Éléments de Géométrie Algébrique — English

§21. Differentials in characteristic-$p$ rings

The results of the present section, of a more special and technical nature than those of §§19, 20 and 22, will be used only exceptionally in the course of Chap. IV. Their principal role here is in the proof of three theorems of §22 (22.3.3, 22.5.8 and 22.7.3), the first and the last of which intervene in an essential way in the "fine" theory of Noetherian local rings of Chap. IV, §7.

21.1. Systems of $p$-generators and $p$-bases

(21.1.1) Given a number $p$ which is either 0 or a prime number, we shall say that a ring $A$ is of characteristic $p$ if there exists a ring homomorphism $P\to A$, where $P$ is the prime field of characteristic $p$; note that this homomorphism is then unique, the composite $\mathbb{Z}\to P\to A$ being the unique homomorphism of $\mathbb{Z}$ into $A$. If $A\ne 0$, this amounts to saying that $A$ contains a field of characteristic $p$, the image of $P$ being necessarily a field isomorphic to $P$ (and moreover the only subfield of $A$ isomorphic to $P$).

(21.1.2) If $p>0$, saying that $A$ is of characteristic $p$ is equivalent to saying that, in $A$, one has $p\cdot 1=0$, or again $pA=0$. If $p=0$, saying that $A$ is of characteristic $p$ is equivalent to saying that for every integer $n\ne 0$, $n\cdot 1$ is invertible in $A$. If $A\ne 0$, there can exist only one $p$ (prime or 0) such that $A$ is of characteristic $p$; this follows from the preceding and from Bezout's identity $ap+bq=1$ for two distinct primes p, q. By contrast the zero ring is of characteristic $p$ for every $p$.

(21.1.3) If $A$ is of characteristic $p$, so is every algebra over $A$. In particular, for every prime ideal $\mathfrak{p}$ of $A$, the residue field of $A$ at $\mathfrak{p}$ is of characteristic $p$. Conversely, if $p=0$ and if for every maximal ideal $\mathfrak{m}$ of $A$ the residue field of $A$ at $\mathfrak{m}$ is of characteristic 0, the same holds for $A$, for every integer $n\ne 0$ is then invertible in all the $A_{\mathfrak{m}}$, hence also in $A$. By contrast, if $p>0$, a local ring may have its residue field of characteristic $p$ without itself being of characteristic $p$, as is shown by the example of the prime ring (integral) ${\mathbb{Z}}_{(p)}$ (19.8.3) or of the Artinian local ring $\mathbb{Z}/p^2\mathbb{Z}$ for $p\ge 2$, which do not contain a field.

Let us finally note that for a ring (even reduced), the residue fields at its prime ideals may have different characteristics, as is shown by the example of $\mathbb{Z}$.

(21.1.4) In all the rest of this section, we suppose fixed a prime number $p$ and all rings will be supposed of characteristic $p$, unless expressly mentioned otherwise. For such a ring $A$, the map $x\mapsto x^p$ is an endomorphism of $A$, which we denote F_A; if $A$ is reduced, F_A is injective. One sets $A^p=F_A(A)$ (subring of $A$ consisting of the $x^p$ for $x\in A$); one can naturally consider $A$ as an $A^p$-algebra.

One can also consider $A$ as an $A$-algebra by means of the homomorphism $F_A:x\mapsto x^p$ of $A$ into $A$; in other words, this is the $A$-algebra $A_{F_A}$ for which the product $\lambda \cdot x$ of $x\in A$ by a scalar $\lambda \in A$ is the product ${\lambda }^{p}x$ in the ring $A$; we shall denote this $A$-algebra $A^{[p]}$. It is clear that for every ring homomorphism $u:A\to B$ the pair $(u,u)$ is a di-homomorphism of $A$-algebras $A^{[p]}\to B^{[p]}$. For every $A$-module $E$, one sets $E^{[p]}=E{\otimes }_A{A}^{[p]}$ where $A^{[p]}$ is considered as an $(A,A)$-bimodule, the left $A$-module structure being the one just defined, and the right $A$-module structure defined by multiplication in $A$; $E^{[p]}$ is equipped with the $A$-module structure coming from the right $A$-module structure of $A^{[p]}$, so that for $x\in E$, $\alpha ,\beta$ in $A$, one has

  α(x ⊗ β) = x ⊗ (αβ)    and    (αx) ⊗ β = α(x ⊗ β) = x ⊗ α^p β.

Setting $x^{[p]}=x\otimes 1$, one therefore has $(\alpha x{)}^{[p]}={\alpha }^p{x}^{[p]}$. When F_A is injective, one can identify $A^{[p]}$ with the ring $A$ considered as algebra over its subring $A^p$.

Proposition (21.1.5).

Let $A$, $B$ be two rings, $u:A\to B$ a homomorphism.

(i) Every $A$-derivation of $B$ into a $B$-module $L$ is zero on $A[B^p]$ (and is therefore an $A[B^p]$-derivation).

(ii) For every sub-$A$-algebra $A^{\prime }$ of $A[B^p]$, if $j:A^{\prime }\to B$ is the canonical injection, the canonical homomorphism

$${\gamma }_{B/A^{\prime }/A}:{\Omega }_{B/A}\to {\Omega }_{B/A^{\prime }}$$

is bijective.

(iii) Suppose there exists an integer $s\ge 0$ such that $u(A)\supset B^{p^s}$. Then, in the ring $B{\otimes }_{A}B$, ${\mathfrak{J}}_{B/A}$ is a nilideal.

(i) By induction starting from (20.1.1.1), one deduces for every $A$-derivation $D:B\to L$ that one has $D(x^k)=k{x}^{k-1}D(x)$, whence in particular $D(x^p)=0$.

(ii) By (20.4.8), assertion (ii) is only a translation of (i), the latter being written ${\mathrm{Der}}_A(B,L)={\mathrm{Der}}_{A[B^p]}(B,L)$ for every $B$-module $L$.

(iii) For every $x\in B$, one has $(x\otimes 1-1\otimes x{)}^{p^s}=x^{p^s}\otimes 1-1\otimes x^{p^s}=x^{p^s}(1\otimes 1-1\otimes 1)=0$, since $x^{p^s}\in u(A)$. The conclusion follows from the fact that the elements $1\otimes x-x\otimes 1$ generate ${\mathfrak{J}}_{B/A}$ (20.4.4).

It follows at once from (21.1.5) that for every pair of ring homomorphisms $A\to B\to C$,

$${\gamma }_{C/B/A}={\gamma }_{C/B/A}(\mathrm{21.1.5.1})$$

for every sub-$A$-algebra $A^{\prime }\subset A[B^p]$ of $B$.

On the other hand, (21.1.5) also shows in particular that for "absolute" modules of differentials,

$${\Omega }_A={\Omega }_{A/A^p}.(\mathrm{21.1.5.2})$$

Corollary (21.1.6).

Suppose that $B$ is an $A$-algebra of finite type and that there exists an integer $s\ge 0$ such that $u(A)\supset B^{p^s}$. If ${\Omega }_{B/A}=0$, then $u$ is surjective.

