :A\to B$ a local homomorphism, $k$ the residue field of $A$, and let $B_0=B{\otimes }_{A}k$; equip $A$, $B$ with their preadic topologies and $k$ with the discrete topology. Then, for $B$ to be a formally smooth $A$-algebra, it is necessary and sufficient that $B$ be a flat $A$-module and that B_0 be a formally smooth $k$-algebra (19.7.1). This theorem thus reduces formal smoothness, for Noetherian local rings, to the same question for Noetherian local rings which are algebras over a field.

II. — Let $k$ be a field, $A$ a Noetherian local ring which is a $k$-algebra. For $A$ to be formally smooth, it is necessary and sufficient that $A$ be geometrically regular over $k$, that is, that for every finite extension $k^{\prime }$ of $k$, the semi-local ring $A{\otimes }_k{k}^{\prime }$ be

regular (22.5.8); the completion Â of $A$ is then a ring isomorphic to a formal power series ring $K[[T_1,\cdots ,T_n]]$ (19.6.5). Moreover, the structure of $k$-algebra of Â, when $A$ is assumed to be complete and formally smooth over $k$, is entirely determined by the residue field $K$ of $A$ and by the dimension of $A$; the latter can moreover be arbitrary provided it satisfies the inequality $\dim (A)⩾r{g}_K({{\rm Y}}_{K/k})$, where ${{\rm Y}}_{K/k}$ is the "imperfection module" of $K$ (22.2.6).

In particular, for an extension $K$ of $k$ to be formally smooth, it is necessary and sufficient that $K$ be a separable extension of $k$ (19.6.1).

III. — Let $A$ be a Noetherian local ring, $\mathfrak{J}$ an ideal of $A$ distinct from $A$, $A_0=A/\mathfrak{J}$, B_0 a complete Noetherian local ring, $A_0\to B_0$ a local homomorphism making B_0 a formally smooth A_0-algebra. Then there exists a complete Noetherian local ring $B$, a local homomorphism $A\to B$ making $B$ a flat $A$-module, and an A_0-isomorphism $B{\otimes }_A{A}_0\cong B_0$ (so that, by I, $B$ is a formally smooth $A$-algebra); furthermore $B$ is determined by these properties up to isomorphism (19.7.2). This theorem contains in particular the theorems of Cohen on the structure of complete Noetherian local rings (19.8), which will play an important role in §§6 and 7 of Chapter IV.

IV. — The interest of the study of formally smooth Noetherian local rings over another arises from the following "pointwise" characterization of smooth morphisms: if $X$ and $Y$ are locally Noetherian preschemes, $f:X\to Y$ a morphism locally of finite type, then, for $f$ to be smooth, it is necessary and sufficient that for every $x\in X$, the ring ${\mathcal{O}}_x$ be formally smooth over ${\mathcal{O}}_{f(x)}$. In particular, if $Y=\mathrm{Spec}(k)$, where $k$ is a perfect field, to say that $f$ is smooth is equivalent to saying that $X$ is a regular prescheme.

V. — Finally, we shall see in §§20, 21, and 22 that the notion of formally smooth algebra arises naturally in the theory of Kähler differentials, the two theories illuminating each other.

(19.0.3) In all this section and the following ones, the topological rings and modules will be assumed to be linearly topologized $(0_I,\mathrm{7.1.1})$; the topological rings considered will be assumed commutative, except when explicitly stated otherwise. Recall that if $A$ and $B$ are two topological rings, $\rho :A\to B$ a ring homomorphism defining on $B$ a structure of $A$-algebra, one says that $B$ is a topological $A$-algebra if $\rho$ is continuous for the topologies in question.

To abbreviate, in a topological ring $A$ (resp. a topological $A$-module $M$), we shall say "fundamental system of open ideals (resp. submodules)" instead of "fundamental system of neighbourhoods of 0 formed of ideals (resp. submodules)".

Given a topological ring $A$ and an $A$-module $M$, the sets $\mathfrak{J}M$, where $\mathfrak{J}$ runs through a fundamental system of open ideals, form a fundamental system of open submodules for a topology on $M$ making $M$ a topological $A$-module, which is said to be deduced from the topology of $A$.

Let $M$ be a topological $A$-module whose topology is coarser than the topology deduced from that of $A$; then, if $N$ is an open submodule of $M$, the discrete $A$-module $M/N$ is annulled by an open ideal of $A$, for by hypothesis there exists such an ideal $\mathfrak{K}$ with $\mathfrak{K}\cdot M\subset N$.

If $M$ and $N$ are two topological $A$-modules whose topologies are both deduced from that of $A$, then every $A$-homomorphism $u:M\to N$ is continuous, for, for every neighbourhood $V$ of 0 in $N$, there exists by hypothesis an open ideal $\mathfrak{J}$ of $A$ such that $\mathfrak{J}N\subset V$, and one therefore has $u(\mathfrak{J}M)\subset \mathfrak{J}N\subset V$.

When $B$ is a topological $A$-algebra, the topology on $B$ deduced from that of $A$ is finer than the given topology, for, for every open ideal $\mathfrak{K}$ of $B$, there is by hypothesis an open ideal $\mathfrak{J}$ of $A$ such that $\mathfrak{J}B\subset \mathfrak{K}$.

19.1. Formal epimorphisms and monomorphisms

Proposition (19.1.1).

Let $A$ be a topological ring, $M$, $N$ two topological $A$-modules, $(W_{\lambda })$ a fundamental system of open submodules of $N$, $u:M\to N$ a continuous $A$-homomorphism, $\hat{M}$, $\hat{N}$ the separated completions of $M$ and $N$, $\hat{u}:\hat{M}\to \hat{N}$ the continuous extension of $u$ to the separated completions.

(i) The following conditions are equivalent:

a) $u(M)$ is dense in $N$.

a') $\hat{u}(\hat{M})$ is dense in $\hat{N}$.

a") For every $\lambda$, the composite homomorphism $M\stackrel{u}{\to }N\to N/W_{\lambda }$ is surjective.

(ii) The following conditions are equivalent:

b) The inverse image by $u$ of the topology of $N$ is equal to the topology of $M$.

b') û is an isomorphism of the topological Â-module $\hat{M}$ onto the topological sub-Â-module $\hat{u}(\hat{M})$ of $\hat{N}$ (which is necessarily closed).

b") The $u^{-1}(W_{\lambda })$ form a fundamental system of neighbourhoods of 0 in $M$.

This follows immediately from the definition of the separated completions and from the fact that $N/W_{\lambda }$ is discrete.

(19.1.2) When the equivalent conditions of (i) (resp. (ii)) in (19.1.1) are satisfied, one says that $u$ is a formal epimorphism (resp. a formal monomorphism). One says that $u$ is a formal bimorphism if it is at once a formal monomorphism and a formal epimorphism; this amounts, by virtue of (19.1.1), to saying that û is an isomorphism of the topological Â-module $\hat{M}$ onto the topological Â-module $\hat{N}$.

Proposition (19.1.3).

Let $A$ be a topological ring, $M$, $N$ two topological $A$-modules, $u:M\to N$ a continuous $A$-homomorphism. Assume that there exist two fundamental systems of open submodules $(V_{\lambda })$, $(W_{\lambda })$ in $M$ and $N$ respectively, having the same set of indices and such that $u(V_{\lambda })\subset W_{\lambda }$ for every $\lambda$; let $u_{\lambda }:M/V_{\lambda }\to N/W_{\lambda }$ be the homomorphism deduced from $u$ by passage to the quotients. Then:

(i) For $u$ to be a formal epimorphism, it is necessary and sufficient that, for every $\lambda$, $u_{\lambda }$ be surjective.

(ii) For $u$ to be a formal monomorphism, it is necessary and sufficient that, for every $\lambda$, there exist a $\mu$ such that $V_{\mu }\subset V_{\lambda }$ and that $Ker(u_{\mu })$ be contained in the kernel $V_{\lambda }/V_{\mu }$ of the canonical map $M/V_{\mu }\to M/V_{\lambda }$.

