Generators of rings of constants for some diagonal derivations in polynomial rings

Generators of rings of constants for some diagonal derivations

in polynomial rings

by

Andrzej Nowicki

Institute of Mathematics, N.Copernicus University 87–100 Toru´n, Poland (e-mail: anow@pltumk11.bitnet).

and

Jean – Marie Strelcyn

D´epartement de Math´ematiques, Universit´e de Rouen,

URA CNRS 1378, B.P.118, 76134 Mont Saint Aignan Cedex, France and

Laboratorie Analyse, G´eometrie et Applications, URA CNRS 742, Institut Galil´ee, Departement de Math´ematiques, Avenue J.–B. Cl´ement,

93430 Villetaneuse, France (e-mail: strelcyn@math.univ-paris13.fr).

Abstract

Let K be a field of characteristic zero. We show that if n ≥ 3, given r ≥ 0 there exists a diagonal K–derivation of K[x₁, . . . , x_n] such that the minimal number of generators over K of the ring of constants is equal to r.

1.Introduction. Let K be a field of characteristic zero and R = K[x₁, . . . , x_n] the polynomial ring over K.

Let d be a K-derivation of R and R^d its ring of constants, that is, R^d= {w ∈ R; d(w) = 0}.

If R^d 6= K and R^d is finitely generated over K then we denote by γ(d) the minimal number of polynomials from R \ K which generate R^d over K. Moreover we assume that γ(d) = 0 iff R^d = K, and γ(d) = ∞ iff R^d is not finitely generated over K.

In a recent paper [1] Derksen show that the Nagata’s counterexample [3] to the fourteenth problem of Hilbert can be put in the form R^d for some derivation d with

n = 32. So there exists a K-derivation d of R (for n = 32) such that γ(d) = ∞ (see also [4]).

If n = 1 and d 6= 0 then of course γ(d) = 0. If n = 2 and d 6= 0 then, by a result of Zaks [6] and Eakin [2], we see that γ(d) ≤ 1. If n = 3 then it is known, by a result of Zariski [7] , that γ(d) < ∞ (see [5] for details).

In this note we show that, for n ≥ 3, the set {γ(d); d is a K-derivation of R} is unbounded.

In our proof we use only diagonal K-derivations, that is, such K-derivations d that d(x₁) = a₁x₁, . . . , d(x_n) = a_nx_n, for some a₁, . . . , a_n ∈ K.

The aim of this note is to prove the following

Theorem 1 If n ≥ 3 and r ≥ 0 then there exists a diagonal K-derivation of R = K[x₁, . . . , x_n] such that γ(d) = r.

2.Proof. Let us start with the following two simple remarks:

(a) If d(x₁) = x₁, . . . , d(x_n) = x_n then γ(d) = 0.

(b) Let 1 ≤ r < n and d(x₁) = . . . = d(x_r) = 0, d(x_r+1) = x_r+1, . . . , d(x_n) = x_n. Then γ(d) = r.

Consequently, in the remaining part of the proof we will assume that r ≥ n.

Define m = n − 3 and p = r − n + 2 = r − m − 1; p ≥ 2.

Let x = x₁, y = x₂, z = x₃ and if n > 3, y₁ = x₄, . . . , y_m = x_n.

To prove our theorem let us consider a K-derivation of R defined as follows:

d(x) = x, d(y) = y, d(z) = −pz and, if n > 3,

d(y₁) = . . . = d(y_m) = 0.

Consider now r polynomials f₀, f₁, . . . f_p+m:

f₀ = x^pz, f₁ = x^p−1y¹z, . . . , f_i = x^p−iyⁱz, . . . , f_p−1 = x¹y^p−1z, f_p = y^pz which are defined for every n ≥ 3 and

f_p+1= y₁, f_p+2= y₂, . . . , f_p+m = y_m when n > 3.

If n = 3 then R = K[x, y, z] and p + 1 = r. If n > 3 then R = K[x, y, z, y₁, . . . , y_m] and p + 1 + m = r.

Let us observe that these polynomials belong to R^d. Indeed; if 0 ≤ i ≤ p then d(f_i) = d(x^p−iyⁱz) = (p − i + i − p)x^p−iyⁱz = 0

and if i = p + j ≥ p, then d(fi) = d(yj) = 0.

Lemma 1 The polynomials f₀, . . . , f_r−1 generate R^d over K.

