Polynomial graded subalgebras of polynomial algebras

Polynomial graded subalgebras of polynomial algebras

Piotr J edrzejewicz, Andrzej Nowicki

Faculty of Mathematics and Computer Science Nicolaus Copernicus University

Toru´n, Poland

Abstract

Let k[x1, . . . , xn] be the polynomial algebra over a field k. We describe polynomial graded subalgebras of k[x1, . . . , xn], containing k[x^p₁¹, . . . , x^pnⁿ], where p₁, . . . , p_n are prime numbers.

Key Words: polynomial algebra, graded algebra, homogeneous derivation, noetherian ring, height of ideal, transcendence degree.

2000 Mathematics Subject Classification: Primary 13F20, Secondary 13A02, 13N15.

Introduction

Throughout this paper k is a field of arbitrary characteristic, n is a pos- itive integer and k[x1, . . . , xn] is a polynomial k-algebra with the natural Z-grading. By hv1, . . . , v_ni we denote the k-linear space spanned by the ele- ments v₁, . . . , v_n and by hv_i; i ∈ T i, where T is a set, we denote the k-linear space spanned by the set {v_i; i ∈ T }.

The following theorem was proved by Ganong in [3] in the case of alge- braically closed field k of characteristic p > 0, and generalized to the case of arbitrary field of characteristic p > 0 by Daigle in [2].

0Corresponding author: Piotr Jedrzejewicz, Nicolaus Copernicus University, Faculty of_, Mathematics and Computer Science, ul. Chopina 12/18, 87–100 Toru´n, Poland. E-mail:

pjedrzej@mat.uni.torun.pl.

Theorem (Ganong, Daigle). If k is a field of characteristic p > 0, A and B are polynomial k-algebras in two variables such that A^p $ B $ A, where A^p = {a^p; a ∈ A}, then there exist x, y ∈ A such that A = k[x, y] and B = k[x^p, y].

Let n be a positive integer and let p be a prime number. Consider the following statement:

”For every field k of arbitrary characteristic and arbitrary homo- geneous polynomials f₁, . . . , f_n ∈ k[x₁, . . . , x_n], such that

k[x^p₁, . . . , x^p_n] ⊆ k[f₁, . . . , f_n], the following holds:

k[f₁, . . . , f_n] =







k[x^l₁¹, . . . , x^l_nⁿ] if char k 6= p, k[y₁^l1, . . . , y_n^ln] if char k = p,

for some l₁, . . . , l_n ∈ {1, p} and some k-linear basis y₁, . . . , y_n of hx₁, . . . , x_ni.”

The above statement was proved in [5] in four particular cases:

1. p = 2, 3 and arbitrary n, 2. n = 2 and arbitrary p, 3. n = 3 and p 6 19, 4. n = 4 and p 6 7.

In this paper we prove that the above statement holds for arbitrary n and p. Moreover, we consider more general situation. Assume that

k[x^p₁¹, . . . , x^p_nⁿ] ⊆ k[f1, . . . , fn]

for some homogeneous polynomials f₁, . . . , f_n∈ k[x₁, . . . , x_n] and some prime numbers p₁, . . . , p_n. We show in Theorem 2.1 that:

a) if char k 6= pi for every i, then

k[f₁, . . . , f_n] = k[x^l₁¹, . . . , x^l_nⁿ] for some l₁, . . . , l_n such that l_i ∈ {1, p_i};

b) if char k = p_i for some i, then

k[f₁, . . . , f_n] = k[y₁^l1, . . . , y^l_nⁿ]

for some l₁, . . . , l_n such that l_i ∈ {1, p_i} and some k-linear basis y₁, . . . , y_n of hx₁, . . . , x_ni with certain properties.