(iii) For $u$ to be a formal bimorphism, it is necessary and sufficient that, for every $\lambda$, $u_{\lambda }$ be surjective and that there exist a $\mu$ such that $V_{\mu }\subset V_{\lambda }$ and that the canonical homomorphism $M/V_{\mu }\to M/V_{\lambda }$

factor as $M/V_{\mu }\stackrel{u_{\mu }}{\to }N/W_{\mu }\to M/V_{\lambda }$ (where the homomorphism $N/W_{\mu }\to M/V_{\lambda }$ is necessarily unique).

Assertion (i) is immediate; on the other hand, $Ker(u_{\mu })=u^{-1}(W_{\mu })/V_{\mu }$, and the kernel of the canonical map $M/V_{\mu }\to M/V_{\lambda }$ is $V_{\lambda }/V_{\mu }$; the condition in (ii) thus amounts to saying that $u^{-1}(W_{\mu })\subset V_{\lambda }$, and assertion (ii) follows at once; finally, when $u_{\mu }$ is surjective, it amounts to the same to say that $Ker(u_{\mu })$ is contained in $V_{\lambda }/V_{\mu }$, or to say that $M/V_{\mu }\to M/V_{\lambda }$ factors as $v\circ u_{\mu }$, where $v$ is a homomorphism $N/W_{\mu }\to M/V_{\lambda }$, since $N/W_{\mu }$ is then identified with $(M/V_{\mu })/Ker(u_{\mu })$.

Corollary (19.1.4).

Let $A$ be a topological ring, $M$, $N$ two topological $A$-modules whose topologies are deduced from that of $A$, $u:M\to N$ a formal epimorphism. Let $({\mathfrak{J}}_{\lambda })$ be a fundamental system of open ideals of $A$. For $u$ to be a formal bimorphism, it is necessary and sufficient that for every $\lambda$, the homomorphism $u_{\lambda }:M/{\mathfrak{J}}_{\lambda }M\to N/{\mathfrak{J}}_{\lambda }N$ deduced from $u$ by passage to the quotients be bijective.

Indeed, one has $u({\mathfrak{J}}_{\lambda }M)\subset {\mathfrak{J}}_{\lambda }N$ and one may apply the criterion (19.1.3, (iii)); but if one has a factorization $M/{\mathfrak{J}}_{\lambda }M\stackrel{u_{\mu }}{\to }N/{\mathfrak{J}}_{\mu }N\stackrel{v}{\to }M/{\mathfrak{J}}_{\lambda }M$, the image of ${\mathfrak{J}}_{\lambda }M/{\mathfrak{J}}_{\mu }M$ under $u_{\mu }$ is ${\mathfrak{J}}_{\lambda }N/{\mathfrak{J}}_{\mu }N$, hence the image of ${\mathfrak{J}}_{\lambda }N/{\mathfrak{J}}_{\mu }N$ under $v$ is 0 and $v$ factors as

                            N/𝔍_μ N → N/𝔍_λ N ─v'→ M/𝔍_λ M;

but then the identity automorphism of $M/{\mathfrak{J}}_{\lambda }M$ factors as

                            M/𝔍_λ M ─u_λ→ N/𝔍_λ N ─v'→ M/𝔍_λ M,

which shows that $u_{\lambda }$ is injective, and since one already knows it is surjective, this proves the corollary.

Proposition (19.1.5).

Let $A$ be a topological ring, $M$, $N$ two topological $A$-modules, $u:M\to N$ a continuous $A$-homomorphism. The following conditions are equivalent:

a) For every discrete topological $A$-module $E$ and every continuous $A$-homomorphism $v:M\to E$, there exists a continuous $A$-homomorphism $w:N\to E$ such that $v=w\circ u$.

b) For every open submodule $V$ of $M$, there exist an open submodule W'' of $N$, an open submodule $V^{\prime \prime }\subset V\cap u^{-1}(W^{\prime \prime })$ and a homomorphism $h:N/W^{\prime \prime }\to M/V$ such that the canonical homomorphism $M/V^{\prime \prime }\to M/V$ factors as

                              M/V'' ─u''→ N/W'' ─h→ M/V

where u'' is deduced from $u$ by passage to the quotients.

To see that a) implies b), it suffices to apply a) to $E=M/V$ and to the canonical homomorphism $v:M\to M/V$; then $W^{\prime \prime }=w^{-1}(0)$ is an open submodule of $N$ and $V^{\prime \prime }=u^{-1}(W^{\prime \prime })$ an open submodule of $M$ contained in $V$; these submodules and the homomorphism $h:N/W^{\prime \prime }\to M/V$ deduced from $w$ by passage to the quotient answer the question. Conversely, to show that b) implies a), one may restrict to the case where $v$ is surjective, by replacing $E$ by $v(M)$; then $V^{\prime }=Ker(v)$ is an open submodule of $M$, hence $E$ is isomorphic to the discrete $A$-module $M/V^{\prime }$, and by applying b) one obtains

the desired continuous $A$-homomorphism $w$ as the composite $N\to N/W^{\prime \prime }\stackrel{h}{\to }M/V^{\prime }$, the diagram

                              M ─────u───→ N
                              │            │
                              ↓            ↓
                            M/V'' ──────→ N/W''
                                    u''

being commutative.

When the equivalent conditions of (19.1.5) are satisfied, we shall say that $u$ is formally left-invertible; since condition b) of (19.1.5) implies that $u^{-1}(W^{\prime \prime })\subset V$, $u$ is then a formal monomorphism. The terminology is further justified by the following corollaries:

Corollary (19.1.6).

If there exists a continuous Â-homomorphism $s:\hat{N}\to \hat{M}$ such that $s\circ \hat{u}=1_{\hat{M}}$, then $u$ is formally left-invertible.

One observes that û is then a topological isomorphism of $\hat{M}$ onto a topological direct factor of $\hat{N}$. To prove (19.1.6), it suffices to note that with the notations of (19.1.5, a)), $v:M\to E$ extends to a continuous Â-homomorphism $\hat{v}:\hat{M}\to E$ since $E$ is discrete; let $j:M\to \hat{M}$ and $j^{\prime }:N\to \hat{N}$ be the canonical homomorphisms; then, setting $w=\hat{v}\circ s\circ j^{\prime }$, one has $w\circ u=\hat{v}\circ s\circ \hat{u}\circ j=\hat{v}\circ j=v$.

Corollary (19.1.7).

Suppose that the topologies of $M$ and $N$ are deduced from that of $A$, and let $({\mathfrak{J}}_{\lambda })$ be a fundamental system of open ideals of $A$. For $u$ to be formally left-invertible, it is necessary and sufficient that, for every $\lambda$, the homomorphism $u_{\lambda }:M/{\mathfrak{J}}_{\lambda }M\to N/{\mathfrak{J}}_{\lambda }N$, deduced from $u$ by passage to the quotients, be left-invertible (in other words, be an isomorphism of $M/{\mathfrak{J}}_{\lambda }M$ onto a direct factor of $N/{\mathfrak{J}}_{\lambda }N$).