By (21.1.5, (iii)), ${\mathfrak{J}}_{B/A}$ is a nilideal; moreover, by (20.4.4) and the hypothesis, ${\mathfrak{J}}_{B/A}$ is an ideal of finite type, so it is nilpotent; as the relation ${\Omega }_{B/A}=0$ means that ${\mathfrak{J}}_{B/A}^2={\mathfrak{J}}_{B/A}$, one concludes that ${\mathfrak{J}}_{B/A}=0$, that is, that $\pi :B{\otimes }_{A}B\to B$ is bijective. Moreover, every element of $B$ having its $p^s$-th power in $u(A)$, $B$ is integral over $A$, and being of finite type, it is a finite $A$-algebra. One is thus reduced to proving the following lemma (in which the rings considered are not supposed of characteristic $p$):

Lemma (21.1.6.1).

Let $R$ be a ring, $S$ a finite $R$-algebra; if the canonical homomorphism $\pi :S{\otimes }_{R}S\to S$ is bijective, then the structural homomorphism $u:R\to S$ is surjective.

It suffices to show that for every maximal ideal $\mathfrak{m}$ of $R$, if one sets $T=R-\mathfrak{m}$, the homomorphism $u_{\mathfrak{m}}:R_{\mathfrak{m}}\to T^{-1}S$ is surjective (Bourbaki, Alg. comm., chap. II, §3, n° 3, th. 1); now the hypothesis entails that the homomorphism $(T^{-1}S){\otimes }_{R_{\mathfrak{m}}}(T^{-1}S)\to T^{-1}S$ is bijective $(0_I,\mathrm{1.3.4})$, and since $T^{-1}S$ is a finite $R_{\mathfrak{m}}$-algebra,

one sees that one can restrict to the case where $R$ is a local ring. Denoting still by $\mathfrak{m}$ its maximal ideal, it suffices to prove that $u\otimes 1:R/\mathfrak{m}\to S/\mathfrak{m}S$ is surjective, by Nakayama's lemma ($S$ being an $R$-module of finite type); as the canonical homomorphism $(S/\mathfrak{m}S){\otimes }_{R/\mathfrak{m}}(S/\mathfrak{m}S)\to S/\mathfrak{m}S$ is bijective and $S/\mathfrak{m}S$ is a finite $(R/\mathfrak{m})$-algebra, one is finally reduced to the case where $R$ is a field; but then the ranks of $S{\otimes }_{R}S$ and of $S$ over $R$ can only be equal if $S$ is of rank 0 or 1, that is, if $u:R\to S$ is surjective.

Proposition (21.1.7).

Let $A$ be a ring, $B$ an $A$-algebra, $(x_{\alpha }{)}_{\alpha \in I}$ a family of elements of $B$. Consider the following properties:

a) $B=A[B^p,(x_{\alpha })]$, in other words the $A[B^p]$-algebra $B$ is generated by the family $(x_{\alpha }{)}_{\alpha \in I}$.

b) The $A[B^p]$-module $B$ is generated by the monomials $\prod x_{\alpha }^{n(\alpha )}$, where $(n(\alpha ){)}_{\alpha \in I}$ is a family of integers of finite support such that $0\le n(\alpha )<p$ for every $\alpha \in I$.

c) The $B$-module ${\Omega }_{B/A}$ is generated by the $d_{B/A}(x_{\alpha })$ $(\alpha \in I)$.

Then properties a) and b) are equivalent and entail c); if moreover $B$ is an $A[B^p]$-algebra of finite type, c) is equivalent to a) and b).

It is clear that b) entails a), and conversely a) entails b), for every monomial ${\prod }_{\alpha }{x}_{\alpha }^{m(\alpha )}$, where the $m(\alpha )$ are integers $\ge 0$ (forming a family of finite support), can be written $({\prod }_{\alpha }{x}_{\alpha }^{q(\alpha )}{)}^p\cdot {\prod }_{\alpha }{x}_{\alpha }^{r(\alpha )}$, by dividing each $m(\alpha )$ by $p$, which gives $m(\alpha )=p\cdot q(\alpha )+r(\alpha )$ with $0\le r(\alpha )<p$. The fact that a) entails c) follows from (21.1.5, (ii)) and from (20.4.7). Conversely suppose c) verified and that $B$ is an $A[B^p]$-algebra of finite type; let $B^{\prime }$ be the sub-$A[B^p]$-algebra of $B$ generated by the $x_{\alpha }$; in the exact sequence (20.5.7.1)

  Ω_{B'/A[B^p]} ⊗_{B'} B → Ω_{B/A[B^p]} → Ω_{B/B'} → 0

the hypothesis entails that the left arrow is surjective (taking account of (21.1.5, (ii))); one therefore has ${\Omega }_{B/B^{\prime }}=0$, and as $B^p\subset B^{\prime }\subset B$ and $B$ is a $B^{\prime }$-algebra of finite type, one necessarily has $B^{\prime }=B$ by (21.1.6).

Remark (21.1.8).

When $B$ is a field, we shall prove the equivalence of properties a), b) and c) without hypothesis of finiteness (21.4.5).

Definition (21.1.9).

Let $A$ be a ring (of characteristic $p$), $B$ an $A$-algebra. One says that a family $(x_{\alpha }{)}_{\alpha \in I}$ of elements of $B$ is $p$-free over $A$ (resp. a system of $p$-generators of $B$ over $A$, resp. a $p$-basis of $B$ over $A$) if the family of monomials ${\prod }_{\alpha }{x}_{\alpha }^{n(\alpha )}$ ($0\le n(\alpha )<p$, $(n(\alpha ){)}_{\alpha \in I}$ of finite support) is a free family (resp. a system of generators, resp. a basis) in the $A[B^p]$-module $B$.

One has corresponding definitions for a set $M$ of elements of $B$, by considering the family defined by the canonical injection $M\to B$. When one takes for $A$ the prime field ${\mathbb{F}}_p$ (in which case $A[B^p]=B^p$), one omits the mention of $A$ in the preceding definition (or one says further that a family is "absolutely" $p$-free, resp. an "absolute" system of $p$-generators, resp. an "absolute" $p$-basis).

It is clear that the notions defined in (21.1.9) do not change when one replaces $A$ therein by the subring $A[B^p]$ of $B$; in other words one can always suppose that one has $B^p\subset A\subset B$.

If $M$ is a $p$-free part of $B$ over $A$, it is clear that every subset of $M$ is $p$-free over $A$. Moreover:

Lemma (21.1.10).

Let $C$ be a sub-$A$-algebra of $B$ such that $B^p\subset C$, $M$ a subset of $C$, $N$ a subset of $B$.

(i) If $M$ is a system of $p$-generators of $C$ over $A$ and $N$ a system of $p$-generators of $B$ over $C$, then $M\cup N$ is a system of $p$-generators of $B$ over $A$.

(ii) Suppose that $M$ is a $p$-basis of $C$ over $A$; then, for $N$ to be $p$-free over $C$, it is necessary and sufficient that $M\cup N$ be $p$-free over $A$.