If char k = p > 0, then the ring of constants of a k-derivation of the polynomial algebra k[X] = k[x1, . . . , xn] always contains the subalgebra k[x^p₁, . . . , x^p_n]. If the derivation is homogeneous (in, particular, linear) then its ring of constants is a graded subalgebra of k[X] (necessary definitions can be found in [8], [9]). So, if we are interested in rings of constants, which are polynomial algebras, in the case of homogeneous derivations and char k = p > 0 we ask about polynomial graded subalgebras of k[X], con- taining k[x^p₁, . . . , x^p_n]. Note that every polynomial graded subalgebra of k[X], containing k[x^p₁, . . . , x^p_n], is of the form k[f₁, . . . , f_n] for some homogeneous polynomials f₁, . . . , f_n (Proposition 1.1).

The general problem, when the ring of constants of a k-derivation of k[X]

is a polynomial subalgebra is known to be difficult. The second author in [7] characterized linear derivations with trivial rings of constants and trivial fields of constants, in the case of char k = 0. The first author in [4] described linear derivations, which rings of constants are polynomial algebras generated by linear forms. Now we can deduce in Theorem 4.1 that in the case of char k = p > 0, if the ring of constants of a homogeneous derivation is a polynomial k-algebra, then it is generated over k[x^p₁, . . . , x^p_n] by linear forms.

1 Preparatory facts

In this paper we are interested in polynomial graded subalgebras of the polynomial algebra k[X] = k[x1, . . . , xn], containing k[x^p₁¹, . . . , x^p_nⁿ], where p₁, . . . , p_n are prime numbers. The following proposition is an easy general- ization of Lemma 2.1 in [5], which was motivated by Lemma II.3.2 in [6].

Proposition 1.1 If B is a polynomial graded k-subalgebra of k[X] such that k[x^p₁¹, . . . , x^p_nⁿ] ⊆ B, where p₁, . . . , p_n are prime numbers, then

B = k[f₁, . . . , f_n]

for some homogeneous polynomials f₁, . . . , f_n∈ k[X].

Hence, we can consider only subalgebras of the form k[f₁, . . . , f_n], where f₁, . . . , f_nare homogeneous polynomials. Note that the transcendence degree of a subalgebra of k[X], containing k[x^p₁¹, . . . , x^p_nⁿ], equals n, so in this case the polynomials f₁, . . . , f_n are algebraically independent. Note also that the above proposition is true even if p₁, . . . , p_nare arbitrary positive integers, but we are interested only in the case when they are prime numbers.

Recall the following well known general form of Krull Theorem (see [10]

IV.14, Theorem 30, [1] (1.2.10)).

Theorem 1.2 (Krull) Let R be a noetherian commutative ring with unity, and let P be a minimal prime ideal of an ideal generated by n elements. Then the height of P is not greater than n.

Note the following useful lemma.

Lemma 1.3 Let h₁, . . . , h_s be homogeneous polynomials belonging to k[X] = k[x1, . . . , xn], of degrees r1, . . . , rs, respectively. Assume that they are alge- braically independent over k. Let w ∈ k[X] r k be a homogeneous polynomial of degree m such that w ∈ k[h₁, . . . , h_s], and let

A = n

(α1, . . . , αs) ∈ N^s; α1r1+ · · · + αsrs= m o

. Then for each α ∈ A there exists a unique element a_α ∈ k such that

w =X

α∈A

a_αh^α,

where if α = (α₁, . . . , α_s) then h^α means h^α₁¹· · · h^α_s^s. Moreover, if j ∈ {1, . . . , s}, and α_j = 0 for all α ∈ A, then w ∈ k[h₁, . . . , h_j−1, h_j+1, . . . , h_s].

Proof. Every element h^α = h^α₁¹· · · h^α_s^s, with α = (α₁, . . . , α_s) ∈ A, is a homogeneous polynomial of degree m. Since w ∈ k[h1, . . . , hs] and w is homogeneous, we have an equality of the form w = P

α∈Aa_αh^α, where each a_α belongs to k. The uniqueness follows from the assumption that h₁, . . . , h_s are algebraically independent over k. The remain part of this lemma is obvious. 

The following proposition will be useful in our proof of Theorem 2.1.

