Bourbaki

Exercise 12, § 2

Commutative Algebra, III, 235

Let K be a commutative field, A the polynomial ring K[Xi]i∈IK[X_i]_{i \in I} in an arbitrary infinite family of indeterminates and m the (maximal) ideal of A generated by the XiX_i. If we set Ai=A/mi+1A_i = A/m^{i+1}, the AiA_i, and the canonical homomorphisms hij:A/mj+1→A/mi+1h_{ij} : A/m^{j+1} \to A/m^{i+1} for i<ji < j satisfy the conditions of Proposition 14; the ring lim⁡←A\lim \leftarrow A, is the completion A^\hat{A} of AA with respect to the m-adic topology and the kernel of the canonical homomorphism A^→A\hat{A} \to A, is equal to (mt)∗(m^t)^*, the closure of mtm^t in AA.

(a) Show that A^\hat{A} is canonically identified with the ring of formal power series in the XnX_n, each of which has only a finite number of terms of given degree (Algebra, Chapter IV, § 5, Exercise 1).

(b) From now on take I=NI = \mathbf{N}. Show that m≠A^.mm \neq \hat{A}.m (consider the formal power series ∑n=1∞Xnn\sum_{n=1}^\infty X_n^n).

(c) Suppose that KK is a finite field. Show that (m^)2≠(m2)∗(\hat{m})^2 \neq (m^2)^*. (Show first the following result: for every integer k>0k > 0, there exists an integer nkn_k and for every integer n≥nkn \geq n_k a homogeneous polynomial FF, of degree nn in n2n^2 indeterminates with coefficients in KK such that FF, cannot be the sum of the terms of degree nn in any polynomial of the form P1Q1+…+PkQkP_1 Q_1 + \ldots + P_k Q_k, where the PiP_i and QiQ_i are polynomials without constant term in the same n2n^2 indeterminates).

Deduce that A^\hat{A} is not complete with respect to the m-adic topology.

Solution

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Transcribed from the printed volume. Not translated.