I am just studying Ext and Tor with my self from Basic homological algebra by Scott. Why we need to use flat resolution and injective resolution in order to compute the Tor and Ext. what is wrong with projective resolution.
Any help will be appreciated
Best Answer
For a ring $R$ and $R$-modules $A$ and $B$, projective resolutions can be used to compute both $\operatorname{Tor}_n^R(A,B)$ and $\operatorname{Ext}_R^n(A,B)$. Let $$\cdots\xrightarrow{\ \ \ }P_2\xrightarrow{\ \ \ }P_1\xrightarrow{\ \ \ }P_0\xrightarrow{\ \varepsilon \ }A\xrightarrow{\ \ \ }0$$ be a projective resolution of $A$. Then, applying $-\otimes_RB$ to the above gives you a chain complex: $$\cdots\xrightarrow{\ \ \ }P_2\otimes_RB\xrightarrow{\ \ \ }P_1\otimes_RB\xrightarrow{\ \ \ }P_0\otimes_RB\xrightarrow{\ \ \ }0$$ and the homology of this complex gives you $\operatorname{Tor}_n^R(A,B)$. You can also choose a projective resolution of $B$, apply $A\otimes_R-$, and compute the homology of the resulting complex.
Similarly, applying $\operatorname{Hom}_R(-,B)$ to the resolution above gives rise to a cochain complex: $$0\xrightarrow{\ \ \ }\operatorname{Hom}_R(P_0,B)\xrightarrow{\ \ \ }\operatorname{Hom}_R(P_1,B)\xrightarrow{\ \ \ }\operatorname{Hom}_R(P_2,B)\xrightarrow{\ \ \ }\cdots$$ and the cohomologyof this complex is $\operatorname{Ext}_R^n(A,B)$.
Edit: As Tobias mentions below the fact that you can compute these groups using both methods is non trivial. A proof of this can be found in Weibel's book, in the "Balancing Tor and Ext" section.