Lemma 10.159.3. Let $R$ be a ring. Let $\mathfrak p \subset R$ be a prime and let $\kappa (\mathfrak p) \subset L$ be a finite extension of fields. Then there exists a finite free ring map $R \to S$ such that $\mathfrak q = \mathfrak pS$ is prime and $\kappa (\mathfrak p) \subset \kappa (\mathfrak q)$ is isomorphic to the given extension $\kappa (\mathfrak p) \subset L$.

Proof. By induction of the degree of $\kappa (\mathfrak p) \subset L$. If the degree is $1$, then we take $R = S$. In general, if there exists a sub extension $\kappa (\mathfrak p) \subset L' \subset L$ then we win by induction on the degree (by first constructing $R \subset S'$ corresponding to $L'/\kappa (\mathfrak p)$ and then construction $S' \subset S$ corresponding to $L/L'$). Thus we may assume that $L \supset \kappa (\mathfrak p)$ is generated by a single element $\alpha \in L$. Let $X^ d + \sum _{i < d} a_ iX^ i$ be the minimal polynomial of $\alpha$ over $\kappa (\mathfrak p)$, so $a_ i \in \kappa (\mathfrak p)$. We may write $a_ i$ as the image of $f_ i/g$ for some $f_ i, g \in R$ and $g \not\in \mathfrak p$. After replacing $\alpha$ by $g\alpha$ (and correspondingly replacing $a_ i$ by $g^{d - i}a_ i$) we may assume that $a_ i$ is the image of some $f_ i \in R$. Then we simply take $S = R[x]/(x^ d + \sum f_ ix^ i)$. $\square$

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