Bourbaki

Exercise 5, § 3

Commutative Algebra, VI, 448

Bourbaki marks this one as harder than the rest.

Let kk be a field and K=k(X,Y)K = k(X, Y) the field of rational functions in two indeterminates over kk. We write x1=X,y1=Yx_1 = X, y_1 = Y and for n≥1n \geq 1 define inductively the elements xn+,ynx_n + , y_n and yn+,=xnyn−1y_n + , = x_n y_n^{-1} of K. (a) Write An=k[xn,yn]A_n = k[x_n, y_n]; the sequence (An)(A_n) is increasing and, if

mn=Anxn+Anyn, m_n = A_n x_n + A_n y_n,
mnm_n is a maximal ideal of A, and mn=An∩mn+1m_n = A_n \cap m_{n+1}; in the ring A=⋂nAnA = \bigcap_n A_n, m=⋂nmnm = \bigcap_n m_n is a maximal ideal. (b) Let vv be the valuation on K defined by v(X)=ρ=(1+5)/2v(X) = \rho = (1 + \sqrt{5})/2, v(Y)=1v(Y) = 1; let B and p be the ring and ideal of vv; show that m=p∩Am = p \cap A (note that ρ2−ρ−1=0\rho^2 - \rho - 1 = 0 and use this fact to calculate v(yn)v(y_n)). Show that for a monomial XiYjX^i Y^j (i,ji, j positive or negative integers) to belong to A, it is necessary and sufficient that (1+52)i+j>0\left( \frac{1 + \sqrt{5}}{2} \right)^i + j > 0; for every monomial z=XiYjz = X^i Y^j, therefore, either z∈Az \in A or 1/z∈A1/z \in A. (c) Show that B=AmB = A_m (write an element t∈Bt \in B in the form of a quotient of two polynomials in X and Y and show in each of them the monomials at which vv takes the least value). (d) Show that, for all nn, (An+1)mn+1(A_{n+1})_{m_{n+1}} is the prime quadratic transform of (An)mn(A_n)_{m_n} in the direction of vv (Exercise 4).

Solution

No solution has been written yet.

Transcribed from the printed volume. Not translated.