Bourbaki

Exercise 15, § 7

Commutative Algebra, VIII, p. 106

Let AA be a local, noetherian, complete ring, with maximal ideal m. Put d=dim⁡(A)d = \dim(A). Suppose that A contains a subfield k such that A=k+mA = k + m.

Prove that one can find x1,...,xdx_1, ..., x_d in m such that: α)\alpha) the morphism φ:k[[X1,...,Xd]]→A\varphi : k[[X_1, ..., X_d]] \to A defined by φ(Xi)=xi\varphi(X_i) = x_i makes A an integral k[[X1,...,Xd]]k[[X_1, ..., X_d]]-algebra ; β)\beta) denoting by m0=(X1,...,Xd)m_0 = (X_1, ..., X_d) the maximal ideal of k[[X1,...,Xd]]k[[X_1, ..., X_d]], one has m0A‾=m\overline{m_0A} = m, and hence em0A(A)=em(A)e_{m_0A}(A) = e_m(A). Prove that if x1,...,xdx_1, ..., x_d satisfy condition α)\alpha), condition β)\beta) is satisfied if and only if the initial forms ξ1,...,ξd\xi_1, ..., \xi_d of the xix_i in grm(A)\mathrm{gr}_m(A) form a maximal transversal sequence.

Solution

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