One can restrict to the case where $B^p\subset A\subset C\subset B$. Then (i) is a particular case of the fact that if $P$ (resp. $Q$) is a system of generators of the $A$-module $C$ (resp. of the $C$-module $B$), the set of xy, where $x\in P$ and $y\in Q$, is a system of generators of the $A$-module $B$. Keeping the same notation, if $P$ is a basis of the $A$-module $C$, saying that $Q$ is a free family over $C$ means that the relation ${\sum }_{\lambda ,\mu }{a}_{\lambda \mu }{x}_{\lambda }{y}_{\mu }=0$, where $a_{\lambda \mu }\in A$, $x_{\lambda }\in P$, $y_{\mu }\in Q$ (the $x_{\lambda }$ (resp. $y_{\mu }$) being pairwise distinct), is equivalent to ${\sum }_{\lambda }{a}_{\lambda \mu }{x}_{\lambda }=0$ for every $\mu$, or again to $a_{\lambda \mu }=0$ for every pair $(\lambda ,\mu )$; whence assertion (ii).

21.2. $p$-bases and formal smoothness

Theorem (21.2.1).

Let $B$ be a ring, $A$ a subring of $B$ such that $B^p\subset A\subset B$, $(x_{\alpha }{)}_{\alpha \in I}$ a $p$-basis of $B$ over $A$. Let $E$ be an $A$-algebra, $u:A\to E$ the structural homomorphism, $q:E\to B$ a surjective $A$-homomorphism, $\mathfrak{K}$ its kernel, and suppose that ${\mathfrak{K}}^2=0$. Then:

(i) For there to exist an $A$-homomorphism $v:B\to E$ right inverse to the homomorphism $q:E\to B$, it is necessary and sufficient that one have $u(q(z{)}^p)=z^p$ for every $z\in E$.

(ii) When the condition of (i) is satisfied, for every family $(z_{\alpha }{)}_{\alpha \in I}$ of elements of $E$ such that $q(z_{\alpha })=x_{\alpha }$ for every $\alpha \in I$, there exists one and only one $A$-homomorphism $v:B\to E$ such that $v(x_{\alpha })=z_{\alpha }$ for every $\alpha \in I$, and $v$ is right inverse to $q$.

If there exists an $A$-homomorphism $v:B\to E$, one must have $v(a)=u(a)$ for every $a\in A\subset B$, hence $v(q(z{)}^p)=u(q(z{)}^p)$ for every $z\in E$, since $B^p\subset A$. But one has $v(q(z{)}^p)=(v(q(z)){)}^p$ and by definition $v(q(z))\equiv z(mod\mathfrak{K})$ if $q\circ v=1_B$, hence $(v(q(z)){)}^p=z^p$ since ${\mathfrak{K}}^2=0$; whence the necessity of (i). The sufficiency of condition (i) will follow from (ii). Now, under the hypotheses of (ii), the uniqueness of $v$ is evident since the monomials ${\prod }_{\alpha }{x}_{\alpha }^{n(\alpha )}$

generate the $A$-module $B$; as these monomials moreover form a basis of the $A$-module $B$, there exists an $A$-linear map $v$ of $B$ into $E$ such that $v({\prod }_{\alpha }{x}_{\alpha }^{n(\alpha )})={\prod }_{\alpha }{z}_{\alpha }^{n(\alpha )}$ for every family $(n(\alpha ){)}_{\alpha \in I}$ of finite support with $0\le n(\alpha )<p$ for every $\alpha$. It remains to see that $v$ is a ring homomorphism. Now, one can write $({\prod }_{\alpha }{x}_{\alpha }^{m(\alpha )})({\prod }_{\alpha }{x}_{\alpha }^{n(\alpha )})=a\cdot {\prod }_{\alpha }{x}_{\alpha }^{r(\alpha )}$, where $r(\alpha )=m(\alpha )+n(\alpha )$ if $m(\alpha )+n(\alpha )<p$, $r(\alpha )=m(\alpha )+n(\alpha )-p$ in the contrary case, and $a\in A$ is the product of the $x_{\alpha }^p$ for those $\alpha \in I$ such that $m(\alpha )+n(\alpha )\ge p$; we have to see that $u(a)={\prod }_{\alpha }{z}_{\alpha }^p$ (the product over the same set of $\alpha$); but as $x_{\alpha }^p=q(z_{\alpha }{)}^p$, this follows from the hypothesis. Q.E.D.

Corollary (21.2.2).

Let $B$ be a ring, $A$ a subring of $B$ such that $B^p\subset A\subset B$. If there exists a $p$-basis of $B$ over $A$, $B$ is an $A$-algebra formally smooth relative to $B^p$ (for the discrete topologies).

In fact, let $E$ be an $A$-extension of $B$ by a $B$-module $L$, $q:E\to B$ the augmentation, $u:A\to E$ the structural homomorphism. Saying that $E$ is $B^p$-trivial means that there exists a ring homomorphism $v:B\to E$ such that $q(v(b))=b$ and $v(b^p)=u(b^p)$ for every $b\in B$. One deduces from this that, for $z\in E$, one has $u(q(z{)}^p)=v(q(z{)}^p)=(v(q(z)){)}^p$; but by virtue of the relation $L^2=0$, one has ${\mathfrak{K}}^2=0$, and as $v(q(z))-z\in L$, one has $(v(q(z)){)}^p=z^p$; the condition of (21.2.1, (i)) is therefore satisfied, and $E$ is $B$-trivial.

Corollary (21.2.3).

Let $B$ be a ring, $A$ a subring of $B$ such that $B^p\subset A\subset B$, and $(x_{\alpha }{)}_{\alpha \in I}$ a $p$-basis of $B$ over $A$. Let $L$ be a $B$-module. Then:

(i) For a derivation $D$ of $A$ into $L$ to extend to a derivation of $B$ into $L$, it is necessary and sufficient that $D$ vanish on $B^p$.

(ii) If $D$ vanishes on $B^p$, then, for every family $(y_{\alpha }{)}_{\alpha \in I}$ of elements of $L$, there exists one and only one derivation $D^{\prime }$ of $B$ into $L$ extending $D$ and such that $D^{\prime }(x_{\alpha })=y_{\alpha }$ for every $\alpha \in I$.

Given a derivation $D$ of $A$ into $L$, consider the ring $E=D_B(L)$ and the homomorphism $u:A\to E$ defined by $u(a)=(a,D(a))$; an $A$-homomorphism $v:B\to E$ right inverse to the canonical homomorphism $q:E\to B$ is then of the form $x\mapsto (x,D^{\prime }(x))$, where $D^{\prime }$ is a derivation of $B$ into $L$ extending $D$ (20.1.5); as $u(q(z{)}^p)=(q(z{)}^p,pq(z{)}^{p-1}{D}^{\prime }(q(z)))$ for $z\in E$, the corollary follows immediately from (21.2.1) applied to $E$.

It follows from (21.2.3) that the sequence

  0 → Der_A(B, L) → Der_{B^p}(B, L) → Der_{B^p}(A, L) → 0                            (21.2.3.1)

(cf. (20.2.2.1)) is exact, and that the map

$$D^{\prime }\mapsto (D^{\prime }(x_{\alpha }){)}_{\alpha \in I}(\mathrm{21.2.3.2})$$

is an isomorphism of the $B$-module ${\mathrm{Der}}_A(B,L)$ onto the $B$-module product $L^I$ (by taking $D=0$ in (21.2.3)).

Corollary (21.2.4).

Under the hypotheses of (21.2.3), the sequence

  0 → Ω_{A/B^p} ⊗_A B → Ω_{B/B^p} → Ω_{B/A} → 0                                      (21.2.4.1)

is exact and split, and the family $(d_{B/A}(x_{\alpha }){)}_{\alpha \in I}$ is a basis of the $B$-module ${\Omega }_{B/A}$.

