Lemma 10.159.3. Let R be a ring. Let \mathfrak p \subset R be a prime and let L/\kappa (\mathfrak p) 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 q)/\kappa (\mathfrak p) is isomorphic to the given extension L/\kappa (\mathfrak p).
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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