2 The derivation ∆

Algebras of constants

for some extensions of derivations

Andrzej J. Maciejewski Andrzej Nowicki March 6, 2002

Abstract

Let A = B[t] be the polynomial ring in one variable over a unique factorization domain B, where k is a field of characteristic zero. Let δ be a derivation of B, ϕ ∈ B with δ(ϕ) 6= 0, and let s be a positive integer. Denote by d the derivation of A such that d(b) = t^sδ(b) for b ∈ B and d(t) = _s+1¹ δ(ϕ). We describe the rings of constants of d.

Several applications of our description are also given.

1 Introduction

Throughout this paper k is a field of characteristic zero. Let R be a commutative algebra over k with identity. A k-linear mapping d : R → R is called a k-derivation of R if it satisfies the Leibniz rule: d(ab) = ad(b) + bd(a) for all a, b ∈ R.

Let d be a k-derivation of R. We denote by R^d the kernel of d, that is, R^d = Ker d = {a ∈ R; d(a) = 0}.

This set is a k-subalgebra of R which we call the ring of constants of d. If R is a domain and k is a field, then we denote by R₀ the field of quotients of R and we denote also by d the unique extension of d to R₀. In this case R^d₀ is a subfield of R₀ containing k.

Assume that R = k[X] := k[x₁, . . . , x_n] is the polynomial ring over k in n ≥ 1 variables. We know a description of all k-derivations of R. If d is a k- derivation of k[X], then we have the polynomials f₁ := d(x₁), . . . , f_n := d(x_n), belonging to k[X], and then

d = f₁ ∂

∂x₁ + · · · + f_n ∂

∂x_n.

Every k-derivation d of k[X] is uniquely determined by a sequence (f₁, . . . , f_n) of polynomials from k[X].

Let d be a k-derivation of k[X] and consider the ring of constants k[X]^d. In this case the ring k[X]^dcoincides with the k-algebra of all polynomial first integrals of the following system of ordinary differential equations:

dx_i

dt = f_i(x₁(t), . . . , x_n(t)), i = 1, . . . , n,

where f₁ = d(x₁), . . . , f_n = d(x_n). It is known ([17], [15]) that if n ≤ 3, then k[X]^dis finitely generated over k. If n ≥ 5, then there exists a k-derivation of k[X] for which the ring of constants is not finitely generated over k ([2], [1]).

For n = 4 the problem is open. Every example of k-derivation of k[X] with nonfinitely generated ring of constants is a counterexample to the famous fourteenth problem of Hilbert (see [12], [14], [3], [6]). We present two new such examples for n = 6 (see Examples 5.9 and 5.10).

The problem of a description of the ring k[X]^d is known to be difficult even for n = 2. The importance of the study of these rings comes from the fact that many important mathematical problems can be transformed into problems concerning kernels of certain derivations of k[X] (see for example [3], [14]). In this paper we describe the rings of constants of some classes of derivations of the polynomial ring k[X].

Let A = B[t] be the polynomial algebra in one variable t over a unique factorization k-domain B. We will consider the following four derivations δ,

∆, D and d.

The derivation δ : B → B is an arbitrary nonzero derivation of B.

The next derivation ∆ : A → A is the unique derivation such that ∆(t) = 0 and ∆(b) = δ(b) for b ∈ B..

Let ϕ ∈ B and δ(ϕ) 6= 0. The derivation D : A → A is defined as:

D(t) = δ(ϕ) and D(b) = δ(b) for b ∈ B.

If s is a positive integer, then d : A → A is the unique derivation such that

d(t) = 1

s + 1δ(ϕ) and d(b) = t^sδ(b) for all b ∈ B.

Throughout this article all algebras are commutative k-algebras with identity and all derivations are k-derivations.

The aim of this note is to describe the algebras of constants of the above derivations. The main result is Theorem 4.5 which states that A^d_o = B_o^δ(h) and A^d= B^δ[h], where h = t^s+1− ϕ.