Proof. Let w ∈ R be such that d(w) = 0. First let us consider the case when w ∈ K[x, y, z]. Thus

w =^Xa_ijkxⁱy^jz^k

with all a_ijk∈ K. Since d is diagonal then all monomials xⁱy^jz^kfrom the above sum are such that d(xⁱy^jz^k) = 0. Let us pick such a monomial w0 = xⁱy^jz^k. Then i + j = pk.

Let a, b, u, v be the nonnegative integers such that

i = ap + u, j = bp + v, u < p, v < p.

Then either u + v = 0, or u + v = p. If u + v = 0, then k = a + b and consequently w₀ = xⁱy^jz^k= x^apy^bpz^a+b = (x^pz)^a(y^pz)^b = f₀^af_p^b.

If u + v = p, then k = a + b + 1 which implies

w₀ = xⁱx^jz^k= x^ap+uy^bp+vz^a+b+1 = (x^uy^vz)(x^pz)^a(y^pz)^b = f_vf₀^af_p^b. Therefore, if w ∈ K[x, y, z] and d(w) = 0, then w ∈ K[f₀, . . . , f_p].

Assume now that w ∈ R = K[x, y, z, y₁, . . . , y_m]. Then w =^Xa_i1...imy₁ⁱ¹. . . yⁱ_m^m

where all coefficients a_i1...im belong to K[x, y, z]. Since d(w) = 0 we have 0 = d(w) =^Xd(a_i1...im)y₁ⁱ¹. . . y_m^im,

and hence d(a_i1...im) = 0. From the first step of our proof, we know that

a_i1...im ∈ K[f₀, . . . , f_p] and therefore w ∈ K[f₀, . . . , f_p, y₁, . . . , y_m] = K[f₀, . . . , f_r−1].

Now we will prove that {f₀, . . . , f_r−1} is a minimal set of generators of R^d.

For this aim suppose that for some s < r there exist polynomials g₁, . . . , g_s such that R^d = K[g1, . . . , gs]. Then K[f0, . . . , fr−1] = K[g1, . . . , gs] so that there exist polynomials α₀, . . . , α_r−1 ∈ K[u₁, . . . u_s] such that

f_i = α_i(g₁, . . . , g_s)

for i = 0, 1, . . . , r − 1. Moreover there exist polynomials β₁, . . . , β_s ∈ K[v₀, . . . v_r−1] such that

g_j = β_j(f₀, . . . , f_r−1) for j = 1, . . . , s.

Denote F = (f₀, . . . , f_r−1) and G = (g₁, . . . , g_s) = (β₁(F ), . . . , β_s(F )). Then in the ring R the following identities are satisfied:

fi = αi(G) = αi(β1(F ), . . . , βs(F )),

for i = 0, 1, . . . , r − 1, that is,

F = (α ◦ β)(F ), where α = (α₀, . . . , α_r−1) and β = (β₁, . . . , β_s).

Let us introduce the notations:

ω = (0, 0, 1, 0, . . . , 0) ∈ Kⁿ, A_ik = ∂α_i

∂uk

(G), B_kq = ∂β_k

∂vq

(F ), a_ik = A_ik(ω), b_kq = B_kq(ω), for any i, q = 0, 1, . . . , r − 1 and k = 1, . . . , s.

Moreover define

Ciq =

s

X

k=1

AikBkq, ciq = Ciq(ω) =

s

X

k=1

aikbkq, where i, q = 0, 1, . . . , r − 1.

Finally let us introduce the matrices:

A = [a_ik], B = [b_kq], C = [c_iq], where 0 ≤ i ≤ r − 1, 1 ≤ k ≤ s and 0 ≤ q ≤ r − 1.

By δ_iq we denote usual Kronecker delta.

Now, using the above notations we will prove the following two lemmas.

Lemma 2 If 0 ≤ i ≤ r − 1 and p + 1 ≤ q ≤ r − 1 then c_iq = δ_iq. Proof. Differentiating the identity F = (α ◦ β)(F ) one obtains that

∂f_i

∂x_j =

r−1

X

l=0

C_il∂f_l

∂x_j (1)

for i = 0, 1, . . . , r − 1 and j = 1, . . . , n.

Let q be as in our lemma and denote

t = q − p, j₀ = t + 3.

Then t > 0, j₀ > 3 and by virtue of our choice of f₀, . . . , f_r−1 we see that f_q = y_t= x_j0, so _∂x^∂fq

j0 = 1 and, if i 6= q, then _∂x^∂fi

j0 = 0, that is,

∂f_i

∂xj0

= δ_iq. (2)

Therefore, by (1) and (2):

C_iq = C_iq · 1 = C_iq ∂f_q

∂xj0

=

r−1

X

l=0

C_il ∂f_l

∂xj0

= ∂f_i

∂xj0

= δ_iq, and consequently c_iq = C_iq(ω) = δ_iq.

