Proposition 11

— Suppose that $A$ is an integral domain.

(i) The ring $A[[I]]$ is again an integral domain.

(ii) If $u, v$ are non-zero elements of $A[[I]]$, then $\omega(uv) = \omega(u) + \omega(v)$.

For each $J \subset I$ let $\varphi_J$ be the homomorphism of $A[[I]]$ into $A[[J]]$ obtained by substituting in each element of $A[[I]]$, $X_i$ for $X_i$ when $i \in J$ and 0 for $X_i$ when $i \in I - J$. Let $u, v$ be non-zero elements of $A[[I]]$, $p = \omega(u)$, $q = \omega(v)$; there exists a finite subset $J$ of $I$ such that

Let a (resp. b) be the homogeneous component of degree $p$ (resp. $q$) of $\varphi_J(u)$ (resp. $\varphi_J(v)$). Since $J$ is finite, a and b are polynomials. We have $a \neq 0, b \neq 0$, hence $ab \neq 0$ (IV, p. 9, Prop. 8). Hence $\varphi_J(u)\ \varphi_J(v)$ is non-zero, of order $p + q$. It follows that $uv \neq 0$ and $\omega(uv) \leq p + q$; but clearly $\omega(uv) \geq p - q$.