Equation (0MVH) · ScalingStacks

Write $M=M_{0}$ . Let $n=1$ ; we construct an object $M_{1}$ and a morphism $\mu_{0}:M_{0}\rightarrow M_{1}$ such that $\textnormal{Hom}(S,\Sigma M_{1})=0$ . If $\textnormal{Hom}(S,\Sigma M_{0})=0$ then set $M_{1}=M_{0}$ and $\mu_{0}=1_{M_{0}}$ , the identity map on $M_{0}$ . If not, we can choose a, possibly infinite, coproduct $S^{(m_{1})}$ of copies of $S$ and a nonzero morphism $S^{(m_{1})}\rightarrow\Sigma M_{0}$ which becomes a surjection under the functor $\textnormal{Hom}(S,-)$ . Since the endomorphism ring $\textnormal{End}(S)$ is a division ring we can, moreover, choose $m_{1}$ so that this morphism becomes an isomorphism under $\textnormal{Hom}(S,-)$ . We now extend this morphism to a distinguished triangle:

(3.1)

S^{(m_{1})}\rightarrow\Sigma M_{0}\rightarrow\Sigma M_{1}\rightarrow\Sigma S^{% (m_{1})}.

Applying $\textnormal{Hom}(S,-)$ to (3.1) gives the exact sequence:

\textnormal{Hom}(S,S^{(m_{1})})\stackrel{{\scriptstyle\sim}}{{\longrightarrow}% }\textnormal{Hom}(S,\Sigma M_{0})\rightarrow\textnormal{Hom}(S,\Sigma M_{1})% \rightarrow\textnormal{Hom}(S,\Sigma S^{(m_{1})}).

Since $\textnormal{Hom}(S,\Sigma S^{(m_{1})})=0$ , we get $\textnormal{Hom}(S,\Sigma M_{1})=0$ .

David Pauksztello

Original source: arXiv:0705.0102v2