Source: Weibel, Page 94. Given an ideal $I$ in a ring $R$, we have the exact sequence: $$0\rightarrow I \rightarrow R \rightarrow R/I \rightarrow 0,$$ so if we apply the contravariant left exact functor $Hom(-,B)$, we have the next exact(why?) sequence: $$Hom(R,B)\rightarrow Hom(I,B)\rightarrow Ext^1(R/I)\rightarrow 0$$ But seems I'm missing something here because I don't see why the $Ext$ functor appears here. I don't know either why the sequence finishes at 0, but doesn't start at 0.
If I get it correctly, applying the functor we would have $$0\rightarrow Hom(R/I,B)\rightarrow Hom(R,B)\rightarrow Hom(I,B),$$ not the sequence above.
Thanks.
He uses the derived functors of Hom, giving rise to the long exact sequence: \begin{multline*}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Ext{Ext} 0\to\Hom(R/I,B)\to\Hom(R,B)\to\Hom(I,B)\to\Ext^1(R/I,B)\to\Ext^1(R,B)\\ \to\Ext^1(I,B)\to \Ext^2(R/I,B)\to\Ext^2(R,B)\to\Ext^2(I,B)\to \dots \end{multline*} Note that $\Ext^i(P,B)=0$ for all projective modules $P$, in particular for $R$.