§ 10. Derivations

In this paragraph, and unless otherwise mentioned, the algebras considered are not assumed to be associative nor necessarily to possess a unit element; K denotes a commutative ring.

1. Commutation factors

When in this paragraph we speak of graduations without specifying them, we shall always mean graduations of type $\Delta$ , where $\Delta$ is a commutative group written additively. In this paragraph, a commutation factor over $\Delta$ with values in $\mathbf{Z}$ is called a commutation factor over $\Delta$ (§ 4, no. 7, Definition 6). A commutation factor over $\Delta$ is therefore identified with a mapping $\varepsilon : (\alpha, \beta) \mapsto \varepsilon_{\alpha \beta} = \varepsilon(\alpha, \beta)$ of $\Delta \times \Delta$ into the multiplicative group $\{-1, 1\}$ such that for $\alpha, \alpha', \beta, \beta'$ in $\Delta$ ,

\begin{cases} \varepsilon(\alpha + \alpha', \beta) = \varepsilon(\alpha, \beta)\varepsilon(\alpha', \beta) \\ \varepsilon(\alpha, \beta + \beta') = \varepsilon(\alpha, \beta)\varepsilon(\alpha, \beta') \\ \varepsilon(\beta, \alpha) = \varepsilon(\alpha, \beta). \end{cases}

(1)

It follows that $\varepsilon(2\alpha, \beta) = \varepsilon(\alpha, 2\beta) = 1$ .

When $\Delta = \mathbf{Z}$ , every commutation factor $\varepsilon$ is determined by giving $\varepsilon(1, 1)$ ; there are therefore only two such factors, the first defined by

(2)

\varepsilon(p, q) = 1 \quad \text{for } p, q \text{ in } \mathbf{Z}

and the second by (3)

\varepsilon(p, q) = (-1)^{pq} \quad \text{for } p, q \text{ in } \mathbf{Z}.

2. General definition of derivations

Consider a commutative ring $K$ , six graded $K$ -modules of type $\Delta : A, A', A'', B, B', B''$ , and three $K$ -linear mappings

\mu : A \times A' \to A'', \qquad \lambda_1 : B \times A' \to B'', \qquad \lambda_2 : A \times B' \to B''

such that the corresponding

K

-linear mappings

A \otimes_K A' \to A'', \qquad B \otimes_K A' \to B'', \qquad A \otimes_K B' \to B''

are graded of degree 0. The image

\mu(a, a')

for

a \in A, a' \in A'

is simply denoted by

a.a'

or even

aa'

, and similarly for the two other bilinear mappings. The degree of

a.a'

is therefore the sum of the degrees of

a

and

a'

Definition 1

Given the above and a commutation factor $\varepsilon$ on $\Delta \times \Delta$ , an $\varepsilon$ -derivation (or $(K, \varepsilon)$ -derivation) of degree $\delta \in \Delta$ of $(A, A', A'')$ into $(B, B', B'')$ is a triple of graded $K$ -linear mappings of degree $\delta$ :*

d : A \to B, \qquad d' : A' \to B', \qquad d'' : A'' \to B''

such that, for every homogeneous element $a \in A$ and every element $a' \in A'$ (4)

d''(a.a') = (da).a' + \varepsilon_{\delta, \deg(a)}a.(d'a').

Cited by 3

Cited by Exercise 1, § 10, Exercise 2, § 10, Example 5

It obviously suffices by linearity to verify relation (4) when $a$ and $a'$ run through respective generating systems of $A$ and $A'$ .

It is often convenient to denote the three mappings $d, d', d''$ by the same letter $d$ (which can be justified by denoting equally by $d$ the graded $K$ -linear mapping of degree $\delta$ )

(a, a', a'') \mapsto (da, d'a', d''a'')

of $A \oplus A' \oplus A''$ into $B \oplus B' \oplus B''$ ). Relation (4) can then be written more simply

d(a.a') = (da).a' + \varepsilon_{\delta,\deg(a)} a.(da').

The

\varepsilon

-derivations of

(A, A', A'')

into

(B, B', B'')

of given degree form a sub-K-module of the K-module of graded linear mappings

\operatorname{Hom}_{\mathbf{K}}(A \oplus A' \oplus A'', B \oplus B' \oplus B'').

When

\varepsilon(\alpha, \beta) = 1

for all

\alpha, \beta

\Delta

, we say simply derivation (or K-derivation) instead of

\varepsilon

-derivation. Derivations form a sub-K-module of

\operatorname{Hom}_{\mathbf{K}}(A \oplus A' \oplus A'', B \oplus B' \oplus B'').

When

\Delta = \mathbf{Z}

and

\varepsilon(p.q) = (-1)^{pq}

, every

\varepsilon

-derivation of even degree is a derivation; every

\varepsilon

-derivation of odd degree is often called an antiderivation (or K-antiderivation); an antiderivation

d

therefore satisfies

d(a.a') = (da).a' + (-1)^{\deg(a)} a.(da')

for a homogeneous element

a \in A

Remark

(1) The notion of derivation can be defined for non-graded modules by agreeing to give these modules the trivial graduation.

(2) If only $\varepsilon$ -derivations of given degree $\delta$ are considered, the commutation factor $\varepsilon$ may be disposed of as follows: the bilinear mapping $\lambda_2 : A \times B' \to B''$ is modified by replacing it by the bilinear mapping $\lambda'_2 : A \times B' \to B''$ such that, for every homogeneous $a$ in $A$ and all $b' \in B'$ ,

\lambda'_2(a, b') = \varepsilon_{\delta, \deg(a)} \lambda_2(a, b').

Then

d

is a derivation relative to the bilinear mappings

\mu, \lambda_1, \lambda'_2

The general definition of $\varepsilon$ -derivations given above is especially used in two cases:

Case (I)*: $A = B, A' = B', A'' = B''$ and the three bilinear mappings $\mu, \lambda_1, \lambda_2$ are equal to the same mapping.

Case (II)*: $A = A' = A'', B = B' = B''$ , so that (for $\mu$ ) $A$ is a graded algebra and the two K-bilinear mappings.

\lambda_1 : B \times A \to B, \quad \lambda_2 : A \times B \to B

are such that the corresponding K-linear mappings

B \otimes_K A \to B, A \otimes_K B \to B

are graded of degree 0. An

\varepsilon

-derivation of degree

\delta

A

into

B

is then a graded K-linear mapping

d : A \to B

of degree

\delta

, such that for every homogeneous

x

A

and every

y \in A

, we have the relation

d(xy) = (dx)y + \varepsilon_{\delta, \deg(a)} x(dy).

Consider in particular in case (II) the case where A is a unital associative K-algebra and $\lambda_1$ and $\lambda_2$ are the external laws of an (A, A)-bimodule (§ 4, no. 3, Definition 3). This holds notably when A and B are two unital associative K-algebras, a unital homomorphism of graded K-algebras $\rho : A \to B$ is given and an (A, A)-bimodule structure is considered on B defined by the two external laws

\lambda_1 : (b, a) \mapsto b \rho(a), \qquad \lambda_2 : (a, b) \mapsto \rho(a)b

for $a \in A, b \in B$ .

Cases (I) and (II) have the following case in common: consider a graded K-algebra A, take $B = A$ , mappings (7) both being multiplication on A. We then speak of an $\varepsilon$ -derivation (or (K, $\varepsilon$ )-derivation) of the graded algebra A: it is a graded K-linear mapping of A into itself, of degree $\delta$ , satisfying (8) for every homogeneous $x$ in A and all $y \in A$ . In particular if A is a graded ring, considered as an (associative) $\mathbf{Z}$ -algebra, we speak of the $\varepsilon$ -derivation of the ring A.

Let A be a unital commutative associative K-algebra and B an A-module; when we speak of a derivation of A into B, it will always be understood that we mean with the A-bimodule structure on B derived from its A-module structure; then the formula

d(xy) = x(dy) + y(dx) \quad \text{for} \quad x \in A, y \in A

holds for such a derivation $d : A \to B$ .

Cites 1

Cites Definition 3, III § 4

3. Examples of derivations

Example 1

Let A be an $\mathbf{R}$ -algebra of differentiable mappings of $\mathbf{R}$ into $\mathbf{R}$ and let $x_0$ be a point of $\mathbf{R}$ ; $\mathbf{R}$ can be considered as an A-module with the external law $(f, a) \mapsto f(x_0)a$ . Then the mapping $f \mapsto Df(x_0)$ is a derivation, since (Functions of a Real Variable, I, § 1, no. 3)

(D(fg))(x_0) = (Df(x_0))g(x_0) + f(x_0)(Dg(x_0)).*

Cited by 1

Cited by Corollary

Example 2

Let X be a differentiable manifold of class $C^\infty$ and let A be the graded $\mathbf{R}$ -algebra of differential forms on X. The mapping which associates with every differential form $\omega$ on X its exterior differential $d\omega$ is an anti-derivation of degree +1 (Differentiable and Analytic Manifolds, R, § 8).

Cited by 1

Cited by § 10. Derivations

Example 3

Let A be an associative K-algebra. For all $a \in A$ , the mapping $x \mapsto ax - xa$ is a derivation of the algebra A (cf. no. 6).

Example 4

Let M be a K-module and A the exterior algebra $\bigwedge(M^*)$ with its usual graduation (§ 7, no. 1). *It will be seen in § 11, no. 9 that, for all $x \in M$ , the right interior product $i(x)$ is an antiderivation of A of degree $-1$ .

