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3

Ok Slup, here goes. Let $R$ be any commutative ring and let $A$ be a polynomial ring over $R$. Let $P$ be any projective module over $R$. Then Quillen (and Suslin a bit later in this generality) proved that if for every maximal ideal $\mathfrak{m}$ of $R$, $P_{\mathfrak{m}}$ is of the form $Q\otimes_{R_{\mathfrak{m}}} A_{\mathfrak{m}}$ for some projective $...


3

Ok, I will take the chance to enlighten you a little bit. First of all, if you are really willing to fully understand why $\operatorname{Ext}^i(k,R^s) \to \operatorname{Ext}^i(k,R^r)$ is the zero map, you should not care that much about the explanations given in textbooks. These short explanations actually phrase the following: Checking that this map is ...


3

Given a morphism of short exact sequences, you get a morphism of long exact sequences, and this map respects composition. $\delta$ itself is, most precisely, a natural transformation between the functors $H_n\circ t$ and $H_{n-1}\circ f$ from short exact sequences of chain complexes to abelian groups, where $t$ sends an s.e.s. to its third complex and $f$, ...


3

The proof can be corrected like this: We have the long exact sequence $$\operatorname{Ext}^i(k,F) \to \operatorname{Ext}^i(k,B)\to \operatorname{Ext}^{i+1}(k,\Omega) \to \operatorname{Ext}^{i+1}(k,F).$$ Let $d$ be the depth of $R$, which coincides with the depth of $F$. We have $\operatorname{Ext}^i(k,F)=0$ for all $i<d$. Thus $\operatorname{Ext}^i(k,B)...


3

In $D(R)$, $\operatorname{Hom}(R,R[t])=0$ for $t\neq0$, so you definitely need the condition that $\operatorname{Hom}(E_i,E_j[t])=0$ for $t\neq0$. In my paper "Morita theory for derived categories" (J. London Math. Soc. (2) 39 (1989), no. 3, 436-456) I proved a result for derived categories of rings with a proof along the lines of the idea in your final ...


3

In the same vein as zyx's answer, if you want to learn about cdgas and their applications to rational homotopy theory (before trying to use them in derived AG, which would make sense IMO), I can recommend these two books: Yves Félix, Stephen Halperin, and Jean-Claude Thomas. Rational homotopy theory. Graduate Texts in Mathematics 205. New York: Springer-...


2

Hint: Use induction to reduce to $\text{pdim}_A(M)\leq 1$. In this case pick a presentation $0\to P\to Q\to M\to 0$ with $P,Q$ finitely generated and projective and consider the associated extension class. Remark: Dropping the finiteness of the global dimension, but assuming instead that the injective dimension of $_AA$ is finite, one can still deduce that $...


2

Start with $q=2$. We actually have $H^2(G,A)=0$ for every $G$-module $A$. This follows from the fact that $H^2(G,A)$ is in bijection with the set of isomorphism classes of extensions $$1 \to A \to E \to G \to 1.$$ But all such extensions are trivial (i.e. split) by the universal property of $G$. Now use dimension shifting as you suggested before. Prove by ...


2

Let $\mathbf{2}$ be the category with two objects and one morphism between them. Then the arrow category of any category $\mathbf{C}$ is isomorphic to the functor category $\text{Funct}(\mathbf{2}, \mathbf{C})$; consequently, limits and colimits can be computed from the general facts about limits of functors. In particular, if $\mathbf{C}$ has all (co)...


2

Any additive functor between abelian categories preserves mapping cones, and a map is a quasi-isomorphism iff its mapping cone is acyclic, or equivalently quasi-isomorphic to $0$.


2

Finding a free resolution can be a difficult task by hand, because the simple linear algebra task of finding the kernel of a map given by a matrix is no longer simple when the base ring isn't a field. That said, modulo the problem of being able to say for sure that you have found a complete set of generators for the kernel of a map (which the computer does ...


2

For reasonable rings, one way to do this would be to use the fact that a module is projective if and only if it is locally so and thus we may assume $A$ is local with residue field $k$. Now, $M$ is projective if and only if $\operatorname{Tor}^1_A(k,M)=0$. Then one has $0=\operatorname{Tor}^1_B(k\otimes_A B, M\otimes_A B) =\operatorname{Tor}^1_A(k,M)\...


2

Finite dimensional algebras have integrals, and using one you can show selfinjectivity at once —in fact, they are Frobenius. See for example the Lectures on Hopf algebras by Schneider, which Google will find for you, or Susan Montgomery's book in Hopf algebras.