Indeed, the condition is sufficient, for, taking $V^{\prime }={\mathfrak{J}}_{\lambda }M$ in condition b) of (19.1.5), one answers the question by taking $W^{\prime \prime }={\mathfrak{J}}_{\lambda }N$, $V^{\prime \prime }=V^{\prime }$ and $h$ such that $h\circ u_{\lambda }$ is the identity on $M/{\mathfrak{J}}_{\lambda }M$. Conversely, if $u$ is formally left-invertible, then, for every $\lambda$, there is, by virtue of (19.1.5, b)), a $\mu$ such that ${\mathfrak{J}}_{\mu }\subset {\mathfrak{J}}_{\lambda }$ and a homomorphism $h:N/{\mathfrak{J}}_{\mu }N\to M/{\mathfrak{J}}_{\lambda }M$ such that the canonical homomorphism $M/{\mathfrak{J}}_{\mu }M\to M/{\mathfrak{J}}_{\lambda }M$ factors as $M/{\mathfrak{J}}_{\mu }M\stackrel{u_{\mu }}{\to }N/{\mathfrak{J}}_{\mu }N\stackrel{h}{\to }M/{\mathfrak{J}}_{\lambda }M$; but since $h({\mathfrak{J}}_{\lambda }(N/{\mathfrak{J}}_{\mu }N))={\mathfrak{J}}_{\lambda }\cdot h(N/{\mathfrak{J}}_{\mu }N)\subset {\mathfrak{J}}_{\lambda }(M/{\mathfrak{J}}_{\lambda }M)=0$, $h$ factors canonically as $N/{\mathfrak{J}}_{\mu }N\to N/{\mathfrak{J}}_{\lambda }N\stackrel{s}{\to }M/{\mathfrak{J}}_{\lambda }M$, and it is immediate that $s$ is a left inverse of $u_{\lambda }$.

Proposition (19.1.8).

Let $A$ be a topological ring admitting a countable decreasing fundamental system $({\mathfrak{J}}_n)$ of open ideals. Let $M$, $N$ be two topological $A$-modules whose topologies are deduced from that of $A$, $u:M\to N$ an $A$-homomorphism. Set $M_n=M/{\mathfrak{J}}_{n}M$, $N_n=N/{\mathfrak{J}}_{n}N$ and let $u_n:M_n\to N_n$ be the $(A/{\mathfrak{J}}_n)$-homomorphism deduced from $u$ by passage to the quotients. Suppose that, for every $n$, $P_n=Coker(u_n)$ is a projective $(A/{\mathfrak{J}}_n)$-module and that $M$ is separated and complete. Then, for $u$ to be formally left-invertible, it is necessary and sufficient that $u$ be left-invertible (and $u$ is then a topological isomorphism of $M$ onto a topological direct factor of $N$).

By virtue of (19.1.7), one has the commutative diagrams

                    u_n         p_n
       0 ───→ M_n ─────→ N_n ────────→ P_n ──→ 0
              │↑          │↑           │↑
              │f          │g           │h
              ↓│          ↓│           ↓│
       0 ──→ M_{n+1} ──→ N_{n+1} ──→ P_{n+1} ──→ 0
                  u_{n+1}      p_{n+1}

where the rows are split exact sequences; in other words, there exists for each $n$ a homomorphism $s_n:P_n\to N_n$ such that $p_n\circ s_n=1_{P_n}$. We shall show that one can, by induction on $n$, define a homomorphism $s_n^{\prime }:P_n\to N_n$ such that $p_n\circ s_n^{\prime }=1_{P_n}$ and that the diagrams

                              s'_n
                       P_n ──────→ N_n
                       │↑          │↑
                       │h          │g
                       ↓│          ↓│
                       P_{n+1} ──→ N_{n+1}
                              s'_{n+1}

be commutative. Indeed, $g\circ s_{n+1}^{\prime }-s_n^{\prime }\circ h$ is a homomorphism of $P_{n+1}$ into $u_{n+1}(M_{n+1})=u_{n+1}(M_{n+1})/{\mathfrak{J}}_n{u}_{n+1}(M_{n+1})$; since $P_{n+1}$ is a projective $(A/{\mathfrak{J}}_{n+1})$-module, this homomorphism factors as

                P_{n+1} ─t_{n+1}→ u_{n+1}(M_{n+1}) → u_{n+1}(M_{n+1})/𝔍_n u_{n+1}(M_{n+1})

whence one concludes at once that $s_{n+1}^{\prime }=s_{n+1}-t_{n+1}$ answers the question. This being so, from the decomposition of $N_n$ as the direct sum of $u_n(M_n)$ and of $s_n^{\prime }(P_n)$, one deduces at once a homomorphism $w_n:N_n\to M_n$ left inverse of $u_n$ and such that the diagrams

                              w_n
                       N_n ──────→ M_n
                       │↑          │↑
                       │g          │f
                       ↓│          ↓│
                       N_{n+1} ──→ M_{n+1}
                              w_{n+1}

be commutative. The projective system $(w_n)$ then admits a projective limit $\hat{w}:\hat{N}\to \hat{M}=M$, whence by composition with the canonical homomorphism $N\to \hat{N}$, a homomorphism $w:N\to M$ such that, for every $n$, the endomorphism $(w\circ u{)}_n$ of $M_n=M/{\mathfrak{J}}_{n}M$ deduced from $w\circ u$ by passage to the quotients is the identity; this entails that $w\circ u=1_M$ since $M$ is separated and complete.

Proposition (19.1.9).

Let $A$ be a preadmissible topological ring $(0_I,\mathrm{7.1.2})$, $\mathfrak{L}$ an ideal of definition of $A$, $({\mathfrak{J}}_{\lambda })$ a fundamental system of open ideals of $A$. Let $M$, $N$ be two topological $A$-modules whose topologies are deduced from that of $A$, and such that for every $\lambda$, $N/{\mathfrak{J}}_{\lambda }N$ is a projective $(A/{\mathfrak{J}}_{\lambda })$-module (cf. (19.2.3)). Let $u:M\to N$ be an $A$-homomorphism. Then the following conditions are equivalent:

a) $u$ is formally left-invertible.

b) The homomorphism $u_0:M/\mathfrak{L}M\to N/\mathfrak{L}N$ deduced from $u$ by passage to the quotients is left-invertible.

We have seen (19.1.7) that condition a) is equivalent to saying that $u_{\lambda }$ is left-invertible for every $\lambda$; since $\mathfrak{L}$ is an open ideal, hence contains a ${\mathfrak{J}}_{\lambda }$, one deduces at once that $u_0$ is left-invertible. To show conversely that b) implies a), note that for every $\lambda$, $\mathfrak{L}/{\mathfrak{J}}_{\lambda }$ is by hypothesis a nilpotent ideal of $A/{\mathfrak{J}}_{\lambda }$. Our assertion will follow from the next proposition:

Proposition (19.1.10).

Let $A$ be a ring, $M$, $N$ two $A$-modules, with $N$ projective, $u:M\to N$ an $A$-homomorphism. Let $\mathfrak{J}$ be an ideal of $A$; suppose that one of the following conditions is satisfied:

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

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

Then, for $u$ to be left-invertible, it is necessary and sufficient that the homomorphism $u_0:M/\mathfrak{J}M\to N/\mathfrak{J}N$ of $(A/\mathfrak{J})$-modules, deduced from $u$ by passage to the quotients, be left-invertible.

The condition being obviously necessary, let us prove that it is sufficient. Let $v_0$ be a left inverse of $u_0$; the composite homomorphism $N\to N/\mathfrak{J}N\stackrel{v_0}{\to }M/\mathfrak{J}M$ factors as $N\stackrel{v}{\to }M\to M/\mathfrak{J}M$ since $N$ is projective; then $w=v\circ u$ is an endomorphism of $M$ such that the endomorphism $w_0$ of $M/\mathfrak{J}M$ deduced by passage to the quotients is the identity, and it suffices to prove that $w$ is itself bijective (for then $w^{-1}v$ will be a left inverse of $u$). Let us now distinguish the two cases.

(i) For every $n$ one has the commutative diagram

            (𝔍ⁿ/𝔍ⁿ⁺¹) ⊗_{A/𝔍} (M/𝔍M)  ──→  𝔍ⁿM/𝔍ⁿ⁺¹M
                   │                            │
                   │1 ⊗ gr⁰(w)                  │gr^n(w)
                   ↓                            ↓
            (𝔍ⁿ/𝔍ⁿ⁺¹) ⊗_{A/𝔍} (M/𝔍M)  ──→  𝔍ⁿM/𝔍ⁿ⁺¹M

where the horizontal arrows are surjective, and since $g{r}^0(w)=w_0$ is the identity, so is $g{r}^n(w)$, which a fortiori is bijective. The $\mathfrak{J}$-preadic filtration on $M$ being finite since $\mathfrak{J}$ is nilpotent, one concludes that $w$ is bijective (Bourbaki, Alg. comm., chap. III, §2, n° 8, cor. 3 of th. 1).

