Lemma 10.115.1. Let n \in \mathbf{N}. Let N be a finite nonempty set of multi-indices \nu = (\nu _1, \ldots , \nu _ n). Given e = (e_1, \ldots , e_ n) we set e \cdot \nu = \sum e_ i\nu _ i. Then for e_1 \gg e_2 \gg \ldots \gg e_{n-1} \gg e_ n we have: If \nu , \nu ' \in N then
10.115 Noether normalization
In this section we prove variants of the Noether normalization lemma. The key ingredient we will use is contained in the following two lemmas.
Proof. Say N = \{ \nu _ j\} with \nu _ j = (\nu _{j1}, \ldots , \nu _{jn}). Let A_ i = \max _ j \nu _{ji} - \min _ j \nu _{ji}. If for each i we have e_{i - 1} > A_ ie_ i + A_{i + 1}e_{i + 1} + \ldots + A_ ne_ n then the lemma holds. For suppose that e \cdot (\nu - \nu ') = 0. Then for n \ge 2,
We may assume that (\nu _1 - \nu '_1) \ge 0. If (\nu _1 - \nu '_1) > 0, then
This contradiction implies that \nu '_1 = \nu _1. By induction, \nu '_ i = \nu _ i for 2 \le i \le n. \square
Lemma 10.115.2. Let R be a ring. Let g \in R[x_1, \ldots , x_ n] be an element which is nonconstant, i.e., g \not\in R. For e_1 \gg e_2 \gg \ldots \gg e_{n-1} \gg e_ n = 1 the polynomial
where d > 0 and a \in R is one of the nonzero coefficients of g.
Proof. Write g = \sum _{\nu \in N} a_\nu x^\nu with a_\nu \in R not zero. Here N is a finite set of multi-indices as in Lemma 10.115.1 and x^\nu = x_1^{\nu _1} \ldots x_ n^{\nu _ n}. Note that the leading term in
Hence the lemma follows from Lemma 10.115.1 which guarantees that there is exactly one nonzero term a_\nu x^\nu of g which gives rise to the leading term of g(x_1 + x_ n^{e_1}, x_2 + x_ n^{e_2}, \ldots , x_{n - 1} + x_ n^{e_{n - 1}}, x_ n), i.e., a = a_\nu for the unique \nu \in N such that e \cdot \nu is maximal. \square
Lemma 10.115.3. Let k be a field. Let S = k[x_1, \ldots , x_ n]/I for some proper ideal I. If I \not= 0, then there exist y_1, \ldots , y_{n-1} \in k[x_1, \ldots , x_ n] such that S is finite over k[y_1, \ldots , y_{n-1}]. Moreover we may choose y_ i to be in the \mathbf{Z}-subalgebra of k[x_1, \ldots , x_ n] generated by x_1, \ldots , x_ n.
Proof. Pick f \in I, f\not= 0. It suffices to show the lemma for k[x_1, \ldots , x_ n]/(f) since S is a quotient of that ring. We will take y_ i = x_ i - x_ n^{e_ i}, i = 1, \ldots , n-1 for suitable integers e_ i. When does this work? It suffices to show that \overline{x_ n} \in k[x_1, \ldots , x_ n]/(f) is integral over the ring k[y_1, \ldots , y_{n-1}]. The equation for \overline{x_ n} over this ring is
Hence we are done if we can show there exists integers e_ i such that the leading coefficient with respect to x_ n of the equation above is a nonzero element of k. This can be achieved for example by choosing e_1 \gg e_2 \gg \ldots \gg e_{n-1}, see Lemma 10.115.2. \square
Lemma 10.115.4.slogan Let k be a field. Let S = k[x_1, \ldots , x_ n]/I for some ideal I. If I \neq (1), there exist r\geq 0, and y_1, \ldots , y_ r \in k[x_1, \ldots , x_ n] such that (a) the map k[y_1, \ldots , y_ r] \to S is injective, and (b) the map k[y_1, \ldots , y_ r] \to S is finite. In this case the integer r is the dimension of S. Moreover we may choose y_ i to be in the \mathbf{Z}-subalgebra of k[x_1, \ldots , x_ n] generated by x_1, \ldots , x_ n.
Proof. By induction on n, with n = 0 being trivial. If I = 0, then take r = n and y_ i = x_ i. If I \not= 0, then choose y_1, \ldots , y_{n-1} as in Lemma 10.115.3. Let S' \subset S be the subring generated by the images of the y_ i. By induction we can choose r and z_1, \ldots , z_ r \in k[y_1, \ldots , y_{n-1}] such that (a), (b) hold for k[z_1, \ldots , z_ r] \to S'. Since S' \to S is injective and finite we see (a), (b) hold for k[z_1, \ldots , z_ r] \to S. The last assertion follows from Lemma 10.112.4. \square
Lemma 10.115.5. Let k be a field. Let S be a finite type k algebra and denote X = \mathop{\mathrm{Spec}}(S). Let \mathfrak q be a prime of S, and let x \in X be the corresponding point. There exists a g \in S, g \not\in \mathfrak q such that \dim (S_ g) = \dim _ x(X) =: d and such that there exists a finite injective map k[y_1, \ldots , y_ d] \to S_ g.
Proof. Note that by definition \dim _ x(X) is the minimum of the dimensions of S_ g for g \in S, g \not\in \mathfrak q, i.e., the minimum is attained. Thus the lemma follows from Lemma 10.115.4. \square
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
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
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
solves the problem. \square
Lemma 10.115.7. Let R \to S be an injective finite type ring map. Assume R is a domain. Then there exists an integer d and a factorization
by injective maps such that S' is finite over R[y_1, \ldots , y_ d] and such that S'_ f \cong S_ f for some nonzero f \in R.
Proof. Pick x_1, \ldots , x_ n \in S which generate S over R. Let K be the fraction field of R and S_ K = S \otimes _ R K. By Lemma 10.115.4 we can find y_1, \ldots , y_ d \in S such that K[y_1, \ldots , y_ d] \to S_ K is a finite injective map. Note that y_ i \in S because we may pick the y_ j in the \mathbf{Z}-algebra generated by x_1, \ldots , x_ n. As a finite ring map is integral (see Lemma 10.36.3) we can find monic P_ i \in K[y_1, \ldots , y_ d][T] such that P_ i(x_ i) = 0 in S_ K. Let f \in R be a nonzero element such that fP_ i \in R[y_1, \ldots , y_ d][T] for all i. Then fP_ i(x_ i) maps to zero in S_ K. Hence after replacing f by another nonzero element of R we may also assume fP_ i(x_ i) is zero in S. Set x_ i' = fx_ i and let S' \subset S be the R-subalgebra generated by y_1, \ldots , y_ d and x'_1, \ldots , x'_ n. Note that x'_ i is integral over R[y_1, \ldots , y_ d] as we have Q_ i(x_ i') = 0 where Q_ i = f^{\deg _ T(P_ i)}P_ i(T/f) which is a monic polynomial in T with coefficients in R[y_1, \ldots , y_ d] by our choice of f. Hence R[y_1, \ldots , y_ d] \subset S' is finite by Lemma 10.36.5. Since S' \subset S we have S'_ f \subset S_ f (localization is exact). On the other hand, the elements x_ i = x'_ i/f in S'_ f generate S_ f over R_ f and hence S'_ f \to S_ f is surjective. Whence S'_ f \cong S_ f and we win. \square
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