Proposition 1.4 Let f₁, . . . , f_s∈ k[x₁, . . . , x_n] be homogeneous polynomials, algebraically independent over k, of degrees r₁ 6 . . . 6 rs, respectively. Let g₁, . . . , g_s ∈ k[x₁, . . . , x_n] be homogeneous polynomials of degrees t₁ 6 . . . 6 t_s, respectively. Assume that

k[f₁, . . . , f_s] ⊆ k[g₁, . . . , g_s].

Then k[f₁, . . . , f_s] = k[g₁, . . . , g_s] if and only if r_i = t_i for all i = 1, . . . , s.

Proof. Note first that the polynomials g₁, . . . , g_s are algebraically indepen- dent over k, because tr deg_kk[g₁, . . . , g_s] = s. For each i = 1, . . . , s, consider the set A_i defined by

A_i =n

α⁽ⁱ⁾=

α⁽ⁱ⁾₁ , . . . , α⁽ⁱ⁾_s 

∈ N^s; α⁽ⁱ⁾₁ t₁+ · · · + α⁽ⁱ⁾_s t_s = r_io .

Since every f_i belongs to k[g₁, . . . , g_s], we have (by Lemma 1.3) the equalities of the form

(a) f_i = X

α⁽ⁱ⁾∈Ai

a_α(i)g^α(i) = X

α⁽ⁱ⁾∈Ai

a_α(i)g^α

(i) 1

1 · · · g_s^α(i)s , with unique elements a_α(i) belonging to k.

Part 1. Let us assume that k[f₁, . . . , f_s] = k[g₁, . . . , g_s]. For each i = 1, . . . , s, consider the set B_i defined by

B_i =n

β⁽ⁱ⁾ =

β₁⁽ⁱ⁾, . . . , β_s⁽ⁱ⁾

∈ N^s; β₁⁽ⁱ⁾r₁ + · · · + β_s⁽ⁱ⁾r_s = t_io .

Since every g_i belongs to k[f₁, . . . , f_s], we have (by Lemma 1.3) the equalities of the form

(b) g_i = X

β⁽ⁱ⁾∈Bi

b_β(i)f^β(i) = X

β⁽ⁱ⁾∈Bi

b_β(i)f^β

(i) 1

1 · · · f_s^βs(i), with unique elements b_β⁽ⁱ⁾ belonging to k.

We will show that rs = ts. Suppose that rs > ts. Then rs > ti for all i = 1, . . . , s, because t_s > ts−1 > · · · > t1. In each equality of the form

β₁⁽ⁱ⁾r₁+ · · · + β_s⁽ⁱ⁾r_s= t_i,

for every β⁽ⁱ⁾ ∈ Bi, the element βs⁽ⁱ⁾ is equal to zero. This means, by Lemma 1.3, that all the polynomials g₁, . . . , g_s belong to the ring k[f₁, . . . , f_s−1]. But it is a contradiction with transcendence degrees. Hence, r_s 6 ts. By the same way we show that ts6 r^s. Therefore, rs= ts.

We will show that r1 = t1. Suppose that r1 > t1. Then all the numbers r₁, . . . , r_s are strictly greater than t₁, and so, in each equality of the form

β₁⁽ⁱ⁾r₁+ · · · + β_s⁽ⁱ⁾r_s= t₁,

for all β⁽ⁱ⁾ ∈ B₁, the numbers β₁⁽ⁱ⁾, . . . , βs⁽ⁱ⁾ are equal to zero. This means that t₁ = 0, that is, g₁ ∈ k. But the polynomials g₁, . . . , g_s are algebraically

independent over k, so we have a contradiction. Hence, r₁ 6 t1. By the same way we show that t₁ 6 r1. Therefore, r₁ = t₁.

