Proposition 1

Let A be a filtered ring, $(A_n)_{n \in \mathbf{Z}}$ its filtration and v its orderfunction. Suppose that $\mathrm{gr}(A)$ is a ring with no divisor of zero. Then, for every ordered pair of elements a, b of the ring $B = \bigcup_{n \in \mathbf{Z}} A_n$, $v(ab) = v(a) + v(b)$.

As $n = \bigcap_{n \in \mathbf{Z}} A_n$ is a two-sided ideal of the ring B, the formula holds if $v(a)$ or $v(b)$ is equal to $+\infty$. If not, $v(a) = r$ and $v(b) = s$ are integers; the classes $\alpha$ of a mod. A, +, and $\beta$ of b mod. A, +_1 are $\neq 0$ by definition, whence by hypothesis $\alpha \beta \neq 0$ in $\mathrm{gr}(A)$ and therefore $ab \notin A_{r+s+1}$, as $ab \in A_{r+s}$,