I'm reading IRREDUCIBLE COMPONENTS OF RIGID SPACES (by Conrad). In this paper he defines the irreducible component of a rigid variety $X$ to be reduced image of a connected component of $\tilde X$ (the normalization of $X$). I wonder why not just define it as the irreducible component of zariski topology? It seems it's the same as the definition of this paper...
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You can make that definition, but then it's not obvious that it has nice properties when you take an open subset. See section 1.2 of Coleman-Mazur's Eigencurve paper: one specific thing you'd want to know that if $X'\to X$ is an open immersion of rigid spaces, then there is an induced map on irreducible components. But the Zariski topology of $X'$ is not in general the topology induced by the Zariski topology of $X$ - it is finer - so constructing said induced map seems a bit mysterious. Conrad's definition is one way to bypass this issue. (It is worth noting that exactly the same issue comes up in complex analytic geometry, and you can deal with it in roughly the same way; see Grauert-Remmert, Coherent Analytic Sheaves, 9.2. There are extra complications for rigid analytic spaces due to usual admissibility issues and because proving excellence of local rings is a bit trickier.)
If you're willing to work with Berkovich spaces, there is an alternative method due to Ducros in Les espaces de Berkovich sont excellents that avoids normalization, although a lot of the intermediate steps turn out to be of a similar nature.