In Section 5 we present several applications of this theorem. Further ap- plications we present in Section 6, where we discuss the algebras of constants

of the derivation d_b = y^sz_∂x^∂ − x^sz_∂y^∂ + (by^sz − xy^s)_∂z^∂, where s is a positive integer and b ∈ k. If s = 2 then the derivations of the form d_b are asso- ciated with systems of differential equations which appear in descriptions of 3-dimensional homogeneous sub-Riemannian spaces which are Lie groups (see [4] for details). Some investigations concerning d_b one may find also in [7] and [9].

2 The derivation ∆

It is obvious that A^∆ is equal to B^δ[t]. In this case we do not need to assume that B is a UFD. However, this assumption is important in our proof of the following proposition.

Proposition 2.1. A^∆_o = B_o^δ(t).

Proof. Let F ∈ A^∆₀ . Put F = P/Q, where P and Q are nonzero relatively prime polynomials from A = B[t]. Let

P = p_ntⁿ+ · · · + p₁t + p₀, Q = q_mt^m+ · · · + q₁t + q₀, with p₀, . . . , p_n, q₀, . . . , q_m ∈ B and p_n6= 0, q_m 6= 0. Then

∆(P ) = δ(p_n)tⁿ+ · · · + δ(p₁)t + δ(p₀), ∆(Q) = δ(q_m)t^m+ · · · + δ(q₁)t + δ(q₀).

Since F ∈ A^∆₀, we have

(i) P ∆(Q) = ∆(P )Q.

Suppose that ∆(P ) = 0. Then ∆(Q) = 0. Hence δ(p_n) = · · · = δ(p₁) = δ(p0) = 0 and δ(qm) = · · · = δ(q1) = δ(q0) = 0, and hence P, Q ∈ B^δ[t], so F ∈ B_o^δ(t). If ∆(Q) = 0, then ∆(P ) = 0 and again F ∈ B_o^δ(t).

Now we may assume that ∆(P ) 6= 0 and ∆(Q) 6= 0.

Since A is a unique factorization domain and the polynomials P and Q are relatively prime, it follows from the equality (i) that

∆(P ) = M P and ∆(Q) = M Q

for some nonzero M ∈ A. Comparing the degrees with respect to t we deduce that M ∈ B. Thus,

δ(p_i) = M p_i, δ(q_j) = M q_j,

for all i = 0, 1, . . . , n and j = 0, 1, . . . , m. In particular, all the fractions

pn

qm, ^p_qⁿ⁻¹

m , . . . ,_q^p0

m and ^q_p^m

n, ^qm−1_p

n , . . . ,_p^q0

n belong to B_o^δ. Hence p_n = a_nq_m,

p_n−1 = a_n−1q_m, . . . , p₀ = a₀q_m and q_m = b_mp_n, q_m−1 = b_m−1p_n, . . . , q₀ = b₀p_n, for some a₀, . . . , a_n, b₀, . . . , b_m ∈ B^δ_o. This means that P = q_mP₁ and Q = p_nQ₁, where P₁, Q₁ ∈ B_o^δ[t] r {0}. Thus we have

F = _Q^P = ^q_p^m

n

P1

Q1 = bmP1

Q1 ∈ B_o^δ(t).

Therefore, A^∆_o ⊆ B_o^δ(t). The opposite inclusion is obvious. 

3 The derivation D

Proposition 3.1. A^D = B^δ[ω], A^D_o = B^δ_o(ω), where ω = t − ϕ.

Proof. Consider the automorphism σ : A → A such that σ(b) = b for all b ∈ B, and σ(t) = ω = t − ϕ. Then D = σ∆σ⁻¹ and we have

A^D = σ A^∆
= σ B^δ[t]
= B^δ[σ(t)] = B^δ[ω].

Similarly, by Proposition 2.1, A^D_o = B_o^δ(ω). 

In the next section we will use the following lemma.

Lemma 3.2. If f ∈ A_o satisfies the equality tD(f ) = λδ(ϕ)f for some λ ∈ Q r Z, then f = 0.

Proof. Suppose that f 6= 0 and put f = ^u_v, where u and v are nonzero relatively prime polynomials from A = B[t]. Since tD(^u_v) = λδ(ϕ)^u_v, we have t_v¹2(D(u)v − uD(v)) = λδ(ϕ)^u_v and so

(1) t(D(u)v − uD(v)) = λδ(ϕ)uv.

This implies that t divides u or v.

