Bourbaki

Example 1

Tag 0DSZ, permanent. Cite it as [Bourbaki, Tag 0DSZ].

Algebra, chapter X, § 2. Complexes de A-modules

Take AA to be the ring k[X1,...,Xn]k[X_1, ..., X_n]. Then ΩA/k1\Omega_{A/k}^1 is a free AA-module with basis dX1,...,dXndX_1, ..., dX_n (III, p. 134, example). Consequently, if, for every subset I={i1,...,ip}I = \{ i_1, ..., i_p \} of {1,n}\{ 1, n \}, one sets dXI=dXi1∧...∧dXipdX_I = dX_{i_1} \wedge ... \wedge dX_{i_p} (with i1<...<ipi_1 < ... < i_p), the AA-module ΩA/kp\Omega_{A/k}^p admits as basis the elements dXIdX_I, where II runs through the set of subsets of {1,n}\{ 1, n \} of cardinal pp. One has

d(PdXI)=dP∧dXI=∑i∉I(−1)n(I,i)∂P∂XidXI∪{i}, d(P dX_I) = dP \wedge dX_I = \sum_{i \notin I} (-1)^{n(I,i)} \frac{\partial P}{\partial X_i} dX_{I \cup \{ i \}} ,
where n(I,i)n(I, i) denotes the number of elements of II strictly less than ii.

The cycles of Zp(ΩA/k)Z^p(\Omega_{A/k}) are therefore the elements ω=∑Card⁡(I)=pPIdXI\omega = \sum_{\operatorname{Card}(I) = p} P_I dX_I such that, for every subset JJ with (p+1)(p + 1) elements of {1,n}\{ 1, n \}, one has:

∑i∈J(−1)n(J,i)∂PJ−{i}∂Xi=0. \sum_{i \in J} (-1)^{n(J,i)} \frac{\partial P_{J-\{i\}}}{\partial X_i} = 0 .

The element ω\omega is a boundary if one can choose, for every subset J⊂{1,n}J \subset \{1, n\} with (p−1)(p - 1) elements, a polynomial QJ∈AQ_J \in A in such a way that:

PI=∑j∈I(−1)n(I,j)∂QI−{j}∂Xj. P_I = \sum_{j \in I} (-1)^{n(I,j)} \frac{\partial Q_{I-\{j\}}}{\partial X_j}.

We shall see in § 9 that the de Rham complex of AA over kk is acyclic in degrees >0> 0 if kk is a Q\mathbf{Q}-algebra (X, p. 159, remark 4).

Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.

Cited by

Nothing.

Cites

Nothing.

Translated with machine help, not yet reviewed by a person.