Dimension definition Consider the definition of the topological dimension of a topological space $X$ as the maximum of an increasing chain of closed irreducible subsets of $X$.
Is dimension a topological concept?
I mean here, If I define the dimension of a ring to be its Krull dimension, and then use this in schemes, am I still talking about the topological concept?
Now, if I have a manifold, and declare its dimension to be the dimension of the Euclidian spaces it resembles on a small scale, is this equivalent to its dimension as a topological space?
Hopefully, my questions make sense.

For $X$ a topological space, let's define $\dim_{\text{top}} X$ to be the maximal $n$ so that there exists a chain of subsets
$$X_0 \subsetneq X_1 \subsetneq \dots \subsetneq X_n$$
where each $X_i$ are irreducible and closed. One then proves that 
$$\dim R = \dim_{\text{top}}X$$ when for $X$ an algebraic variety over a field $K$ with coordinate ring $R$ (more generally, if $X = \text{Spec} (R)$ is an affine scheme).   
The fact that the Krull dimension and topological dimension coincide boils down to the observation that chains of prime ideals in $R$ correspond to chains of irreducible closed subsets in $X$. Said otherwise, the structure of ideals in the coordinate ring gives topological information about the associated variety -- perhaps not terribly surprising given that they are used to define the topology on the space :)
In the case of a manifold, the topological dimension is badly behaved and does not agree with the Euclidean dimension. Any closed set that is not a point is reducible, so the Krull dimension is always zero!
