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Question

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

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

In particular, if

M=M1MnM=M_{1} \oplus \cdots \oplus M_{n}

prove that

Supp(M)=Supp(M1)Supp(Mn).\operatorname{Supp}(M)=\operatorname{Supp}\left(M_{1}\right) \cup \cdots \cup \operatorname{Supp}\left(M_{n}\right).

Solution

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

Assume that at least one of the LPL_P and NPN_P is non-zero, then by the exactness of the sequence would imply that MPM_P is non-zero.

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

We have that the obvious morphisms 0i=1n1Mii=1nMiMn00\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 nn.

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