# Talk:Ext functor

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Field: Algebra

## Left or right derived functors?

According to Osborne it is Left derived functors that are defined using projective resolutions and right derived functors that use injective resolutions. Doesn't this make Ext left derive.

IE start with a covariant functor Hom(X,-) and some object A, take a projective resolution of X, P_n --> ... --> P_0 --> X, then apply Hom, cut the end off: Hom(P_n,A) --> ... --> Hom(P_0,A), then take cohomology of the complex?

Fixed that. And a few things more. Michiexile 15:18, 9 February 2007 (UTC)
Isn't that Hom(P_n,A) <-- ... <-- Hom(P_0,A) ? Charles Matthews 15:53, 9 February 2007 (UTC)
It is. But in my rewrite, things are tweaked the right way around, regardless of what's in the discussion. Michiexile 21:51, 10 February 2007 (UTC)

Hom(A,-) is [covariant] left-exact, but contrary to what is in the article, Hom(-,B) is also left-exact. (See exact functor.) This is why both have right-derived functors. (Changed.) Tesseran 20:08, 11 February 2007 (UTC)

In answer to the original question: left-exact covariant functors, as well as right-exact contravariant functors, have their derived functors (right-derived and left-derived respectively) defined using injective resolutions. Right-exact covariant and left-exact contravariant functors have their derived functors (left-derived and right-derived resp.) defined using projective resolutions. Your statement is only true if all your original functors are covariant. Tesseran 20:12, 11 February 2007 (UTC)
Hom(-,B) is contravariant though. Which, upon a reread, meshes perfectly with what you wrote. Thanks for the correction. Michiexile 16:29, 12 February 2007 (UTC)

## examples have dangling references

The examples have dangling references. Should R and M be the same? --MarSch 13:53, 2 May 2007 (UTC)

No, the first paragraph of the article introduces $\mathcal{C}$ as the category of modules over some ring $R$. So, the statements in the example sequence just fix the category in which the example is exhibited, and $M$ is the module of coefficients for the functor.
I'm not certain how to make this distinction clearer - suggestions are very welcome. Michiexile 22:33, 3 May 2007 (UTC)
Mmmh, at least this solves where the R dependency should go, although it currently makes no sense. The solution is to be explicit about when C is some particular category and when it is merely _a_ category (which is the usual meaning) and especially not suppress R-dependence. Any idea what M is? --MarSch 10:38, 5 May 2007 (UTC)
Ayup, M is some R-module. Any R-module.

The interesting examples subsection reads fine as English (I can't follow the math at this level). I made a couple of minor edits in earlier subsections for clarity; for instance Ring Structure... subsection above. I didn't change any math/markup/symbols/ , so someone might like to check the argument remains sound.Newbyguesses 23:13, 3 May 2007 (UTC)

## Definition and computation

The claims of this chapter should be made more explicit. Quote:

Let $R$ be a ring and let $\mathrm{Mod}_R$ be the category of modules over R. Let $B$ be in $\mathrm{Mod}_R$ and set $T(B) = \operatorname{Hom}_{\mathrm{Mod}_R}(A,B)$, for fixed $A$ in $\mathrm{Mod}_R$. This is a left exact functor and thus has right derived functors $R^nT$. Define

$\operatorname{Ext}_R^n(A,B)=(R^nT)(B),$

i.e., take an injective resolution

$J(B)\leftarrow B\leftarrow 0,$

compute

$\operatorname{Hom}_{\mathrm{Mod}_R}(A,J(B))\leftarrow\operatorname{Hom}_{\mathrm{Mod}_R}(A,B)\leftarrow0,$

and take the cohomology of this complex.

So is Extn supposed to be the n-th cohomology group of this complex? And why is the right derived functor the same as this cohomology? This should be explained. --Roentgenium111 (talk) 03:25, 5 July 2009 (UTC)

Wouldn't such explanations belong in the Derived functor article? I would think so. And they are there. RobHar (talk) 05:15, 5 July 2009 (UTC)
By this logic the whole "i.e." part should be deleted. But since it's there, it should also be stated that R^n is the n-th co(?)homology group of the complex, if this is the case (the Derived functor article says homology). --Roentgenium111 (talk) 14:32, 5 July 2009 (UTC)

## Interesting examples

This needs some correction or clarification:

"For Fp the finite field on p elements, we also have that H*(G,M) = Ext* Fp[G](Fp, M), and it turns out that the group cohomology doesn't depend on the base ring chosen." — Preceding unsigned comment added by 80.42.253.109 (talk) 17:59, 7 May 2013 (UTC)