Let $R$ be a finitely generated $k$-algebra of dimension greater than $1$, let $Q$ be any maximal ideal of $R$. It is claimed by my lecturer that one can find a homogeneous, non-unit, non-zero divisor in the associated graded ring gr$R$ corresponding to the $Q$-adic filtration of $R$. Explicitly gr$R$=$\bigoplus_{i\geq 0} Q^i/Q^{i+1}$. I do not see why this is the case.
In particular I tried to show $(Q,x+Q^2,Q^3,..)$ in gr$R$ has these properties where $x\in Q-Q^2$. Note that $Q\neq Q^2$ else gr$R$ is a field which is a contradiction since know dim(gr$R$)$=$dim($R)\gt1$. For this example to work I really want something of the form that if $x.y\in Q^n $ where $y\in Q^{n-2}$ then $y\in Q^{n-1}$. This is true for $n=2$ by using $R/Q$ field.
Another thing is that this result is a step to prove another result which implied $Q^i \neq Q^{i+1}$ for all $i$. It is clear that what I claimed in the last paragraph would be false without this.
I might be mixing up things, but intuitively I think you need stronger assumptions on $R$: The dimension of $R$ concerns the minimal primes of $R$ (geometrically, the irreducible components of $\text{Spec}(R)$) while the (non-)presence of zero-divisors concerns the associated primes of $R$ (geometrically, the 'virtual' irreducible components of $\text{Spec}(R)$), not all of which need to be minimal.
Example: Begin with the $n$-plane $S=k[x_1,...,x_n]$ and build $R$ from $S$ by adding an infinitesimal line at the origin, i.e. $R := S[t]/(t^2, tx_1,...,tx_n)$. Then $R$ stil has dimension $n$, but any homogeneous element of positive degreee annihilates $t$, hence is a zero-divisor. So, in the associated graded of $R$ with respect to ${\mathfrak m} = (t,x_1,...,x_n)$, any homogeneous non-unit is a zero-divisor.