# Injective dimension of graded-injective modules

In "Existence theorems..." Van den Bergh proposes the following "pleasant excercise in homological algebra":

Let $A$ be a connected graded noetherian $k$-algebra (that is, $\mathbb N$-graded with $A_0 = k$ a field). A graded module $I$ which is injective in the category of left graded modules has injective dimension at most $1$ in the category of left modules.

There is a proof of this fact for commutative graded algebras in this paper by Fossum and Foxby, but I don't really see how to transfer this to the non-commutative setting. Can anyone provide any pointers or a reference to a proof? Thanks in advance.

PS: In a later paper, Yekutieli and Zhang state that the only proof they know of this fact is "quite involved", which eased my anxiety at being unable to solve a pleasant excercise in the area I'm supposed to be PhD-ing in...

• If $0 \to I \to E \to C \to 0$ is exact with $E$ injective, then we can make an exact sequence of graded modules $0 \to \oplus_{\mathbb Z} I \to \oplus_{\mathbb Z} E \to \oplus_{\mathbb Z} C \to 0$. This sequence must split since $\oplus_{\mathbb Z} I$ is graded-injective, so at least $\oplus_{\mathbb Z} C$ is injective. – Dag Oskar Madsen Feb 25 '12 at 18:22
• $C$ is not necessarily an $A$-direct summand of $\oplus_{\mathbb Z} E$, that's why I deleted my answer. The injectivity of $\oplus_{\mathbb Z} E$ also questionable since is not really a direct sum of $A$-modules. Actually, it cannot always be injective since that would imply $I$ injective, which we know usually isn't the case. – Dag Oskar Madsen Feb 25 '12 at 21:10
• I don't see what the graded module structure on $\oplus_{\mathbb Z} E$ is, or how to make it compatible with the obvious morphism $\bigoplus_{\mathbb Z} I \rightarrow \bigoplus_{\mathbb Z} E$. – Pablo Zadunaisky Feb 26 '12 at 18:52
• You make the graded module in the following way: In each degree put a copy of $E$. If $a \in A n$ and $x$ is an element in the degree $m$ copy of $E$, then the product $ax$ in the graded modules structure is the element $ax$ in the degree $n+m$ copy of $E$. – Dag Oskar Madsen Feb 29 '12 at 22:40

In the following all gradings are over $\mathbb Z$.

Theorem: Let $R$ be a graded left-Noetherian ring and $N$ a graded $R$-module of finite graded injective dimension. Then $\operatorname{injdim}(N)\le \operatorname{gr-injdim}(N) + 1$.

Remark: In  Ekström showed $\operatorname{injdim}(R)\le \operatorname{gr-injdim}(R) + 1$ if $R$ is left- and right-Noetherian. The proof below is a generalization of Ekström's proof. The crucial observation is that in a spectral sequence she used, finite generation of the coefficient-module can be replaced by boundedness below (compare part I).

Proof of the theorem: The notation is taken from . In particular, $S:=R[t]$ is a poynomial ring in the indeterminate $t$ and if $M$ is an $R$-module then $M[t]:= R[t]\otimes_R M = \sum_{i\ge 0} Mt^i$. Note that $R[t]$ is left-Noetherian by the Hilbert basis theorem.

Set $v:= \operatorname{gr-injdim}(N)$. In order to prove the theorem, by [2, 3.1.10] it suffices to show $Ext^i_R(M,N)= 0$ for all f.g. $R$-modules $M$ and all $i > v+1$.

By [1, Prop. 1.3] there is a graded f.g $R[t]$-module $L$ such that $M \cong L/(1-t)L=:\epsilon L$ as $R$-module. Since $1-t$ is regular on $S=R[t]$ and on $L$ (see the remark in the proof of [1, Lemma 1.4]), the short exact sequence $0 \to L \xrightarrow[]{1-t} L \to \epsilon{L} \to 0$ induces the exact sequence

$$Ext_S^i(L,N[t]) \to Ext^{i+1}_S(\epsilon L,N[t]) \to Ext^{i+1}_S(L,N[t]).$$

Now suppose the Proposition below has already been shown. Hence $Ext^i_S(L,N[t])=0$ for $i > v+1$ holds and $Ext^{i+1}_S(\epsilon L,N[t])=0$ follows.

