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If 0 $\rightarrow$ L $\rightarrow$ M $\rightarrow$ N $\rightarrow$ 0 is an exact sequence of R-modules, prove that the localization $M_{P}$ is nonzero if and only if one of the localizations $N_{P}$ and $L_{P}$ is nonzero and deduce that

$\operatorname{Supp}(M)=\operatorname{Supp}(L) \cup \operatorname{Supp}(N).$

In particular, if

$M=M_{1} \oplus \cdots \oplus M_{n}$

prove that

$\operatorname{Supp}(M)=\operatorname{Supp}\left(M_{1}\right) \cup \cdots \cup \operatorname{Supp}\left(M_{n}\right).$

Solution

VerifiedSuppose $M_P$ is non-zero. If $L_P$ and $N_P$ are both zero the exactness of the sequence and the preservation of exactness by localization ($\textbf{proposition 42}$) would imply that $M_P=0$, a contradiction. Therefore at least one of the $L_P$ and $N_P$ must be non-zero.

Assume that at least one of the $L_P$ and $N_P$ is non-zero, then by the exactness of the sequence would imply that $M_P$ is non-zero.

The first implication implies that $Supp(M)\subseteq Supp(L)\cup Supp(N)$, and the second implication implies $Supp(M)\supseteq Supp(L)\cup Supp(N)$. Therefore $Supp(M)= Supp(L)\cup Supp(N)$.

We have that the obvious morphisms $0\rightarrow \oplus_{i=1}^{n-1}M_i\rightarrow \oplus_{i=1}^nM_i\rightarrow M_n\rightarrow 0$ is an exact sequence. Therefore the conclusion follows by the previous observation andinduction on $n$.

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