This follows at once from (21.2.3) and from the formula ${\mathrm{Der}}_A(B,L)={\mathrm{Hom}}_B({\Omega }_{B/A},L)$ (20.4.8).

Corollary (21.2.5).

Let $A$ be a ring, $B$ an $A$-algebra, $(x_{\alpha }{)}_{\alpha \in I}$ a $p$-basis of $B$ over $A$. Then $(d_{B/A}(x_{\alpha }){)}_{\alpha \in I}$ is a basis of the $B$-module ${\Omega }_{B/A}$.

Using (21.1.5, (ii)), one reduces to the case where $A=A[B^p]$, and it then suffices to apply (21.2.4).

(21.2.6) Let $A$, $B$ be two rings, $u:A\to B$ a ring homomorphism; one has (with the notation of (21.1.4)) a commutative diagram

  A^{[p]} —u^{[p]}→ B^{[p]}
    │                │
    F_A              F_B                                                             (21.2.6.1)
    │                │
    ↓                ↓
    A   ———u———→    B.

One therefore deduces canonically a homomorphism of $B$-algebras

  u ⊗ 1 : A^{[p]} ⊗_A B → B^{[p]}                                                    (21.2.6.2)

whose image is the ring $A[B^p]$ (which is a $B$-algebra by the homomorphism $F_B:B\to B^{[p]}$).

Theorem (21.2.7).

Let $A$ be a ring, $B$ an $A$-algebra, $u:A\to B$ the structural homomorphism. Suppose verified the following conditions:

(i) The homomorphism (21.2.6.2) deduced from $u$ is injective.

(ii) $B$ admits a $p$-basis $(x_{\alpha }{)}_{\alpha \in I}$ relative to $A$.

Then $B$ is an $A$-algebra formally smooth (for the discrete topologies). More precisely, equip $A$ and $B$ with the discrete topologies; let $E$ be an admissible topological $A$-algebra $(0_I,\mathrm{7.1.2})$, $\mathfrak{K}$ an ideal of definition of $E$, $C=E/\mathfrak{K}$, $v:B\to C$ an $A$-homomorphism, $q:E\to C$ the augmentation. Then, for every family $(z_{\alpha }{)}_{\alpha \in I}$ of elements of $E$ such that $v(x_{\alpha })=q(z_{\alpha })$ for every $\alpha \in I$, there exists one and only one $A$-homomorphism $w:B\to E$ such that $v=q\circ w$ and $w(x_{\alpha })=z_{\alpha }$ for every $\alpha \in I$.

Consider first the case where $E$ is discrete, hence $\mathfrak{K}$ nilpotent. One can restrict to the case where $v$ is surjective; moreover, the reasoning of (19.4.3) permits supposing ${\mathfrak{K}}^2=0$ and consequently ${\mathfrak{K}}^2=0$. Finally, by considering the inverse image by $v$ of the extension $E$ of $C$ by $\mathfrak{K}$, one can restrict to the case where $C=B$ and $v$ is the identity (19.4.4). Since the ring homomorphism $F_E:E\to E$ vanishes on $\mathfrak{K}$, it factors as $E\to B\to E$, and $F^{\prime }$, considered as homomorphism of $B$ into $E^{[p]}$, is an $A$-homomorphism by definition of the $A$-algebra structure of $E^{[p]}$ by means of the composite homomorphism $A\to E\to E^{[p]}$, where $r:A\to E$ is the structural homomorphism of the $A$-algebra $E$; moreover $r:A^{[p]}\to E^{[p]}$ is also an $A$-homomorphism. There is therefore a unique $A$-homomorphism $f:A^{[p]}{\otimes }_{A}B\to E^{[p]}$ such that the composites of $f$ with the canonical homomorphisms $A^{[p]}\to A^{[p]}{\otimes }_{A}B$ and $B\to A^{[p]}{\otimes }_{A}B$ are respectively $r$ and $F^{\prime }$. Now, by hypothesis, one can identify $A^{[p]}{\otimes }_{A}B$ with $A[B^p]$, the canonical homomorphisms $A^{[p]}\to A^{[p]}{\otimes }_{A}B$ and $B\to A^{[p]}{\otimes }_{A}B$ identifying respectively with $u$ and F_B. One can therefore now consider $E$ as an $A[B^p]$-algebra by means of the homomorphism $f$, and, by construction, one has $f(q(z{)}^p)=z^p$ for every $z\in E$; one is thus in the conditions of application of (21.2.1), whence the theorem in this case.

To pass to the general case, consider a fundamental system $({\mathfrak{J}}_{\lambda })$ of open ideals of $E$, and set $E_{\lambda }=E/{\mathfrak{J}}_{\lambda }$, ${\mathfrak{K}}_{\lambda }=(\mathfrak{K}+{\mathfrak{J}}_{\lambda })/{\mathfrak{J}}_{\lambda }$, $C_{\lambda }=E_{\lambda }/{\mathfrak{K}}_{\lambda }=E/(\mathfrak{K}+{\mathfrak{J}}_{\lambda })$; as $E$ is admissible, one has $E=\lim E_{\lambda }$ $(0_I,\mathrm{7.2.2})$. For every pair $(\alpha ,\lambda )$, denote by $z_{\alpha ,\lambda }$ the canonical image of $z_{\alpha }$ in $E_{\lambda }$, by $p_{\lambda }:C\to C_{\lambda }$ the canonical homomorphism,

by $q_{\lambda }$ the canonical homomorphism $E_{\lambda }\to C_{\lambda }$. As by hypothesis ${\mathfrak{K}}_{\lambda }$ is nilpotent in $E_{\lambda }$, the first part of the proof proves the existence and uniqueness of an $A$-homomorphism $w_{\lambda }:B\to E_{\lambda }$ such that $p_{\lambda }\circ v=q_{\lambda }\circ w_{\lambda }$ and $w_{\lambda }(x_{\alpha })=z_{\alpha ,\lambda }$ for every $\alpha$. The uniqueness of the $w_{\lambda }$ then shows immediately that $(w_{\lambda })$ is a projective system of homomorphisms, and $w=\lim w_{\lambda }$ answers the question.

Remark (21.2.8).

The verification of the existence and uniqueness of the homomorphism $w:B\to E$ such that $w(x_{\alpha })=z_{\alpha }$ for every $\alpha$ can be deduced directly from the fact that $B$ is an $A$-algebra formally smooth and from the fact that the $d_{B/A}(x_{\alpha })$ form a basis of the $B$-module ${\Omega }_{B/A}$ (21.2.4), without bringing in the fact that one deals with a $p$-basis (so that the result is valid without supposing the rings of characteristic $p$).

In fact, one can restrict to the case where ${\mathfrak{K}}^2=0$; as ${\mathrm{Der}}_A(B,\mathfrak{K})={\mathrm{Hom}}_B({\Omega }_{B/A},\mathfrak{K})$ by (20.4.8), there exists one and only one $A$-derivation $D$ of $B$ into $\mathfrak{K}$ such that $D(x_{\alpha })=t_{\alpha }$ for every family $(t_{\alpha })$ of elements of $\mathfrak{K}$; the conclusion follows from (20.1.1).

