Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 4 down vote accepted

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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.