Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 have a question regarding an exercise I found in Eisenbud's Commutative Algebra with a view towards Algebraic Geometry:

Exercise 17.6: Any ideal of a Noetherian ring generated by a regular sequence can be generated by a regular sequence in any order (i.e. permutations of the regular sequence still form a regular sequence).

In the section "Hints and Solutions for Selected Exercises", there are hints on how to solve the exercise, which break the proof down to 4 separate steps. Steps 1, 2 and 4 are manageable, but I am kind of stuck at step 3:

if $x_1,\ldots,x_r$ is a regular sequence $\Rightarrow \exists j \in (x_1,\ldots,x_{r-1})$ such that $x_r+j$ is a nonzerodivisor modulo every subset of $\{x_1,\ldots,x_{r-1}\}$.

The hint he stated ("use prime avoidance in the form given in Exercise 3.8b") is not that useful, since there is no Exercise 3.8b. I tried to find out what else he might have referred to but I couldn't find anything applicable, so I am stuck with the hint that some sort of prime avoidance is necessary.

One can write the set of zero divisors as a union of prime ideals, the radicals of the primary ideals which decompose $(0)$, but I also fail to see how it might help.

I hope somebody can help me with this problem that has been bugging me for the past week.

Thanks in advance!

share|cite|improve this question
Maybe Exercise 3.8b was meant to be Exercise 3.19b? – Andrew Jul 13 '12 at 17:23
Exercise 3.19b solves Step 3 immediately. One can also look for the exercise 1.2.21 from Bruns & Herzog, Cohen-Macaulay Rings. – user26857 Jul 13 '12 at 20:54

Your Answer


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

Browse other questions tagged or ask your own question.