Bourbaki

Exercise 21, § 2

Commutative Algebra, VII, 562

Bourbaki marks this one as harder than the rest.

(a) Let A\mathbf{A} be an integral domain, f,gf, g two polynomials in A[X]\mathbf{A}[X] and h=fgh = fg. Let a,b,ca, b, c denote the ideals of A\mathbf{A} generated respectively by the coefficients of f,g,hf, g, h. Show that, if deg⁡(g)=n\deg(g) = n, then an+1b=anca^{n+1}b = a^n c. (Let f(X)=∑i=1maiXi,g(X)=∑j=1nbjXjf(X) = \sum_{i=1}^m a_i X^i, g(X) = \sum_{j=1}^n b_j X^j; for every increasing sequence

σ=(ik)1≤k≤n+1 \sigma = (i_k)_{1 \leq k \leq n+1}
of integers ≤m\leq m and all jj, let
uσ,j=ai1ai2⋯aij+1bjaij+2⋯ain+1; u_{\sigma,j} = a_{i_1} a_{i_2} \cdots a_{i_j+1} b_j a_{i_j+2} \cdots a_{i_{n+1}};
take on the set of uσ,ju_{\sigma,j} a total ordering such that, for j<j′j < j', uσ,j<uτ,j′u_{\sigma,j} < u_{\tau,j'} for all σ,τ\sigma, \tau, and, for all jj, uσ,j≤uτ,ju_{\sigma,j} \leq u_{\tau,j} if and only if σ≤τ\sigma \leq \tau in the lexicographical ordering on [0,m]n+1[0, m]^{n+1}. Then argue by induction on this totally ordered set.)

(b) Deduce from (a) that, if A\mathbf{A} is Pruferian (Exercise 12), then c=abc = ab.

(c) Taking A\mathbf{A} to be the polynomial ring Z[Y]\mathbf{Z}[Y], give an example of two polynomials f,gf, g of the first degree in A[X]\mathbf{A}[X] such that c≠abc \neq ab.

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.