If s = 2, then we are done. Assume now that s > 3. Let l ∈ {2, 3, . . . , s − 1}. We will show that r_l= t_l. Suppose that r_l > t_l. Then we have

t₁ 6 t2 6 · · · 6 tl< r_l 6 rl+1 6 · · · 6 rs. In each equality of the form

β₁⁽ⁱ⁾r₁+ · · · + β_l⁽ⁱ⁾r_l+ β_l+1⁽ⁱ⁾ r_l+1+ · · · + β_s⁽ⁱ⁾ = t_i,

for i = 1, . . . , l, the numbers β_l⁽ⁱ⁾, β_l+1⁽ⁱ⁾ , . . . , βs⁽ⁱ⁾are equal to zero. This means, that in the equalities (b) for i = 1, . . . , l, there are not the polynomials f_l, f_l+1, . . . , f_s. Thus, in this case k[g₁, . . . , g_l] ⊆ k[f₁, . . . , f_l−1], but it is a contradiction with transcendence degrees. Hence, r_l 6 tl. By the same way we show that t_l 6 rl. Therefore, r_l = t_l. Therefore, we proved that if k[f₁, . . . , f_s] = k[g₁, . . . , g_s], then r_i = t_i for all i = 1, . . . , s.

Part 2. Assume that r_i = t_i for all i = 1, . . . , s. We will show that k[f₁, . . . , f_s] = k[g₁, . . . , g_s]. We arrange the proof in two steps.

Step 1. Let m be the maximal number belonging to {1, . . . , s} such that r₁ = r₂ = · · · = r_m. We will show that k[f₁, . . . , f_m] = k[g₁, . . . , g_m].

Look at the equalities (a) for i = 1, . . . , m. Observe that in these equali- ties there are not polynomials g_j with j > m. Moreover, each equality of the form α⁽ⁱ⁾₁ t₁+ · · · + α⁽ⁱ⁾s t_s = r_i, for every α⁽ⁱ⁾ ∈ A_i with i = 1, . . . , m, reduces to the equality α⁽ⁱ⁾₁ r₁+ · · · + α⁽ⁱ⁾mr₁ = r₁, that is, α⁽ⁱ⁾₁ + · · · + α⁽ⁱ⁾m = 1. Hence, the equalities (a), for i = 1, . . . , m, are of the forms











f₁ = a₁₁g₁+ a₁₂g₂+ · · · + a_1mg_m, f₂ = a₂₁g₁+ a₂₂g₂+ · · · + a_2mg_m,

...

f_m = a_m1g₁+ a_m2g₂+ · · · + a_mmg_m,

where each a_ij belongs to k. Since f₁, . . . , f_m are algebraically independent over k, the determinant of the m × m matrix [a_ij] is nonzero. This implies that k[f₁, . . . , f_m] = k[g₁, . . . , g_m].

If m = s, then we are done; we have the equality k[f₁, . . . , f_s] = k[g₁, . . . , g_s], so we have a proof of Part 2.

Now we may assume that m < s. In this case there exists u ∈ {1, . . . , s − 1} with r_u < r_u+1.

Step 2. Assume that u is a number belonging to {1, . . . , s − 1} such that r_u < r_u+1 and k[f₁, . . . , f_u] = k[g₁, . . . , g_u]. Let v be the maximal natural number satisfying the equalities r_u+1 = r_u+2 = · · · = r_u+v. We will show that k[f₁, . . . , f_u+v] = k[g₁, . . . , g_u+v].

Look at the equalities (a) for i = u + 1, . . . , u + v. Observe that in these equalities there are not polynomials g_j with j > u + v. Moreover, each equality of the form

α⁽ⁱ⁾₁ t₁+ · · · + α_s⁽ⁱ⁾t_s= r_i,

for every α⁽ⁱ⁾ ∈ A_i with i ∈ {u + 1, . . . , u + v}, reduces to the equality α⁽ⁱ⁾₁ r₁+ · · · + α⁽ⁱ⁾_u r_u+