(ii) It suffices to show that for every maximal ideal $\mathfrak{m}$ of $A$, the endomorphism $w_{\mathfrak{m}}$ of $M_{\mathfrak{m}}$ is bijective (Bourbaki, Alg. comm., chap. II, §3, n° 3, th. 1) and since $\mathfrak{J}{A}_{\mathfrak{m}}\subset \mathfrak{m}{A}_{\mathfrak{m}}$ and $A_{\mathfrak{m}}/\mathfrak{J}{A}_{\mathfrak{m}}=(A/\mathfrak{J}{)}_{\mathfrak{m}}$, one is reduced to proving the proposition when $A$ is a local ring. Moreover, one may suppose that $\mathfrak{J}$ is the maximal ideal of $A$, for if $u_0$ is left-invertible,

then so is $u_{00}:M/{\mathfrak{J}}_{0}M\to N/{\mathfrak{J}}_{0}M$ obtained by tensoring $u_0$ with $1_{A/{\mathfrak{J}}_0}$, since one has $M/{\mathfrak{J}}_{0}M=(M/\mathfrak{J}M){\otimes }_{A/\mathfrak{J}}(A/{\mathfrak{J}}_0)$. Suppose then that $\mathfrak{J}$ is maximal, so that $A/\mathfrak{J}$ is a field. It clearly suffices to show that $M$ is a free $A$-module under the conditions of the statement: indeed, $det(w_0)$ is then the canonical image of $det(w)$ in $A/\mathfrak{J}$, hence $det(w)$ does not belong to the ideal $\mathfrak{J}$ and is consequently invertible. Now, the $(A/\mathfrak{J})$-vector space $M/\mathfrak{J}M$ being free of finite type, there is an $A$-module $L$ of finite type and an $A$-homomorphism $f:L\to M$ such that the homomorphism $f_0:L/\mathfrak{J}L\to M/\mathfrak{J}M$ deduced from $f$ by passage to the quotients is bijective. Since $M$ is of finite type, one concludes first of all that $f$ is surjective by Nakayama's lemma (Bourbaki, Alg. comm., chap. II, §3, n° 2, cor. 1 of prop. 4); furthermore, if $g=u\circ f$, the homomorphism $g_0:L/\mathfrak{J}L\to N/\mathfrak{J}N$ deduced by passage to the quotients is left-invertible, and since here $L$ is free, the remark at the beginning proves that $g$ is itself left-invertible; but this clearly entails that $f$ is injective, which completes the proof.

Let us mention in passing the following corollaries of (19.1.10):

Corollary (19.1.11).

Let $A$ be a local ring, $k$ its residue field, $M$ an $A$-module of finite type, $N$ a projective $A$-module, $u:M\to N$ a homomorphism. For $u$ to be left-invertible, it is necessary and sufficient that there exist a system of generators $(x_i{)}_{1⩽i⩽m}$ of $M$ such that the images under $u\otimes 1:M{\otimes }_{A}k\to N{\otimes }_{A}k$ of the $x_i\otimes 1$ be linearly independent in $N{\otimes }_{A}k$; the $x_i$ then form a basis of $M$.

The condition is obviously necessary, for if $u$ is left-invertible, $M$ is a projective $A$-module of finite type, hence free. Conversely, if the condition is satisfied, it is clear that the $x_i\otimes 1$ form a basis of $M{\otimes }_{A}k$ and that $u\otimes 1$ is left-invertible; it then suffices to apply (19.1.10) to the maximal ideal $\mathfrak{J}$ of $A$.

Corollary (19.1.12).

Let $A$ be a ring, $M$ an $A$-module of finite type, $N$ a projective $A$-module, $u:M\to N$ a homomorphism. For every prime ideal $\mathfrak{p}\in \mathrm{Spec}(A)$, the following conditions are equivalent:

a) The homomorphism $u_{\mathfrak{p}}:M_{\mathfrak{p}}\to N_{\mathfrak{p}}$ is left-invertible.

b) The homomorphism $u\otimes 1:M{\otimes }_{A}k(\mathfrak{p})\to N{\otimes }_{A}k(\mathfrak{p})$ is injective (recall that $k(\mathfrak{p})$ denotes the residue field of $A$ at $\mathfrak{p}$).

c) There exists a finite system of elements $x_i\in M$ ($1⩽i⩽m$) such that the images of the $x_i$ in $M_{\mathfrak{p}}$ generate $M_{\mathfrak{p}}$, and a system of $m$ linear forms $y_i^{\ast }$ on $N$ ($1⩽i⩽m$) such that $det(⟨y_i^{\ast },u(x_j)⟩)\notin \mathfrak{p}$.

d) There exists $f\in A-\mathfrak{p}$ such that the homomorphism $u_f:M_f\to N_f$ is left-invertible.

Moreover, the set of $\mathfrak{p}\in \mathrm{Spec}(A)$ satisfying these conditions is open in $\mathrm{Spec}(A)$.

The last condition is a trivial consequence of d). Since $N$ is projective, it is a direct factor of a free $A$-module $A^{(I)}$; furthermore, since $M$ is of finite type, $u(M)$ is contained in a sub-module of $A^{(I)}$ of the form $A^n$ for $n$ finite; since each of the statements a), b), c), d) is equivalent to the corresponding statement where one replaces $N$ by $N\oplus P$ (with $u$ still considered as mapping $M$ into $N$), one is reduced to the case where $N$ is free of finite type. It is trivial that d) implies a), and a) and b) are equivalent by virtue of (19.1.11); moreover a) entails that $M_{\mathfrak{p}}$ is free (19.1.11), hence a) entails c), by taking for the $x_i$ a family whose images in $M_{\mathfrak{p}}$ form a basis of this

$A_{\mathfrak{p}}$-module and noting that (since $N$ is free of finite type) every linear form on the $A_{\mathfrak{p}}$-module $N_{\mathfrak{p}}$ is written $u^{\ast }=v^{\ast }/f$, where $f\in A-\mathfrak{p}$, and $v^{\ast }$ is a linear form on the $A$-module $N$. It is clear that c) implies b), and it remains to see that a) implies d). Now, since $N$ is of finite presentation, $({\mathrm{Hom}}_A(N,M){)}_{\mathfrak{p}}$ is canonically identified with ${\mathrm{Hom}}_{A_{\mathfrak{p}}}(N_{\mathfrak{p}},M_{\mathfrak{p}})$ (Bourbaki, Alg. comm., chap. II, §2, n° 7, prop. 19). If $w_{\mathfrak{p}}$ is a left inverse of $u_{\mathfrak{p}}$, there exists thus a homomorphism $w:N\to M$ and an element $f\in A-\mathfrak{p}$ such that $w_{\mathfrak{p}}=w\otimes (1/f)1_{A_{\mathfrak{p}}}$. The relation $w_{\mathfrak{p}}\circ u_{\mathfrak{p}}=1_{M_{\mathfrak{p}}}$ thus also reads $(w\circ u)\otimes 1_{A_{\mathfrak{p}}}=f\cdot 1_{M_{\mathfrak{p}}}$. But since $M$ is an $A$-module of finite type, there exists $g\in A-\mathfrak{p}$ such that the endomorphism $g((w\circ u)-f\cdot 1_M)$ vanishes on all the generators of $M$, hence is zero. Setting $h=fg$, and $u_h=u\otimes 1_{A_h}$, $w_h=w\otimes 1_{A_h}$, one therefore has $w_h(u_h(z))=(f/1)z$ for every $z\in M_h$; but since $h/1$ is invertible in $A_h$, so is $f/1=f$, and $f^{-1}{w}_h$ is consequently a left inverse of $u_h$, which completes the proof.

Proposition (19.1.13).