Cites 2

Cites III § 11. Cogebras, products of multilinear forms, inner products and duality, III § 7. Exterior algebras

Example 5

Returning to the general situation of Definition 1 of no. 2, let $\overline{K}$ be another commutative ring and $\rho : K \to \overline{K}$ a ring homomorphism; let $\overline{A}, \overline{A}', \overline{A}'', \overline{B}, \overline{B}', \overline{B}''$ denote the graded $\overline{K}$ -modules obtained respectively from $A, A', A'', B, B', B''$ by extending the ring of scalars to $\overline{K}$ (II, § 11, no. 5); we derive from $\mu, \lambda_1$ and $\lambda_2$ $\overline{K}$ -bilinear mappings

\bar{\mu} : \overline{A} \times \overline{A}' \to \overline{A}'', \quad \bar{\lambda}_1 : \overline{B} \times \overline{A}' \to \overline{B}'', \quad \bar{\lambda}_2 : \overline{A} \times \overline{B}' \to \overline{B}''

by considering the tensor products by

l_{\overline{K}}

of the corresponding

K

-linear mappings to

\mu, \lambda_1

and

\lambda_2

(II, § 5, no. 1). Then, if

d

is an

\varepsilon

-derivation of degree

\delta

(A, A', A'')

into

(B, B', B'')

, the mapping

\bar{d} = d \otimes l_{\overline{K}}

\overline{A} \oplus \overline{A}' \oplus \overline{A}''

into

\overline{B} \oplus \overline{B}' \oplus \overline{B}''

is an

\varepsilon

-derivation of degree

\delta

(\overline{A}, \overline{A}', \overline{A}'')

into

(\overline{B}, \overline{B}', \overline{B}'')

Cites 3

Cites II § 11. Graded modules and rings, Definition 1, II § 5. Extension of the ring of scalars

Example 6

Let $A$ be a graded $K$ -algebra of type $\mathbf{Z}$ ; a graded linear $K$ -mapping of degree 0, $d : A \to A$ , is defined by taking, for $x_n \in A_n (n \in \mathbf{Z})$ , $d(x_n) = nx_n$ . This mapping is a derivation of $A$ , since, for $x_p \in A_p, x_q \in A_q$ ,

d(x_p x_q) = (p + q)x_p x_q = d(x_p)x_q + x_p d(x_q).

Cited by 1

Cited by Proposition 6 (*« Euler's identity »*), IV § 1

4. Composition of derivations

We suppose in this no. that case (I) of no. 2 holds, that is that $A, A', A''$ are three graded $K$ -modules of type $\Delta$ and that we are given a $K$ -bilinear mapping $\mu : A \times A' \to A''$ corresponding to a graded $K$ -linear mapping of degree 0, $A \otimes_K A' \to A''$ . The graded endomorphisms $f$ of $A \oplus A' \oplus A''$ such that $f(A) \subset A, f(A') \subset A'$ and $f(A'') \subset A''$ form a graded subalgebra of the graded associative algebra $\mathrm{Endgr}_K(A \oplus A' \oplus A'')$ (§ 3, no. 1, Example 2). In particular two $\varepsilon$ -derivations of $(A, A', A'')$ can be composed, but it should not be thought that the composition of two $\varepsilon$ -derivations is an $\varepsilon$ -derivation.

On every graded algebra $B$ of type $\Delta$ is defined the $\varepsilon$ -bracket (or simply bracket when $\varepsilon = 1$ ) of two homogeneous elements $u, v$ , by the formula (10)

[u, v]_\varepsilon = uv - \varepsilon_{\deg u, \deg v} vu \text{ (denoted simply by } [u, v] \text{ if } \varepsilon = 1).

By extending this mapping by linearity, a

K

-bilinear mapping

(u, v) \mapsto [u, v]_\varepsilon

B \times B

into

B

is obtained. Then, for homogeneous

u

and

v

B

[v, u]_\varepsilon = -\varepsilon_{\deg u, \deg v}[u, v]_\varepsilon.

Applying this definition to the graded algebra

\mathrm{Endgr}_K(A \oplus A' \oplus A'')

, the

\varepsilon

-bracket of two graded endomorphisms is thus defined.

Proposition 1

Let $d_1, d_2$ be two $\varepsilon$ -derivations of $(A, A', A'')$ of respective degrees $\delta_1, \delta_2$ . Then the $\varepsilon$ -bracket*

[d_1, d_2]_\varepsilon = d_1 \circ d_2 - \varepsilon_{\delta_1, \delta_2} d_2 \circ d_1

*is an

\varepsilon

-derivation of degree

\delta_1 + \delta_2

. Moreover, if

d

is an

\varepsilon

-derivation of

(A, A', A'')

of degree

\delta

and if

\varepsilon_{\delta, \delta} = -1

, then

d^2 = d \circ d

is a derivation.

Cited by 1

Cited by Proposition 14

Suppose $x \in A$ is homogeneous of degree $\xi$ ; for all $y \in A'$ ,

d_1(d_2(xy)) = ((d_1 d_2)(x))y + \varepsilon_{\delta_1, \delta_2 + \xi}(d_2 x)(d_1 y) + \varepsilon_{\delta_2, \xi}(d_1 x)(d_2 y) + \varepsilon_{\delta_1 + \delta_2, \xi} x((d_1 d_2)(y))

taking account of formulae (1) of no. 1. Exchanging the roles of

d_1

and

d_2

, we obtain, after simplifications again using (1) (no. 1),

(d_1 d_2)(xy) - \varepsilon_{\delta_1, \delta_2}(d_2 d_1)(xy) = ((d_1 d_2)(x))y - \varepsilon_{\delta_1, \delta_2}((d_2 d_1)(x))y + \varepsilon_{\delta_1 + \delta_2, \xi} x((d_1 d_2)(y)) - \varepsilon_{\delta_1, \delta_2} \varepsilon_{\delta_1 + \delta_2, \xi} x((d_2 d_1)(y))

that is, writing

d = [d_1, d_2]_\varepsilon

and

\delta = \delta_1 + \delta_2

d(xy) = (dx)y + \varepsilon_{\delta, \xi} x(dy)

which proves that

d

is an

\varepsilon

-derivation.

On the other hand, if, in (11), we let $d_1 = d_2 = d$ , $\delta_1 = \delta_2 = \delta$ and $\varepsilon_{\delta, \delta} = -1$ , we obtain, since then $\varepsilon_{\delta, \delta + \xi} = -\varepsilon_{\delta, \xi}$ by (1),

d^2(xy) = (d^2 x)y + \varepsilon_{2\delta, \xi} x(d^2 y)

and as

\varepsilon_{2\delta, \xi} = 1

it is seen that

d^2

is a derivation.

Corollary

Suppose that $\Delta = \mathbf{Z}$ . Then:* (i) The square of an antiderivation is a derivation. (ii) The bracket of two derivations is a derivation. (iii) The bracket of an antiderivation and a derivation of even degree is an antiderivation. (iv) *If $d_1$ and $d_2$ are antiderivations, $d_1 d_2 + d_2 d_1$ is a derivation.

Cites 1

Cites III § 2. Examples of algebras

Under the hypotheses of the beginning of this no., consider now a finite sequence $D = (d_i)_{1 \leq i \leq n}$ of pairwise permutable derivations of $(A, A', A'')$ . For every polynomial $P(X_1, \ldots, X_n)$ in the algebra $K[X_1, \ldots, X_n]$ , the element $P(d_1, \ldots, d_n)$ of $\mathrm{Endgr}_K(A \oplus A' \oplus A'')$ is then defined (§ 2, no. 9); its abbreviated notation is $P(D)$ .

Proposition 2

With the above hypotheses and notation, consider 2n indeterminates $T_1, \ldots, T_n, T'_1, \ldots, T'_n$ and for every polynomial $F \in K[X_1, \ldots, X_n]$ write $F(T) = F(T_1, \ldots, T_n), F(T') = F(T'_1, \ldots, T'_n)$ and*

F(T + T') = F(T_1 + T'_1, \ldots, T_n + T'_n).

Suppose that

P(T + T') = \sum_i Q_i(T) R_i(T')

where the $Q_i$ and $R_i$ belong to $K[X_1, \ldots, X_n]$ . Then, for all $x \in A$ and $y \in A'$ ,

P(D)(xy) = \sum_i (Q_i(D)x)(R_i(D)y).

Cited by 2

Cited by Corollary (Leibniz's formula), Proposition 14

We introduce $n$ other indeterminates $T_1'', \ldots, T_n''$ and consider the polynomial algebra $K[T_1, \ldots, T_n, T_1', \ldots, T_n', T_1'', \ldots, T_n''] = B$ ; on the other hand we consider the $K$ -module $M$ of bilinear mappings of $A \times A'$ into $A''$ ; a $B$ -module structure is defined on $M$ by writing, for every $K$ -bilinear mapping $f \in M$ and $1 \leq i \leq n$ ,

\begin{cases} (T_i f)(a, a') = f(d_i a, a') \\ (T_i' f)(a, a') = f(a, d_i a') \\ (T_i'' f)(a, a') = d_i(f(a, a')) \end{cases}

Since the $d_i$ are permutable with one another, it is seen that, for every polynomial $F \in K[X_1, \ldots, X_n]$ , $(F(T)f)(a, a') = f(F(D)a, a')$ ,

(F(T')f)(a, a') = f(a, F(D)a')

and $(F(T'')f) = F(D)(f(a, a'))$ . Hence, to prove (12) it suffices to show that

(P(T'') - \sum_i Q_i(T)R_i(T')).\mu = 0

or also $(P(T'') - P(T + T')).\mu = 0$ in the $B$ -module $M$ . Now, the hypothesis that the $d_i$ are derivations can also be expressed by saying that, for $1 \leq i \leq n$ ,

(T_i'' - T_i - T_i').\mu = 0

in the $B$ -module $M$ . By considering successively the polynomials

P(T_1'', T_2'', \ldots, T_n'') - P(T_1 + T_1', T_2'', \ldots, T_n'')

P(T_1 + T_1', T_2'', \ldots, T_n'') - P(T_1 + T_1', T_2 + T_2', \ldots, T_n'')

\ldots

P(T_1 + T_1', \ldots, T_{n-1} + T_{n-1}', T_n'') - P(T_1 + T_1', \ldots, T_{n-1} + T_{n-1}', T_n + T_n')

it is seen that the difference $P(T'') - P(T + T')$ can be written in the form

\sum_{i=1}^n (T_i'' - T_i - T_i') G_i(T, T', T'')

where the $G_i$ are elements of $B$ . Relation (14) is therefore an immediate consequence of relations (15).

