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For an ideal $I$ in a commutative ring $R$, define $\operatorname{codim}I=\dim R-\dim R/I$. Does codimension equals height for all ideals in the formal power series ring? Does this hold for complete local domains in general?

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up vote 6 down vote accepted

The standard definition of codimension of $R/I$ is the height of $I$.

Anyway, a first observation is $\dim R - \dim R/I=\operatorname{ht} I$ for all $I$ is equivalent to require the equality for all prime ideals. So let $P$ be a prime ideal in $R$ and you want $$ \dim R = \dim R/P + \operatorname{ht}P.$$ This is true for any local Cohen-Macaulay ring. If you take a formal power series rings with coefficients in a field and finitely many variables, then it is regular, hence Cohen-Macaulay. Another class of rings for which the egality holds are Noetherian local catenary integral domains (this includes the Noetherian complete local domains).

If $R$ is not a domain, there are easy counterexamples (e.g. if the spectrum of $R$ has two irreducible components and if they have distinct dimensions).

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Very nicely explained! – Georges Elencwajg Mar 16 '12 at 10:08
Thank you. This is the definition of dimension used in Bruns-Herzog. – Jake Voigt Mar 16 '12 at 13:35

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