Assume that t | u. Let u = t^mu1, where m ≥ 1, u1 ∈ A, t - u¹. Then, by (1), we get t(t^mD(u₁)v + mt^m−1u₁δ(ϕ)v − t^mu₁D(v)) = λδ(ϕ)t^mu₁v, and hence

t(D(u1)v − u1D(v)) = (λ − m)δ(ϕ)u1v.

This implies, that t | v (since t - u¹ and λ − m 6= 0), but it is impossible because t | u and the polynomials u and v are relatively prime.

A similar argument we use when t | v. In this case we put v = t^mv₁, where m ≥ 1, v1 ∈ A, t - v¹ and, by (1), we obtain

t(D(u)v₁− uD(v₁)) = (λ + m)δ(ϕ)uv₁.

This implies that t | u and again we have a contradiction with the assumption that u and v are relatively prime. 

4 The derivation d

In this section we denote by p the polynomial t^s+1.

Since t is an algebraic element over the field B_o(p), A_o is an algebraic extension of B_o(p) of the degree s + 1. This implies that every element g ∈ Ao has a unique presentation of the form

(∗) g = g_s(p)t^s+ g_s−1(p)t^s−1+ · · · + g₁(p)t + g₀(p),

where g₀, g₁, . . . , g_s∈ A_o. It is clear that if g ∈ A, then g₀, g₁, . . . , g_s ∈ A.

Lemma 4.1. Let g ∈ A_o and g₀, g₁, . . . , g_s ∈ A_o as in (∗). If g ∈ A^d_o, then g_i(p)tⁱ ∈ A^d_o

for every i = 0, 1, . . . , s.

Proof. Replacing A and d by A ⊗_kk and d ⊗_k1, respectively, where k is an algebraic closure of k, we may assume that the field k is algebraically closed.

Denote by ε a primitive root of unity of degree s + 1 and consider the automorphism τ : A_o → A_o such that

τ (t) = εt and ∀_b∈Bo τ (b) = b.

Then τ dτ⁻¹ = ε^sd. In fact, if b ∈ B_o, then τ dτ⁻¹(b) = τ d(b) = τ (t^sδ(ϕ)) = (εt)^sδ(ϕ) = ε^st^sδ(ϕ) = ε^sd(b). Moreover, τ dτ⁻¹(t) = τ d(ε^st) = ε^sτ (d(t)) = ε^sd(t).

Let g = g(t) ∈ A^d_o. The equality τ dτ⁻¹ = ε^sd implies that g(εt) = τ (g(t)) ∈ A^d_o, and consequently g(εⁱt) = τⁱ(g(t)) ∈ A^d_o for any integer i. This means, in particular, that every element of the form

(ε^s)ⁱg_s(p)t^s+ (ε^s−1)ⁱg_s−1(p)t^s−1+ · · · + (ε¹)ⁱg₁(p)t + g₀(p), i ∈ {0, 1, . . . , s}, belongs of A^d_o. Now, using a determinant of Vandermonde we deduce that g_i(p)tⁱ ∈ A^d_o for i = 0, 1, . . . , s. 

Lemma 4.2. If f ∈ A_o, then d(f (p)) = D(f )(p)t^s.

Proof. It is sufficient to show that the equality d(f (p)) = D(f )(p)t^s holds when f = btⁿ with b ∈ B_o and n ≥ 0. In this case we have:

d(f (p)) = d (b(t^s+1)ⁿ)

= d(b)(t^s+1)ⁿ+ (s + 1)nbd(t)t^(s+1)n−1

= t^sδ(b)(t^s+1)ⁿ+ (s + 1)nb_s+1¹ δ(ϕ)t^(s+1)n−1

= (δ(b)tⁿ) (p)t^s+ (nbtⁿ⁻¹δ(ϕ)) (p)t^s

= (D(b)tⁿ+ bD(tⁿ)) (p)t^s

= D(btⁿ)(p)t^s

= D(f )(p)t^s. This completes the proof. 

Lemma 4.3. Let f ∈ A_o and 1 ≤ r ≤ s. If f (p)t^r ∈ A^d_o, then f = 0.

