Example 89.9.3. There exists an excellent 2-dimensional Noetherian local ring and a modification X \to S = \mathop{\mathrm{Spec}}(A) which is not a scheme. We sketch a construction. Let X be a normal surface over \mathbf{C} with a unique singular point x \in X. Assume that there exists a resolution \pi : X' \to X such that the exceptional fibre C = \pi ^{-1}(x)_{red} is a smooth projective curve. Furthermore, assume there exists a point c \in C such that if \mathcal{O}_ C(nc) is in the image of \mathop{\mathrm{Pic}}\nolimits (X') \to \mathop{\mathrm{Pic}}\nolimits (C), then n = 0. Then we let X'' \to X' be the blowing up in the nonsingular point c. Let C' \subset X'' be the strict transform of C and let E \subset X'' be the exceptional fibre. By Artin's results ([ArtinII]; use for example [Mumford-topology] to see that the normal bundle of C' is negative) we can blow down the curve C' in X'' to obtain an algebraic space X'''. Picture
We claim that X''' is not a scheme. This provides us with our example because X''' is a scheme if and only if the base change of X''' to A = \mathcal{O}_{X, x} is a scheme (details omitted). If X''' where a scheme, then the image of C' in X''' would have an affine neighbourhood. The complement of this neighbourhood would be an effective Cartier divisor on X''' (because X''' is nonsingular apart from 1 point). This effective Cartier divisor would correspond to an effective Cartier divisor on X'' meeting E and avoiding C'. Taking the image in X' we obtain an effective Cartier divisor meeting C (set theoretically) in c. This is impossible as no multiple of c is the restriction of a Cartier divisor by assumption.
To finish we have to find such a singular surface X. We can just take X to be the affine surface given by
in \mathbf{A}^3_\mathbf {C} = \mathop{\mathrm{Spec}}(\mathbf{C}[x, y, z]) and singular point (0, 0, 0). Then (0, 0, 0) is the only singular point. Blowing up X in the maximal ideal corresponding to (0, 0, 0) we find three charts each isomorphic to the smooth affine surface
which is nonsingular with exceptional divisor C given by x = 0. The reader will recognize C as an elliptic curve. Finally, the surface X is rational as projection from (0, 0, 0) shows, or because in the equation for the blowup we can solve for x. Finally, the Picard group of a nonsingular rational surface is countable, whereas the Picard group of an elliptic curve over the complex numbers is uncountable. Hence we can find a closed point c as indicated.
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