2

Let us say that a ring $A$ satisfies condition $(S)$ if every finite type projective $A$-module is free. Clearly a necessary condition for $(S)$ to hold is $K_0(A)=\mathbf{Z}$. Example. (i) If $A$ is local Noetherian then $(S)$ holds. (ii) If $A$ is a Dedekind domain then $K_0(A)=\mathbf{Z}\oplus\mathrm{Pic}(A)$, and thus a necessary condition for $(S)$ is $...


2

In an answer to a question you posted recently, a construction was given that satisfies all the conditions you want except that $A$, and therefore all the morphisms from it, are zero. You can easily adapt that to satisfy all your conditions by taking the direct sum with a diagram of the form $$\require{AMScd}\begin{CD} @.0@.0@.0\\ @.@VVV@VVV@VVV\\ 0@>>&...


1

It should be true for any PID. See the book by Lam, Serre's Conjecture


1

Let $V$ be a finite dimensional space and let $W \subseteq V$ be a subspace. Denote by $W^{0}$ the annihilator of $W$. Then we have a natural isomorphism $W^{*} \cong V^{*}/W^{0}$ obtained by mapping $[\varphi] \in V^{*}/W^{0}$ to the linear functional $\varphi|_{W}$. Support we are given $k$ linear functionals $(\varphi)_{i=1}^k$ on $V$. We can define a ...


1

Use the long exact sequence of $\operatorname{Ext}$: if $0 \to A \to B \to C \to 0$ is exact, then it induces a long exact sequence (everything is over $R$): $$0 \to \hom(C,I) \to \hom(B,I) \to \hom(A,I) \to \operatorname{Ext}^1(C,I) \to \operatorname{Ext}^1(B,I) \to \operatorname{Ext}(A,I) \to \dots$$ Now for an injection $i : A \to B$, let $C = \...


1

You can also show there is a length one resolution of the trivial $G$-module $\Bbb Z$ as follows. Let $\varepsilon : \Bbb ZG\longrightarrow \Bbb Z$ be the canonical augmentation that sends eveyr $g\in G\mapsto 1 \in\Bbb Z$, and let $K=\ker\varepsilon$. Show that $K$ is a free $\Bbb ZG$-module with basis $\{ x-1 : x\in X\}$ where $X$ is a basis of $G$, so ...


1

I hate to answer my own question, but I think the answer is yes and here is a nice proof: compute all the first Ext groups of the indecomposable complexes. There's only one nonvanishing Ext group and it "classifies" or "records" the boundary map. Let $P[i]$ denote the indecomposable complex $k \xrightarrow{\sim} k$ in degrees $i,i-1$. Let $S[i]$ denote the ...


1

Yes, by definition the derived functor must be a functor of triangulated categories which verifies a certain universal property. But by definition a functor of triangulated category is additive.


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Suppose the sequence $0\to M\xrightarrow{\phi}N\to\operatorname{coker}\phi\to0$ splits. Then there exists the diagram $$\DeclareMathOperator{\coker}{coker}\require{AMScd} \begin{CD} 0 @>>> M @>\phi>> N @>\pi>> \coker\phi @>>> 0 \\ @. @V{\alpha}V\sim V @V\beta V\sim V @V\gamma V\sim V @. \\ 0 @>>> M' @>f>> ...


1

What is actually true is that $F$ is the limit of the direct system $(R_r, u_{rs})$, where each $R_r$ is copy of the ring $R$ and $u_{rs}\colon R_r\longrightarrow R_s$ is defined as multiplication by $r$ in $R$ whenever $r\mid s$.


1

After doing some more research into the topic, I've noticed that the definition I've given above for an $E$-injective spectrum is not the definition everyone makes in a triangulated category setting. In some 2012 notes by Haynes Miller, the definition of an $E$-injective spectrum $I$ is exactly one where the canonical map $i:I\rightarrow I\wedge E$ is an ...


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Here is a more recent and concise version on this: Jenny August, Matt Booth, Juliet Cooke, Tim Weelinck, "Differential Graded Algebras and Applications," Jan 2016, available at http://www.maths.ed.ac.uk/~mbooth/files/hodgeproject.pdf This also has 31 references, that can help go deep in the sub-part where you need more details.


1

No. For example, let $A=\mathbb{C}[[x]]$, the algebra of power series. Then $A$ is local, so only has one simple module $S=A/(x)$.Then $M=A[x^{-1}]$ is not projective ($x$ does not act invertibly on any non-zero projective module) but $\operatorname{Ext}^1_A(M,S)=0$ ($x$ acts invertibly on $M$, and so invertibly on this $\operatorname{Ext}$-module, but $x$ ...



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