Proof. Assume that d(f (p)t^r) = 0. Then, by Lemma 4.2, we have 0 = d(f (p))t^r+ rt^r−1f (p)d(t)

= D(f )(p)t^st^r+ rt^r−1f (p)_s+1¹ δ(ϕ)

= t^r−1· D(f )t + _s+1^r δ(ϕ)f
(p).

This implies that D(f )t + _s+1^r δ(ϕ)f
(p) = 0, and hence tD(f ) = λδ(ϕ)f

for λ = −_s+1^r . Since 1 ≤ r ≤ s, the coefficient λ belongs to Q r Z. Therefore, by Lemma 3.2, f = 0. 

Using the above lemmas we may prove the following proposition.

Proposition 4.4. If g ∈ A^d_o, then g = f (p) for some f ∈ A^D_o. If g ∈ A^d, then g = f (p) for some f ∈ A^D.

Proof. Assume that g ∈ A^d_o and let g₀, . . . , g_s ∈ A_o be such as in (∗).

Put f = g₀. Then, by Lemmas 4.1 and 4.3, g = f (p). Moreover, Lemma 4.2 implies that f ∈ A^D_o. It is clear that if g ∈ A, then f ∈ A. 

Now we are ready to prove the following theorem which is the main result of this article.

Theorem 4.5. Let A = B[t] be a polynomial algebra in a single variable t over a unique factorization k-domain B, where k is a field of characteristic zero. Let δ : B → B be a nonzero derivation and let ϕ be an element of B such that δ(ϕ) 6= 0. Moreover, let s be a positive integer. Consider the derivation d : A → A defined by

d(t) = 1

s + 1δ(ϕ) and ∀_b∈B d(b) = t^sδ(b).

Then A^d_o = B_o^δ(h) and A^d = B^δ[h], where h = t^s+1− ϕ.

Proof. First observe that h = ω(p) where, as in Proposition 3.1, ω = t − ϕ.

Now let g ∈ A^d. Then, by Proposition 4.4, g = f (p) for some f ∈ A^D. We know, by Proposition 3.1, that A^D = B^δ[ω]. Hence f = γ(ω) for some γ ∈ B^δ[t], and hence g = f (p) = γ(ω(p)) = γ(h) ∈ B^δ[h]. Thus, A^d⊆ B^δ[h].

Since the opposite inclusion is obvious, we have A^d= B^δ[h].

Repeating the same arguments, thanks to Propositions 3.1 and 4.4, we obtain the equality A^d_o = B_o^δ(h). 

5 Examples

Example 5.1. Let d be the derivation of k[x, y, z] defined by







d(x) = yz d(y) = zx d(z) = xy.

Then k[x, y, z]^d = k[x²− y², z²− x²] and k(x, y, z)^d= k(x²− y², z²− x²).

Proof. Consider the derivation δ = y_∂x^∂ + x_∂y^∂ of the polynomial algebra B = k[x, y] and let ϕ = x². It is easy to check that k[x, y]^δ = k[x²− y²] and k(x, y)^δ = k(x²− y²). Moreover, ¹₂δ(ϕ) = xy = d(z). Using Theorem 4.5 for B, δ, ϕ as above and t = z, A = k[x, y, z] and s = 1, we obtain:

k[x, y, z]^d= k[x, y]^δ[z²− ϕ] = k[x²− y²][z²− x²] = k[x²− y², z²− x²].

The same we do for k(x, y, z)^d. 

The next example is a generalization of Example 5.1.

Example 5.2. Let d_n be the derivation of k[X] = k[x₁, . . . , x_n], (n ≥ 2) defined by















d_n(x₁) = x₂x₃· · · x_n dn(x2) = x1x3· · · xn

...

d_n(x_n) = x₁x₂· · · x_n−1.

Then k[X]^dn = k[h₂, h₃, . . . , h_n] and k(X)^dn = k(h₂, h₃, . . . , h_n), where hi = x²₁− x²_i, i = 2, 3, . . . , n. 

Proof. We use an induction with respect to n. This is clear for n = 2. Now assume that it is true for n − 1. Let B = k[x₁, . . . , x_n−1], δ = d_n−1, t = x_n, A = k[x₁, . . . , x_n], s = 1 and ϕ = x²₁. Then A = B[t], t²− ϕ = x²_n− x²₁ = −hn and

1

1+sδ(ϕ) = ¹₂2x₁d_n−1(x₁) = x₁x₂. . . x_n−1= d_n(x_n) = d_n(t).