Corollary (Leibniz's formula)

Let $d_i$ ( $1 \leq i \leq n$ ) be $n$ derivations of $(A, A', A'')$ which are permutable with one another. For all $\alpha = (\alpha_1, \ldots, \alpha_n) \in \mathbf{N}^n$ , we write

Cites 2

Cites I § 8. Rings, Proposition 2

d^\alpha = d_1^{\alpha_1} d_2^{\alpha_2} \ldots d_n^{\alpha_n}.

Then, for $x \in \mathbf{A}$ and $y \in \mathbf{A}'$ ,

d^\alpha(xy) = \sum_{\beta + \gamma = \alpha} ((\beta, \gamma)) d^\beta(x)d^\gamma(y)

where we have written (in the notation introduced at the beginning of the chapter)

((\beta, \gamma)) = (\beta + \gamma)!/(\beta! \gamma!).

This follows immediately from the multinomial formula (I, § 8, no. 2)

(T + T')^\alpha = \sum_{\beta + \gamma = \alpha} ((\beta, \gamma)) T^\beta {T'}^\gamma

and Proposition 2.

5. Derivations of an algebra a into an a-module

We suppose in this no. that Case (II) of no. 2 holds. Then there is a graded K-algebra A and a graded K-module E and also two K-linear mappings of degree 0

E \otimes_K A \to E, \quad A \otimes_K E \to E

denoted by

x \otimes a \mapsto x.a \quad \text{and} \quad a \otimes x \mapsto a.x \quad \text{for } a \in A \text{ and } x \in E.

Proposition 3

Let $d : A \to E$ be an $\varepsilon$ -derivation of degree $\delta$ . Then $\mathrm{Ker}(d)$ is a graded subalgebra of A; if A admits a unit element, it belongs to $\mathrm{Ker}(d)$ .

Cites 1 · Cited by 3

Cites III § 3. Graded algebras

Cited by Proposition 10, Proposition 4, VIII § 1. Artinian Modules and Noetherian Modules

Clearly $\mathrm{Ker}(d)$ is a graded sub-K-module of A; further, relation (8) of no. 2 shows that, if $x$ and $y$ are two homogeneous elements belonging to $\mathrm{Ker}(d)$ , then $d(xy) = 0$ and hence $xy \in \mathrm{Ker}(d)$ . Finally, if A admits a unit element 1 (of degree 0, cf. § 3, no. 1), relation (8) of no. 2, where $x$ and $y$ are replaced by 1, gives $d(1) = d(1) + d(1)$ and hence $d(1) = 0$ .

Corollary

Let $d_1$ and $d_2$ be two $\varepsilon$ -derivations from A to E of the same degree $\delta$ . If $d_1$ and $d_2$ take the same values at each element of a generating system of the algebra A, then $d_1 = d_2$ .

Cited by 1

Cited by Corollary

$d_1 - d_2$ is an $\varepsilon$ -derivation of degree $\delta$ , hence $\mathrm{Ker}(d_1 - d_2)$ is a subalgebra of A which contains a generating system of A and hence is equal to A.

Proposition 4

Let $d : A \to E$ be an $\varepsilon$ -derivation of degree $\delta$ . Suppose that A has a unit element 1 and let $x$ be a homogeneous element of A with an inverse $x^{-1}$ in A. Then

d(x^{-1}) = -\varepsilon_{\delta, \deg(x)} x^{-1}((dx)x^{-1}) = -\varepsilon_{\delta, \deg(x)} (x^{-1}(dx))x^{-1}.

Cites 1

Cites Proposition 3

We have $d(xx^{-1}) = d(1) = 0$ (Proposition 3), whence

(dx)x^{-1} + \varepsilon_{\delta, \deg(x)} x(d(x^{-1})) = 0

which proves the first formula of (19). On the other hand,

x^{-1}

is homogeneous of degree

-\deg(x)

and

\varepsilon_{\delta, \deg(x)} = \varepsilon_{\delta, -\deg(x)}

by formulae (1) of no. 1; writing

d(x^{-1}x) = 0

, the second formula of (19) is obtained similarly.

Proposition 5

Suppose that $\mathbf{A}$ is an integral domain and let $\mathbf{L}$ be its field of fractions. Every derivation of $\mathbf{A}$ into a vector space $\mathbf{E}$ over $\mathbf{L}$ (considered as an $\mathbf{A}$ -module) can be extended uniquely to a derivation of $\mathbf{L}$ into $\mathbf{E}$ .

Cited by 2

Cited by Proposition 23, IV § 3. Rational fractions

Let $d$ be a derivation of $\mathbf{A}$ into $\mathbf{E}$ and $\bar{d}$ a derivation of $\mathbf{L}$ into $\mathbf{E}$ extending $d$ ; then, for $u \in \mathbf{A}, v \in \mathbf{A}, v \neq 0$ , of necessity, by virtue of (19),

\bar{d}(u/v) = v^{-1}du - uv^{-2}dv

which proves the uniqueness of

\bar{d}

. Conversely, we show that

\bar{d}

can be defined by formula (20); it must first be verified that if

u/v = u'/v'

the value of the right hand side of (20) does not change when

u

is replaced by

u'

and

v

v'

. Now,

uv' = vu'

, hence

v'(du) + u(dv') = v(du') + u'(dv)

and therefore

v'(du - uv^{-1}dv) = v(du' - u'{v'}^{-1}dv')

, since

uv'v^{-1} = u'

and

u'{v'}^{-1}v = u

. Thus a mapping

\bar{d}: \mathbf{L} \to \mathbf{E}

has been defined which extends

d

; it is immediately verified that it is

\mathbf{K}

-linear and a derivation.

Proposition 6

Suppose that $\mathbf{A}$ is a unital associative graded $\mathbf{K}$ -algebra and $\mathbf{E}$ a graded $(\mathbf{A}, \mathbf{A})$ -bimodule. If $d : \mathbf{A} \to \mathbf{E}$ is an $\varepsilon$ -derivation of degree $\delta$ , then, for every finite sequence $(x_i)_{1 \leq i \leq n}$ of homogeneous elements of $\mathbf{A}$ , of respective degrees $\xi_i$ ( $1 \leq i \leq n$ ),

d(x_1 x_2 \ldots x_n) = \sum_{i=1}^n \varepsilon_{\delta, \xi_1 + \cdots + \xi_{i-1}} x_1 \cdots x_{i-1} (dx_i) x_{i+1} \cdots x_n.

Formula (21) is trivial for

n = 0

and is proved by induction on

n

, taking account of (4) (no. 2).

Cited by 1

Cited by Proposition 4, IV § 1

Corollary

Suppose that $\mathbf{A}$ is a unital commutative associative algebra and $\mathbf{E}$ an $\mathbf{A}$ -module. If $d : \mathbf{A} \to \mathbf{E}$ is a derivation, then, for every integer $n \geq 0$ ,

d(x^n) = nx^{n-1}(dx) \quad \text{for all } x \in \mathbf{A}.

It suffices to give

\mathbf{A}

the trivial graduation and apply (21) with all the

x_i

equal to

x

We return to the general case of an $\varepsilon$ -derivation $d : \mathbf{A} \to \mathbf{E}$ of degree $\delta$ . Let $Z_\delta$ be the set of $a \in \mathbf{A}$ such that for every homogeneous component $a_\alpha$ of $a$ of degree $\alpha$ , for every homogeneous $x$ in $\mathbf{E}$ ,

xa_\alpha = \varepsilon_{\alpha, \deg(x)} a_\alpha x.

If $A$ is a unital associative graded algebra and $E$ a graded $(A, A)$ -bimodule it follows immediately from this definition that $Z_\varepsilon$ is a graded subalgebra of $A$ containing the unit element.

Proposition 7

Suppose that $A$ is a unital associative graded algebra and $E$ a graded $(A, A)$ -bimodule. Let $d : A \to E$ be an $\varepsilon$ -derivation of degree $\delta$ and $a$ a homogeneous element of $Z_\varepsilon$ of degree $\alpha$ . Then the mapping $x \mapsto a(dx)$ is an $\varepsilon$ -derivation of degree $\delta + \alpha$ .

Cited by 1

Cited by Corollary

We write $d'(x) = a(dx)$ ; for $x$ homogeneous of degree $\xi$ in $A$ and $y \in A$ , by virtue of (23) and (1) (no. 1),

d'(xy) = a((dx)y) + \varepsilon_{\delta, \xi} a(x(dy)) = (a(dx))y + \varepsilon_{\delta+\alpha, \xi}(xa)(dy) = (d'x)y + \varepsilon_{\delta+\alpha, \xi} x(d'y).

Proposition 7 says that the K-module of $\varepsilon$ -derivations of $A$ into $E$ is a graded $Z_\varepsilon$ -module of type $\Delta$ .

6. Derivations of an algebra

Let $A$ be a graded K-algebra; for every homogeneous element $a \in A$ , let $\mathrm{ad}_\varepsilon(a)$ , or simply $\mathrm{ad}(a)$ if no confusion can arise, denote the K-linear mapping of $A$ into $A$

x \mapsto [a, x]_\varepsilon

(no. 4, formula (10)) which is graded of degree

\deg a

Proposition 8

Let $A$ be a graded K-algebra.

