# On the grade of an ideal

I need to prove the following statment (actually a special case of it).

Let $R$ be a Noetherian ring, $M$ a finite $R$-module and $I$ an ideal of $R$. Then $\operatorname{grade}(I,M)\geq 2$ if and only if the homomorphism $M\rightarrow$Hom$_R(I,M)$ given by $m\mapsto(i\mapsto im)$ is an isomorphism.

This is Exercise 1.2.24 in Bruns and Herzog, Cohen-Macaulay Rings.

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What is $\mathrm{grade}(I,M)$? –  Arturo Magidin Jan 26 '11 at 19:24
It's the length of a maximal regular sequence of $M$ contained in $I$. –  Wesley Farrel Jan 26 '11 at 20:27

Use the long exact sequence in Ext from the ses $$0 \to I \to R \to R/I \to 0,$$ leading to $$0 \to \hom(R/I, M) \to \hom(R, M) \to\hom(I,M) \to \mathrm{Ext}^1(R/I, M) \to 0,$$ because $R$ is projective. It is known that depth (or grade) can be measured in terms of Exts, and in particular that it is at least 2 if the two ends of the sequence vanish. But this is equivalent to the condition that $\hom(R, M) \to \hom(I,M)$ be an isomorphism.