Take the 2-minute tour ×
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.

Which of the following inequalities hold for a ring $R$ and an ideal of its $I$?

height $I\leq\mathrm{dim}\;R-\mathrm{dim}\;R/I$

height $I\geq\mathrm{dim}\;R-\mathrm{dim}\;R/I$

share|improve this question

2 Answers 2

up vote 2 down vote accepted

Assume dim R is finite. Let P be a prime containing I such that dim R/I = dim R/p. We have then that

ht(I) + dim (R/I) <= ht(I) + dim (R/p) <= ht(P) + dim(R/p) <= dim(R)

Reference

share|improve this answer
    
I might be wrong, but I think that $\mathrm{dim}\;R/I\geq \mathrm{dim}\;R/\mathfrak{p}$. For example take $I\subset\mathfrak{p}_1\subset\mathfrak{p}_2$ then $\mathrm{dim}\;R/I\geq\mathrm{dim}\;R/\mathfrak{p}_2$, isn't it? –  Jacob Fox Jul 11 '11 at 13:46
    
But choose a prime P such that dim R/I = dim R/p. –  YequalsX Jul 11 '11 at 14:01
    
Right, you're right –  Jacob Fox Jul 11 '11 at 14:18

I think to have it: suppose $\mathrm{height}\;I=n$ and $\mathrm{dim}\;R/I=m$ then we have a chain

$\mathfrak{p}_0\subset\ldots\subset\mathfrak{p}_n\subset I\subset\mathfrak{p}_{n+1}\subset\ldots\subset\mathfrak{p}_{n+m}$

but in general $\mathrm{dim}\;R$ would be greater, so

$\mathrm{height}\;I+\mathrm{dim}\;R/I\leq\mathrm{dim}\;R$ holds

share|improve this answer

Your Answer

 
discard

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.