# Tagged Questions

**1**

vote

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99 views

### Thickness of the category of perfect complexes with finite length homology

Let $R$ be a commutative Noetherian local ring and let $D(R)$ be the derived category of $R$-modules. Recall that a chain complex $C_\bullet$ of modules over $R$ is called perfect if it is isomorphic ...

**3**

votes

**1**answer

238 views

### Reference for comparison of heart cohomology with standard cohomology

I'm looking for a reference for the following fact (which I believe to be true and should be easy for people who understand how spectral sequences arise from filtrations).
Let A,B be two hearts of ...

**10**

votes

**1**answer

385 views

### Fullness of pullback functor in algebraic geometry

Given $f:X\to Y$ a morphism of schemes (or stacks if it's not harder), I am interested in a geometric reformulation of the condition that the functor $f^*:D^b(Coh(Y))\to D^b(Coh(X))$ is full. I can ...

**7**

votes

**1**answer

388 views

### Lifting isomorphisms between derived categories

(Remark: I first asked this question at math.stackexchange. As it received no answer, I'm posting it here).
Suppose $A$ and $B$ are commutative rings. Let $A\to B$ be a surjective ring homomorphism. ...

**6**

votes

**1**answer

585 views

### Do I need to know what an infinity-Gerstenhaber algebra is, and if so, what is it?

I am in the following situation. I have two (rather explicit and specific) dg commutative algebras $R,S$ over a field of characteristic $0$. In fact, $S$ is an $R$-algebra, in that I have a map $R ...

**13**

votes

**0**answers

469 views

### When is every “solid” perfect complex faithful?

Let $R$ be a noetherian commutative ring. Consider $D^{perf}(R)=K^b(R-proj)$ the category of bounded complexes of finitely generated projective $R$-modules, with maps of complexes up to homotopy.
...

**39**

votes

**7**answers

3k views

### What does a projective resolution mean geometrically?

For R a commutative ring and M an R-module, we can always find a projective resolution of M which replaces M by a sequence of projective R-modules. But as R is commutative, we can consider the affine ...