# Suppose we're given a filtered algebra A over a field k with filtration F &#x2

Suppose we're given a filtered algebra $A$ over a field $k$ with filtration ${F}_{\bullet }A$ over the subspaces of $A$:
$\left\{0\right\}\subseteq {F}_{0}A\subseteq \cdots \subseteq {F}_{i}A\subseteq \cdots \subseteq A,$
and suppose that
${\mathrm{g}\mathrm{r}}_{\bullet }^{F}A:=\underset{i\in {\mathbb{N}}_{0}}{⨁}{\mathrm{g}\mathrm{r}}_{i}^{F}A$
is the associated graded algebra of $A$, where ${\mathrm{g}\mathrm{r}}_{i}^{F}A:={F}_{i}A/{F}_{i-1}A$ and ${\mathrm{g}\mathrm{r}}_{0}^{F}A={F}_{0}A$.
If ${\mathrm{g}\mathrm{r}}_{\bullet }^{F}A$ is commutative, does it follow that ${F}_{i+j}A\subseteq {F}_{i}A\cdot {F}_{j}A$ for all $i,j\in {\mathbb{N}}_{0}$?
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Zackery Harvey
Nope. To keep things really simple let's take $A$ itself to be commutative and graded (with the filtration where ${F}_{n}\left(A\right)$ is sums of elements of degree at most n, so the associated graded is just $A$ again): specifically, take
$A=k\left[{x}_{1},{x}_{2}\right]/\left({x}_{1}^{2}={x}_{2}^{2}=0\right)$
where $\mathrm{deg}{x}_{i}=i$. (It doesn't really matter whether we impose ${x}_{2}^{2}=0$ or not, the point is just to have an element of degree $2$ that isn't a sum of products of elements of degree $1$.)