$, one has $\bar{a}\bar{b} = [a+F^{n-1}R][b+F^{m-1}R]
\subseteq [ab+F^{n+m-1}R]$. Since $\bar{a}\bar{b}\not=0$, $ab\in F^{n+m}R\smallsetminus F^{n+m-1}R$, whence $\bar{a}\bar{b}=\sigma(ab)$ is the image of $ab$ in $\operatorname{gr}_F(AB)$.

\(2) Define a map $\rho: \operatorname{gr}_FA\times \operatorname{gr}_FB \to \operatorname{gr}_F(A\otimes_RB)$ by $\rho(\bar{a}, \bar{b} ) = [a\otimes b + F^{n+m-1}(A\otimes B)]$, for $\bar{a}\in \operatorname{gr}_F^nA$, $\bar{b}\in \operatorname{gr}_F^mB$ and where the rest of the notation is the same as for part (1). This clearly defines a ${\mathbb{C}}$-bilinear map that is $\operatorname{gr}_F R$-balanced in the sense that $\rho(\bar{a}\bar{r}, \bar{b}) = \rho(\bar{a}, \bar{r}  \bar{b})$ for $\bar{r}\in \operatorname{gr}^s_FR$. By universality, $\rho$ therefore induces a map $ \operatorname{gr}_FA\otimes_{\operatorname{gr}R}\operatorname{gr}_FB\to \operatorname{gr}_F(A\otimes_RB)$. It is surjective since $F^{n+m}(A\otimes B)/ F^{n+m-1}(A\otimes B)$ is spanned by elements of the given form $[a\otimes b + F^{n+m-1}(A\otimes B)].$

Lemma {#grade-elements}
-----

Let $R=\bigcup_{i\geq 0}F^iR$ be a filtered ring, pick $r\in R$ and let $ I$ be a subset of $R$. Under the induced filtrations, $\operatorname{gr}_F(rI) = \sigma(r)\operatorname{gr}_F(I)$ in the following cases:

1.  $\sigma(r)$ is regular in $\operatorname{gr}_FR$;

2.  $ r=r^2 \in F^0(R)$ and $rI\subseteq I$.

Assume that $r\in F^sR\smallsetminus F^{s-1}R$. We claim that, in both cases, it suffices to prove that $F^n(rI) = rF^{n-s}I$ for all $n\geq s$. Indeed, if this is true then the identity $F^m(rI) = rI\cap F^mR$ implies that the $n^{\mathrm{th}}$ summand of $\operatorname{gr}(rI)$ equals $$\frac{F^n(rI)}{F^{n-1}(rI)} \ = \  \frac{F^n(rI)}{F^n(rI)\cap F^{n-1}R}
      \  \cong\  \frac{ F^n(rI) + F^{n-1}R}{F^{n-1}R}
        \ = \  \frac{rF^{n-s}I + F^{n-1}R}{F^{n-1}R},$$ which is the $n^{\mathrm{th}}$ summand of $\sigma(r)gr(I)$.

\(1) In this case, $rt\in F^n(rI) = rI\cap F^n(R) \Leftrightarrow
 t\in I\  \mathrm