Consider the polynomial ring $\mathbb C[t]$ with $\deg t=0$, that is, we view $\mathbb C[t]$ as ungraded ring concentrated in degree zero. On the other hand there is a decreasing filtration of $\mathbb C[t]$
$$F_0:=\mathbb C[t]\supseteq F_1:=t\mathbb C[t]\supseteq\dots\supseteq F_n:=t^n\mathbb C[t]\supseteq\cdots\supseteq\{0\}.$$
We see from the above decreasing chain that $t$ has a filtration degree $1$ since it lies in $F_1$ but not in $F_2$. On the other hand, the element $t$ has a graduation degree $0$.
- The filtration degree does not assign a weight to the elements? An element might have a non-trivial filtration degree and still have weight (or graduation degree) $0$, right? When does the filtration degree coincide with the graduation degree of an element in a filtered and graded module - when the filtration is induced by the graduation? In all other cases the filtration degree and the graduation degree are independent?
Let $A$ be an associative $\mathbb C$-algebra (which is not necessarily finite dimensional as a complex vector space) with an increasing filtration $A_0\subseteq A_1\subseteq\dots\subseteq A_i\subseteq A_{i+1}\subseteq\dots\subseteq A$. Let us consider the $\mathbb C(t)$-algebra $A(t)$. What is the set of all elements of $n$-th filtration degree of $A[t]\otimes_{\mathbb C[t]}\mathbb C(t)$? Is it $$F_n(A[t]\otimes_{\mathbb C[t]}\mathbb C(t))=\sum_{i+j=n}F_i(A\otimes_{\mathbb C}\mathbb C[t])\otimes_{\mathbb C[t]}t^{-j}\mathbb C[t]=\sum_{i+j=n}A_i[t]\otimes_{\mathbb C[t]}t^{-j}\mathbb C[t]$$