21.3. $p$-bases and imperfection modules

(21.3.1) Let $A$, $B$ be two rings (of characteristic $p$), $i:A\to B$, $j:B\to A$ two ring homomorphisms such that one has

  j(i(a)) = a^p              for every a ∈ A,                                        (21.3.1.1)
  i(j(b)) = b^p              for every b ∈ B.                                        (21.3.1.2)

Most often, $i$ will be injective, so that $A$ will be identified by $i:A\to B$ to a subring of $B$; once this identification is made, the existence of $j$ implies that $B^p\subset A$, and $j$ is then identified with F_B.

(21.3.2) If $h:A^p\to A$ is the canonical injection, one has, by (21.3.1.1), a commutative diagram

  B ——j——→ A
  │         │
  i         h                                                                        (21.3.2.1)
  │         │
  ↓         ↓
  A ——F_A→ A^p

so that the pair $(j,F_A)$ may be considered as a di-homomorphism of the $A$-algebra $B$ (for $i$) into the $A^p$-algebra $A$ (for $h$). One deduces a canonical homomorphism of $A$-modules

  π_{B/A} : Ω_{B/A} ⊗_{B, j} A → Ω_A = Ω^1_A                                         (21.3.2.2)

(cf. (20.5.4); the identification of ${\Omega }_{A/A^p}^1$ and of the module of absolute differentials ${\Omega }_A^1$ comes from (21.1.5, (ii))).

One sets

  Θ_{B/A} = Ω_{B/A} ⊗_{B, j} A                                                       (21.3.2.3)
  Ξ_{B/A} = Ker(π_{B/A})                                                             (21.3.2.4)

so that one has the exact sequence

$$0\to {\Xi }_{B/A}\to {\Theta }_{B/A}\to {\Omega }_A^1(\mathrm{21.3.2.5})$$

(one notes that this notation may lead to confusion since ${\Theta }_{B/A}$ and ${\pi }_{B/A}$ depend not only on $A$ and $B$, but also on $i$ and $j$).

(21.3.3) Since by (21.3.1.2), one has $F_B=i\circ j$, one can write for every $B$-module $M$ (cf. (21.1.4))

  M^{[p]} = M ⊗_B B = (M ⊗_B A) ⊗_A B                                                (21.3.3.1)

whence in particular

  (Ω_{B/A})^{[p]} = Θ_{B/A} ⊗_A B                                                    (21.3.3.2)

and one deduces from (21.3.2.2) a canonical homomorphism of $B$-modules

  ω_{B/A} : (Ω_{B/A})^{[p]} → Ω^1_A ⊗_A B.                                           (21.3.3.3)

Taking account of (20.5.4.2), the image of this homomorphism is the $B$-module generated by the elements $d_A(j(b))\otimes 1$ for $b\in B$. Their image by the canonical homomorphism $i_{B/A}:{\Omega }_A^1{\otimes }_{A}B\to {\Omega }_B^1$ deduced from $i$ (20.5.2) is therefore in the sub-$B$-module of ${\Omega }_B^1$ generated by the $d_B(i(j(b)))$ for $b\in B$; by virtue of (21.3.1.2), this image is zero. In other words, one has a sequence of homomorphisms

  0 → Ξ_{B/A} ⊗_A B → (Ω_{B/A})^{[p]} → Υ_{B/A}                                      (21.3.3.4)

which is not necessarily exact, but in which the composite of two consecutive homomorphisms is zero.

Proposition (21.3.4).

If $B$ is an $A$-module flat (for $i$), the sequence (21.3.3.4) is exact.

This follows from the definition of flatness.

Proposition (21.3.4) applies in particular when one has $B^p\subset A\subset B$, $i$ and $j$ being respectively the canonical injection and F_B, and $B$ admits a $p$-basis over $A$ (so that $B$ is then a free $A$-module). But even in this case the kernel ${\Xi }_{B/A}$ is not necessarily zero. Nevertheless:

Proposition (21.3.5).

Let $B$ be a reduced ring, $A$ a subring of $B$ such that $B^p\subset A\subset B$. Suppose there exists a $p$-basis $(x_{\lambda }{)}_{\lambda \in L}$ of $B$ over $A$ and a $p$-basis $(y_{\mu }{)}_{\mu \in M}$ of $A$ over $A^p$; then the canonical homomorphism (21.3.3.4)

$$({\Omega }_{B/A}{)}^{[p]}\to {{\rm Y}}_{B/A}(\mathrm{21.3.5.1})$$

is bijective, and the elements $d_B(x_{\lambda })\otimes 1$ form a basis of the $B$-module ${{\rm Y}}_{B/A}$.

Since $B$ is reduced, $x\mapsto x^p$ is an isomorphism of $B$ onto $B^p$, and by transport of structure by means of this isomorphism, one sees that $(x_{\lambda }^p{)}_{\lambda \in L}$ is a $p$-basis of $B^p$ over $A^p$; one concludes that the $x_{\lambda }^p$ and the $y_{\mu }$ form a $p$-basis of $A$ over $A^p$ (21.1.10); consequently (21.1.5, (ii) and 21.2.5) the $d_A(x_{\lambda }^p)$ and the $d_A(y_{\mu })$ form a basis of the $A$-module ${\Omega }_A^1$; hence the $d_A(x_{\lambda }^p)\otimes 1$ and $d_A(y_{\mu })\otimes 1$ form a basis of the $B$-module ${\Omega }_A^1{\otimes }_{A}B$. Now, the image of $d_A(x_{\lambda }^p)\otimes 1$ by $i_{B/A}$ is $d_B(x_{\lambda }^p)=0$; on the other hand, the image of $d_A(y_{\mu })\otimes 1$ by $i_{B/A}$

is $d_B(y_{\mu })$, and as the $y_{\mu }$ and the $x_{\lambda }$ form a $p$-basis of $B$ over $B^p$ (21.1.10), the $d_B(y_{\mu })$ are linearly independent (over $B$) in ${\Omega }_B^1$ (21.2.5); one deduces at once that the kernel of $i_{B/A}$ has a basis formed of the $d_A(x_{\lambda }^p)\otimes 1$; as these are the images by (21.3.5.1) of the elements $d_B(x_{\lambda })\otimes 1$ in ${\Omega }_{B/A}{\otimes }_B{B}^{[p]}$, and as the latter form a basis of the $B$-module ${\Omega }_{B/A}{\otimes }_B{B}^{[p]}$ (21.2.5), the homomorphism (21.3.5.1) of $B$-modules is bijective.

Remarks (21.3.6).

(i) Let $A^{\prime }$, $B^{\prime }$ be two rings of characteristic $p$, and suppose one has two homomorphisms $i^{\prime }:A^{\prime }\to B^{\prime }$, $j^{\prime }:B^{\prime }\to A^{\prime }$ verifying (21.3.1.1) and (21.3.1.2), and ring homomorphisms $f:A\to A^{\prime }$, $g:B\to B^{\prime }$ making the diagram

  A'  ——i'——→ B'  ——j'——→ A'
  ↑           ↑           ↑
  f           g           f
  │           │           │
  A   ——i——→  B   ——j——→  A

commute. Then the canonical homomorphism ${\Omega }_{B/A}\to {\Omega }_{B^{\prime }/A^{\prime }}$ (20.5.4.3) gives a canonical di-homomorphism ${\Theta }_{B/A}\to {\Theta }_{B^{\prime }/A^{\prime }}$ making the diagram

$${\pi }_{B/A}0\to {\Xi }_{B/A}\to {\Theta }_{B/A}\to {\Omega }_A^1\downarrow \downarrow \downarrow 0\to {\Xi }_{B^{\prime }/A^{\prime }}\to {\Theta }_{B^{\prime }/A^{\prime }}\to {\Omega }_{A^{\prime }}^1{\pi }_{B^{\prime }/A^{\prime }}$$

commute.

