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One can define $\mathrm{Ext}^n(M,N)$ (where $M,N$ are $R$-modules) in two ways, either by taking an injective resolution of $N$ and applying $\mathrm{Hom}(M,-)$or by taking a projective resolution and applying $\mathrm{Hom}(-,N)$.

I'm struggling to remember which is which and I'm sure there is a(n obvious?) reason. Why does it have to be projective + $\mathrm{Hom}(-,N)$ or injective + $\mathrm{Hom}(M,-)$ but cannot be projective + $\mathrm{Hom}(M,-)$?

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    Which one has the lifting property in which direction?2012-07-18
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    Always use injective resolutions for right derived functors. (The fact that you can use a projective resolution here is because there's an ${}^\textrm{op}$ floating around!)2012-07-18
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    The projectives are the injectives of the category $(\text{$R$-$\mathrm{Mod}$})^{\mathrm{op}}$. Eventually, you will just remember which to use when, independently of these rationalizations :)2012-07-18
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    @QiaochuYuan Mariano Zhen I guess someone should just post an answer to close the question.2012-07-18
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    @ZhenLin I have to go for projective in this case because the other one is not even mentioned in the lecture. The lecture also doesn't mention what a functor or what a category is but I want to understand this properly hence the question.2012-07-18
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    @MTurgeon Well as of know I cannot use any of these comments to answer my question. I have zero knowledge about category theory and the reason I have this question is because $\mathrm{Ext}$ is defined via projective in my commutative algebra lecture.2012-07-18
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    The answer to your question «why cannot it it be...?» is simply because doing that would be more or less useless: the properties that you want from Ext would not hold. This is something you can check yourself: go through the proofs and see. Since you are starting with the subject, you may well still not know what are the properties that you want Ext to have: in that case, just wait until you do and *then* see what happens if you try the "alternative definition".2012-07-18
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    @MattN. You say your definition of $\mathrm{Ext}$ uses projective resolutions. What Mariano and Zhen said is that, due to the "contravariant" nature of $\mathrm{Hom}(-,N)$, your projective resolution is really an injective resolution, but in the opposite category. All that says is that in both cases you're really doing the same thing.2012-07-18
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    @MTurgeon Ok. Thank you. But it doesn't yet help me remember which of the two $\mathrm{Hom}$ functors to apply to the projective resolution.2012-07-18
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    @MattN. By definition, projective modules are the ones that have nice properties when you look at functions out of them, and injectives are the ones that have nice properties when you look at functions into them. That's one way to remember which side to use.2012-07-18

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