Injective dimension of graded-injective modules. - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T02:38:16Z http://mathoverflow.net/feeds/question/89453 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89453/injective-dimension-of-graded-injective-modules Injective dimension of graded-injective modules. Pablo Zadunaisky 2012-02-25T00:33:43Z 2012-04-23T06:57:26Z <p>In "<a href="http://www.ams.org/mathscinet/search/publdoc.html?arg3=&amp;co4=AND&amp;co5=AND&amp;co6=AND&amp;co7=AND&amp;dr=all&amp;pg4=AUCN&amp;pg5=TI&amp;pg6=PC&amp;pg7=ALLF&amp;pg8=ET&amp;r=1&amp;review_format=html&amp;s4=van%2520den%2520Bergh&amp;s5=Existence&amp;s6=&amp;s7=&amp;s8=All&amp;vfpref=html&amp;yearRangeFirst=&amp;yearRangeSecond=&amp;yrop=eq" rel="nofollow">Existence theorems...</a>" Van den Bergh proposes the following "pleasant excercise in homological algebra":</p> <blockquote> <p>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.</p> </blockquote> <p>There is a proof of this fact for commutative graded algebras in <a href="http://www.mscand.dk/article.php?id=2222" rel="nofollow">this paper</a> 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.</p> <p>PS: In a <a href="http://www.math.bgu.ac.il/~amyekut/publications/aus-gor/journal.pdf" rel="nofollow">later paper</a>, 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...</p> http://mathoverflow.net/questions/89453/injective-dimension-of-graded-injective-modules/94910#94910 Answer by Ralph for Injective dimension of graded-injective modules. Ralph 2012-04-23T06:55:03Z 2012-04-23T06:55:03Z <p><strong>Part I:</strong> A spectral sequence </p> <p>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).</p> <p>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. </p> <blockquote> <p><strong>Theorem:</strong> 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. </p> <p><strong>Corollary:</strong> If $Ext^n_{\operatorname{gr}(R)}(\operatorname{gr}(M),\operatorname{gr}(N))=0$ for some integer $n$ then $Ext_R^n(M,N)=0$. </p> </blockquote> <p>Remark: The corollary generalizes [3, Prop. 3.2]. </p> <p>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. </p> <p>Filter $Hom_R(M,N)$ by </p> <p>$$F_kHom_R(M,N) = \lbrace f \mid \forall i: f(F_iM) \subseteq F_{i+k}N\;\rbrace.$$ </p> <p>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. </p> <p>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:<br> $$E^1_{p,q}=H_{p+q}(F_pHom_R(P,N)/F_{p-1}Hom_R(P,N))$$ 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 $p$. </p> <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 $$E^1_{pq} \cong H_{p+q}\; Hom^p_{\operatorname{gr}(R)}(\operatorname{gr}(P),\operatorname{gr}(N)) = Ext^{-(p+q),p}_{\operatorname{gr}(R)}(\operatorname{gr}(M),\operatorname{gr}(N)).$$</p> <p><strong>Remark:</strong> 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 $$Ext_R^i(M,N) = \oplus_p Ext_R^{i,p}(M,N).$$</p> <p><strong>References</strong></p> <ol> <li>Nastasescu, Oystaeyen: Graded and Filtered Rings and Modules</li> <li>Weibel: An Introduction to Homological Algebra </li> <li>Björk: The Auslander Condition on Noetherian Rings, Lect. Notes Math. 1404(1989),137-173</li> <li>Grünenfelder: On the Homology of Filtered and Graded Rings, J. Pure Appl. Alg. 14(1997), 21-37</li> </ol> http://mathoverflow.net/questions/89453/injective-dimension-of-graded-injective-modules/94911#94911 Answer by Ralph for Injective dimension of graded-injective modules. Ralph 2012-04-23T06:57:26Z 2012-04-23T06:57:26Z <p><strong>Part II:</strong> Answering the question </p> <p>In the following all gradings are over $\mathbb Z$. </p> <blockquote> <p><strong>Theorem:</strong> Let $R$ be a graded left-Noetherian ring and $N$ a graded $R$-module of finite graded injective dimension. Then $\operatorname{injdim}(N)\le \operatorname{gr-injdim}(N) + 1$.</p> </blockquote> <p><strong>Remark:</strong> 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). </p> <p>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. </p> <p>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 $i > v+1$. </p> <p>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 </p> <p>$$Ext_S^i(L,N[t]) \to Ext^{i+1}_S(\epsilon L,N[t]) \to Ext^{i+1}_S(L,N[t]).$$</p> <p>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. </p> <p>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. </p> <blockquote> <p><strong>Proposition:</strong> $\operatorname{gr-injdim}_{R[t]}(N[t]) = \operatorname{gr-injdim}_R(N) + 1$.</p> </blockquote> <p>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: </p> <ol> <li><p>Show $Ext_{R[t]}^i(L,N[t])=0$ for all $i > v+1$ and all special graded f.g. $R[t]$-modules $L$. </p></li> <li><p>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 $Ext_{R[t]}^i(L,N[t])=0$. </p></li> </ol> <p><strong>Remark:</strong> 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. </p> <p><strong>References:</strong></p> <ol> <li>Ekström: The Auslander Condition on graded and filtered noetherian rings, Lect. Notes Math. 1404(1989),220-245.</li> <li>Bruns-Herzog: Cohen-Macaulay Rings </li> <li>Rotman: An Introduction to Homological Algebra </li> </ol>