α⁽ⁱ⁾_u+1+ · · · + α⁽ⁱ⁾_u+v

r_u+1= r_u+1 Hence, the equalities (a), for i = u + 1, . . . , u + v, are of the forms











f_u+1 = a₁₁g_u+1+ a₁₂g_u+2+ · · · + a_1vg_u+v + H₁(g₁, . . . , g_u), f_u+2 = a₂₁g_u+1+ a₂₂g_u+2+ · · · + a_2vg_u+v + H₂(g₁, . . . , g_u),

...

f_u+v = a_v1g_u+1+ a_v2g_u+2+ · · · + a_vvg_u+v + H_v(g₁, . . . , g_u), where each a_ij belongs to k, and H₁, . . . , H_v are polynomials in u variables over k. Put

h_j = a_j1g_u+1+ a_j2g_u+2+ · · · + a_jvg_u+v

for all j = 1, . . . , v. Hence, fu+j = hj + Hj(g1, . . . , gu) for j = 1, . . . , v, and hence

k[f₁, . . . , f_u+v] ⊆ k[g₁, . . . , g_u, h₁, . . . , h_v],

because k[f₁, . . . , f_u] = k[g₁, . . . , g_u]. But the polynomials f₁, . . . , f_u+v are algebraically independent over k, so it follows from the above inclusion, then the polynomials

g₁, . . . , g_u, h₁, . . . , h_v

are also algebraically independent over k. In particular, the polynomials h₁, . . . , h_v are linearly independent over k. This means that the determinant of the v × v matrix [a_ij] is nonzero, and we have the equality k[h₁, . . . , h_v] = k[g_u+1, . . . , g_u+v]. Now we have:

k[f₁, . . . , f_u+v] = k[f₁, . . . , f_u][f_u+1, . . . , f_u+v]

= k[g₁, . . . , g_u] [h₁+ H₁(g₁, . . . , g_u), . . . , h_v+ H_v(g₁, . . . , g_u)]

= k[g₁, . . . , g_u][h₁, . . . , h_v] = k[g₁, . . . , g_u][g_u+1, . . . , g_u+v]

= k[g₁, . . . , g_u+v].

Therefore, k[f₁, . . . , f_u+v] = k[g₁, . . . , g_u+v].

Now, starting from Step 1, and using one or several times Step 2, we obtain the equality k[f1, . . . , fs] = k[g1, . . . , gs]. 

2 The main theorem

Now we are ready to prove the following theorem.

Theorem 2.1 Let k be a field and let f₁, . . . , f_n ∈ k[x₁, . . . , x_n] be homoge- neous polynomials such that

k[x^p₁¹, . . . , x^p_nⁿ] ⊆ k[f₁, . . . , f_n] for some prime numbers p₁, . . . , p_n.

a) If char k 6= p_i for every i ∈ {1, . . . , n}, then k[f₁, . . . , f_n] = k[x^l₁¹, . . . , x^l_nⁿ] for some l1 ∈ {1, p1}, . . . , ln ∈ {1, pn}.

b) If char k belongs to the set {p₁, . . . , p_n}, then k[f₁, . . . , f_n] = k[y₁^l1, . . . , y^l_nⁿ]

for some l₁ ∈ {1, p₁}, . . . , l_n ∈ {1, p_n} and some k-linear basis y₁, . . . , y_n of hx₁, . . . , x_ni such that







hy_i; i ∈ T i = hx_i; i ∈ T i, y_i = x_i for i ∈ {1, . . . , n} \ T, where T = {i ∈ {1, . . . , n}; char k = p_i}.

Proof. Let r₁, . . . , r_n be the degrees of f₁, . . . , f_n, respectively. We arrange the proof in three steps.

Step 1. Assume that r₁, . . . , r_n > 2. We will show that there exists a permutation σ of the set {1, . . . , n} such that r_i = p_σ(i) for i = 1, . . . , n.

For each element q ∈ {p₁, . . . , p_n} ∪ {r₁, . . . , r_n} consider two sets:

A_q = {i ∈ {1, . . . , n}; p_i = q}, B_q = {i ∈ {1, . . . , n}; r_i = q}.

We will show that |A_q| > |Bq| for every q.