Suppose the hypotheses of (19.1.9) are satisfied and suppose further that $({\mathfrak{J}}_{\lambda })$ is a decreasing sequence $({\mathfrak{J}}_n)$ and that $M$ is separated and complete. Then conditions a) and b) of (19.1.9) are also equivalent to:

c) $u$ is left-invertible.

One already knows that c) implies a) (19.1.6). Conversely, if a) is satisfied, one knows (with the notations of (19.1.8)) that $M_n$ is a direct factor of the projective $(A/{\mathfrak{J}}_n)$-module $N_n$, hence $P_n$ is also isomorphic to a direct factor of $N_n$, and consequently is projective; it then suffices to apply (19.1.8).

Let us finally note the following proposition:

Proposition (19.1.14).

Let $A$ be a ring, $M$ an $A$-module of finite type, $N$ a projective $A$-module, $u:M\to N$ an $A$-homomorphism.

(i) For $u$ to be left-invertible, it is necessary and sufficient that, for every maximal ideal $\mathfrak{m}$ of $A$, $u_{\mathfrak{m}}:M_{\mathfrak{m}}\to N_{\mathfrak{m}}$ be left-invertible.

(ii) Let $A^{\prime }$ be an $A$-algebra which is a faithfully flat $A$-module. For $u$ to be left-invertible, it is necessary and sufficient that $u\otimes 1:M{\otimes }_A{A}^{\prime }\to N{\otimes }_A{A}^{\prime }$ be left-invertible.

As in (19.1.12), one can restrict to the case where $N$ is free of finite type; to say that $u$ is left-invertible then means that $u$ is injective and that the quotient module $P=N/u(M)$ is projective, for $u(M)$ will then be a direct factor of $N$. Note further that since $M$ is of finite type, $P$ is of finite presentation. This being so:

(i) The condition is obviously necessary. Conversely, if it is satisfied, one knows that $u$ is injective (Bourbaki, Alg. comm., chap. II, §3, n° 3, th. 1) and since $P_{\mathfrak{m}}=N_{\mathfrak{m}}/u_{\mathfrak{m}}(M_{\mathfrak{m}})$ is projective for every $\mathfrak{m}$, one knows that this implies that $P$ is projective (loc. cit., §5, n° 2, th. 1).

(ii) Here again, the condition is trivially necessary. Conversely, if it is satisfied, one knows that $u$ is injective $(0_I,\mathrm{6.4.1})$ and since $P{\otimes }_A{A}^{\prime }=Coker(u\otimes 1)$ is projective, hence flat, one deduces that $P$ is a flat $A$-module $(0_I,\mathrm{6.6.3})$, hence projective since it is of finite presentation (Bourbaki, Alg. comm., chap. II, §5, n° 2, cor. 2 of th. 1).

Remark (19.1.15).

The notion "dual" to that of formally left-invertible homomorphism is that of formally right-invertible homomorphism; such a continuous $A$-homomorphism $u:M\to N$ verifies, by definition, the following condition: for every open submodule $W$ of $N$, there exist an open submodule $V^{\prime }\subset u^{-1}(W^{\prime })$ of $M$, an open submodule $W^{\prime \prime }\subset W$ of $N$ and a homomorphism $h:N/W^{\prime \prime }\to M/V^{\prime }$ such that the canonical homomorphism $N/W^{\prime \prime }\to N/W$ factors as

                              N/W'' ─h→ M/V' ─u'→ N/W

where $u^{\prime }$ is deduced from $u$ by passage to the quotients. This implies that $u$ is a formal epimorphism; if there exists a continuous $A$-homomorphism $r:N\to M$ such that $u\circ r=1_N$, one verifies at once that $u$ is formally right-invertible. If the conditions of (19.1.7) are satisfied, for $u$ to be formally right-invertible, it is necessary and sufficient that the $u_{\lambda }$ be right-invertible (that is, that the kernel of $u_{\lambda }$ be a direct factor of $M/{\mathfrak{J}}_{\lambda }M$ and that $u_{\lambda }$ be an isomorphism of a supplement of $Ker(u_{\lambda })$ onto $N/{\mathfrak{J}}_{\lambda }N$). We leave to the reader the task of stating and proving the propositions corresponding to (19.1.8) and (19.1.9) (in the analogue of (19.1.8), one must assume $M$ separated and complete and that $M_n$ is a projective $(A/{\mathfrak{J}}_n)$-module; in the analogue of (19.1.9), one must drop the hypothesis on the $N/{\mathfrak{J}}_{\lambda }N$, but assume on the other hand that, for every $\lambda$, $M/{\mathfrak{J}}_{\lambda }M$ is a projective $(A/{\mathfrak{J}}_{\lambda })$-module).

19.2. Formally projective modules

Definition (19.2.1).

Let $A$ be a topological ring. One says that a topological $A$-module $M$ is formally projective if it satisfies the following condition: for every open ideal $\mathfrak{J}$ of $A$, every pair of (discrete) $(A/\mathfrak{J})$-modules $P$, $Q$, every surjective $A$-homomorphism $u:P\to Q$ and every continuous $A$-homomorphism $v:M\to Q$, there exists a continuous $A$-homomorphism $w:M\to P$ such that $v=u\circ w$.

(19.2.2) To verify the condition of (19.2.1), one may evidently restrict (by replacing $Q$ by $v(M)$ and $P$ by $u^{-1}(v(M))$) to the case where $v$ is itself surjective; then $Q$ is isomorphic to $M/V$, where $V$ is an open submodule of $M$ such that $\mathfrak{J}M\subset V$; condition (19.2.1) is then equivalent to saying that for every discrete $(A/\mathfrak{J})$-module $P$ and every surjective $A$-homomorphism $u:P\to M/V$, there exist an open submodule $V^{\prime }\subset V$ of $M$ and an $A$-homomorphism $w:M/V^{\prime }\to P$ such that the canonical homomorphism $M/V^{\prime }\to M/V$ factors as $M/V^{\prime }\stackrel{w}{\to }P\stackrel{u}{\to }M/V$. We note that it suffices to verify this condition when $\mathfrak{J}$ runs through a fundamental system of neighbourhoods of 0 in $A$ formed of ideals.

(19.2.3) Suppose there exists a fundamental system $({\mathfrak{J}}_{\lambda })$ of open ideals of $A$ and a fundamental system $(V_{\lambda })$ of open submodules of $M$, having the same set of indices as $({\mathfrak{J}}_{\lambda })$ and such that, for every $\lambda$, $M/V_{\lambda }$ is a projective $(A/{\mathfrak{J}}_{\lambda })$-module. Then $M$ is formally projective: it suffices indeed, with the notations of (19.2.2), to take $\lambda$ such that ${\mathfrak{J}}_{\lambda }\subset \mathfrak{J}$ and $V_{\lambda }\subset V$; since $P$ is also an $(A/{\mathfrak{J}}_{\lambda })$-module, the factorization of the canonical homomorphism $M/V_{\lambda }\to M/V$ as $M/V_{\lambda }\to P\to M/V$ follows from the fact that we deal with an $(A/{\mathfrak{J}}_{\lambda })$-homomorphism and that $M/V_{\lambda }$ is assumed to be a projective $(A/{\mathfrak{J}}_{\lambda })$-module.

When the stricter condition of this number is satisfied, one says that $M$ is strictly formally projective.

Proposition (19.2.4).

Let $A$ be a topological ring, $M$ a topological $A$-module whose topology is deduced from that of $A$. For $M$ to be formally projective, it is necessary and sufficient that it be strictly formally projective.

