The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 10.152.11. Let $k \subset K$ be a field extension. Then $K$ is a filtered colimit of global complete intersection algebras over $k$. If $K/k$ is separable, then $K$ is a filtered colimit of smooth algebras over $k$.

Proof. Suppose that $E \subset K$ is a finite subset. It suffices to show that there exists a $k$ subalgebra $A \subset K$ which contains $E$ and which is a global complete intersection (resp. smooth) over $k$. The separable/smooth case follows from Lemma 10.152.10. In general let $L \subset K$ be the subfield generated by $E$. Pick a transcendence basis $x_1, \ldots , x_ d \in L$ over $k$. The extension $k(x_1, \ldots , x_ d) \subset L$ is finite. Say $L = k(x_1, \ldots , x_ d)[y_1, \ldots , y_ r]$. Pick inductively polynomials $P_ i \in k(x_1, \ldots , x_ d)[Y_1, \ldots , Y_ r]$ such that $P_ i = P_ i(Y_1, \ldots , Y_ i)$ is monic in $Y_ i$ over $k(x_1, \ldots , x_ d)[Y_1, \ldots , Y_{i - 1}]$ and maps to the minimum polynomial of $y_ i$ in $k(x_1, \ldots , x_ d)[y_1, \ldots , y_{i - 1}][Y_ i]$. Then it is clear that $P_1, \ldots , P_ r$ is a regular sequence in $k(x_1, \ldots , x_ r)[Y_1, \ldots , Y_ r]$ and that $L = k(x_1, \ldots , x_ r)[Y_1, \ldots , Y_ r]/(P_1, \ldots , P_ r)$. If $h \in k[x_1, \ldots , x_ d]$ is a polynomial such that $P_ i \in k[x_1, \ldots , x_ d, 1/h, Y_1, \ldots , Y_ r]$, then we see that $P_1, \ldots , P_ r$ is a regular sequence in $k[x_1, \ldots , x_ d, 1/h, Y_1, \ldots , Y_ r]$ and $A = k[x_1, \ldots , x_ d, 1/h, Y_1, \ldots , Y_ r]/(P_1, \ldots , P_ r)$ is a global complete intersection. After adjusting our choice of $h$ we may assume $E \subset A$ and we win. $\square$


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