Lemma 3 If 0 ≤ i ≤ p and 1 ≤ q ≤ p then c_iq = δ_iq.

Proof. Let d_x = _∂x^∂ , d_y = _∂y^∂. Given a natural number t ≥ 2 we define M_t as the ideal in R generated by all elements of the form

d^a_xd^b_y(fk), where k = 0, 1, . . . , p and 1 ≤ a + b ≤ t − 1.

It is clear that d_x(M_t) ⊆ M_t+1 and d_y(M_t) ⊆ M_t+1. Moreover every element from M_p is of the form zh, where h ∈ R is such that h(ω) = 0.

By succesive differentiations of the identity F = (α ◦ β)(F ) one easily sees (see (1)) that, for a ≥ 0, b ≥ 0, a + b > 0 and 0 ≤ k ≤ p,

d^a_xd^b_y(fk) =

r−1

X

l=0

Ckld^a_xd^b_y(fl) + Ek,a,b,

where E_k,a,b ∈ M_a+b. From the above identity and Lemma 2 one deduces that d^a_xd^b_y(f_k)(ω) =

p

X

l=0

c_kld^a_xd^b_y(f_l)(ω), (3) because every element of the ideal M_a+bvanishes at ω. Now, observe that if 0 ≤ k, q ≤ p then

d^p−q_x d^q_y(f_k) = (p − q)!q!δ_kqz. (4) Therefore, by (3) and (4), one obtains that

(p − q)!q!δ_kq = d^p−q_x d^q_y(f_k)(ω)

=

p

X

l=0

ckld^p−q_x d^q_y(fl)(ω)

= (p − q)!q!c_kq and consequently c_iq = δ_iq.

Now we can conclude the proof of our theorem. By Lemmas 1 and 2, the matrix C is invertible. Let D = C⁻¹A. Then D is an r × s matrix and I = DB, where I is the r × r identity matrix. Therefore there exist two K-linear mappings B : K^r −→ K^s, D : K^s −→ K^r such that D ◦ B = id. Then B is injective, but it is a contradiction because s < r. This proves that {f₀, . . . , f_r−1} is a minimal set of generators of R^d over K, that is, γ(d) = r.

3.Remark. In the proof we never used the assumption that {g_i}, {f_j}, {α_k} are polynomials. Note that the same proof gives the following

Proposition Let K be the field of real or of complex numbers. Let n, p be natural numbers such that n ≥ 3 and 2 ≤ p ≤ n. Denote x = (x₁, . . . , x_n) and let

f_i(x) = x^p−i₁ xⁱ₂x₃, f or 0 ≤ i ≤ p

f_i(x) = x_n−p+i, f or p + 1 ≤ i ≤ p + n − 3.

If g₁, . . . , g_s ∈ C^∞(Kⁿ) and f_i = α_i(g₁, . . . , g_s), g_j = β_j(f₀, . . . , f_p+n−3), for some functions

α_i ∈ C^∞(K^s), 0 ≤ i ≤ p + n − 3, β_j ∈ C^∞(K^p+n−2), 1 ≤ j ≤ s, then s ≥ r.

If K is the field of real numbers instead of C^∞ functions it suffices to consider functions of class C^p.

Acknowledgnents. The first version of this paper was written when the se-cond author was a guest at Department of Mathematics of Toru´n University. It was com- pleted during the stay of the first author at the Department of Mathematics of Rouen University as visiting professor. Both Departments are acknowledged for their hospi- tality and excellent working conditions.

References

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[2] P. Eakin, A note of finite dimensional subrings of polynomial rings, Proc. Amer.

Math. Soc., 31(1972), 75 – 80.

[3] M. Nagata, On the fourteenth problem of Hilbert, Proc. Intern. Congress Math.,1958, 459 – 462, Cambridge Univ. Press, New York, 1966.

[4] A. Nowicki, Rings and fields of constants for derivations in characteristic zero, to appear.

[5] A. Nowicki, M. Nagata, Rings of constants for k–derivations in k[x₁, . . . , x_n], J. Math. Kyoto Univ., 28(1988), 111 – 118.

[6] A. Zaks, Dedekind subrings of k[x₁, . . . , x_n] are rings of polynomials, Israel J. Math., 9(1971), 285 – 289.

[7] O. Zariski, Interpretations algebraico–geometriques du quatorzieme probleme de Hilbert, Bull. Sci. Math., 78(1954), 155 – 168.