(i) For every $\varepsilon$ -derivation $d : A \to A$ and every homogeneous element $a$ of $A$ ,

(24)

[d, \mathrm{ad}_\varepsilon(a)]_\varepsilon = \mathrm{ad}_\varepsilon(da).

(ii) If the algebra $A$ is associative, $\mathrm{ad}_\varepsilon(a)$ is an $\varepsilon$ -derivation of $A$ of degree $\deg(a)$ .

(i) Suppose that $d$ is of degree $\delta$ , let $\alpha = \deg a$ and write $f = [d, \mathrm{ad}_\varepsilon(a)]_\varepsilon$ . For every homogeneous element $x \in A$ of degree $\xi$ , we have, by (1) (no. 1),

f(x) = d(ax - \varepsilon(\alpha, \xi) xa) - \varepsilon_{\delta, \alpha}(a(dx)) - \varepsilon_{\alpha, \delta+\xi}(dx)a \\ = (da)x + \varepsilon_{\delta, \alpha} a(dx) - \varepsilon_{\alpha, \xi}(dx)a - \varepsilon_{\delta+\alpha, \xi} x(da) \\ - \varepsilon_{\delta, \alpha} a(dx) + \varepsilon_{\alpha, \xi}(dx)a \\ = (da)x - \varepsilon_{\delta+\alpha, \xi} x(da) = [da, x]_\varepsilon.

(ii) For all $x$ homogeneous of degree $\xi$ and all $y$ homogeneous of degree $\eta$ in $A$ ,

\mathrm{ad}_\varepsilon(a)(xy) = a(xy) - \varepsilon_{\alpha, \xi+\eta}(xy)a \\ = (ax - \varepsilon_{\alpha, \xi} xa)y + \varepsilon_{\alpha, \xi} x(ay - \varepsilon_{\alpha, \eta} ya) \\ = \mathrm{ad}_\varepsilon(a)(x).y + \varepsilon_{\alpha, \xi} x.\mathrm{ad}_\varepsilon(a)(y)

taking account of (1) and the associativity of

A

When $A$ is associative, $\mathrm{ad}_\varepsilon(a)$ is called the inner $\varepsilon$ -derivation of $A$ defined by $a$ .

Corollary

Let $A$ be an associative graded algebra. For two homogeneous elements, $a, b$ of $A$ ,*

[\mathrm{ad}_\varepsilon(a), \mathrm{ad}_\varepsilon(b)]_\varepsilon = \mathrm{ad}_\varepsilon([a, b]_\varepsilon).

(25)

It suffices to replace $d$ by $\mathrm{ad}_\varepsilon(a)$ and $\mathrm{ad}_\varepsilon(a)$ by $\mathrm{ad}_\varepsilon(b)$ in (24).

If $\deg a = \alpha, \deg b = \beta$ , formula (25) is equivalent to the following relation for every homogeneous element $c \in A$ of degree $\gamma$

\varepsilon_{\alpha, \gamma}[a, [b, c]_\varepsilon]_\varepsilon + \varepsilon_{\beta, \alpha}[b, [c, a]_\varepsilon]_\varepsilon + \varepsilon_{\gamma, \beta}[c, [a, b]_\varepsilon]_\varepsilon = 0

called the Jacobi identity.

7. Functorial Properties

In this no., all algebras are assumed to be associative and unital and every algebra homomorphism is assumed to be unital.

Proposition 9

Let $A, B$ be two graded $K$ -algebras, $E$ an $(A, A)$ -bimodule and $F$ a graded $(B, B)$ -bimodule; let $\rho : A \to B$ be a graded algebra homomorphism and $\theta : E \to F$ a graded $A$ -homomorphism of $A$ -bimodules (relative to $\rho$ ), of degree 0. Then:

Cited by 6

Cited by Corollary, Proposition 10, Proposition 11, Proposition 17, Proposition 19, Proposition 22

(i) For every $\varepsilon$ -derivation $d' : B \to F$ , $d' \circ \rho : A \to \rho^*(F)$ is an $\varepsilon$ -derivation of the same degree. (ii) For every $\varepsilon$ -derivation $d : A \to E$ , $\theta \circ d : A \to \rho^*(F)$ is an $\varepsilon$ -derivation of the same degree.

The two assertions follow immediately from the relations

d'(\rho(xy)) = d'(\rho(x)\rho(y)) = d'(\rho(x))\rho(y) + \varepsilon_{\delta', \xi}\rho(x)d'(\rho(y))

\theta(d(xy)) = \theta((dx)y + \varepsilon_{\delta, \xi}x(dy)) = \theta(dx)\rho(y) + \varepsilon_{\delta, \xi}\rho(x)\theta(dy)

for

x \in A

homogeneous of degree

\xi

and

y \in A

\delta

and

\delta'

denoting the respective degrees of

d

and

d'

Corollary

Let $S$ be a generating system of the algebra $A$ . In order that $d' \circ \rho = \theta \circ d$ , it is necessary and sufficient that $d'(\rho(x)) = \theta(d(x))$ for all $x \in S$ .

Cites 4 · Cited by 1

Cites Corollary, Proposition 9, II § 1. Modules, Example 1

Cited by Proposition 11, III § 11

This is an immediate consequence of Proposition 9 and no. 5, Corollary to Proposition 3.

Under the conditions of Proposition 9, we know that $B$ has (by means of $\rho$ ) an $(A, A)$ -bimodule structure (II, § 1, no. 14, Example 1).

Proposition 10

Under the conditions of Proposition 9, for an $\varepsilon$ -derivation $d' : B \to F$ to be $A$ -linear for the left (resp. right) $A$ -module structures on $B$ and $\rho^*(F)$ , it is necessary and sufficient that $d'$ be zero on the subalgebra $\rho(A)$ of $B$ .

Cites 2 · Cited by 1

Cites Proposition 9, Proposition 3

Cited by Corollary

We perform the proof for left A-module structures. For $a \in A, b \in B$ ,

d'(\rho(a)b) = d'(\rho(a))b + \rho(a)d'b

and hence if

d' \circ \rho = 0

d'

is linear for the left A-module structures on B and

\rho^*(F)

. Conversely, if this is so, in particular

d'(\rho(a)) = d'(\rho(a).1) = \rho(a)d'(1) = 0

(no. 5, Proposition 3).

In particular let $D_K(B, F)$ denote the K-module of derivations of B into F (no. 2); those among these derivations which are A-linear, in other words those which are zero on $\rho(A)$ , form a sub-K-module of $D_K(B, F)$ , denoted by $D_{A,\rho}(B, F)$ or simply $D_A(B, F)$ (obviously $D_K(B, F) = D_{K,\phi}(B, F)$ , where $\phi : K \to B$ is the homomorphism defining the K-algebra structure on B).

Now let A, B, C be three graded K-algebras, $\rho : A \to B, \sigma : B \to C$ two graded algebra homomorphisms and G a graded (C, C)-bimodule; if $D_A(B, G)$ , $D_B(C, G)$ and $D_A(C, G)$ denote the respective K-modules $D_{A,\rho}(B, \sigma_*(G))$ , $D_{B,\sigma}(C, G)$ and $D_{A,\sigma \circ \rho}(C, G)$ , $D_B(C, G)$ is clearly a sub-K-module of $D_A(C, G)$ since $\sigma(\rho(A)) \subset \sigma(B)$ .

Proposition 11

Under the above conditions, there is an exact sequence of K-homomorphisms*

0 \to D_B(C, G) \xrightarrow{u} D_A(C, G) \xrightarrow{v} D_A(B, G)

where u is the canonical injection and v the homomorphism $d \mapsto d \circ \sigma$ (Proposition 9).

Cites 1 · Cited by 2

Cites Proposition 9

Cited by Corollary, Proposition 21

The kernel of v is the set of derivations $d : C \to G$ such that $d(\sigma(b)) = 0$ for all $b \in B$ , which is precisely the image of u.

8. Relations between derivations and algebra homomorphisms

We suppose again in this no. that Case (II) of no. 2 holds and the graded K-algebra A is not assumed to be associative. Given an element $\delta \in \Delta$ , consider the graded K-module E( $\delta$ ) (II, § 11, no. 2) such that

(E(\delta))_\mu = E_{\mu+\delta}

for all

\mu \in \Delta

. We define on the graded K-module

A \oplus E(\delta)

a graded K-algebra structure by setting, for every homogeneous element

a \in A

and arbitrary elements

a' \in A, x, x'

E(\delta)

(a, x)(a', x') = (aa', x.a' + \varepsilon_{\delta, \deg(a)} a.x');

the verification of the fact that this multiplication defines a graded ring structure is immediate.

The projection $p : (a, x) \mapsto a$ is called the augmentation of the algebra $A \oplus E(\delta)$ and is a graded algebra homomorphism. The graded K-linear mappings $g : A \to A \oplus E(\delta)$ of degree 0 such that the composition

A \xrightarrow{g} A \oplus E(\delta) \xrightarrow{p} A

is the identity $1_A$ are the mappings of the form $x \mapsto (x, f(x))$ , where $f : A \to E$ is a graded K-linear mapping of degree $\delta$ .

Proposition 12

For a graded K-linear mapping $f : A \to E$ of degree $\delta$ to be an $\varepsilon$ -derivation, it is necessary and sufficient that the mapping $x \mapsto (x, f(x))$ of A into $A \oplus E(\delta)$ be a graded K-algebra homomorphism.

Cited by 1

Cited by Proposition 14

Using the fact that for $x$ homogeneous in $A$ and $y \in A$

(xy, f(xy)) = (x, f(x)).(y, f(y)),

we obtain, taking account of (28), the equivalent relation

f(xy) = f(x).y + \varepsilon_{\delta, \deg(x)} x.f(y),

whence the proposition.

