The Stacks project

Proof. To get a contradiction, assume that $U \subset \mathop{\mathrm{Spec}}(R)$ is finite. In this case $(0) \in U$ and $\{ (0)\} $ is an open subset of $U$ (because the complement of $\{ (0)\} $ is the union of the closures of the other points). Thus we may assume $U = \{ (0)\} $. Let $\mathfrak m \subset R$ be the maximal ideal. We can find an $x \in \mathfrak m$, $x \not= 0$ such that $V(x) \cup U = \mathop{\mathrm{Spec}}(R)$. In other words we see that $D(x) = \{ (0)\} $. In particular we see that $\dim (R/xR) = \dim (R) - 1 \geq 1$, see Lemma 10.60.13. Let $\overline{y}_2, \ldots , \overline{y}_{\dim (R)} \in R/xR$ generate an ideal of definition of $R/xR$, see Proposition 10.60.9. Choose lifts $y_2, \ldots , y_{\dim (R)} \in R$, so that $x, y_2, \ldots , y_{\dim (R)}$ generate an ideal of definition in $R$. This implies that $\dim (R/(y_2)) = \dim (R) - 1$ and $\dim (R/(y_2, x)) = \dim (R) - 2$, see Lemma 10.60.14. Hence there exists a prime $\mathfrak p$ containing $y_2$ but not $x$. This contradicts the fact that $D(x) = \{ (0)\} $. $\square$