Exercise 1, § 3

Commutative Algebra, VI, 446

Bourbaki marks this one as harder than the rest.

(a) Let L be a field and T an indeterminate; in the field of formal power series $L((T))$ , consider the series

s = c_0 + c_1 T^{1!} + c_2 T^{2!} + \ldots + c_n T^{n!} + \ldots

where the $c_i \in L$ are all $\neq 0$ . Show that s is transcendental over the subfield $L(T)$ of $L((T))$ . (Argue by reductio ad absurdum by supposing that there exists a relation $a_0(T) + a_1(T)s + \ldots + a_q(T)s^q = 0$ where $a_i(T)$ are polynomials in $L(T)$ such that $a_i \neq 0$ ; construct the equation satisfied by

s' = s - c_0 - c_1 T^{1!} - \ldots - c_{n-1} T^{(n-1)!}

and show that for n sufficiently large its constant term is a polynomial of degree $< n!$ , whence our contradiction.)

(b) Let k be a field, X, Y, Z, T four indeterminates, $K = k(X, Y, Z)$ and E an algebraic closure of $k(X)$ ; in E, for all n, let $x_n$ denote an element such that $x_n^n = X$ . In the field of formal power series $E((T))$ , consider the element

s = x_1 T + x_2 T^{2!} + \ldots + x_n T^{n!} + \ldots

The elements s and T are algebraically independent over E by (a). Consider the homomorphism $f : K \to E((T))$ such that $f(X) = X, f(Y) = T, f(Z) = s$ . If v is the discrete valuation on $E((T))$ defined by the order of formal power series (no. 3, Example 3), $v \circ f$ is a discrete valuation on $K$ ; show that the residue field of $v \circ f$ is the subfield $k(x_1, x_2, \ldots, x_n, \ldots)$ of $E$ and is therefore infinite over $k$ .

Solution

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