Lemma 33.43.7. Let $X$ be a separated scheme of finite type over $k$ with $\dim (X) \leq 1$. Then there exists a commutative diagram
\[ \xymatrix{ \overline{Y}_1 \amalg \ldots \amalg \overline{Y}_ n \ar[rd] & Y_1 \amalg \ldots \amalg Y_ n \ar[r]_-\nu \ar[d] \ar[l]^ j & X_{k'} \ar[r] \ar[d] & X \ar[d]^ f \\ & \mathop{\mathrm{Spec}}(k'_1) \amalg \ldots \amalg \mathop{\mathrm{Spec}}(k'_ n) \ar[r] & \mathop{\mathrm{Spec}}(k') \ar[r] & \mathop{\mathrm{Spec}}(k) } \]
of schemes with the following properties:
$k'/k$ is a finite purely inseparable extension of fields,
$\nu $ is the normalization of $X_{k'}$,
$j$ is an open immersion with dense image,
$k'_ i/k'$ is a finite separable extension for $i = 1, \ldots , n$,
$\overline{Y}_ i$ is smooth, projective, geometrically irreducible dimension $\leq 1$ over $k'_ i$.
Proof.
As we may replace $X$ by its reduction, we may and do assume $X$ is reduced. Choose $X \to \overline{X}$ as in Lemma 33.43.6. If we can show the lemma for $\overline{X}$, then the lemma follows for $X$ (details omitted). Thus we may and do assume $X$ is projective.
Choose $k'/k$ finite purely inseparable such that the normalization of $X_{k'}$ is geometrically normal over $k'$, see Lemma 33.27.4. Denote $Y = (X_{k'})^\nu $ the normalization; for properties of the normalization, see Section 33.27. Then $Y$ is geometrically regular as normal and regular are the same in dimension $\leq 1$, see Properties, Lemma 28.12.6. Hence $Y$ is smooth over $k'$ by Lemma 33.12.6. Let $Y = Y_1 \amalg \ldots \amalg Y_ n$ be the decomposition of $Y$ into irreducible components. Set $k'_ i = \Gamma (Y_ i, \mathcal{O}_{Y_ i})$. These are finite separable extensions of $k'$ by Lemma 33.9.3. The proof is finished by Lemma 33.9.4.
$\square$
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