Proposition 13

For the algebra $A \oplus E(\delta)$ to be associative and unital, it is necessary and sufficient that $A$ be associative and unital, and that the mappings $(a, x) \mapsto a.x$ and $(a, x) \mapsto x.a$ define on $E$ an $(A, A)$ -bimodule structure; the unit element of $A \oplus E(\delta)$ is then $(1, 0)$ .

Cited by 1

Cited by Proposition 14

If an element $(u, m) \in A \oplus E(\delta)$ is written as unit element of this algebra, it is immediately found that $u$ must be the unit element of $A$ ; writing $(u, m).(0, x) = (0, x).(u, m) = (0, x)$ , we obtain $u.x = x.u = x$ for $x \in E$ and, writing $(u, m).(u, 0) = (u, 0).(u, m) = (u, 0)$ , we obtain $m = 0$ . The fact that $A$ is associative when $A \oplus E(\delta)$ is follows from the fact that the augmentation is a surjective homomorphism. The condition $(x.a').a'' = x.(a'.a'')$ is then equivalent to $((0, x)(a', 0))(a'', 0) = (0, x)((a', 0)(a'', 0))$ and similarly the condition $a.(a'.x) = (a.a').x$ is equivalent to

(a, 0)((a', 0)(0, x)) = ((a, 0)(a', 0))(0, x);

finally the condition $a.(x.a') = (a.x).a'$ is equivalent to

(a, 0)((0, x)(a', 0)) = ((a, 0)(0, x))(a', 0).

9. Extension of derivations

Proposition 14

Let $A$ be a commutative ring, $M$ an $A$ -module, $B$ the $A$ -algebra $T(M)$ (resp. $S(M)$ , resp. $\Lambda(M)$ ) and $E$ a $(B, B)$ -bimodule. Let $d_0 : A \to E$ be a derivation of the ring $\mathbf{A}$ into the $\mathbf{A}$ -module $\mathbf{E}$ and $d_1 : M \to \mathbf{E}$ an additive group homomorphism such that, for all $a \in \mathbf{A}$ and all $x \in M$ ,

Cites 4 · Cited by 2

Cites Proposition 1, Proposition 12, Proposition 13, Proposition 2

Cited by Example, Corollary

(29) \quad d_1(ax) = ad_1(x) + d_0(a) \cdot x,

and further, when $B = S(M)$ ,

(30) \quad x \cdot d_1(y) + d_1(x) \cdot y = y \cdot d_1(x) + d_1(y) \cdot x

for all $x, y$ in $M$ , and, when $B = \bigwedge(M)$ ,

(31) \quad x \cdot d_1(x) + d_1(x) \cdot x = 0

for all $x \in M$ . Then there exists one and only one derivation $d$ of $B$ (considered as a $\mathbf{Z}$ -algebra) into the $(B, B)$ -bimodule $\mathbf{E}$ such that $d|_{\mathbf{A}} = d_0$ and $d|_{M} = d_1$ .

We take on the $\mathbf{Z}$ -module $B \oplus \mathbf{E}$ the associative $\mathbf{Z}$ -algebra structure defined by

(b, t)(b', t') = (bb', bt' + tb')

which has $(1, 0)$ as unit element (no. 8, Proposition 13). Under the canonical injection $t \mapsto (0, t)$ , $\mathbf{E}$ is identified with a two-sided ideal of $B \oplus \mathbf{E}$ such that $\mathbf{E}^2 = \{0\}$ . On the other hand, the mapping $h_0 : B \oplus \mathbf{E}$ defined by $h_0(a) = (a, d_0(a))$ is a unital ring homomorphism (no. 8, Proposition 12); under this mapping, $B \oplus \mathbf{E}$ then becomes an $\mathbf{A}$ -algebra. Moreover, if, for all $x \in M$ , we write $h_1(x) = (x, d_1(x))$ , it follows from the definition of $h_0$ and (29) that $h_1(ax) = h_0(a)h_1(x)$ ; in other words $h_1$ is an $\mathbf{A}$ -linear mapping of $M$ into $B \oplus \mathbf{E}$ . Then there exists one and only one $\mathbf{A}$ -algebra homomorphism, $h : B \to B \oplus \mathbf{E}$ such that $h|_M = h_1$ (and necessarily $h|_{\mathbf{A}} = h_0$ ): for, if $B = T(M)$ , this follows from $§ 5$ , no. 1, Proposition 1; if $B = S(M)$ , condition (30) shows that $h(x)h(y) = h(y)h(x)$ for all $x, y$ in $M$ and the conclusion follows from $§ 6$ , no. 1, Proposition 2; finally if $B = \bigwedge(M)$ , condition (31) shows that $(h(x))^2 = 0$ for all $x \in M$ , since $x \wedge x = 0$ and the conclusion follows from $§ 7$ , no. 1, Proposition 1. The homomorphism $h$ is such that the composition $p \circ h : B \to B$ with the augmentation $\rho : B \oplus \mathbf{E} \to B$ is the identity $l_B$ , for $p \circ h$ and $l_B$ coincide by definition for the elements of $\mathbf{A}$ and those of $M$ and the set of these elements is a generating system of $B$ . We can therefore write $h(b) = (b, d(b))$ for all $b \in B$ and the mapping $b \mapsto d(b)$ of $B$ into $\mathbf{E}$ is a derivation with the required properties, by virtue of Proposition 12 of no. 8.

Corollary

Let $M$ be a graded $K$ -module of type $\Delta$ ; the $K$ -algebras $T(M)$ , $S(M)$ and $\bigwedge(M)$ are given the corresponding graduations of type $\Delta' = \Delta \times \mathbf{Z}$ ( $§ 5$ , no. 5, Proposition 7, $§ 6$ , no. 6, Proposition 10 and $§ 7$ , no. 7, Proposition 11). On the other hand $M$ is given the graduation of type $\Delta'$ such that $M_{\alpha, 1} = M_{\alpha}$ for all $\alpha \in \Delta$ and $M_{\alpha, n} = \{0\}$ for $\alpha \in \Delta$ and $n \neq 1$ . Let $\varepsilon'$ be a commutation factor over $\Delta'$ .

Cites 4 · Cited by 1

Cites Proposition 10, Proposition 11, Proposition 7, Proposition 14

Cited by Proposition 15

(i) Let E be a graded (left and right) $\mathbf{T}(M)$ -bimodule of type $\Delta'$ ; for all $\delta \in \Delta$ and every integer $n \in \mathbf{Z}$ , every graded K-linear mapping $f : M \to E$ of degree $\delta_1' = (\delta, n)$ can be extended uniquely to an $\varepsilon'$ -derivation $d : \mathbf{T}(M) \to E$ of degree $\delta'$ .

(ii) Let E be a graded $S(M)$ -module of type $\Delta'$ ; for a graded K-linear mapping $f : M \to E$ of degree $\delta'$ to be extendable to an $\varepsilon'$ -derivation $d : S(M) \to E$ of degree $\delta'$ , it is necessary and sufficient that, for every ordered pair $(x, y)$ of homogeneous elements of M,

x . f(y) + \varepsilon_{\delta', (\deg(y), 1)} y . f(x) = y . f(x) + \varepsilon_{\delta', (\deg(x), 1)} x . f(y).

The $\varepsilon'$ -derivation d is then unique.

(iii) Let E be a graded (left and right) $\Lambda(M)$ -bimodule of type $\Delta'$ ; for a graded K-linear mapping $f : M \to E$ of degree $\delta'$ to be extendable to an $\varepsilon'$ -derivation

d : \Lambda(M) \to E

of degree $\delta'$ , it is necessary and sufficient that, for every homogeneous element x of M,

x . f(x) + \varepsilon_{\delta', (\deg(x), 1)} f(x) . x = 0.

The $\varepsilon'$ -derivation d is then unique.

Remark 2 of no. 2 is applied with one of the external B-module laws on E (with B equal to $\mathbf{T}(M)$ , $S(M)$ or $\Lambda(M)$ ) modified; the external law thus modified is still, by (1) (no. 1), a B-module law and the B-module law thus obtained on E is still compatible with the other B-module structure. It then suffices to apply Proposition 14 with $A = K$ and $d_0 = 0$ .

Example (1). In the application of Proposition 14 note that if $d_0 = 0$ condition (29) means simply that $d_1$ is A-linear. If we take in particular $E = B$ and the (B, B)-bimodule structure derived from the ring structure on B, conditions (30) and (31) are automatically satisfied when $d_1$ is taken to be the composition of an endomorphism s of M and the canonical injection $M \to B$ ; this is obvious for (30) since $S(M)$ is commutative and for (31) this follows from the fact that x and $s(x)$ are of degree 1 in $\Lambda(M)$ . It is therefore seen that every endomorphism s of M can be extended uniquely to a derivation $D_s$ of $\mathbf{T}(M)$ (resp. $S(M)$ , resp. $\Lambda(M)$ ), which is of degree 0. Moreover, for two endomorphisms s, t of M,

[D_s, D_t] = D_{[s, t]}

for both sides are derivations of $\mathbf{T}(M)$ (resp. $S(M)$ , resp. $\Lambda(M)$ ) which are equal to $[s, t]$ on M.

The expression for $D_s$ is obtained using formula (21) of no. 5, which gives respectively, for $x_1, x_2, \ldots, x_n$ in $M$ ,

\begin{cases} D_s(x_1 \otimes x_2 \otimes \cdots \otimes x_n) \\ \phantom{D_s(x_1 \otimes x_2 \otimes \cdots \otimes x_n)} = \sum_{i=1}^n x_1 \otimes \cdots \otimes x_{i-1} \otimes s(x_i) \otimes x_{i+1} \otimes \cdots \otimes x_n \\ D_s(x_1 x_2 \ldots x_n) = \sum_{i=1}^n x_1 \ldots x_{i-1} s(x_i) x_{i+1} \ldots x_n \\ D_s(x_1 \wedge x_2 \wedge \cdots \wedge x_n) \\ \phantom{D_s(x_1 \wedge x_2 \wedge \cdots \wedge x_n)} = \sum_{i=1}^n x_1 \wedge \cdots \wedge x_{i-1} \wedge s(x_i) \wedge x_{i+1} \wedge \cdots \wedge x_n. \end{cases}

In the case of the algebra $\Lambda(M)$ , there is the following interpretation of $D_s$ :

Proposition 15

If $M$ is a free $K$ -module of finite rank $n$ , then, for every endomorphism $s$ of $M$ , the restriction to $\Lambda^n(M)$ of the derivation $D_s$ is the homothety of ratio $\operatorname{Tr}(s)$ .

Cites 2

Cites Corollary, III § 11. Cogebras, products of multilinear forms, inner products and duality

Let $(e_j)_{1 \leq j \leq n}$ be a basis of $M$ and write $e = e_1 \wedge e_2 \wedge \cdots \wedge e_n$ . If

s(e_j) = \sum_{k=1}^n \alpha_{jk} e_k,

the third formula in (36) gives

D_s(e) = \sum_{i=1}^n e_1 \wedge \cdots \wedge e_{i-1} \wedge s(e_i) \wedge e_{i+1} \wedge \cdots \wedge e_n = \left( \sum_{j=1}^n \alpha_{jj} \right) e.

Example (2). In the Corollary to Proposition 14, part (iii), let $\Delta = \{0\}$ , the graduation on $\Lambda(M)$ then being the usual graduation of type $\mathbf{Z}$ ; on the other hand take $\varepsilon(p, q) = (-1)^{pq}$ . Then, for every linear form $x^* \in M^*$ on $M$ , $x \mapsto \langle x, x^* \rangle$ is a graded $K$ -linear mapping of degree $-1$ of $M$ into $\Lambda(M)$ satisfying relation (34); then there exists an antiderivation $i(x^*)$ of $\Lambda(M)$ , of degree $-1$ , such that (by virtue of formula (21) of no. 5)

i(x^*)(x_1 \wedge \cdots \wedge x_n) = \sum_{i=1}^n (-1)^{i-1} \langle x_i, x^* \rangle x_1 \wedge \cdots \wedge x_{i-1} \wedge x_{i+1} \wedge \cdots \wedge x_n

and which is a special case of the inner product to be defined in § 11, no. 9, formula (68).

Proposition 16

Let $A$ be a commutative $K$ -algebra, $M_i$ ( $1 \leq i \leq n$ ) and $P$ $A$ -modules and $H$ the $A$ -module of $A$ -multilinear mappings of $M_1 \times M_2 \times \cdots \times M_n$ * into P. Suppose that there is given a K-derivation $d_0 : A \to A$ of the algebra A, for each i, a K-linear mapping $d_i : M_i \to M_i$ and a K-linear mapping $D : P \to P$ , so that, for $1 \leq i \leq n$ , $(d_0, d_i, d_i)$ is a K-derivation of $(A, M_i, M_i)$ into itself and $(d_0, D, D)$ is a K-derivation of $(A, P, P)$ into itself. Then there exists a K-linear mapping $D' : H \to H$ such that $(d_0, D', D')$ is a K-derivation of $(A, H, H)$ into itself and

Cited by 1

Cited by Example

(37) $D(f(x_1, \ldots, x_n))$

= (D'f)(x_1, \ldots, x_n) + \sum_{i=1}^n f(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)

for all $x_i \in M_i$ for $1 \leq i \leq n$ and $f \in H$ .

We show that for $f \in H$ , the mapping $D'f$ of $M_1 \times M_2 \times \cdots \times M_n$ into P defined by (37) is A-multilinear. For $a \in A$ ,

(D'f)(ax_1, x_2, \ldots, x_n) = D(af(x_1, \ldots, x_n)) - f(d_1(ax_1), x_2, \ldots, x_n)

-a \sum_{i=2}^n f(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)

and by hypothesis

D(af(x_1, \ldots, x_n)) = (d_0a)f(x_1, \ldots, x_n) + aD(f(x_1, \ldots, x_n))

and $d_1(ax_1) = (d_0a)x_1 + a.d_1x_1$ , which gives

(D'f)(ax_1, x_2, \ldots, x_n) = a.(D'f)(x_1, \ldots, x_n)

and linearity in each of the $x_i$ is proved similarly. On the other hand,

(D'(af))(x_1, \ldots, x_n) = D(af(x_1, \ldots, x_n))

- \sum_{i=1}^n af(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)

= (d_0a)f(x_1, \ldots, x_n)) + aD(f(x_1, \ldots, x_n))

- \sum_{i=1}^n af(x_1, \ldots, x_{i-1}, d_ix_i, x_{i+1}, \ldots, x_n)

= (d_0a)f(x_1, \ldots, x_n) + a(D'f)(x_1, \ldots, x_n)

in other words

D'(af) = (d_0a)f + a(D'f)

which establishes the proposition.

Example

(3) Applying Proposition 16 to the case $n = 1$ , $M_1 = M$ , $P = A$ , then $H = M^*$ , the dual of M, and it is seen that for a K-derivation $(d_0, d, d')$ of $(A, M, M)$ is derived a K-derivation $(d_0, d^*, d^*)$ of $(A, M^*, M^*)$ such that

(38) $d_0\langle m, m^*\rangle = \langle dm, m^*\rangle + \langle m, d^*m^*\rangle$

for $m \in M$ and $m^* \in M^*$ . The K-linear mapping of $M \oplus M^*$ into itself which is equal to $d$ on $M$ and to $d^*$ on $M^*$ then satisfies condition (29) and there is therefore a $K$ -derivation $D$ of the $A$ -algebra $T(M \oplus M^*)$ , which reduces to $d_0$ on $A$ , to $d$ on $M$ and to $d^*$ on $M^*$ . The restriction $d_J^I$ of $D$ to the sub- $A$ -module $T_J^I(M)$ of $T(M \oplus M^*)$ ( $§ 5$ , no. 6) is a $K$ -endomorphism of $T_J^I(M)$ such that $(d_0, d_J^I, d_J^I)$ is a $K$ -derivation of $(A, T_J^I(M), T_J^I(M))$ . Moreover, for $i \in I, j \in J$ , if we write $I' = I - \{i\}, J' = J - \{j\}$ , it is immediately verified that for the contraction $c_j^i$ ( $§ 5$ , no. 6)

c_j^i(d_J^I(z)) = d_{J'}^{I'}(c_j^i(z)) \quad \text{for all } z \in T_J^I(M).

(4) Let $M_i$ ( $1 \leq i \leq 3$ ) be three $A$ -modules and, for each $i$ , suppose that $(d_0, d_i, d_i)$ is a derivation of $(A, M_i, M_i)$ ; applying Proposition 16 again for $n = 1$ , a derivation $(d_0, d_{ij}, d_{ij})$ of $(A, H_{ij}, H_{ij})$ , where $H_{ij} = \operatorname{Hom}_A(M_i, M_j)$ , is derived for each ordered pair $(i, j)$ . With this notation, for $u \in \operatorname{Hom}_A(M_1, M_2)$ and $v \in \operatorname{Hom}_A(M_2, M_3)$ ,

d_{13}(v \circ u) = (d_{23}v) \circ u + v \circ (d_{12}u)

as is immediately verified from the definitions.

Cites 1 · Cited by 1

Cites Proposition 16

Cited by LIE III App. 0. Operations on linear representations

10. Universal problem for derivations; non-commutative case

Throughout the rest of $§ 10$ all the algebras are assumed to be associative and unital and all the algebra homomorphisms are assumed to be unital.

Let $A$ be a $K$ -algebra; the tensor product $A \otimes_K A$ has canonically an $(A, A)$ -bimodule structure under which

x . (u \otimes v) . y = (xu) \otimes (vy)

for all

x, y, u, v

A

(

§ 4

, no. 3, Example 2). The

K

-linear mapping

m : A \otimes_K A \to A

corresponding to multiplication in

A

(and hence such that

m(x \otimes y) = xy

) is an

(A, A)

-bimodule homomorphism; its kernel

I

is therefore a sub-bimodule of

A \otimes_K A

Lemma 1

The mapping $\delta_A : x \mapsto x \otimes 1 - 1 \otimes x$ is a $K$ -derivation of $A$ into $I$ and $I$ is generated, as a left $A$ -module, by the image of $\delta_A$ .

Cited by 2

Cited by § 10. Derivations, Proposition 17

The first assertion follows from the fact that

(xy) \otimes 1 - 1 \otimes (xy) = (x \otimes 1 - 1 \otimes x) . y + x . (y \otimes 1 - 1 \otimes y)

by (40). On the other hand, if the element

\sum_i x_i \otimes y_i

(for

x_i, y_i

A

) belongs to

I

, by definition

\sum_i x_i y_i = 0

and hence

\sum_i (x_i \otimes y_i) = \sum_i x_i (1 \otimes y_i - y_i \otimes 1)

by (40), which completes the proof of the lemma.

Proposition 17

The derivation* $\delta_A$ has the following universal property: for every $(A, A)$ -bimodule $E$ and every $K$ -derivation $d : A \to E$ , there exists one and only one $(A, A)$ -bimodule homomorphism $f : I \to E$ such that $d = f \circ \delta_A$ .