Because of $R[t]/(t-1) \cong R$ and $N[t]/(1-t)N[t] \cong N$, Rees' theorem [3, Theorem 8.34] yields $Ext^{i+1}_S(\epsilon(L),N[t])=Ext^i_R(\epsilon(L),N)=Ext^i_R(M,N)$. Thus $Ext^i_R(M,N)=0$ for $i > v+1$ and we are done.

Proposition: $\operatorname{gr-injdim}_{R[t]}(N[t]) = \operatorname{gr-injdim}_R(N) + 1$.

The proof is the same as the proof of [1, Lemma 2.1] with $N[t]$ in place of $A[t]$. The two basic steps are:

1. Show $Ext_{R[t]}^i(L,N[t])=0$ for all $i > v+1$ and all special graded f.g. $R[t]$-modules $L$.

2. Let $L$ be a f.g. graded $R[t]$-module. Take the (positive) $\Sigma$-filtrations $F_iR := \oplus_{j\le i}Rt^i$, $F_iN[t]:= \oplus_{j\le i}Rt^i$ and choose a good $\Sigma$-filtration $FL$ of $L$. Note that $FL$ is bounded below (and hence Hausdorff) since $FR$ is bounded below and $L$ is f.g. By [1, 1.11], $\operatorname{gr}(L)$ is special graded and by the previous step, $Ext_{R[t]}^i(\operatorname{gr}(L),N[t])=0$. Applying the Corollary from Part I to $(R[t],L,N[t])$ finally gives $Ext_{R[t]}^i(L,N[t])=0$.

Remark: From 1.6 on,  assumes $R$ to be left- and right-Noetherian. But I checked the proofs of the results from [1, sect. 1] about the filtration stuff that I used in the proof the theorem above. As far as I can see everything works fine for left-Noetherian rings. In fact, the only place I see where the left-right comes into play is in the referenced [Bj:2] (that's  in Part I) where the $R$-bimodule structure of $R$ is used to obtain an $R$-right-module structure on $Ext_R(-,R)$ that is used to prove the convergence of their spectral sequence. But that's not of relevance here since the convergence in Part I is obtained in a completely different way.

References:

1. Ekström: The Auslander Condition on graded and filtered noetherian rings, Lect. Notes Math. 1404(1989),220-245.
2. Bruns-Herzog: Cohen-Macaulay Rings
3. Rotman: An Introduction to Homological Algebra
• Wow! Thanks again Ralph. I will look into the details tomorrow, but on a first reading everything seems to work :). – Pablo Zadunaisky Apr 24 '12 at 1:51
• BTW: Do you know how the solutions of van den Bergh or Yekutieli/Zhang look like ? – Ralph Apr 24 '12 at 23:30
• The best I can tell you is that I sent an email to vdB but received no answer. I did not try to contact Yekutieli or Zhang. – Pablo Zadunaisky Apr 25 '12 at 18:54

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 (, 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

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

Here is a proof of Van den Bergh's theorem, just posted on the arxiv: http://arxiv.org/abs/1407.5916 .

My proof is pretty much self contained, and very detailed. The proof uses the same strategy as the original proof of Van den Bergh (which was very similar to the one given above by Ralph), namely passing through the Rees ring. But it is somewehat more conceptual, in that it isolates the precise "reason" for the dimension jump. This is possible since I work with derived categories (and not with the more cumbersome method of spectral sequences).

Your problem can be solved from the inequalities: $$inj.dim M_A = gr.inj.dim M[t^{\pm 1}]_{A[t^{\pm 1}]} \leq gr.inj.dim M[t]_{A[t]} \leq 1 + gr.inj.dim M_A.$$

• How do you prove the last inequality? I don't see how to relate the graded structure of $M$ over $A$ with that of $M[t]$ over $A[t]$ – Pablo Zadunaisky Oct 2 '13 at 11:35