Let Q = {p₁, . . . , p_n} ∪ {r₁, . . . , r_n}, and let q ∈ Q. Put s = |A_q| and t = |B_q|; s, t ∈ {0, . . . , n}. For i ∈ {1, . . . , n} we may assume that







p_i = q if i 6 s, p_i 6= q if i > s, and we may also assume that







r_i = q if i 6 t, r_i 6= q if i > t.

If s = n or t = 0, then obviously |A_q| > |Bq|. Assume that t > 0 and s < n.

The polynomials f₁, . . . , f_n are homogeneous of degrees r₁, . . . , r_n, re- spectively, so for every i ∈ {1, . . . , n} we have

x^p_iⁱ = X

α1,...,αn>0 α1r1+...+αnrn=pi

a⁽ⁱ⁾_α f₁^α1. . . f_n^αn,

where a⁽ⁱ⁾α ∈ k, α = (α₁, . . . , α_n). Note that if t = n, then each equality α₁r₁+ . . . + α_nr_n = p_i yields that q divides p_i, what gives a contradiction for i > s. Therefore t < n.

For i = s + 1, . . . , n we can present x^p_iⁱ in the following way:

x^p_iⁱ = g_n⁽ⁱ⁾f_n+ . . . + g_t+2⁽ⁱ⁾ f_t+2+ h⁽ⁱ⁾

(if t = n − 1, we just put x^p_iⁱ = h⁽ⁱ⁾), where g_j⁽ⁱ⁾ ∈ k[f1, . . . , fj] for j = t + 2, . . . , n and h⁽ⁱ⁾ ∈ k[f₁, . . . , f_t+1],

h⁽ⁱ⁾ = X

α1,...,αt+1>0 α1r1+...+αt+1rt+1=pi

b⁽ⁱ⁾_α f₁^α1. . . f_t+1^αt+1,

where b⁽ⁱ⁾α ∈ k, α = (α1, . . . , αt+1). Since r1, . . . , rt = q, q > 1 and q 6= pi, the equality α₁r₁+ . . . + α_t+1r_t+1 = p_i implies that α_t+1 6= 0. This means that h⁽ⁱ⁾ belongs to the principal ideal generated by f_t+1, so x^p_iⁱ belongs to the ideal I = (ft+1, . . . , fn).

Let P be a minimal prime ideal of the ideal I. By Krull Theorem 1.2, since the ideal I is generated by n − t elements, the height of the ideal P is not greater than n − t. On the other side, we have x^p_iⁱ ∈ P , so x_i ∈ P for

every i ∈ {s + 1, . . . , n}. Since (x_s+1, . . . , x_n) ⊆ P , the height of the ideal P is not lesser than n − s. Therefore s > t, that is, |Aq| > |Bq|.

Now observe that the sets from the family {A_q}_q∈Q are pairwise disjoint and their union is {1, . . . , n}. The same we can tell about the family {B_q}_q∈Q. We have proved that |Aq| > |B^q| for every q ∈ Q, so in fact |Aq| = |Bq| for every q ∈ Q. Bijections from B_q to A_q for every q taken together form a required permutation σ of the set {1, . . . , n} such that r_i = p_σ(i) for i = 1, . . . , n.

Step 2. We may assume that there exists m ∈ {0, 1, . . . , n} such that for i ∈ {1, . . . , n} we have







r_i = 1 if i 6 m, r_i > 1 if i > m.

We will show that if 0 < m < n, then

k[f₁, . . . , f_n] = k[f₁, . . . , f_m, x_j^p_m+1^jm+1, . . . , x^p_jn^jn] for some j_m+1 < . . . < j_n.

Assume that 0 < m < n. Choose j_m+1, . . . , j_n ∈ {1, . . . , n}, j_m+1 < . . . <

j_n, such that the elements

f₁, . . . , f_m, x_jm+1, . . . , x_jn

form a basis of hx₁, . . . , x_ni. For a simplicity denote z_i = x_ji and q_i = p_ji for i = m + 1, . . . , n.

