Bourbaki

Exercise 26, § 3

Commutative Algebra, VII, 570

Let AA be a factorial domain and KK its field of fractions. For elements fif_i (1≤i≤r1 \leq i \leq r) of B=A[X1,…,Xn]B = A[X_1, \ldots, X_n] to be such that the ideal ∑i=1rBfi\sum_{i=1}^r Bf_i is equal to B, it is necessary and sufficient that there exist polynomials viv_i (1≤i≤r1 \leq i \leq r) in K[X1,…,Xn]K[X_1, \ldots, X_n] such that ∑i=1rvifi=1\sum_{i=1}^r v_i f_i = 1 and which satisfy further the following condition: if we write vi=wi/dv_i = w_i / d, where d∈Ad \in \mathbf{A}, the polynomials wiw_i belong to A[X1,…,Xn]\mathbf{A}[X_1, \ldots, X_n] and the g.c.d. of the set of coefficients of all the wiw_i (1≤i≤r1 \leq i \leq r) is equal to 1, then, for every extremal element a\mathfrak{a} of A\mathbf{A} dividing dd, the ideal generated by the classes of theft in the ring (A/Ap)[X1,…,Xn](\mathbf{A}/\mathfrak{A}\mathfrak{p})[X_1, \ldots, X_n] is the whole of this ring.

Solution

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Transcribed from the printed volume. Not translated.