Given a filtered vector space (or module over a ring) $0=V\_{0}\subseteq V\_{1}\subseteq\cdots\subseteq V$, you can construct the *associated graded* vector space $\mathrm{gr}\left(V\right)=\oplus\_{i}V\_{i+1}/V\_{i}$. Does $\mathrm{gr}\left(V\right)$ satisfy a universal property? What is it?


Before anybody hastily says, "it's the universal graded vector space with a filtered map from $V$," let me point out that it's not so simple. A map of filtered vector spaces is a map of vector spaces which respects the filtration. It's clear what the map $V\_{i+1}\rightarrow V\_{i+1}/V\_{i}$ should be, but what would the map $\cup\_{i}V\_{i+1}\rightarrow\oplus\_{i}V\_{i+1}/V\_{i}$ be?



What is the universal property of associated graded?
$\DeclareMathOperator{\Rees}{Rees}\DeclareMathOperator{\gr}{gr}$ The associated graded of a filtered $\mathbf R$-module $M$ is the universal $\mathbf R$-module with a map of the Rees module of $M$ over $\mathbf R[t]$ to $\gr M$.


Let me explain what the Rees module $\operatorname{Rees}(M)$ is: it's the submodule of $M[t,t^{-1}]$ which is generated as a $\mathbf{R}[t]$ module by $t^iM\_i$. Give this the obvious grading by degree of $t$. So $\Rees(M)/t\Rees(M)=gr M$, whereas $\Rees(M)/(t-1)\Rees(M)=M$ with the induced filtration. This is the thing that has a map to $\gr M$.


