On the height of an ideal

Question

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

YequalsX · Accepted Answer · 2011-07-11 12:52:21Z

up vote 1 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)

answered Jul 11 '11 at 12:52

YequalsX
962

	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

Jacob Fox · Answer 2 · 2011-07-11 13:53:48Z

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

2 Answers