(ii) Let $A^{\prime }$ be any $A$-algebra, and set $B^{\prime }=B{\otimes }_A{A}^{\prime }$; then $i^{\prime }=i\otimes 1:A^{\prime }\to B^{\prime }$ and $j^{\prime }=j\otimes 1:B^{\prime }\to A^{\prime }$ verify (21.3.1.1) and (21.3.1.2), and one has

  Ω_{B'/A'} = Ω_{B/A} ⊗_A A' = Ω_{B/A} ⊗_B B'

(20.5.5); one deduces a canonical $A^{\prime }$-isomorphism

  Θ_{B/A} ⊗_A A' ⥲ Θ_{B'/A'}                                                         (21.3.6.1)

and also a canonical $B^{\prime }$-homomorphism

  Υ_{B/A} ⊗_A A' → Υ_{B'/A'}.                                                        (21.3.6.2)

21.4. Case of field extensions

(21.4.0) Let $K$ be a field of characteristic $p>0$, $k$ a subfield of $K$; then the ring $k[K^p]$ is equal to the field $k(K^p)$ since $k$ is algebraic over $K^p$. One can therefore apply the results of the preceding numbers by replacing throughout $A$, $B$ and $A[B^p]$ by $k$, $K$ and $k(K^p)$.

Lemma (21.4.1).

Let $k$ be a field of characteristic $p>0$, $K$ an extension of $k$. For an element $x\in K$ to be $p$-free over $k$, it is necessary and sufficient that $x\notin k(K^p)$.

In fact, $x$ is a root of the polynomial $X^p-x^p$ of $k(K^p)[X]$, and one knows (Bourbaki, Alg., chap. V, §8, n° 1, prop. 1) that if $x\notin k(K^p)$, this polynomial is irreducible, so that the elements $1,x,\cdots ,x^{p-1}$ form a basis of the $k(K^p)$-module $k(K^p)(x)$.

Theorem (21.4.2).

Let $k$ be a field of characteristic $p>0$, $K$ an extension of $k$, $S$ a system of $p$-generators of $K$ over $k$, $L\subset S$ a part $p$-free over $k$. Then there exists a $p$-basis $B$ of $K$ over $k$ such that $L\subset B\subset S$. In particular, every extension of $k$ admits a $p$-basis over $k$.

One can restrict to the case where $K^p\subset k$. By Zorn's theorem, there exists in $K$ a subset $B$ such that $L\subset B\subset S$, $p$-free over $k$ and maximal among the subsets of $S$ having these properties. It suffices to see that the subfield $K^{\prime }$ of $K$ generated by $k$ and $B$ is equal to $K$. In the contrary case, there would exist $x\in S$ not in $K^{\prime }$; as $K^p\subset k$, one has $K^{\prime }\supset k(K^p)$; hence $x$ would be $p$-free over $K^{\prime }$ (21.4.1), and consequently $B\cup x$ would be $p$-free over $k$ (21.1.10), contrary to the definition of $B$. The last assertion of (21.4.2) is obtained by taking $L=\varnothing$, $S=K$. Q.E.D.

Corollary (21.4.3).

Let $k$ be a field of characteristic $p>0$, $K$ an extension of $k$. For a family $(x_{\alpha })$ of elements of $K$ to be $p$-free over $k$, it is necessary and sufficient that, for every $\alpha$, $x_{\alpha }$ not belong to the field $K_{\alpha }$ generated by $k(K^p)$ and by the $x_{\beta }$ of index $\beta \ne \alpha$.

The condition is necessary by (21.1.10). Conversely, suppose it satisfied; one can restrict to the case where $(x_{\alpha })$ is a system of $p$-generators of $K$. There then exists a sub-family of $(x_{\alpha })$ which is a $p$-basis of $K$ (21.4.2), but this family cannot be distinct from $(x_{\alpha })$, otherwise, by the hypothesis, it would not be a family of $p$-generators of $K$.

Corollary (21.4.4).

Let $k$ be a field of characteristic $p>0$, $K$ an extension of $k$. For ${\Omega }_{K/k}^1=0$, it is necessary and sufficient that $K=k(K^p)$. In particular, for ${\Omega }_K^1=0$, it is necessary and sufficient that $K$ be a perfect field.

In fact, if $(x_{\alpha })$ is a $p$-basis of $K$ over $k$, the $d_K(x_{\alpha })$ form a basis of the $K$-vector space ${\Omega }_{K/k}^1$ (21.2.5).

Theorem (21.4.5).

Let $k$ be a field of characteristic $p>0$, $K$ an extension of $k$, $(x_{\alpha })$ a family of elements of $K$. For $(x_{\alpha })$ to be $p$-free over $k$ (resp. a family of $p$-generators of $K$ over $k$, resp. a $p$-basis of $K$ over $k$), it is necessary and sufficient that the family $(d_{K/k}(x_{\alpha }))$ be a free family (resp. a system of generators, resp. a basis) in the $K$-vector space ${\Omega }_{K/k}^1$.

Let $K^{\prime }$ be the subfield (equal to the subring) of $K$ generated by $k(K^p)$ and the $x_{\alpha }$; taking account of (20.4.7), one sees that the $d_{K^{\prime }/k}(x_{\alpha })$ generate ${\Omega }_{K^{\prime }/k}^1={\Omega }_{K^{\prime }/k(K^p)}^1$. If the $d_{K/k}(x_{\alpha })$ generate ${\Omega }_{K/k}^1={\Omega }_{K/k(K^p)}^1$, the left arrow in the exact sequence (20.5.7.1)

  Ω^1_{K'/k(K^p)} ⊗_{K'} K → Ω^1_{K/k(K^p)} → Ω^1_{K/K'} → 0

is surjective, hence ${\Omega }_{K/K^{\prime }}^1=0$, which implies $K^{\prime }=K$ by (21.4.4).

If now $(x_{\alpha })$ is a $p$-free family, it is a sub-family of a $p$-basis of $K$ over $k$ (21.4.2), hence the $d_K(x_{\alpha })$ form part of a basis of the $K$-vector space ${\Omega }_{K/k}^1$ (21.2.5), and are consequently linearly independent over $K$. Conversely let us prove that if the $d_{K/k}(x_{\alpha })$ are linearly independent over $K$, the family $(x_{\alpha })$ is $p$-free over $k$.

Taking account of (21.4.3), it suffices to see that, for each $\alpha$, $x_{\alpha }$ does not belong to $K_{\alpha }$. Now, in the exact sequence

  Ω^1_{K_α/k(K^p)} ⊗_{K_α} K → Ω^1_{K/k(K^p)} → Ω^1_{K/K_α} → 0,

the images by the left arrow of the $d_{K/k}(x_{\beta })$ for $\beta \ne \alpha$ are the $d_{K_{\alpha }/k}(x_{\beta })$, which therefore generate a sub-vector space of ${\Omega }_{K_{\alpha }/k(K^p)}^1$ not containing $d_{K/k}(x_{\alpha })$, and as this sub-vector space is the kernel of the right arrow, one sees that $d_{K/K_{\alpha }}(x_{\alpha })\ne 0$, hence $x_{\alpha }\notin K_{\alpha }$.

