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In the context of Krull dimension, given any commutative ring $R$ and $\mathfrak{p}\subset R$ a prime ideal, we have (almost by definition) $$ \mathrm{ht}\:\mathfrak{p}+\mathrm{coht}\:\mathfrak{p} \leq \dim R $$ In the special circumstance of the coordinate ring of an (irreducible) affine variety, i.e. $R$ being a finitely generated $K$-algebra which is an integral domain we have the equality $$ \mathrm{ht}\:\mathfrak{p}+\mathrm{coht}\:\mathfrak{p} = \dim R\quad (*) $$

I was curious on when (which type of rings) this equality $(*)$ happens? By the way, does this equality/inequality have a name?

First of all seems like we better assume $\dim R<\infty$ for preventing things to go haywire. Then if $\mathfrak{p}_0\subsetneq\cdots\subsetneq \mathfrak{p}_m=\mathfrak{p}$ is a chain the length of which is $\mathrm{ht} \: \mathfrak{p}$ and $\mathfrak{p}\subsetneq\cdots\subsetneq \mathfrak{p}_n=\mathfrak{m}$ a chain the length of which is $\mathrm{coht}\: \mathfrak{p}$, then defining $n$ as the length of the two chains combined: $$n\leq \mathrm{ht}\:\mathfrak{m}\leq \dim R.$$ So we have two inequalities we need to satisfy. The second inequality can be satisfied by assuming $R$ is equicodimensional, i.e. $\operatorname{ht}\mathfrak{m}=\dim R$ for any maximal ideal $\mathfrak m$.

The first one is trickier. It is happening because it is possible to have another maximal chain of primes inside $\mathfrak{m}$ which doesn't pass through $\mathfrak{p}$ and its length is bigger than $n$. I know of one way to make this imposible (which may not be the most general condition):

Fact: If $R$ is a equicodimensional Cohen-Macaulay ring then every maximal chain of prime ideals is of the same length $\dim R$.

So what I have come up with is if $R$ is a equicodimensional CM-ring then the equality (*) holds (right?).

How much further can one go in stating the most general condition for the onset of equality $(*)$? Is it even useful to go further?

user26857
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Hamed
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  • Related: http://math.stackexchange.com/questions/49136/operatornameheight-mathfrakp-dim-a-mathfrakp-dim-a – user26857 May 17 '16 at 08:33
  • You are right about the equicodimensional CM rings. (In particular, it holds for finitely generated CM algebras over an algebraically closed field.) – user26857 May 17 '16 at 09:06
  • One more thing: the first chain condition for prime ideals (f.c.c. for short) says that every maximal chain of prime ideals in $R$ has length equal to $\dim R$. This clearly implies $\mathrm{ht}:\mathfrak{p}+\mathrm{coht}:\mathfrak{p} = \dim R$, but in general the converse can be false. As a side remark, equicodimensional CM rings satisfy f.c.c. – user26857 May 18 '16 at 07:55
  • Btw, Ratliff has shown that f.c.c. is equivalent to $\mathrm{ht}:\mathfrak{p}+\mathrm{coht}:\mathfrak{p} = \dim R$ for local domains. – user26857 May 18 '16 at 09:39

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