Proposition 6

— Let $u \in A[[I]]$; then for $u$ to be invertible in $A[[T]]$ it is necessary and sufficient that its constant term should be invertible in $A$.

Suppose that there exists $v \in A[[I]]$ such that $uv = 1$. Let $a, \beta$ be the constant terms of $u$ and $v$, then $\alpha \beta = 1$, so $a$ is invertible.

Conversely, suppose that the constant term $a$ of $u$ is invertible. Then there exists a formal power series $t \in A[[I]]$ such that $u = \alpha (1 - t)$ and $\omega(t) > 0$. Now there is a ring homomorphism $\varphi : A[[T]] \to A[[I]]$ such that $\varphi(T) = t$, and $1 - T$ is invertible in $A[[T]]$ (Prop. 5); consequently $1 - t$ is invertible in $A[[I]]$, and hence so is $u$.