We have just seen that the condition is sufficient. Conversely, suppose $M$ is formally projective and let $({\mathfrak{J}}_{\lambda })$ be a fundamental system of open ideals of $A$. For every $\lambda$, let $P$ be a free $(A/{\mathfrak{J}}_{\lambda })$-module and $u:P\to M/{\mathfrak{J}}_{\lambda }M$ a surjective $(A/{\mathfrak{J}}_{\lambda })$-homomorphism. There exists therefore a ${\mathfrak{J}}_{\mu }\subset {\mathfrak{J}}_{\lambda }$ such that the canonical homomorphism $M/{\mathfrak{J}}_{\mu }M\to M/{\mathfrak{J}}_{\lambda }M$ factors as $M/{\mathfrak{J}}_{\mu }M\stackrel{w}{\to }P\stackrel{u}{\to }M/{\mathfrak{J}}_{\lambda }M$; but since $w({\mathfrak{J}}_{\mu }(M/{\mathfrak{J}}_{\mu }M))\subset {\mathfrak{J}}_{\mu }P=0$, $w$ factors as $M/{\mathfrak{J}}_{\mu }M\to M/{\mathfrak{J}}_{\lambda }M\stackrel{v}{\to }P$, and it is clear that $u\circ v$ is the identity on $M/{\mathfrak{J}}_{\lambda }M$, which proves that this $(A/{\mathfrak{J}}_{\lambda })$-module is projective.

Proposition (19.2.5).

Let $A$ be a topological ring, $M$ a topological $A$-module.

(i) For $M$ to be formally projective (resp. strictly formally projective), it is necessary and sufficient that the topological Â-module $\hat{M}$ be so.

(ii) Let $A^{\prime }$ be a topological $A$-algebra. If $M$ is formally projective (resp. strictly formally projective), then $M{\otimes }_A{A}^{\prime }$ (equipped with the tensor product topology) is a formally projective (resp. strictly formally projective) topological $A^{\prime }$-module.

(i) It suffices to remark that when $\mathfrak{J}$ (resp. $V$) runs through the set of open ideals of $A$ (resp. the set of open submodules of $M$), the separated completion $\hat{\mathfrak{J}}$ (resp. $\hat{V}$) runs through the set of open ideals of Â (resp. the set of open submodules of $\hat{M}$), and $\hat{A}/\hat{\mathfrak{J}}=A/\mathfrak{J}$ (resp. $\hat{M}/\hat{V}=M/V$) up to a canonical isomorphism; since the notion of formally projective (resp. strictly formally projective) module only involves the $A/\mathfrak{J}$ and the $M/V$, one deduces (i) at once.

(ii) Suppose first $M$ is formally projective and set $M^{\prime }=M{\otimes }_A{A}^{\prime }$; let ${\mathfrak{J}}^{\prime }$ be an open ideal of $A^{\prime }$, $P^{\prime }$, $Q^{\prime }$ two discrete $(A^{\prime }/{\mathfrak{J}}^{\prime })$-modules, $u^{\prime }:P^{\prime }\to Q^{\prime }$ a surjective $A^{\prime }$-homomorphism, $v^{\prime }:M^{\prime }\to Q^{\prime }$ a continuous $A^{\prime }$-homomorphism. There is an open ideal $\mathfrak{J}$ of $A$ such that $\mathfrak{J}{A}^{\prime }\subset {\mathfrak{J}}^{\prime }$, hence $P^{\prime }$ and $Q^{\prime }$ are also discrete $(A/\mathfrak{J})$-modules. If one considers the composite $A$-homomorphism $v:M\stackrel{j}{\to }{M}^{\prime }\stackrel{v^{\prime }}{\to }{Q}^{\prime }$, which is continuous, the hypothesis implies that there exists a continuous $A$-homomorphism $w:M\to P^{\prime }$ such that $v=u^{\prime }\circ w$; but since $P^{\prime }$ is an $A^{\prime }$-topological module, $w$ factors as $M\stackrel{j}{\to }{M}^{\prime }\stackrel{w^{\prime }}{\to }{P}^{\prime }$, where $w^{\prime }$ is a continuous $A^{\prime }$-homomorphism, and since $v^{\prime }\circ j=(u^{\prime }\circ w^{\prime })\circ j$, one concludes that $v^{\prime }=u^{\prime }\circ w^{\prime }$.

Suppose next that $M$ is strictly formally projective; let $({\mathfrak{J}}_{\lambda })$ (resp. $(V_{\lambda })$) be a fundamental system of open ideals of $A$ (resp. of open submodules of $M$) such that $(M/V_{\lambda })$ is a projective $(A/{\mathfrak{J}}_{\lambda })$-module, and let $({\mathfrak{J}}_{\mu }^{\prime })$ be a fundamental system of open ideals of $A^{\prime }$. One has a fundamental system of neighbourhoods of 0 in $M^{\prime }$ by taking the submodules $Im(M{\otimes }_A{\mathfrak{J}}_{\mu }^{\prime })+Im(V_{\lambda }{\otimes }_A{A}^{\prime })=W_{\lambda \mu }$, where $\lambda$ and $\mu$ are such that ${\mathfrak{J}}_{\lambda }{A}^{\prime }\subset {\mathfrak{J}}_{\mu }^{\prime }$. Since $(M{\otimes }_A{A}^{\prime })/W_{\lambda \mu }=(M/V_{\lambda }){\otimes }_{A/{\mathfrak{J}}_{\lambda }}(A^{\prime }/{\mathfrak{J}}_{\mu }^{\prime })$ and since $M/V_{\lambda }$ is a projective $(A/{\mathfrak{J}}_{\lambda })$-module, $M^{\prime }/W_{\lambda \mu }$ is a projective $(A^{\prime }/{\mathfrak{J}}_{\mu }^{\prime })$-module.

19.3. Formally smooth algebras

Definition (19.3.1).

Let $A$ be a topological ring, $B$ a topological $A$-algebra. One says that $B$ is a formally smooth $A$-algebra if, for every discrete topological $A$-algebra $C$

and every nilpotent ideal $\mathfrak{J}$ of $C$, every continuous $A$-homomorphism $u:B\to C/\mathfrak{J}$ factors as $B\stackrel{v}{\to }C\stackrel{\varphi }{\to }C/\mathfrak{J}$, where $v$ is a continuous homomorphism and $\varphi$ is the canonical homomorphism.

Definition (19.3.1) amounts to saying that the following property holds:

(P) For every open ideal $\mathfrak{K}$ of the $A$-algebra $B$ and every $A$-homomorphism $u^{\prime }:B/\mathfrak{K}\to C/\mathfrak{J}$, there is an open ideal ${\mathfrak{K}}^{\prime }\subset \mathfrak{K}$ of $B$ such that the homomorphism $B\to B/\mathfrak{K}\stackrel{u^{\prime }}{\to }C/\mathfrak{J}$ factors as $B\to B/{\mathfrak{K}}^{\prime }\stackrel{v^{\prime }}{\to }C\to C/\mathfrak{J}$, where $v^{\prime }$ is an $A$-homomorphism.

Indeed, if $u:B\to C/\mathfrak{J}$ is a continuous $A$-homomorphism, it has for kernel $\mathfrak{K}$ an open ideal of $B$, hence $u$ factors as $B\to B/\mathfrak{K}\stackrel{u^{\prime }}{\to }C/\mathfrak{J}$, and if (P) is satisfied, it suffices to apply it to $u^{\prime }$ and to take for $v$ the composite $B\to B/{\mathfrak{K}}^{\prime }\stackrel{v^{\prime }}{\to }C$ to satisfy the conditions of (19.3.1). Conversely, suppose that $B$ is a formally smooth $A$-algebra; let us give an open ideal $\mathfrak{K}$ of $B$ and an $A$-homomorphism $u^{\prime }:B/\mathfrak{K}\to C/\mathfrak{J}$ and apply definition (19.3.1) to $u:B\to B/\mathfrak{K}\to C/\mathfrak{J}$; if $v:B\to C$ is a continuous $A$-homomorphism such that $u$ factors as $B\stackrel{v}{\to }C\to C/\mathfrak{J}$, the ideal ${\mathfrak{K}}^{\prime }=Ker(v)\cap \mathfrak{K}$ of $B$ is open and contained in $\mathfrak{K}$; consequently $v$ factors as $B\to B/{\mathfrak{K}}^{\prime }\stackrel{v^{\prime }}{\to }C$, and $v^{\prime }$ indeed satisfies condition (P).

Proposition (19.3.2).