Corollary (21.4.6).

Let $k$ be a field of characteristic $p>0$, $K$ an extension of $k$, $x$ an element of $K$. The three following conditions are equivalent:

a) $x\notin k(K^p)$.

b) $d_{K/k}(x)\ne 0$.

c) The element $x$ is $p$-free over $k$.

Proposition (21.4.7).

Let $L$ be a field of characteristic $p>0$, $K$ a subfield of $L$ such that $L^p\subset K\subset L$. Then the sequence of $L$-vector spaces

  0 → Ω^1_{K/L^p} ⊗_K L → Ω^1_{L/L^p} → Ω^1_{L/K} → 0                                (21.4.7.1)

is exact; in other words, one has ${{\rm Y}}_{L/K/L^p}=0$.

This is a particular case of (21.2.4), taking account of (21.4.2).

Proposition (21.4.8).

Under the hypotheses of (21.4.7), the canonical homomorphism

$$({\Omega }_{L/K}{)}^{[p]}\to {{\rm Y}}_{L/K}(\mathrm{21.4.8.1})$$

is bijective; if $(x_{\alpha })$ is a $p$-basis of $L$ over $K$, the elements $d_L(x_{\alpha })\otimes 1$ form a basis of the $L$-vector space ${{\rm Y}}_{L/K}$.

This is a particular case of (21.3.5).

21.5. Application: separability criteria

In this number and the two following, we no longer suppose that the rings considered are of characteristic $p>0$.

(21.5.1) Note first that the criterion (21.2.7) permits proving a part of Cohen's theorem on separable extensions (19.6.1), namely that if $k$ is of characteristic $p>0$ and if $K$ is a separable extension of $k$, then $K$ is an $A$-algebra formally smooth. In fact, $K$ admits a $p$-basis over $k$ (21.4.2), and on the other hand, it follows from MacLane's criterion (Bourbaki, Alg., chap. V, §8, n° 2, prop. 3) that in an algebraic closure of $K$, $k^{1/p}$ and $K$ are linearly disjoint over $k$, and consequently the canonical homomorphism $k^{1/p}{\otimes }_{k}K\to k^{1/p}(K)$ is bijective, which is precisely condition (i) of (21.2.7), after transport of structure by the isomorphism $k^{1/p}\stackrel{\sim }{\to }k$.

Proposition (21.5.2) (MacLane).

Let $A$, $B$ be two complete discrete valuation rings, $\mathfrak{m}$, $\mathfrak{n}$ their respective maximal ideals, $K=A/\mathfrak{m}$ and $L=B/\mathfrak{n}$ their residue fields, $u:A\to B$ a homomorphism such that $Bu(\mathfrak{m})=\mathfrak{n}$, $u_0=u\otimes 1:K\to L$ the corresponding homomorphism for residue fields. Consider the following conditions:

a) $L$ is a separable extension of $K$ (for $u_0$).

b) For every complete discrete valuation ring $B^{\prime }$ of maximal ideal ${\mathfrak{n}}^{\prime }$ and residue field $L^{\prime }$,

every homomorphism $u^{\prime }:A\to B^{\prime }$ such that $B^{\prime }{u}^{\prime }(\mathfrak{m})={\mathfrak{n}}^{\prime }$, and every $K$-isomorphism $\sigma :L\to L^{\prime }$ (relative to $u_0$ and $u_0^{\prime }=u^{\prime }\otimes 1:K\to L^{\prime }$), there exists an isomorphism $w:B\to B^{\prime }$ such that $u^{\prime }=w\circ u$ and that $w_0=w\otimes 1$ be equal to $\sigma$.

b') For every homomorphism $u^{\prime }:A\to B$ such that $B{u}^{\prime }(\mathfrak{m})=\mathfrak{n}$ and such that $u_0^{\prime }=u^{\prime }\otimes 1:K\to L$ be equal to $u_0$, there exists an automorphism $w$ of $B$ such that $u^{\prime }=w\circ u$ and that $w_0=w\otimes 1:L\to L$ be the identity.

c) Denoting by $u_1:A/{\mathfrak{m}}^2\to B/{\mathfrak{n}}^2$ the homomorphism deduced from $u$ by passage to quotients, then, for every local homomorphism $u_1^{\prime }:A/{\mathfrak{m}}^2\to B/{\mathfrak{n}}^2$ such that $g{r}_0(u_1^{\prime })=g{r}_0(u_1)(=u_0)$, there exists an automorphism $w_1$ of $B/{\mathfrak{n}}^2$ such that $g{r}_0(w_1)$ be the identity and that $u_1^{\prime }=w_1\circ u_1$.

c') For every local homomorphism $u_1^{\prime }:A/{\mathfrak{m}}^2\to B/{\mathfrak{n}}^2$ such that $g{r}_0(u_1^{\prime })=g{r}_0(u_1)$ and $g{r}_1(u_1^{\prime })=g{r}_1(u_1)$ (homomorphism of $\mathfrak{m}/{\mathfrak{m}}^2$ into $\mathfrak{n}/{\mathfrak{n}}^2$), there exists an automorphism $w_1$ of $B/{\mathfrak{n}}^2$ such that $g{r}_0(w_1)$ be the identity and that $u_1^{\prime }=w_1\circ u_1$.

Then one has the implications c) ⇔ c') ⇔ a) ⇒ b) ⇒ b').

If moreover $A$ is a Cohen ring, the five preceding conditions are equivalent.

The implications b) ⇒ b') and c) ⇒ c') are trivial. Let us show that a) implies b). The homomorphism $u$ makes $B$ an $A$-module without torsion, since it transforms by hypothesis a uniformizer of $A$ into a uniformizer of $B$; it follows that $B$ is a flat $A$-module $(0_I,\mathrm{6.3.4})$, hence a Cohen $A$-algebra (19.8.1); the fact that a) implies b) is then a consequence of (19.8.2, (i)) applied with $C=B^{\prime }$, $\mathfrak{J}={\mathfrak{n}}^{\prime }$, $B^{\prime }$ being considered as $A$-algebra for $u^{\prime }$ and the homomorphism $B\to B^{\prime }/{\mathfrak{n}}^{\prime }$ being the composite $B\to B/\mathfrak{n}\to B^{\prime }/{\mathfrak{n}}^{\prime }$, which is an $A$-homomorphism by virtue of the hypothesis $u_0^{\prime }=\sigma \circ u_0$. The same reasoning proves that a) entails c), by taking this time $C=B/{\mathfrak{n}}^2$, $\mathfrak{J}=\mathfrak{n}/{\mathfrak{n}}^2$, $C$ being considered as $A$-algebra for $u_1^{\prime }$.

