Bourbaki

Exercise 7, § 3

Commutative Algebra, VII, 565

Bourbaki marks this one as harder than the rest.

Let KK be a Noetherian factorial domain and FF 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 attributed to each XiX_i, FF is isobaric and of weight q>0q > 0. Let AA be the domain generated, in an algebraic closure Ω\Omega of the field of fractions of K(X1,…,Xn)K(X_1, \ldots, X_n), by KK, 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. Show that AA is factorial in the two following cases: (1) c≡1(modq)c \equiv 1 \pmod{q}; (2) every finitely generated projective KK-module is free (which holds for example when KK 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 a factorial domain and apply Proposition 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 such that z=z′dz = {z'}^d; the domain B=A[z′]B = A[z'] is factorial by virtue of the first case and is a free AA-module. Consider BB as a graded ring taking z′z' to be of weight qq, each of the XiX_i of weight cdq(i)cdq(i) and reduce it to proving that for two homogeneous elements u,vu, v of AA the ideal Au∩AvAu \cap Av is principal, using Exercise 16 of § 1; consider finally the ideal Bu∩BvBu \cap Bv in BB and use Chapter I, § 3, no. 6, Proposition 12.) In particular, if KK is a field and a,b,ca, b, c are three integers >0> 0 which are relatively prime in pairs, the domain

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.

Transcribed from the printed volume. Not translated.