Take A to be the ring k[X1,...,Xn]. Then ΩA/k1 is a free A-module with basis dX1,...,dXn (III, p. 134, example). Consequently, if, for every subset I={i1,...,ip} of {1,n}, one sets dXI=dXi1∧...∧dXip (with i1<...<ip), the A-module ΩA/kp admits as basis the elements dXI, where I runs through the set of subsets of {1,n} of cardinal p. One has
d(PdXI)=dP∧dXI=i∈/I∑(−1)n(I,i)∂Xi∂PdXI∪{i}, where
n(I,i) denotes the number of elements of
I strictly less than
i.
The cycles of Zp(ΩA/k) are therefore the elements ω=∑Card(I)=pPIdXI such that, for every subset J with (p+1) elements of {1,n}, one has:
i∈J∑(−1)n(J,i)∂Xi∂PJ−{i}=0.
The element ω is a boundary if one can choose, for every subset J⊂{1,n} with (p−1) elements, a polynomial QJ∈A in such a way that:
PI=j∈I∑(−1)n(I,j)∂Xj∂QI−{j}.
We shall see in § 9 that the de Rham complex of A over k is acyclic in degrees >0 if k is a 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.
Nothing.
Nothing.