Bourbaki

Exercise 1, § 4

Algebra, IV, p. 90

Let u(X)=∑n=0mαnXnu(X) = \sum_{n=0}^m \alpha_n X^n be a formal power series over a commutative field KK. a) For uu to be a rational fraction of K(X)K(X) it is necessary and sufficient that there should exist a finite sequence of elements (λi)1≤i≤q(\lambda_i)_{1 \leq i \leq q} of elements of KK, not all zero, and an integer d≥0d \geq 0 such that for all n≥dn \geq d we have

λ1αn+λ2αn+1+⋯+λqαn+q−1=0. \lambda_1 \alpha_n + \lambda_2 \alpha_{n+1} + \cdots + \lambda_q \alpha_{n+q-1} = 0 .
b) Put
Hn(k)=∣αnαn+1⋯αn+k−1αn+1αn+2⋯αn+kαn+2αn+3⋯αn+k+1αn+k−1αn+k⋯αn+2k−2∣ H_n^{(k)} = \begin{vmatrix} \alpha_n & \alpha_{n+1} & \cdots & \alpha_{n+k-1} \\ \alpha_{n+1} & \alpha_{n+2} & \cdots & \alpha_{n+k} \\ \alpha_{n+2} & \alpha_{n+3} & \cdots & \alpha_{n+k+1} \\ \alpha_{n+k-1} & \alpha_{n+k} & \cdots & \alpha_{n+2k-2} \end{vmatrix}
(« Hankel determinant ») show that if Hd+i(q+1)=0H_{d+i}^{(q+1)} = 0 and Hd+i(q)≠0H_{d+i}^{(q)} \neq 0 for every integer j≥0j \geq 0, then u(X)u(X) is a rational fraction (use a)). c) Prove the identity
Hn(k)Hn+2(k)−Hn(k+1)Hn+2(k−1)=(Hn+1(k))2 H_n^{(k)} H_{n+2}^{(k)} - H_n^{(k+1)} H_{n+2}^{(k-1)} = (H_{n+1}^{(k)})^2
(cf. III, p. 639, Exercise 10). Deduce that if Hm+j(k+1)=0H_{m+j}^{(k+1)} = 0 for 0≤j≤r−10 \leq j \leq r-1, then the rr determinants Hm+j(k)H_{m+j}^{(k)} where 1≤j≤r1 \leq j \leq r are either all zero or all non-zero. d) Deduce from b) and c) that for u(X)u(X) to be a rational fraction it is necessary and sufficient that there exist two integers dd and qq such that Hd+i(q+1)=0H_{d+i}^{(q+1)} = 0 for all integers j≥0j \geq 0. e) Let kk be a subfield of KK, and identify K(X)K(X), k(X)k(X) and k[[X]]k[[X]] with subrings of K((X))K((X)). Prove that k[[X]]∩K(X)=k[[X]]∩k(X)k[[X]] \cap K(X) = k[[X]] \cap k(X).

Solution

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