Let $A$ be a discrete ring, $V$ a projective $A$-module; the symmetric algebra $B=S_A^{\bullet }(V)$, equipped with the discrete topology, is a formally smooth $A$-algebra.

Indeed, with the notations $C$ and $\mathfrak{J}$ having the same meaning as in (19.3.1), let $u:B\to C/\mathfrak{J}$ be a homomorphism of $A$-algebras, which by restriction to $V=S_A^1(V)$ gives a homomorphism of $A$-modules $u_1:V\to C/\mathfrak{J}$; since $V$ is projective, $u_1$ factors as $V\stackrel{v_1}{\to }C\to C/\mathfrak{J}$, and $v_1$ extends to a homomorphism of $A$-algebras $v:S_A^{\bullet }(V)\to C$, such that the composite $\varphi \circ v$ coincides with $u$ on $V$, hence $\varphi \circ v=u$, which proves the proposition.

Corollary (19.3.3).

If $A$ is a discrete ring, every polynomial algebra $B=A[T_{\alpha }{]}_{\alpha \in I}$, equipped with the discrete topology, is a formally smooth $A$-algebra.

Proposition (19.3.4).

Let $A$ be a topological ring, and let $B=A[[T_{\alpha }]{]}_{\alpha \in I}={\sum }_{\lambda \in {\mathbb{N}}^{(I)}}{c}_{\lambda }{T}^{\lambda }$ be a formal power series ring (broad algebra over $A$ of the monoid ${\mathbb{N}}^{(I)}$, identified as an $A$-module with the product $A^{{\mathbb{N}}^{(I)}}$). If one equips $B$ with the product topology, $B$ is a formally smooth $A$-algebra.

Let $({\mathfrak{L}}_{\mu }{)}_{\mu \in M}$ be a fundamental system of open ideals of $A$. For every finite part $H$ of $I$, every $\mu \in M$ and every integer $n$, denote by ${\mathfrak{K}}_{H,\mu ,n}$ the neighbourhood of 0 in $B$ consisting of the $(c_{\lambda })$ such that for every $\lambda =({\lambda }_{\alpha }{)}_{\alpha \in I}$ with ${\lambda }_{\alpha }=0$ for $\alpha \notin H$ and ${\sum }_{\alpha \in H}{\lambda }_{\alpha }⩽n$, one has $c_{\lambda }\in {\mathfrak{L}}_{\mu }$. One verifies immediately that the ${\mathfrak{K}}_{H,\mu ,n}$ form a fundamental system of neighbourhoods of 0 in $B$ and are ideals of $B$, hence the product topology is compatible with the $A$-algebra structure of $B$.

Let us first note, with the same notations, the

Lemma (19.3.4.1).

Let $E$ be a discrete $A$-algebra.

(i) If $f:B\to E$ is a continuous $A$-homomorphism, there exists a finite part $H$ of $I$ such that $f(T_{\alpha })=0$ for $\alpha \notin H$, and $f(T_{\alpha })$ is nilpotent in $E$ for every $\alpha \in H$.

(ii) Conversely, let $H$ be a finite part of $I$, $(z_{\alpha }{)}_{\alpha \in H}$ a family of nilpotent elements of $E$. There exists a continuous $A$-homomorphism $g:B\to E$ and only one such that $g(T_{\alpha })=z_{\alpha }$ for $\alpha \in H$ and $g(T_{\alpha })=0$ for $\alpha \notin H$.

(i) follows at once from the fact that $f^{-1}(0)$ is a neighbourhood of 0 in the product $A$-module $A^{{\mathbb{N}}^{(I)}}$, whence $f(T^{\lambda })=0$ except for a finite number of values of $\lambda \in {\mathbb{N}}^{(I)}$. To prove (ii), it suffices to remark that the polynomial ring $B^{\prime }=A[T_{\alpha }{]}_{\alpha \in I}$ is dense in $B$; the existence and uniqueness of the restriction $g|B^{\prime }$ are trivial and its continuity follows from the hypothesis that the $f(T_{\alpha })$ for $\alpha \in H$ are nilpotent, for if $(f(T_{\alpha }){)}^n=0$ for every $\alpha \in H$, one has $g(T^{\lambda })=0$ for every $\lambda =({\lambda }_{\alpha }{)}_{\alpha \in I}$ of finite support except those for which ${\lambda }_{\alpha }=0$ for $\alpha \notin H$ and ${\lambda }_{\alpha }<n$ for $\alpha \in H$, that is, except for a finite number of values of $\lambda$.

This lemma being established, and the notations $C$ and $\mathfrak{J}$ having the same meaning as in (19.3.1), one has $u(T_{\alpha })=0$ except for $\alpha \in H$, $H$ being a finite part of $I$, and the $z_{\alpha }=u(T_{\alpha })$ for $\alpha \in H$ are nilpotent in $C/\mathfrak{J}$; since $\mathfrak{J}$ is nilpotent, there exists a family $(x_{\alpha }{)}_{\alpha \in H}$ of nilpotent elements of $C$ whose canonical images in $C/\mathfrak{J}$ are the $z_{\alpha }$; if $v$ is the continuous $A$-homomorphism of $B$ into $C$ such that $v(T_{\alpha })=0$ for $\alpha \notin H$, $v(T_{\alpha })=x_{\alpha }$ for $\alpha \in H$, it is clear that $u$ factors as $B\stackrel{v}{\to }C\to C/\mathfrak{J}$.

Proposition (19.3.5).

Let $A$ be a topological ring.

(i) $A$ is a formally smooth $A$-algebra.

(ii) If $B$ is a formally smooth $A$-algebra and $C$ a formally smooth $B$-algebra, then $C$ is a formally smooth $A$-algebra.

(iii) Let $B$ be a formally smooth $A$-algebra, $A^{\prime }$ a topological $A$-algebra; then the topological $A^{\prime }$-algebra $B{\otimes }_A{A}^{\prime }$ $(0_I,\mathrm{7.7.5}and\mathrm{7.7.6})$ is formally smooth.

(iv) Let $B$ be a topological $A$-algebra, $S$ (resp. $T$) a multiplicative part of $A$ (resp. $B$) such that the canonical image of $S$ in $B$ is contained in $T$. If $B$ is a formally smooth $A$-algebra, then $T^{-1}B$ is a formally smooth $S^{-1}A$-algebra.

(v) Let $B_i$ ($1⩽i⩽n$) be topological $A$-algebras. For ${\prod }_{i=1}^n{B}_i$ to be a formally smooth $A$-algebra, it is necessary and sufficient that each of the $B_i$ be so.

(i) If $C$ is a discrete $A$-algebra, $\varphi :C\to C/\mathfrak{J}$ the canonical homomorphism of $C$ onto an arbitrary quotient $A$-algebra of $C$, the only $A$-homomorphism of $A$ into $C/\mathfrak{J}$ is the composite homomorphism $A\stackrel{\psi }{\to }C\stackrel{\varphi }{\to }C/\mathfrak{J}$, where $\psi$ is the homomorphism defining the $A$-algebra structure on $C$; since $\psi$ is continuous, the condition of (19.3.1) is trivially satisfied.

(ii) Let $\alpha :A\to B$, $\beta :B\to C$ be the continuous homomorphisms defining respectively the $A$-algebra structure on $B$ and the $B$-algebra structure on $C$, so that $\beta \circ \alpha$ defines the $A$-algebra structure on $C$. Let $E$ be a discrete $A$-algebra, $\mathfrak{L}$ a nilpotent ideal of $E$, $u:C\to E/\mathfrak{L}$ a continuous $A$-homomorphism, so that $u\circ \beta \circ \alpha$ is the homomorphism defining the $A$-algebra structure of $E/\mathfrak{L}$. Since $B$ is a formally smooth $A$-algebra, the continuous $A$-homomorphism $u\circ \beta :B\to E/\mathfrak{L}$ factors as $B\stackrel{v}{\to }E\to E/\mathfrak{L}$, where $v$ is a continuous $A$-homomorphism; $v$ and $u\circ \beta$ then define on $E$ and $E/\mathfrak{L}$ respectively structures of topological $B$-algebra, for which $E/\mathfrak{L}$ is

still the (discrete) quotient algebra of the $B$-algebra $E$. Furthermore, $u$ is a continuous $B$-homomorphism, hence factors as $C\stackrel{w}{\to }E\to E/\mathfrak{L}$, where $w$ is a continuous $B$-homomorphism; since $v\circ \alpha$ is the $A$-homomorphism defining the $A$-algebra structure on $E$, $w$ is indeed a continuous $A$-homomorphism, whence our assertion.