Hence, by induction and Theorem 4.5, we get

k[X]^dn = A^dn = B^δ[t²− ϕ] = k[h₂, . . . , h_n−1][h_n] = k[h₂, h₃, . . . , h_n].

The same we do for k(X)^dn.  By a similar way we obtain:

Example 5.3. Let d_n be the derivation of k[X] = k[x₁, . . . , x_n], (n ≥ 2) defined by



















dn(x1) = +x2x3· · · xn

d_n(x₂) = −x₁x₃· · · x_n d_n(x₃) = −x₁x₂· · · x_n

...

d_n(x_n) = −x₁x₂· · · x_n−1.

Then k[X]^dn = k[h₂, h₃, . . . , h_n] and k(X)^dn = k(h₂, h₃, . . . , h_n), where h_i = x²₁+ x²_i, i = 2, 3, . . . , n. 

Using Theorem 4.5 it is easy to prove the following examples which de- scribe the algebras of constants of a derivation d of a polynomial algebra over k. In Examples 5.4 and 5.5 the mapping d is a derivation of k[x, y, z].

Example 5.4. If d = xy_∂x^∂ + zy_∂y^∂ + (x² + y²)_∂z^∂ , then k[x, y, z]^d = k[f ], k(x, y, z)^d= k(^x_y, f ), where f = z²− x²− y². 

Example 5.5. Let s be a positive integer. If d = z^sy_∂x^∂ + z^sx_∂y^∂ + x^sy_∂z^∂ , then k[x, y, z]^d= k[f, g], k(x, y, z)^d = k(f, g), where f = z^s+1− x^s+1, g = x²− y².



In the next examples d is a derivation of the polynomial algebra k[x, y, z, t].

Example 5.6 is connected with a diagonal derivation described in [16].

Example 5.6. Let d = tx_∂x^∂ + ty_∂y^∂ + −ntz_∂x^∂ + (x²+ y²− nz²)_∂t^∂, where n ≥ 1. Then k[x, y, z, t]^d = k[f, xⁿz, xⁿ⁻¹yz, xⁿ⁻²y²z, . . . , yⁿz], k(x, y, z, t)^d = k(x/y, x/z, f ), with f = t²− x²− y²− z². 

The next two examples concern a Jouanolou derivation (see [8], [11], [10]

or [14]).

Example 5.7. Let d = ty^{2 ∂}_∂x+ tz^{2 ∂}_∂y+ +tx^{2 ∂}_∂z + (xy²+ yz²+ zx²)_∂t^∂. Then k[x, y, z, t]^d= k[f ], k(x, y, z, t)^d= k(f ), with f = t²− x²− y²− z².  Example 5.8. If d = t²y^{2 ∂}_∂x + t²z^{2 ∂}_∂y + +t²x^{2 ∂}_∂z + (x²y² + y²z² + z²x²)_∂t^∂, then k[x, y, z, t]^d = k[f ], k(x, y, z, t)^d= k(f ), where f = t³− x³− y³− z³.  Note also two examples of derivations of k[x₁, . . . , x₆] with nonfinitely generated ring of constants.

Example 5.9. Let d be the derivation of k[X] := k[x₁, x₂, . . . , x₆] defined by d = x²₁x₆ ∂

∂x₃ + (x₁x₃ + x₄)x₆ ∂

∂x₄ + x₄x₆ ∂

∂x₅ + x²₁ ∂

∂x₆. Then k[X]^d is not finitely generated over k.

Proof. Consider the derivation δ = x²₁ ∂

∂x₃ + (x₁x₃+ x₄) ∂

∂x₄ + x₄ ∂

∂x₅

of the polynomial ring B = k[x₁, . . . , x₅] and let ϕ = 2x₃, t = x₆ and s = 1. Then k[X] = B[t] and, by Theorem 4.5, k[X]^d = B^δ[x6 − 2x3].

Daigle and Freudenburg ([1], [5]) proved that B^δ is not finitely generated over k. This implies that k[X]^d is also not finitely generated over k, because k[X]^d/(x6− 2x3) ≈ B^δ. 

