Lemma 10.115.6. Let k be a field. Let \mathfrak q \subset k[x_1, \ldots , x_ n] be a prime ideal. Set r = \text{trdeg}_ k\ \kappa (\mathfrak q). Then there exists a finite ring map \varphi : k[y_1, \ldots , y_ n] \to k[x_1, \ldots , x_ n] such that \varphi ^{-1}(\mathfrak q) = (y_{r + 1}, \ldots , y_ n).
Proof. By induction on n. The case n = 0 is clear. Assume n > 0. If r = n, then \mathfrak q = (0) and the result is clear. Choose a nonzero f \in \mathfrak q. Of course f is nonconstant. After applying an automorphism of the form
k[x_1, \ldots , x_ n] \longrightarrow k[x_1, \ldots , x_ n], \quad x_ n \mapsto x_ n, \quad x_ i \mapsto x_ i + x_ n^{e_ i}\ (i < n)
we may assume that f is monic in x_ n over k[x_1, \ldots , x_ n], see Lemma 10.115.2. Hence the ring map
k[y_1, \ldots , y_ n] \longrightarrow k[x_1, \ldots , x_ n], \quad y_ n \mapsto f, \quad y_ i \mapsto x_ i\ (i < n)
is finite. Moreover y_ n \in \mathfrak q \cap k[y_1, \ldots , y_ n] by construction. Thus \mathfrak q \cap k[y_1, \ldots , y_ n] = \mathfrak pk[y_1, \ldots , y_ n] + (y_ n) where \mathfrak p \subset k[y_1, \ldots , y_{n - 1}] is a prime ideal. Note that \kappa (\mathfrak p) \subset \kappa (\mathfrak q) is finite, and hence r = \text{trdeg}_ k\ \kappa (\mathfrak p). Apply the induction hypothesis to the pair (k[y_1, \ldots , y_{n - 1}], \mathfrak p) and we obtain a finite ring map k[z_1, \ldots , z_{n - 1}] \to k[y_1, \ldots , y_{n - 1}] such that \mathfrak p \cap k[z_1, \ldots , z_{n - 1}] = (z_{r + 1}, \ldots , z_{n - 1}). We extend the ring map k[z_1, \ldots , z_{n - 1}] \to k[y_1, \ldots , y_{n - 1}] to a ring map k[z_1, \ldots , z_ n] \to k[y_1, \ldots , y_ n] by mapping z_ n to y_ n. The composition of the ring maps
k[z_1, \ldots , z_ n] \to k[y_1, \ldots , y_ n] \to k[x_1, \ldots , x_ n]
solves the problem. \square
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