(iii) Let $C$ be a discrete topological $A^{\prime }$-algebra, $\mathfrak{J}$ a nilpotent ideal of $C$, $u:B{\otimes }_A{A}^{\prime }\to C/\mathfrak{J}$ a continuous $A^{\prime }$-homomorphism. If one composes $u$ with the canonical homomorphism $\rho :B\to B{\otimes }_A{A}^{\prime }$, one obtains (since $A\to B\stackrel{\rho }{\to }B{\otimes }_A{A}^{\prime }$ is equal to the composite $A\to A^{\prime }\stackrel{\sigma }{\to }B{\otimes }_A{A}^{\prime }$) a continuous $A$-homomorphism which, by hypothesis, therefore factors as $B\stackrel{v}{\to }C\to C/\mathfrak{J}$, where $v$ is a continuous $A$-homomorphism (for the topological $A$-algebra structure on $C$ defined by the composite $A\to A^{\prime }\stackrel{u}{\to }C$ of the canonical homomorphisms). The equality of the composites $A\to B\stackrel{v}{\to }C$ and $A\to A^{\prime }\stackrel{u}{\to }C$ thus entails the existence of a continuous ring homomorphism $f:B{\otimes }_A{A}^{\prime }\to C$ such that $v=f\circ \rho$ and $u=f\circ \sigma$ $(0_I,\mathrm{7.7.6})$; since the composite homomorphisms $B\stackrel{\rho }{\to }B{\otimes }_A{A}^{\prime }\stackrel{f}{\to }C\to C/\mathfrak{J}$ and $B\stackrel{\rho }{\to }B{\otimes }_A{A}^{\prime }\stackrel{u}{\to }C/\mathfrak{J}$ (resp. $A^{\prime }\stackrel{\sigma }{\to }B{\otimes }_A{A}^{\prime }\stackrel{f}{\to }C\to C/\mathfrak{J}$ and $A^{\prime }\stackrel{\sigma }{\to }B{\otimes }_A{A}^{\prime }\stackrel{u}{\to }C/\mathfrak{J}$) are equal, one indeed has the factorization $u:B{\otimes }_A{A}^{\prime }\to C\to C/\mathfrak{J}$, which establishes (iii).

(iv) The topology considered on $S^{-1}A$ (resp. $T^{-1}B$) is naturally that for which a fundamental system of neighbourhoods of 0 is formed of the $S^{-1}\mathfrak{J}$ (resp. $T^{-1}\mathfrak{K}$), where $\mathfrak{J}$ (resp. $\mathfrak{K}$) runs through a fundamental system of open ideals of $A$ (resp. $B$) $(0_I,\mathrm{7.6.1})$. If $\alpha :A\to B$ is the canonical homomorphism, it is clear that the canonical homomorphism ${\alpha }^{\prime }:S^{-1}A\to T^{-1}B$ deduced from $\alpha$ (and whose existence results from $\alpha (S)\subset T$ by hypothesis) is continuous $(0_I,\mathrm{7.6.6})$. Let then $C$ be a discrete topological $S^{-1}A$-algebra, $\mathfrak{J}$ a nilpotent ideal of this algebra, $u:T^{-1}B\to C/\mathfrak{J}$ a continuous $S^{-1}A$-homomorphism; then the composite $B\to T^{-1}B\to C/\mathfrak{J}$ is a continuous $A$-homomorphism which, by hypothesis, factors as $B\to C\to C/\mathfrak{J}$, where $v$ is a continuous $A$-homomorphism. Since for every $t\in T$, $t/1$ is invertible in $T^{-1}B$, $u(t/1)$ is invertible in $C/\mathfrak{J}$. Since $\mathfrak{J}$ is nilpotent, every element of the class $u(t/1)$ in $C$, and in particular $v(t)$, is invertible in $C$ $(0_I,\mathrm{7.1.12})$, and consequently $v$ factors as $B\to T^{-1}B\stackrel{w}{\to }C$; since $v$ is continuous, so is $w$ $(0_I,\mathrm{7.6.6})$, and it is an $S^{-1}A$-homomorphism because the composite $A\to B\to T^{-1}B\to C$ is equal to

                              A → S⁻¹ A → T⁻¹ B → C,

hence $S^{-1}A\to T^{-1}B\to C$ is the canonical homomorphism defining on $C$ the $S^{-1}A$-algebra structure. Finally, the composite homomorphisms $B\to T^{-1}B\to C\to C/\mathfrak{J}$ and $B\to T^{-1}B\stackrel{u}{\to }C/\mathfrak{J}$ being equal, the same reasoning shows that $u$ is indeed equal to the composite $T^{-1}B\stackrel{w}{\to }C\to C/\mathfrak{J}$, whence (iv).

Finally, (v) is immediate, the data of a continuous $A$-homomorphism of $B={\prod }_{i=1}^n{B}_i$ into $C$ (resp. $C/\mathfrak{J}$) being equivalent to that of $n$ continuous $A$-homomorphisms $B_i\to C$

(resp. $B_i\to C/\mathfrak{J}$), and any continuous $A$-homomorphism $B_j\to C$ (resp. $B_j\to C/\mathfrak{J}$) giving by composition a continuous $A$-homomorphism $B\to B_j\to C$ (resp. $B\to B_j\to C/\mathfrak{J}$).

Proposition (19.3.6).

Let $A$ be a topological ring, $B$ a topological $A$-algebra, Â and $\hat{B}$ the respective separated completions of $A$ and $B$, so that $\hat{B}$ is a topological Â-algebra. The following conditions are equivalent:

a) $B$ is a formally smooth $A$-algebra.

b) $\hat{B}$ is a formally smooth $A$-algebra.

c) $\hat{B}$ is a formally smooth Â-algebra.

Of course, the Â-algebra structure on $\hat{B}$ is defined by the homomorphism $\hat{\varphi }$, if $\varphi :A\to B$ is the homomorphism defining the $A$-algebra structure on $B$. Since every discrete $A$-algebra $C$ is separated and complete, it is also an Â-algebra (by canonical extension of the homomorphism from $A$ into $C$), and a continuous $A$-homomorphism from $B$ into $C$ gives by canonical extension an $A$-homomorphism (and a fortiori an Â-homomorphism) from $\hat{B}$ into $C$ (in other words, every $A$-homomorphism $B\to C$ factors as $B\to \hat{B}\to C$ in a unique way). These remarks and definition (19.3.1) entail the proposition at once.

Corollary (19.3.7).

Under the hypotheses of (19.3.5, (iv)), the topological $A{S}^{-1}$-algebra $B{T}^{-1}$ is formally smooth.

This follows from the definitions $(0_I,\mathrm{7.6.1}and\mathrm{7.6.7})$ and from (19.3.5, (iv)) and (19.3.6).

Proposition (19.3.8).

Let $A$ be a topological ring, $B$ a topological $A$-algebra, and suppose that for every open ideal $\mathfrak{K}$ of $B$, ${\mathfrak{K}}^2$ is also open. Let $A^{\prime }$, $B^{\prime }$ be topological rings obtained by equipping $A$ and $B$ with topologies finer than the initial topologies, and suppose that the canonical homomorphism $\varphi :A\to B$ is still a continuous homomorphism from $A^{\prime }$ into $B^{\prime }$. Then, if $B^{\prime }$ is a formally smooth $A^{\prime }$-algebra, $B$ is a formally smooth $A$-algebra.