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 étale. 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 29.57). Moreover, as $p$ is étale 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. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (0)
There are also: