## 88.9 Examples

Some examples related to the results earlier in this chapter.

Example 88.9.1. Let $k$ be a field. The ring $A = k[x, y, z]/(x^ r + y^ s + z^ t)$ is a UFD for $r, s, t$ pairwise coprime integers. Namely, since $x^ r + y^ s + z^ t$ is irreducible $A$ is a domain. The element $z$ is a prime element, i.e., generates a prime ideal in $A$. On the other hand, if $t = 1 + ers$ for some $e$, then

$A[1/z] \cong k[x', y', 1/z]$

where $x' = x/z^{es}$, $y' = y/z^{er}$ and $z = (x')^ r + (y')^ s$. Thus $A[1/z]$ is a localization of a polynomial ring and hence a UFD. It follows from an argument of Nagata that $A$ is a UFD. See Algebra, Lemma 10.120.7. A similar argument can be given if $t$ is not congruent to $1$ modulo $rs$.

Example 88.9.2. The ring $A = \mathbf{C}[[x, y, z]]/(x^ r + y^ s + z^ t)$ is not a UFD when $1 < r < s < t$ are pairwise coprime integers and not equal to $2, 3, 5$. For example consider the special case $A = \mathbf{C}[[x, y, z]]/(x^2 + y^5 + z^7)$. Consider the maps

$\psi _\zeta : \mathbf{C}[[x, y, z]]/(x^2 + y^5 + z^7) \to \mathbf{C}[[t]]$

given by

$x \mapsto t^7,\quad y \mapsto t^3,\quad z \mapsto -\zeta t^2(1 + t)^{1/7}$

where $\zeta$ is a $7$th root of unity. The kernel $\mathfrak p_\zeta$ of $\psi _\zeta$ is a height one prime, hence if $A$ is a UFD, then it is principal, say given by $f_\zeta \in \mathbf{C}[[x, y, z]]$. Note that $V(x^3 - y^7) = \bigcup V(\mathfrak p_\zeta )$ and $A/(x^3 - y^7)$ is reduced away from the closed point. Hence, still assuming $A$ is a UFD, we would obtain

$\prod \nolimits _\zeta f_\zeta = u(x^3 - y^7) + a(x^2 + y^5 + z^7) \quad \text{in}\quad \mathbf{C}[[x, y, z]]$

for some unit $u \in \mathbf{C}[[x, y, z]]$ and some element $a \in \mathbf{C}[[x, y, z]]$. After scaling by a constant we may assume $u(0, 0, 0) = 1$. Note that the left hand side vanishes to order $7$. Hence $a = - x \bmod \mathfrak m^2$. But then we get a term $xy^5$ on the right hand side which does not occur on the left hand side. A contradiction.

Example 88.9.3. 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 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

$\xymatrix{ & X'' \ar[ld] \ar[rd] \\ X' \ar[rd] & & X''' \ar[ld] \\ & X }$

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

$x^3 + y^3 + z^3 + x^4 + y^4 + z^4 = 0$

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

$1 + s^3 + t^3 + x(1 + s^4 + t^4) = 0$

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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