**Part I:** A spectral sequence

For better readability the answer is subdivided into two parts: In part I a spectral sequence is constructed that is of interest in its own right. Part II then solves the original problem by applying the spectral sequence (among others).

If $R$ is a filtered module then $\operatorname{gr}(R) = \oplus_{i \in \mathbb Z}F_iR/F_{i-1}R$ denotes the associated graded ring and if $M$ is a filtered $R$-module, $\operatorname{gr}(M)$ is the associated graded $\operatorname{gr}(R)$-module.

**Theorem:** Let $R$ be a filtered complete ring, $M$ a filtered f.g. $R$-module those filtration is Hausdorff and $N$ a filtered $R$-module those filtration is bounded below. Suppose that all filtrations are exhaustive and that $\operatorname{gr}(R)$ is left-Noetherian. Then there is a convergent spectral sequence
$$E^1_{pq}= Ext^{-(p+q),p}_{\operatorname{gr}(R)}(\operatorname{gr}(M),\operatorname{gr}(N))
\Rightarrow Ext_R^{-(p+q)}(M,N).$$
The filtration of the Ext-group on the r.h.s. is bounded below and exhaustive.

**Corollary:** If $Ext^n_{\operatorname{gr}(R)}(\operatorname{gr}(M),\operatorname{gr}(N))=0$ for some integer $n$ then $Ext_R^n(M,N)=0$.

Remark: The corollary generalizes [3, Prop. 3.2].

Similar spectral sequences can be found in [3, sect. 3] and [4, sect. 3]. But either requires the coefficients $N$ to be finitely generated over $R$ what is too restrictive for our purposes. Therefore I prefer to give a separate set up of the spectral sequence here.

Filter $Hom_R(M,N)$ by

$$F_kHom_R(M,N) = \lbrace f \mid \forall i: f(F_iM) \subseteq F_{i+k}N\;\rbrace.$$

Since $M$ is filtered f.g. (for a definition of filtered f.g. see [1 before 5.2]) and $FN$ is exhaustive, this filtration is also exhaustive. Moreover, since $M$ is filtered f.g. and $FN$ is bounded below, this filtration is also bounded below.

According to [1, 5.1, 5.3] there is a filtered free resolution
$$\cdots P_{-2} \to P_{-1} \to P_0 \to M$$
with each $P_i$ filtered f.g. By the same reason as above, we therefore have a filtered
complex
$$\cdots \to F_kHom_R(P_{-i},N) \to F_kHom_R(P_{-i-1},N) \to \cdots$$
those filtration is exhaustive and bounded below. Thus there is a converging spectral sequence ([2], 5.5.1.2)
$$E^1_{pq} \Rightarrow H_{p+q}Hom_R(P,N)=Ext_R^{-(p+q)}(M,N)$$
such that the filtration on the r.h.s. is again bounded below and exhaustive. The $E^1$-term is:

$$E^1_{p,q}=H_{p+q}(F_pHom_R(P,N)/F_{p-1}Hom_R(P,N))$$
and since $P$ is filtered projective, by [1, 6.14]
$$F_pHom_R(P_i,N)/F_{p-1}Hom_R(P_i,N) \cong Hom^{\;p}_{\operatorname{gr}(R)}(\operatorname{gr}(P_i),\operatorname{gr}(N))$$
where the r.h.s. are the graded $\operatorname{gr}(R)$-module homomorphisms of degree $p$.

Finally, by [1, 5.14, 4.4], $\operatorname{gr}(P) \to \operatorname{gr}(M)$ is a projective resolution in the category of graded $\operatorname{gr}(R)$-modules. Hence
$$E^1_{pq} \cong H_{p+q}\; Hom^p_{\operatorname{gr}(R)}(\operatorname{gr}(P),\operatorname{gr}(N)) = Ext^{-(p+q),p}_{\operatorname{gr}(R)}(\operatorname{gr}(M),\operatorname{gr}(N)).$$

**Remark:** The double grading of Ext in the spectral sequence arises because the Hom-functor in the graded module category is graded: $Hom_R(M,N) = \oplus_p Hom^p_R(M,N)$, whence $$Ext_R^i(M,N) = \oplus_p Ext_R^{i,p}(M,N).$$

**References**

- Nastasescu, Oystaeyen: Graded and Filtered Rings and Modules
- Weibel: An Introduction to Homological Algebra
- Björk: The Auslander Condition on Noetherian Rings, Lect. Notes Math. 1404(1989),137-173
- Grünenfelder: On the Homology of Filtered and Graded Rings, J. Pure Appl. Alg. 14(1997), 21-37