The Stacks project

Lemma 53.9.4. Let $Z \subset \mathbf{P}^2_ k$ be as in Lemma 53.9.1 and let $I(Z) = (F)$ for some $F \in k[T_0, T_1, T_2]$. If $Z \to \mathop{\mathrm{Spec}}(k)$ is smooth in at least one point and $k$ is infinite, then there exists a closed point $z \in Z$ contained in the smooth locus such that $\kappa (z)/k$ is finite separable of degree at most $d$.

Proof. Suppose that $z' \in Z$ is a point where $Z \to \mathop{\mathrm{Spec}}(k)$ is smooth. After renumbering the coordinates if necessary we may assume $z'$ is contained in $D_+(T_0)$. Set $f = F(1, x, y) \in k[x, y]$. Then $Z \cap D_+(X_0)$ is isomorphic to the spectrum of $k[x, y]/(f)$. Let $f_ x, f_ y$ be the partial derivatives of $f$ with respect to $x, y$. Since $z'$ is a smooth point of $Z/k$ we see that either $f_ x$ or $f_ y$ is nonzero in $z'$ (see discussion in Algebra, Section 10.137). After renumbering the coordinates we may assume $f_ y$ is not zero at $z'$. Hence there is a nonempty open subscheme $V \subset Z \cap D_{+}(X_0)$ such that the projection

\[ p : V \longrightarrow \mathop{\mathrm{Spec}}(k[x]) \]

is ├ętale. Because the degree of $f$ as a polynomial in $y$ is at most $d$, we see that the degrees of the fibres of the projection $p$ are at most $d$ (see discussion in Morphisms, Section 29.56). Moreover, as $p$ is ├ętale the image of $p$ is an open $U \subset \mathop{\mathrm{Spec}}(k[x])$. Finally, since $k$ is infinite, the set of $k$-rational points $U(k)$ of $U$ is infinite, in particular not empty. Pick any $t \in U(k)$ and let $z \in V$ be a point mapping to $t$. Then $z$ works. $\square$


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