Corollary

Let A be a filtered ring and $(A_n)_{n \in \mathbf{Z}}$ its filtration; let us set $B = \bigcup_{n \in \mathbf{Z}} A_n$, $n = \bigcap_{n \in \mathbf{Z}} A_n$. If the ring $\mathrm{gr}(A)$ has no divisors of zero, neither has the ring $B/n$. If a and b are elements of B not belonging to n, then $v(a) \neq +\infty$ and $v(b) \neq +\infty$, whence $v(ab) \neq +\infty$ and therefore $ab \notin n$.

Note that the ring A can be an integral domain and the filtration (A_n) exhaustive and separated without $\mathrm{gr}(A)$ being an integral domain (Exercise 2).