Bourbaki

Exercise 7, § 3

Commutative Algebra, VII

Bourbaki marks this one as harder than the rest.

Soient K a noetherian factorial ring, F an extremal element of the polynomial ring K[X1,…,Xn]K[X_1, \ldots, X_n]; suppose that when a weight q(i)>0q(i) > 0 (1≤i≤n1 \leq i \leq n) is assigned to each XiX_i, F is isobaric and of weight q>0q > 0. Let A be the ring generated, in an algebraic closure Ω\Omega of the field of fractions of K(X1,…,Xn)K(X_1, \ldots, X_n), by K, the XiX_i (1≤i≤n1 \leq i \leq n), and a root zz of the polynomial Zc−FZ^c - F, where cc is an integer prime to qq. Now prove that A is factorial in the following two cases: 1) one has c≡1(modq)c \equiv 1 \pmod{q}; 2) every finitely generated projective K-module is free (which is the case for example when K is a field, or a principal ideal domain, or a local ring). (In the first case, consider the ring of fractions A[1/z]A[1/z]; show that it is factorial and apply prop. 3 of No. 4. In the second case, consider an integer dd such that cd≡1(modq)cd \equiv 1 \pmod{q}, and let z′∈Ωz' \in \Omega be such that z=z′dz = {z'}^d; the ring B=A[z′]B = A[z'] is factorial by virtue of the first case, and is a free a-module. Consider B as a graded ring by taking z′z' of weight qq, each of the XiX_i of weight cdq(i)cdq(i), and reduce to proving that for two homogeneous elements u,vu, v of A, the ideal Au∩AvAu \cap Av is principal, using Exercise 16 of § 1; finally consider the ideal Bu∩BvBu \cap Bv in B and use Chapter I, § 3, No. 6, prop. 12). In particular, if K is a field, a,b,ca, b, c three integers >0> 0 prime to each other pairwise, the ring

A=K[X,Y,Z]/(Za−Xb−Yc) A = K[X, Y, Z]/(Z^a - X^b - Y^c)
is factorial.

Solution

No solution has been written yet.

Machine translation of the English, by gpt-5-mini. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.