Lifting graded ideal to ideal in filtered ring Let $R$ be a Noetherian filtered ring. Let $P\subset \operatorname{gr} R$ be a homogeneous ideal. Does there exists an ideal $\tilde{P}\subset R$ such that $\operatorname{gr}\tilde{P}$ (with the induced filtration) is isomorphic, as subobject of $\operatorname{gr}R$, to $P$?
We could try taking homogeneous generators $p_1,\dots,p_m$ for $P$, and lifting them to $\tilde{p}_1,\dots,\tilde{p}_m$ in $R$. Then we would hope that the ideal generated by the $\tilde{p}_i$ satisfies the requirements, but I cannot prove this...

Inspired by the answer to this question, the answer is no.
As an example, consider the ideal $(x,y)\subset \mathbb{C}[x,y]\cong \operatorname{gr} D_{\mathbb A^1}$, where $D_{\mathbb A^1}=\mathbb{C}[x,\partial]$, with $\partial x=x\partial+1$. We work with the standard filtration
$$F^kD_{\mathbb A^1}=\left\{\sum_{i=0}^k P_i(x)\partial^k\mid P_i\in \mathbb{C}[x]\right\}.$$
Then if $\tilde{P}$ would be an ideal lifting $(x,y)$, there must be an element $f(x,\partial)\in F^0D_{\mathbb A^1}\cap \tilde P=\mathbb{C}[x]\cap\tilde P$ equal to $x$ modulo $F^{-1}D_{\mathbb A^1}=0$. In other words, $x\in \tilde{P}$. 
There must also be an element $f(x,\partial)\in F^1D_{\mathbb A^1}\cap \tilde{P}$ such that $f(x,\partial)=\partial +f(x)$ for some $f(x)\in \mathbb{C}[x]=F^0D_{\mathbb A^1}$.
This means that $\tilde{P}$ contains both $x$ and an element of the form $\partial+f(x)$. So $\tilde{P}$ contains $x$ and $\partial$. But then it contains $\partial x + x\partial=1$, so $\tilde{P}=D_{\mathbb A^1}$. This is not what we want.
