Exercise 17, § 1

Suppose that $K$ is a field. Let $T$ be an indeterminate and $L = K((T))$; $L$ has canonically a valuation (order of the formal power series) which makes it a complete valued field.

(a) Let $V$ be a finite-dimensional vector space over $K$. Then $V_{(L)}$ has canonically a complete metrizable topological vector space structure (Topological Vector Spaces, Chapter I, § 2, no. 3, Theorem 2). Let $(x_p)_{p \in \mathbf{Z}}$ be a family of elements of $V$ such that $x_p = 0$ for $p$ less than some rational integer. Then the series $\sum_{-\infty}^{+\infty} x_p T^p$ converges in $V_{(L)}$ and every element of $V_{(L)}$ can be expressed uniquely in this way. (Take a basis of $V$.)

(b) The vector space $\mathscr{L}_L(V_{(L)}) = \mathscr{L}_K(V)_{(L)}$ has a canonical complete metrizable topological vector space structure. Every endomorphism of $V_{(L)}$ can be expressed uniquely in the form $\sum_{-\infty}^{+\infty} u_p T^p$, where $u_p \in \mathscr{L}_K(V)$ and the $u_p$ are zero for $p$ less than some rational integer.

(c) If $K$ is of characteristic 0 and $u \in \mathscr{L}_K(V)$, we write

If $x \in V$ is such that $ux = 0$, then $e^{T u} . x = x$.

(d) If $W$ is another finite-dimensional vector space over $K$ and $v \in \mathscr{L}_K(W)$, then $e^{T(u \otimes 1 + 1 \otimes v)} = e^{T u} \otimes e^{T v}$.

(e) Deduce from (c) and (d) that, if $V$ has a (not necessarily associative) algebra structure and $u$ is a derivation of $V$, then $e^{T u}$ is an automorphism of the algebra $V_{(L)}$.