2
votes
1answer
17 views

Behaviour of $\operatorname{Ext}$ with left exact sequences.

Maybe is a trivial question but I am not so good in derived functors. Assume we are in the category of abelian groups and we have an exact sequence $$0\longrightarrow A\longrightarrow B\longrightarrow ...
0
votes
0answers
26 views

Exactness of derived functor [duplicate]

Is the right derived functor of a left exact functor left exact also? If now, can anything be said about its exactness in general?
3
votes
1answer
52 views

Fully faithful and essentially surjective is an equivalence

The question asks to prove the statement in the subject. So assume the functor is $F: \mathcal{C} \rightarrow \mathcal{D}$ is fully faithful and essentially surjective. We need to construct a map ...
1
vote
0answers
48 views

Where is the mistake? (derived functors )

Assume $pd(M) =n \leq \infty$ for a left $R$-module. I then have to show there exists a free module $F$ such that $Ext_{R}^{n}(M,F) \neq 0 $. I have tried these steps and obtained a contradiction: ...
17
votes
1answer
357 views

Homological methods in algebraic geometry

This question will probably seem quite silly to those well-versed in algebraic geometry (about which I admittedly hardly know anything); in the preface of Atiyah-Macdonald's book on commutative ...
11
votes
1answer
186 views

What is a concrete example of why one wants to have a *derived category* in algebraic geometry?

My question asks for a concrete (and hopefully easy) example, why one wants to derive things in algebraic geometry. I heard, that a resolution of an object by free ones behaves much better than the ...
5
votes
3answers
172 views

What does it mean to have exact derived functors?

Let $F:\mathcal A\to \mathcal B$ be a functor between abelian categories. Suppose $F$ is, say, left exact (plus additive and covariant). We have built its right derived functors $R^iF$. I see no ...
3
votes
2answers
79 views

Derived functors definition

I´m searching for a reference that defines $n^{th}$derived functors in an analogous way to the definition given in Mitchell´s "Theory of Categories" for the $0^{th}$ derived functor of $T$ covariant ...
2
votes
2answers
126 views

Equivalence of categories and derived functors.

Don't know if this kind of a dumb question but let $A$ and $B$ be abelian categories and suppose they're equivalent: there are two functors $P: A \rightarrow B$ and $Q: B \rightarrow A$ satisfying the ...
4
votes
1answer
186 views

Why are these two functors isomorphic?

Let $A$ be a local noetherian ring, $M$ an $A$-module finitely generated. Let $f$ be an $A$-regular and $M$-regular element (i.e. $f$ is not a zero divisors on $A$ nor on $M$). Then inside the ...
0
votes
0answers
66 views

What is the cohomological dimension of a functor?

Let $F:\mathcal{C}\rightarrow \mathcal{D}$ be a functor between abelian categories. Could anyone explain what the cohomological dimension the functor $F$ is? We may need some additional condition to ...
2
votes
0answers
147 views

Left-derived functors

Let $F:\mathcal{A}\to\mathcal{B}$ be a covariant right-exact functor between two abelian categories. Suppose $\mathcal{A}$ has enough projectives. Then we define the left derived functors of $F$ by ...
2
votes
2answers
289 views

Derived functors are Kan extensions

In this short paper by G. Maltsiniotis derived functors are presented as Kan extensions along the localization functor. I began studying derived categories only a couple of months ago, so I'm not at ...