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Let $0\rightarrow N \rightarrow M\rightarrow P \rightarrow 0$ be an exact sequence of finitely generated $\mathbb{N}$ graded module over a commutative ring $R$.

The vanishing degree of a $\mathbb{N}$ graded module $M$ is defined to be the maximal number $m$ such that $M_m \neq 0$ and denoted by $v(M)$.

I guess that : $v(M)=\text{max}\lbrace v(N),v(P)\rbrace$ but I could not prove it. Could you please help me ? If I am wrong please show me a counter example.

Thanks.

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1 Answer 1

For all $m \in \mathbb{N}$ we have

$m > v(M) \Leftrightarrow M_m = 0 \Leftrightarrow N_m=0 \wedge P_m = 0\\ ~ ~~~~~~~~~~~~~~~~~\Leftrightarrow m>v(N) \wedge m > v(P) \Leftrightarrow m > \max(v(N),v(P)).$

It follows $v(M)=\max(v(N),v(P))$. [can you see Yoneda?]

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Could you please explain the relation with Yoneda in your proof ? Did you get it Axy ? –  knot Dec 1 '12 at 15:57

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