
Lemma 10.141.15. Let $R$ be a ring. Let $\mathfrak p$ be a prime of $R$. Let $\kappa (\mathfrak p) \subset L$ 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 $\kappa (\mathfrak p) \subset \kappa (\mathfrak p')$ 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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