The Krull dimension, as defined by Gabriel and Rentschler, of not-necessarily commutative rings is an ordinal. See, for example, [John C. McConnell, James Christopher Robson, Lance W. Small, Noncommutative Noetherian rings].
More generally, they define the deviation of a poset $A$ as follows. If $A$ does not have comparable elements, $\mathrm{dev}\;A=-\infty$; if $A$ is has comparable elements but satisfies the d.c.c., then $\mathrm{dev}\;A=0$. In general, if $\alpha$ is an ordinal, we say that $\mathrm{dev}\;A=\alpha$ if (i) the deviation of $A$ is not an ordinal strictly less that $\alpha$, and (ii) in any descending sequence of elements in $A$ all but finitely many factors (ie, the intervals of $A$ determined by the successive elements in the sequence) have deviation less that $\alpha$.
Then the Gabriel-Rentschler left Krull dimension $\mathcal K(R)$ of a ring $R$ is the deviation of the poset of left ideals of $R$. A poset does not necessarily have a deviation, but if $R$ is left nötherian, then $\mathcal K(R)$ is defined.
A few examples: if a ring is nötherian commutative (or more generally satisfies a polynomial identity), then its G-R Krull dimension coincides with the combinatorial dimension of its prime spectrum, so in this definition extends classical one when these dimensions are finite. A non commutative example is the Weyl algebra $A_{n}(k)$: if $k$ has characteristic zero, then $\mathcal K(A_n(k))=n$, and if $k$ has positive characteristic, $\mathcal K(A_n(k))=2n$. The book by McConnel and Robson has lots of information and references.
Dear Amitsur,
It might help you to think geometrically. For example, $k[x,x^{-1}]$ is the ring of functions
on a hyperbola $xy = 1$, and the projection from this hyperbola to the line $x = y$ is a finite projection. This corresponds to the fact that $k[x,x^{-1}]$ is finitely generated as a module over $k[x + x^{-1}].$ (If we write $f = x + x^{-1}$, then $x^2 - f x +1 = 0$
and $x^{-2} - f x^{-1} + 1 = 0$.)
Best Answer
For a Noetherian ring R, the Krull dimension of its $I$-adic completion, $\hat{R}$ is given by $\sup h(J)$, where $J$ ranges over all maximal ideals of $R$ containing $I$ and $h(J)$ is the height of $J$. Therefore $\dim \hat R\le \dim R$ with equality only when $I\subset \operatorname{rad} R$. A reference is "Topics in $\mathfrak m$-adic topologies" by S.Greco, P.Salmon