Cites 2 · Cited by 1

Cites Proposition 9, Lemma 1

Cited by § 10. Derivations

Note first that, for every $(A, A)$ -bimodule homomorphism $f : I \to E$ , $f \circ \delta_A$ is a derivation (no. 7, Proposition 9). Conversely, let $d : A \to E$ be a K-derivation; then we prove first that if there exists an $(A, A)$ -bimodule homomorphism $f : I \to E$ such that $d = f \circ \delta_A$ , $f$ is uniquely determined by this condition for the definition of $\delta_A$ gives

f(x \otimes 1 - 1 \otimes x) = dx

and our assertion follows from the fact that the image of $\delta_A$ already generates I as a left A-module (Lemma 1): hence of necessity

f\left( \sum_i x_i \otimes y_i \right) = \sum_i x_i \cdot f(1 \otimes y_i - y_i \otimes 1) = -\sum_i x_i \cdot dy_i.

Conversely, as the mapping $(x, y) \mapsto -x \cdot dy$ of $A \times A$ into $E$ is $K$ -bilinear, there exists one and only one $K$ -linear mapping $g : A \otimes_K A \to E$ such that $g(x \otimes y) = -x \cdot dy$ ; it suffices to verify that the restriction $f$ of $g$ to $I$ is A-linear for the left and right A-module structures. The first assertion is obvious since $(xx') \cdot dy = x \cdot (x' \cdot dy)$ ; to prove the second, note that, if $\sum_i x_i \otimes y_i \in I$ and $x \in A$ , then

\sum_i x_i \cdot d(y_i x) = \sum_i x_i \cdot dy_i \cdot x + \sum_i (x_i y_i) \cdot dx

but since $\sum_i x_i y_i = 0$ by definition of $I$ , this completes the proof.

We have thus defined a canonical $K$ -module isomorphism $f \mapsto f \circ \delta_A$

\operatorname{Hom}_{(A, A)}(I, E) \to D_K(A, E)

the left hand side being the $K$ -module of $(A, A)$ -bimodule homomorphisms of $A$ into $E$ .

11. Universal problem for derivations; commutative case

Suppose now that $A$ is a commutative $K$ -algebra and $E$ an $A$ -*module*; $E$ can be considered as an $(A, A)$ -bimodule whose two external laws are identical with the given $A$ -module law. On the other hand the $(A, A)$ -bimodule structure on $A \otimes_K A$ is identical with its $(A \otimes_K A)$ -module structure arising from the commutative ring structure on $A \otimes_K A$ , since in this case, for $x, y, u, v$ in $A$ ,

x \cdot (u \otimes v) \cdot y = (xu) \otimes (vy) = (xu) \otimes (yv) = (x \otimes y)(u \otimes v).

The kernel $\mathfrak{J}$ of $m$ is therefore in this case an ideal of the ring $A \otimes_K A$ and, as $m : A \otimes_K A \to A$ is surjective, $(A \otimes_K A)/\mathfrak{J}$ is isomorphic to $A$ ; if also $E$ is considered as an $(A \otimes_K A)$ -module by means of $m$ (in other words the $(A \otimes_K A)$ -module $m_*(E)$ ), the $(A, A)$ -*bimodule* homomorphisms $\mathfrak{J} \to E$ are identified with the $(A \otimes_K A)$ -*module* homomorphisms $\mathfrak{J} \to E$ (§ 4, no. 3), in other words there is a canonical $K$ -module isomorphism.

\operatorname{Hom}_{(A, A)}(\mathfrak{J}, E) \to \operatorname{Hom}_{A \otimes_K A}(\mathfrak{J}, E).

On the other hand, $\mathfrak{J}E = \{0\}$ , for the elements $1 \otimes x - x \otimes 1$ generate $\mathfrak{J}$ as an $(A \otimes_K A)$ -module (no. 10, Lemma 1) and, for all $z \in E$ ,

(1 \otimes x - x \otimes 1)z = 0

by virtue of the definition of the $(A \otimes_K A)$ -module structure on $E$ . Since $\mathfrak{J}$ is contained in the annihilator of the $(A \otimes_K A)$ -module $E$ and the $((A \otimes_K A)/\mathfrak{J})$ -module structure on $E$ is by definition just the initial $A$ -module structure given on $E$ , there is, taking account of the canonical isomorphism of

\mathfrak{J} \otimes_K ((A \otimes_K A)/\mathfrak{J})

onto $\mathfrak{J}/\mathfrak{J}^2$ (§ 4, no. 1, Corollary 1 to Proposition 1), a canonical $K$ -module isomorphism

\operatorname{Hom}_{A \otimes_K A}(\mathfrak{J}, E) \to \operatorname{Hom}_A(\mathfrak{J}/\mathfrak{J}^2, E).

Taking account of Proposition 17 of no. 10, it is seen that we have proved the following proposition:

Proposition 18

Let $A$ be a commutative $K$ -algebra and $\mathfrak{J}$ the ideal the kernel of the surjective canonical homomorphism $m : A \otimes_K A \to A$ , so that $A$ is isomorphic to $(A \otimes_K A)/\mathfrak{J}$ and $\mathfrak{J}/\mathfrak{J}^2$ has canonically an $A$ -module structure. Let $d_{A/K} : A \to I/I^2$ be the $K$ -linear mapping which associates with every $x \in A$ the class of $x \otimes 1 - 1 \otimes x$ modulo $\mathfrak{J}^2$ . The mapping $d_{A/K}$ is a $K$ -derivation and, for every $A$ -module $E$ and every $K$ -derivation $D : A \to E$ , there exists one and only one $A$ -linear mapping $g : \mathfrak{J}/\mathfrak{J}^2 \to E$ such that $D = g \circ d_{A/K}$ .

Cited by 2

Cited by Proposition 19, Proposition 20

Example

Let M be a K-module; it follows from Proposition 14 of no. 9 that for every S(M)-module E, the mapping D ↦ D | M defines an S(M)-module isomorphism of D_K(S(M), E) onto Hom_K(M, E); on the other hand, since E is an S(M)-module, Hom_K(M, E) is canonically isomorphic to

\operatorname{Hom}_{S(M)}(M \otimes_K S(M), E),

every K-homomorphism of M into E being uniquely expressible in the form $x \mapsto h(x \otimes 1)$ , where

h \in \operatorname{Hom}_{S(M)}(M \otimes_K S(M), E)

(II, § 5, no. 1). Let D_0 be the K-derivation $S(M) \to M \otimes_K S(M)$ whose restriction to M is the canonical homomorphism $x \mapsto x \otimes 1$ ; every K-derivation D : S(M) → E can therefore be written uniquely as $h \circ D_0$ with

h \in \operatorname{Hom}_{S(M)}(M \otimes_K S(M), E).

By the uniqueness of a solution of a universal mapping problem, it is seen that there exists a unique S(M)-module isomorphism

\omega : M \otimes_K S(M) \to \Omega_K(S(M))

such that $D_0 \circ \omega = d_{S(M)/K}$ ; in other words, for all $x \in M$ , $\omega(x \otimes 1) = dx$ .

In particular, if M is a free K-module with basis $(e_\lambda)_{\lambda \in L}$ , $\Omega_K(S(M))$ is a free S(M)-module with basis the set of differentials $de_\lambda$ . Consider in particular the case where $L = \{1, n\}$ , so that S(M) is identified with the polynomial algebra $K[X_1, \ldots, X_n]$ (§ 6, no. 6); for every polynomial $P \in K[X_1, \ldots, X_n]$ , we can write uniquely

dP = \sum_{i=1}^n D_i P . dX_i

with $D_i P \in K[X_1, \ldots, X_n]$ and, by virtue of the above, the mappings $P \mapsto D_i P$ are the K-derivations of $K[X_1, \ldots, X_n]$ corresponding to the coordinate forms on $\Omega_K(S(M))$ for the basis $(dX_i)$ ; we also write $\frac{\partial P}{\partial X_i}$ instead of $D_i P$ and this is called the partial derivative of P with respect to $X_i$ .

Cites 3

Cites Proposition 14, II § 5. Extension of the ring of scalars, III § 6. Symmetric algebras

12. Functiorial properties of k-differentials

Proposition 19

Let

Cites 2

Cites Proposition 18, Proposition 9

\begin{array}{ccc} K & \xrightarrow{\rho} & K' \\ \downarrow \eta & & \downarrow \eta' \\ A & \xrightarrow{u} & A' \end{array}

be a commutative diagram of commutative ring homomorphisms, $\eta$ (resp. $\eta'$ ) making

A (resp. $A'$ ) into a $K$ -algebra (resp. $K'$ -algebra). There exists one and only one $A$ -linear mapping $v : \Omega_K(A) \to \Omega_{K'}(A')$ rendering commutative the diagram

\begin{array}{ccc} A & \xrightarrow{u} & A' \\ d_{A/K} \downarrow & & d_{A'/K'} \downarrow \\ \Omega_K(A) & \xrightarrow{v} & \Omega_{K'}(A') \end{array}

$d_{A'/K'} \circ u$ is a $K$ -derivation of $A$ with values in the $A$ -module $\Omega_{K'}(A')$ ; the existence and uniqueness of $v$ then follow from Proposition 18 of no. 11.

The mapping $v$ of Proposition 19 will be denoted by $\Omega(u)$ ; if there is a commutative diagram of commutative ring homomorphisms

\begin{array}{cccccc} K & \xrightarrow{\sigma} & K' & \xrightarrow{\sigma'} & K'' \\ \eta \downarrow & & \eta' \downarrow & & \eta'' \downarrow \\ A & \xrightarrow{u} & A' & \xrightarrow{u'} & A'' \end{array}

it follows immediately from the uniqueness property of Proposition 19 that

\Omega(u' \circ u) = \Omega(u') \circ \Omega(u).