Let us prove in the second place that c') implies a). The two homomorphisms $u_1$ and $u_1^{\prime }$ are such that $u_1^{\prime }=u_1+D$, where $D$ is a derivation of $A/{\mathfrak{m}}^2$ into $\mathfrak{n}/{\mathfrak{n}}^2$ (20.1.1); moreover, the hypothesis $g{r}_1(u_1^{\prime })=g{r}_1(u_1)$ means that $D$ vanishes on $\mathfrak{m}/{\mathfrak{m}}^2$, and can consequently be considered as a derivation of $K=A/\mathfrak{m}$ into $\mathfrak{n}/{\mathfrak{n}}^2$, and $\mathfrak{n}/{\mathfrak{n}}^2$ is identified (by choice of a uniformizer of $B$) with the $K$-module $L$ (for $u_0$). On the other hand, the conditions imposed on $w_1$ entail that $g{r}_1(w_1)$ is also the identity (since $u_1(\mathfrak{m}/{\mathfrak{m}}^2)$ generates $\mathfrak{n}/{\mathfrak{n}}^2$); $w_1$ is therefore of the form $z\mapsto z+D^{\prime }(z)$, where $D^{\prime }$ is this time a derivation of $B/{\mathfrak{n}}^2$ into the $(B/{\mathfrak{n}}^2)$-module $\mathfrak{n}/{\mathfrak{n}}^2$; as $w_1$ is the identity on $\mathfrak{n}/{\mathfrak{n}}^2$, $D^{\prime }$ can again (by the preceding identification of $\mathfrak{n}/{\mathfrak{n}}^2$ and of $L$) be considered as a derivation of $L$ into $L$; finally, the relation $u_1^{\prime }=w_1\circ u_1$ means that $D^{\prime }$ extends $D$. Note on the other hand that every derivation $D$ of $K$ into $L$ corresponds to a homomorphism $u_1^{\prime }$ verifying the conditions of c') (20.1.1); condition c') means therefore that every derivation of $K$ into $L$ extends to a derivation of $L$ into itself, that is to say (20.6.5) that $L$ is separable over $K$.

Let us prove finally that, when $A$ is a Cohen ring, b') entails c'). Let us prove first that under the hypotheses of c'), there exists a homomorphism $u^{\prime }:A\to B$ which verifies the hypotheses of b') and, by passage to quotients, yields $u_1^{\prime }:A/{\mathfrak{m}}^2\to B/{\mathfrak{n}}^2$. In fact, this follows from (19.8.6, (i)) applied to the composite homomorphism $v^{\prime }:A\to A/{\mathfrak{m}}^2\to B/{\mathfrak{n}}^2$;

this last factors as $A\to B\to B/{\mathfrak{n}}^2$ and the hypotheses on $g{r}_0(u_1^{\prime })$ and $g{r}_1(u_1^{\prime })$ entail $g{r}_0(u^{\prime })=g{r}_0(u)=u_0$ and $g{r}_1(u^{\prime })=g{r}_1(u)$, hence the image by $u^{\prime }$ of a uniformizer of $A$ is a uniformizer of $B$, and one has consequently $B{u}^{\prime }(\mathfrak{m})=\mathfrak{n}$. It then suffices, in order to obtain an automorphism $u_1^{\prime }$ answering the question, to take the automorphism deduced by passage to quotients from the automorphism $w$ of $B$ furnished by the application of b') to the homomorphism $u^{\prime }$.

Remarks (21.5.3).

(i) The differential properties of fields permit solving the question of uniqueness of the field of representatives in a complete Noetherian local ring $C$ (19.8.7, (ii)). Let in fact $\mathfrak{J}$ be the maximal ideal of $C$, $K=C/\mathfrak{J}$ the residue field of $C$; one can restrict to the case $\mathfrak{J}\ne 0$. Suppose there exists a homomorphism $u:K\to C$ which, composed with the augmentation $C\to K$, yields the identity; then, for this homomorphism to be unique, it is necessary and sufficient that ${\Omega }_K^1=0$. In fact, the condition ${\Omega }_K^1=0$ entails that $K$ is formally unramified over its prime field (20.7.4); if there existed a second homomorphism $v\ne u$ answering the question, there would be a greatest integer $n$ such that ${\mathfrak{J}}^n$ contains the set of $u(x)-v(x)$ for $x\in K$; by passage to quotient, $u$ and $v$ would yield two distinct homomorphisms u', v' of $K$ into $C/{\mathfrak{J}}^{n+1}$, whose composites with $C/{\mathfrak{J}}^{n+1}\to C/{\mathfrak{J}}^n$ would be equal, which contradicts the definition (19.10.2). Conversely, suppose ${\Omega }_K^1\ne 0$; there then exists a derivation $D\ne 0$ of $K$ into $\mathfrak{J}/{\mathfrak{J}}^2$ (20.4.8), hence a homomorphism $v_1:K\to C/{\mathfrak{J}}^2$ such that, if $u_1:K\to C/{\mathfrak{J}}^2$ is obtained by passage to quotient from $u$, one has $v_1=u_1+D$ (20.1.1). If $k$ is the prime field of $K$, $C$ is a $k$-algebra and $K$ a $k$-algebra formally smooth (19.6.1), and the restrictions of $u_1$ and $v_1$ to $k$ coincide, hence $v_1$ factors as $K\to C\to C/{\mathfrak{J}}^2$ and one has $v\ne u$.

Let us recall ((20.6.20) and (21.4.4)) that the condition ${\Omega }_K^1=0$ means that $K$ is perfect if it is of characteristic $\ne 0$, and an algebraic extension of $\mathbb{Q}$ if it is of characteristic 0.

(ii) In the same manner, let $W$ be a Cohen ring, $C$ a complete Noetherian local ring, $\mathfrak{J}\ne 0$ an ideal of $C$ contained in the maximal ideal; for the factorization $W\to C\to C/\mathfrak{J}$ in (19.8.6, (i)) to be unique, it is necessary and sufficient that the residue field $K$ of the Cohen ring $W$ satisfy ${\Omega }_K^1=0$. In fact, if ${\Omega }_K^1\ne 0$ and if $\mathfrak{m}$ denotes the maximal ideal of $W$, it suffices to compose a derivation $D\ne 0$ of $K$ into $\mathfrak{J}/\mathfrak{mJ}$ with the augmentation $W\to K$ to obtain a derivation $D^{\prime }\ne 0$ of $W$ into $\mathfrak{J}/\mathfrak{mJ}$, and one finishes the reasoning as in (i) by forming with the aid of $D^{\prime }$ a homomorphism $v_1:W\to C/\mathfrak{mJ}$ distinct from the homomorphism $u_1:W\to C/\mathfrak{mJ}$ deduced from $u$ by passage to quotient; one completes the reasoning by invoking this time (19.8.6, (i)). If on the contrary ${\Omega }_K^1=0$, the uniqueness of $v$ follows already from (i) when $W=K$ is a field of characteristic 0. In the contrary case, one has ${\Omega }_W^1=0$; in fact, one then has $\mathfrak{m}=pW$, $p$ being the characteristic of $K$ (19.8.5), and the canonical homomorphism $\mathfrak{m}/{\mathfrak{m}}^2\to {\Omega }_W^1/\mathfrak{m}{\Omega }_W^1$ (20.5.11.2) is consequently zero. The exact sequence (20.5.12.1) applied to $W$ and to $K=W/\mathfrak{m}$ then entails that ${\Omega }_W^1=\mathfrak{m}{\Omega }_W^1$, whence our assertion. But then (20.7.4) $W$ is formally unramified (for its adic topology) over $\mathbb{Z}$, and the uniqueness of $v$ is proved as in (i).