Example 5.10. Let d be the derivation of k[X] := k[x₁, x₂, . . . , x₆] defined by

d = x²₁x²₆ ∂

∂x₃ + (x₁x₃+ x₄)x²₆ ∂

∂x₄ + x₄x²₆ ∂

∂x₅ + x₄x²₅ ∂

∂x₆. Then k[X]^d is not finitely generated over k.

Proof. We repeat the proof of Example 5.9 with ϕ = x³₅ and s = 2. 

6 Constants of the derivation d

Let s ≥ 1 be a fixed integer. If b ∈ k, then we denote by d_b the derivation of k[x, y, z] defined by







d_b(x) = y^sz d_b(y) = −x^sz

d_b(z) = −xy^s+ by^sz.

For any b ∈ k, the derivation d_b has a nontrivial constant polynomial, namely db(h) = 0 where

h = x^s+1+ y^s+1.

Thus, k[h] ⊆ k[x, y, z]^d and k(h) ⊆ k(x, y, z)^d. We will characterize all the cases in which k[x, y, z]^d= k[h] and all the cases in which k(x, y, z)^d= k(h).

For this aim, we consider the following derivation δ_b : k[x, z] → k[x, z] defined by

 δ_b(x) = z

δ_b(z) = −x + bz.

The derivation δ_b is linear with matrix M =

 0 1

−1 b

 .

Let α, β ∈ k be the eigenvalues of M . Then α and β are roots of the polynomial x² − xb + 1. It is known (see for example [13] or [14]) that k[x, z]^d6= k if and only if the eigenvalues α and β are linearly dependent over N.

We know also (see [13] or [14]) that k(x, z)^d 6= k if and only if the eigen- values α and β are linearly dependent over Z and the Jordan matrix of M is not of the form

 c 1 0 c



for some nonzero c ∈ k (in our case this means that b 6= ±2).

Lemma 6.1. The following conditions are equivalent.

(1) The eigenvalues α and β are linearly dependent over N.

(2) There exist positive integers n, m such that −b² = ^(m−n)_mn ².

Proof. It is clear if b = 0. Let us assume that b 6= 0. Observe that α + β = b and αβ = 1. In particular, α 6= 0 and β 6= 0.

(1) ⇒ (2). Assume that nα + mβ = 0 for some n, m ∈ N with (n, m) 6=

(0, 0). Then n 6= 0, m 6= 0 and n 6= m (because b 6= 0). Moreover, 0 = nα + m(b − α) and 0 = n(b − β) + mβ. Hence α = _m−n^m b, β = _n−mⁿ b and we have 1 = αβ = _m−n^m b · _n−mⁿ b = −_(m−n)^nm 2, that is, −b² = ^(m−n)_mn ².

(2) ⇒ (1). Assume that −b² = ^(m−n)_mn ², where m, n ∈ N, m > 0, n > 0.

Since b 6= 0, we have n 6= m. Put α⁰ = _m−n^m b, β⁰ = _n−mⁿ b. Then α⁰+ β⁰ = b, α⁰β⁰ = 1, hence α⁰ and β⁰ are the roots of x²−bx+1, that is, {α⁰, β⁰} = {α, β}.

Now the linear dependence over N of α and β follows from the equality nα⁰+ mβ⁰ = 0. 

Repeating the above proof we get

Lemma 6.2. The following conditions are equivalent.

(1) The eigenvalues α and β are linearly dependent over Z.

(2) There exist nonzero integers n, m such that −b² = ^(m−n)_mn ². 

Now, using the above facts and Theorem 4.5 (for B = k[x, z], t = y, A = k[x, y, z], δ = δ_b, d = d_b and ϕ = −x^s+1), we get the following proposition.

Proposition 6.3. (1) k[x, y, z]^db 6= k[h] if and only if there exist positive integers n, m such that −b² = ^(m−n)_mn ².

(2) k(x, y, z)^db 6= k(h) if and only if b 6= ±2 and there exist nonzero integers n, m such that −b² = ^(m−n)_mn ². 