Since $\Omega_{K'}(A')$ is an $A'$ -module, from $\Omega(u)$ we derive canonically an $A'$ -linear mapping

(41)

\Omega_0(u) : \Omega_K(A) \otimes_A A' \to \Omega_{K'}(A')

such that $\Omega(u)$ is the composition of $\Omega_0(u)$ and the canonical homomorphism $i_A : \Omega_K(A) \to \Omega_K(A) \otimes_A A'$ . For every $A'$ -module $E'$ , there is a commutative diagram

\begin{array}{ccc} \operatorname{Hom}_{A'}(\Omega_{K'}(A'), E') & \xrightarrow{\operatorname{Hom}(\Omega_0(u), 1_{E'})} & \operatorname{Hom}_{A'}(\Omega_K(A) \otimes_A A', E') \\ \phi_{A'} \downarrow & & \phi_A \circ r_A \downarrow \\ D_{K'}(A', E') & \xrightarrow{C(u)} & D_K(A, E) \end{array}

where $C(u)$ is the mapping $D \mapsto D \circ u$ (no. 7, Proposition 9) and $r_A$ is the canonical isomorphism

\operatorname{Hom}(i_A, 1_{E'}) : \operatorname{Hom}_{A'}(\Omega_K(A) \otimes_A A', E') \to \operatorname{Hom}_A(\Omega_K(A), E');

this follows immediately from Proposition 19 and the definition of the isomorphisms $\phi_A$ and $\phi_{A'}$ .

Proposition 20

Suppose that $A' = A \otimes_K K'$ , with $\eta': K' \to A'$ and $u: A \to A'$ the canonical homomorphisms. Then the $A'$ -linear mapping

Cites 3 · Cited by 1

Cites Theorem 1, II § 2, Proposition 1, II § 5, Proposition 18

Cited by Theorem 3, V § 6

\Omega_0(u): \Omega_K(A) \otimes_A A' \to \Omega_{K'}(A')

is an isomorphism.

By virtue of the fact that in diagram (42) the vertical arrows are bijective, it reduces to proving that, for every $A'$ -module $E'$ , the homomorphism $C(u): D \mapsto D \circ u$ in diagram (42) is bijective (II, § 2, no. 1, Theorem 1). Now $\mathrm{Hom}(u, 1_{E'}): \mathrm{Hom}_{K'}(A \otimes_K K', E') \to \mathrm{Hom}_K(A, E')$ is an isomorphism (II, § 5, no. 1, Proposition 1) and $C(u)$ is its restriction to $D_{K'}(A', E')$ and hence is injective; moreover, if $f: A' \to E'$ is a $K'$ -linear mapping such that

f \circ u: A \to E'

is a $K$ -derivation, it follows immediately from the fact that $f$ is $K'$ -linear and the fact that $f((x \otimes 1)(y \otimes 1)) = (y \otimes 1)f(x \otimes 1) + (x \otimes 1)f(y \otimes 1)$ for $x, y$ in $A$ , that $f$ is a $K'$ -*derivation*, the elements $x \otimes 1$ for $x \in A$ forming a generating system of the $K'$ -module $A'$ ; this completes the proof that $C(u)$ is bijective.

From now on we confine our attention to the case where $\rho: K \to K'$ is the identity mapping of $K$ ; every $K$ -algebra homomorphism $u: A \to B$ is therefore mapped to a $B$ -linear mapping

\Omega_0(u): \Omega_K(A) \otimes_K B \to \Omega_K(B).

On the other hand, we can consider the $B$ -module of $A$ -*differentials* $\Omega_A(B)$ since $B$ is an $A$ -algebra by means of $u$ ; the canonical derivation $d_{B/A}: B \to \Omega_A(B)$ being a fortiori a $K$ -derivation, it factorizes uniquely into

B \xrightarrow{d_{B/K}} \Omega_K(B) \xrightarrow{\Omega_u} \Omega_A(B)

where $\Omega_u$ is a $B$ -linear mapping (no. 11, Proposition 18). For every $B$ -module $E$ , there is a commutative diagram

\begin{array}{ccc} \mathrm{Hom}_B(\Omega_A(B), E) & \xrightarrow{\mathrm{Hom}(\Omega_u, 1_E)} & \mathrm{Hom}_B(\Omega_K(B), E) \\ \downarrow \phi_{A,B} & & \downarrow \phi_{K,B} \\ D_A(B, E) & \xrightarrow{j_u} & D_K(B, E) \end{array}

where $j_u$ is the canonical injection (no. 7); this follows immediately from Proposition 18 of no. 11.

Proposition 21

The sequence of $B$ -linear mappings

Cites 2 · Cited by 1

Cites Theorem 1, II § 2, Proposition 11

Cited by V § 16. Differential criteria of separability

\Omega_K(A) \otimes_A B \xrightarrow{\Omega_0(u)} \Omega_K(B) \xrightarrow{\Omega_u} \Omega_A(B) \to 0

is exact.

It reduces to verifying that, for every B-module E, the sequence

0 \to \operatorname{Hom}_B(\Omega_A(B), E) \xrightarrow{\operatorname{Hom}(\Omega_u, 1_E)} \operatorname{Hom}_B(\Omega_K(B), E) \xrightarrow{\operatorname{Hom}(\Omega_0(u), 1_E)} \operatorname{Hom}_B(\Omega_K(A) \otimes_A B, E)

is exact (II, § 2, no. 1, Theorem 1); but by virtue of the fact that in the commutative diagrams (42) and (44) the vertical arrows are isomorphisms, it suffices to show that the sequence

0 \longrightarrow D_A(B, E) \xrightarrow{j_u} D_K(B, E) \xrightarrow{C(u)} D_K(A, E)

is exact, which is just Proposition 11 of no. 7.

We consider now the case where the K-algebra homomorphism $u : A \to B$ is surjective; if $\mathfrak{J}$ is its kernel, B is then isomorphic to $A/\mathfrak{J}$ . We consider the restriction $d|_{\mathfrak{J}} : \mathfrak{J} \to \Omega_K(A)$ of the canonical derivation $d = d_{A/K}$ and the composite A-linear mapping

d' : \mathfrak{J} \xrightarrow{d|_{\mathfrak{J}}} \Omega_K(A) \xrightarrow{i_A} \Omega_K(A) \otimes_A B.

Then $d'(\mathfrak{J}^2) = 0$ , since, for $x, y$ in $\mathfrak{J}$ ,

d'(xy) = d(xy) \otimes 1 = (x.dy + y.dx) \otimes 1 = dy \otimes u(x) + dx \otimes u(y) = 0

since $u(x) = u(y) = 0$ . Hence we derive from $d'$ , by passing to the quotient, an A-linear mapping

\bar{d} : \mathfrak{J}/\mathfrak{J}^2 \to \Omega_K(A) \otimes_A B

and as $\mathfrak{J}$ annihilates the A-module $\mathfrak{J}/\mathfrak{J}^2$ , $\bar{d}$ is a B-linear mapping.

Proposition 22

Let $\mathfrak{J}$ be an ideal of the commutative K-algebra A, $B = A/\mathfrak{J}$ and $u : A \to B$ the canonical homomorphism. The sequence of B-linear mappings

Cites 2 · Cited by 2

Cites Proposition 9, Exercise 5, § 10

Cited by Lemma 4, V § 6, Proposition 3, V § 7

\mathfrak{J}/\mathfrak{J}^2 \xrightarrow{\bar{d}} \Omega_K(A) \otimes_A B \xrightarrow{\Omega_0(u)} \Omega_K(B) \to 0

is then exact.

Note that $\Omega_K(A) \otimes_A B$ is identified with $\Omega_K(A)/\mathfrak{J}\Omega_K(A)$ and that the image of $\bar{d}$ is the image of $d(\mathfrak{J})$ in this quotient module; the quotient of $\Omega_K(A) \otimes_A B$ by $\operatorname{Im}(\bar{d})$ is therefore identified with the quotient $\Omega_K(A)/N$ , where N is the sub-A-module generated by $\mathfrak{J}\Omega_K(A)$ and $d(\mathfrak{J})$ . Moreover, the composite mapping

A \xrightarrow{d_{A/K}} \Omega_K(A) \longrightarrow \Omega_K(A)/N

is a K-derivation (no. 7, Proposition 9) and, since it is zero on $\mathfrak{J}$ by definition of N, it defines, when passing to the quotient, a K-derivation $D_0 : B \to \Omega_K(A)/N$ .

Taking account of the uniqueness of the solution of a universal mapping problem, it reduces to proving that, for every B-module E and every K-derivation D: B → E, there exists a unique B-linear mapping g: Ω_K(A)/N → E such that D = g ∘ D_0. But, the composite mapping D ∘ u: A → E is a K-derivation (no. 7, Proposition 9) and hence there exists one and only one A-linear mapping f: Ω_K(A) → E such that f ∘ d_{A/K} = D ∘ u. This relation shows already that f is zero on d(Σ); as also ΣE = {0} since E is a B-module, f is zero on ΣΩ_K(A); hence f is zero on N and defines, when passing to the quotient, a B-linear mapping g: Ω_K(A)/N → E such that g ∘ D_0 = D; the uniqueness of g follows from the uniqueness of f.

It must not be thought that, even if u: A → B is an injective homomorphism, Ω_0(u): Ω_K(A) ⊗_A B → Ω_K(B) is injective (Exercise 5). However we have the following proposition:

Proposition 23

Let A be an integral K-algebra, B its field of fractions and u: A → B the canonical injection. Then Ω_0(u): Ω_K(A) ⊗_A B → Ω_K(B) is an isomorphism.

Cites 1

Cites Proposition 5

Using the fact that in diagram (42) the vertical arrows are bijective, it reduces to proving that, for every vector space E over B, the mapping C(u): D_K(B, E) → D_K(A, E) is bijective. But this follows from the fact that every K-derivation of A into E can be extended uniquely to a K-derivation of B into E (no. 5, Proposition 5).

Exercises

See the exercises for § 10.