Lemma 10.158.11. Let $K/k$ 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 10.158.10. 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 $L/k(x_1, \ldots , x_ d)$ 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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