Using the remarks given before Lemma 6.1 it is not difficult to present a description of the algebras of constants of the derivation δ_b. Therefore, as a consequence of Theorem 4.5 we get

Proposition 6.4. (0) If b = 0, then k[x, y, z]^d = k[h, f ] and k(x, y, z)^d = k(h, f ) where f = x^s+1+ z^s+1.

(1) If b 6= 0 and k[x, y, z]^db 6= k[h], then there exist relatively prime positive integers m, n such that b² = −^(m−n)_mn ², and then k[x, y, z]^db = k[h, f ] and k(x, y, z)^db = k(h, f ), where

f = (ax + z)ⁿ(x + az)^m, with a = _m−nⁿ b.

(2) If b 6= 0 and k(x, y, z)^db 6= k(h), then there exist nonzero relatively prime integers m, n such that b² = −^(m−n)_mn ², and then k(x, y, z)^db = k(h, F ), where

F = (ax + z)ⁿ(x + az)^m, with a = _m−nⁿ b. 

Corollary 6.5. If b ∈ R, then C[x, y, z]^db = C[h], where R and C are fields of real and complex numbers, respectively.

Proof. Suppose that C[x, y, z]^db 6= C[h]. Then, by Proposition 6.3, there exist positive integers m, n such that b² = −^(n−m)_nm ² and then b² < 0.  Corollary 6.6. If 0 6= b ∈ Q, then k[x, y, z]^d = k[h] and the following conditions are equivalent

(1) k(x, y, z)^db 6= k(h).

(2) There exist relatively prime positive integers p and q such that (p, q) 6=

(1, 1) and b = ±^p2_pq^+q2.

If p, q are such as in (2), then k(x, y, z)^db = k(h, F ) where F is the rational function defined as follows

F =







(px − qz)^p2(qx − pz)^−q2, if b = ^p2_pq^+q2, (px + qz)^p2(qx + pz)^−q2, if b = −^p2_pq^+q2.

Proof. For a proof that k[x, y, z]^db = k[h] we use the same argument as in the proof of Corollary 6.5.

(2) ⇒ (1). If b = ±^p2_pq^+q2, p, q ∈ N with (p, q) 6= (1, 1), then b 6= ±2 and for n = −p², m = q² we have: −b² = −^(p2_p^+q2q²²⁾² = ^(m−n)_mn ². Hence, by Proposition 6.3, k(x, y, z)^db 6= k(h).

(1) ⇒ (2) The condition (1) implies, by Proposition 6.3, that b 6= ±2 and −b² = ^(m−n)_mn ² for some nonzero integers n, m. We may assume that n i m are relatively prime and that n = −n1 < 0 and m > 0. Let b = ±^u_v, where u and v are relatively prime positive integers. Then ^u_v2² = ^(m+n_mn¹⁾²

1 , and hence u = m + n₁ and v² = mn₁. Since gcd(m, n₁) = 1, there exists relatively prime positive integers p and q such that m = p² and n₁ = q². Thus, b = ±^u_v = ±^m+n_pq ¹ = ±^p2_pq^+q2.

The remaining part of this corollary is a consequence of Proposition 6.4.



Example 6.7. If b = ⁵₂, then k[x, y, z]^db = k[h] and k(x, y, z)^db = k(h, F ), where

F = ^(2x−z)_x−2z⁴. 

Corollary 6.8. If 0 6= b ∈ Z, then k(x, y, z)^db = k(h).

Proof. This is a consequence of Corollary 6.6, because if p and q are relatively prime positive numbers and (p, q) 6= (1, 1), then the number

p²+q²

pq = ^p_q + ^q_p is not integer.  Example 6.9. Let k = C and b =

√2

2 iyz. Then k[x, y, z]^db = k[h, f ] and k(x, y, z)^db = k(h, f ), where

f = 2x³+ 3z²x +√

2iz³. 

Acknowledgement.

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Author’s addresses

Andrzej J. Maciejewski, Institute of Astronomy, University of Zielona G´ora, Lubuska 2, 65-265 Zielona G´ora, Poland; e-mail: maciejka@astri.uni.torun.pl Andrzej Nowicki, Faculty of Mathematics and Computer Science, N. Coper- nicus University, 87–100 Toru´n, Poland, e-mail: anow@mat.uni.torun.pl