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Let X be a surface. (A surface is an excellent integral normal separated 2-dimensional scheme.)

Let $\psi:Y\longrightarrow X$ be a minimal resolution of singularities and let $E$ be an irreducible component of the exceptional locus of $\psi$.

Is $E$ of genus zero? That is, do we have that $E$ is isomorphic to $\mathbf{P}^1_k$ for some field $k$?

Is the exceptional locus of $\psi$ a chain of rational curves? (This means that $(E_i,E_i) <0$, $(E_i, E_{i+1}) = (E_i, E_{i-1}) = 1$ and $(E_i,E_j) = 0$ if $j \neq i-1, i,i+1$. Here $E_i$ denotes an exceptional component.)

I know this is true if $X$ has "cyclic quotient singularities".

I also know that this is true, by Lipman's theorem and what is stated on Wikipedia about it, that this is true if $X$ has pseudo-rational singularities.

What about the general case?

What else can we say in general about the "shape" of the exceptional locus. If it's not a chain, does it have a loop? Can things get arbitrarily complicated?

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This answer probably comes a little late. The exceptional locus can be artitrarily complicated. It can have smooth curve of any genus and any number of loops.

Here is a may to create such situations. Let $R$ be a complet DVR. Let $Z$ be the minimal regular model over $R$ of some smooth proper curve of positive genus over $\mathrm{Frac}(R)$. Let $Z_s$ be the close fiber of $Z$. Blow-up $Y\to Z$ along a close point of $Z_s$. Denote by $E$ the strict transform of $Z_s$ in $Y$. It is essentially isomorphic to $Z_s$. Now there exists a contraction morphism $Y\to X$ which maps $E$ to a single point in $X$ and is an isomorphism away from $E$ (see Bosch-Lütkebohmert-Raynaud's book "Néron model", § 6.7). As $Y$ is regular, it is the minimal resolution of $X$, and the exceptional locus of $Y\to X$ is $E$. As $Z_s$ can be as "complicated" as you can, so is $E$.

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    Unless there is a very good reason, we prefer not to arbitrarily delete good content: the answers could well help other users in the future.2013-08-09