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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