Remark

— Let $A$ be the set of all formal power series with constant term 1. By Prop. 6, $M$ is a commutative group under multiplication; the multiplicative group of $A[[I]]$ is thus the direct product of $M$ and the multiplicative group of $A$. We shall equip $A$ with the topology induced from that of $A[[I]]$. For each $\beta \in \mathbf{N}^{(I)}$ we have in IV, p. 26 defined the ideal $a_\beta$ of $A[[I]]$; then $1 + a_\beta$ is a subgroup of $M$ and the family $(1 + a_{\beta'})$ is a fundamental system of neighbourhoods of 1 in $M$. Since the multiplication in $M$ is continuous, we see that $A$ is a topological group (Gen. Top., III, p. 223); in other words, the mapping $f \mapsto f^{-1}$ is continuous in $M$.

Let $K$ be a commutative field and $\mathscr{O}$ the subring of the field of rational fractions $K((X_i)_{i \in I})$ formed of rational fractions in which the element 0 of $K^1$ is substitutable. If $f \in \mathscr{O}$, we have $f = \frac{u}{v}$, where $u$ and $v$ are polynomials such that the constant term of $v$ is $\neq 0$, hence $v$ is invertible in $K[[I]]$. We can verify at once that the element $uv^{-1}$ of $K[[I]]$ depends only on $f$; we say that the formal power series $uv^{-1}$ is the expansion at the origin of the rational fraction $\frac{u}{v}$. The mapping $f \mapsto uv^{-1}$ is an injective homomorphism of $\mathscr{O}$ into $K[[I]]$; we shall often identify $\mathscr{O}$ with its image under this mapping.