In "[Existence theorems...](http://www.ams.org/mathscinet/search/publdoc.html?arg3=&co4=AND&co5=AND&co6=AND&co7=AND&dr=all&pg4=AUCN&pg5=TI&pg6=PC&pg7=ALLF&pg8=ET&r=1&review_format=html&s4=van%2520den%2520Bergh&s5=Existence&s6=&s7=&s8=All&vfpref=html&yearRangeFirst=&yearRangeSecond=&yrop=eq)" Van den Bergh proposes the following "pleasant excercise in homological algebra":



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> 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.
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There is a proof of this fact for commutative graded algebras in [this paper](http://www.mscand.dk/article.php?id=2222) 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](http://www.math.bgu.ac.il/~amyekut/publications/aus-gor/journal.pdf), 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...



Injective dimension of graded-injective modules
**Part II:** Answering the question 


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



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



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


