Lemma 10.144.3. Let R be a ring. Let \mathfrak p be a prime of R. Let L/\kappa (\mathfrak p) be a finite separable field extension. There exists an étale ring map R \to R' together with a prime \mathfrak p' lying over \mathfrak p such that the field extension \kappa (\mathfrak p')/\kappa (\mathfrak p) is isomorphic to \kappa (\mathfrak p) \subset L.
Proof. By the theorem of the primitive element we may write L = \kappa (\mathfrak p)[\alpha ]. Let \overline{f} \in \kappa (\mathfrak p)[x] denote the minimal polynomial for \alpha (in particular this is monic). After replacing \alpha by c\alpha for some c \in R, c\not\in \mathfrak p we may assume all the coefficients of \overline{f} are in the image of R \to \kappa (\mathfrak p) (verification omitted). Thus we can find a monic polynomial f \in R[x] which maps to \overline{f} in \kappa (\mathfrak p)[x]. Since \kappa (\mathfrak p) \subset L is separable, we see that \gcd (\overline{f}, \overline{f}') = 1. Hence there is an element \gamma \in L such that \overline{f}'(\alpha ) \gamma = 1. Thus we get a R-algebra map
\begin{eqnarray*} R[x, 1/f']/(f) & \longrightarrow & L \\ x & \longmapsto & \alpha \\ 1/f' & \longmapsto & \gamma \end{eqnarray*}
The left hand side is a standard étale algebra R' over R and the kernel of the ring map gives the desired prime. \square
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