Consider a homomorphism of k-algebras

ϕ : k[x₁, . . . , x_n] → k[z_m+1, . . . , z_n]

such that ϕ(f_i) = 0 for i = 1, . . . , m and ϕ(z_i) = z_i for i = m + 1, . . . , n. By the assumption of the theorem we have the inclusion

k[z_m+1^qm+1, . . . , z^q_nⁿ] ⊆ k[f₁, . . . , f_n].

Applying the homomorphism ϕ we obtain

k[z_m+1^qm+1, . . . , z_n^qn] ⊆ k[g_m+1, . . . , g_n],

where g_m+1 = ϕ(f_m+1), . . . , g_n = ϕ(f_n). Note that the polynomials g_m+1, . . . , g_n are nonzero, because tr deg_kk[g_m+1, . . . , g_n] = n − m. Moreover, g_m+1, . . . , g_n are homogeneous polynomials of degrees r_m+1, . . . , r_n > 2, re- spectively, because ϕ is a homogeneous homomorphism. Then, if we assume,

for example, p₁ 6 . . . 6 pn and r₁ 6 . . . 6 rn, by Step 1 we obtain r_i = q_i for i = m + 1, . . . , n. Therefore, since

k[f₁, . . . , f_m, z_m+1^qm+1, . . . , z_n^qn] ⊆ k[f₁, . . . , f_n], by Proposition 1.4 we have

k[f₁, . . . , f_n] = k[f₁, . . . , f_m, z^q_m+1^m+1, . . . , z_n^qn].

Step 3. Now we finish the proof.

Note that in the case m = 0, from Step 1 by Proposition 1.4 we obtain k[f₁, . . . , f_n] = k[x^p₁¹, . . . , x^p_nⁿ].

Note also that if m = n, then obviously hf₁, . . . , f_ni = hx₁, . . . , x_ni, so k[f₁, . . . , f_n] = k[x₁, . . . , x_n].

Now, assume that 0 < m < n. Recall that f₁, . . . , f_m, x_jm+1, . . . , x_jn is a basis of hx₁, . . . , x_ni and j_m+1 < . . . < j_n. Arrange all elements of the set {1, . . . , n} \ {jm+1, . . . , jn} in the increasing order: j1 < . . . < jm. For a simplicity denote z_i = x_ji and q_i = p_ji for 1, . . . , m.

For i = 1, . . . , m we present z_i in the form z_i = v_i + w_i, where v_i ∈ hf1, . . . , fmi and wi ∈ hzm+1, . . . , zni. Observe that vi 6= 0 for every i ∈ {1, . . . , m}, because z_i 6∈ hz_m+1, . . . , z_ni. We may assume that there exists t ∈ {0, . . . , m} such that for i ∈ {1, . . . , m} we have







qi = char k if i 6 t, q_i 6= char k if i > t.

Let i ∈ {t + 1, . . . , m}. Put K = k[f1, . . . , fm]. Then k[x1, . . . , xn] = K[z_m+1, . . . , z_n] is a polynomial algebra over K. By Step 2 we have k[f₁, . . . , f_n] = K[z_m+1^qm+1, . . . , z_n^qn]. By the assumption of the theorem, z_i^qi ∈ k[f1, . . . , fn], so z_i^qi as a polynomial in zm+1, . . . , zn over K has zero compo- nent of degree 1. On the other hand, z_i = v_i+ w_i, where v_i ∈ K and w_i, if nonzero, is a homogeneous polynomial of degree 1 in K[z_m+1, . . . , z_n]. Thus qiv^q_iⁱ⁻¹wi is the component of degree 1 of z_i^qi = (vi+ wi)^qi. Since char k 6= qi, we have w_i = 0, so z_i = v_i and z_i ∈ hf₁, . . . , f_mi. We obtain that

hz_t+1, . . . , z_mi ⊆ hf₁, . . . , f_mi.

Now, let i₀ ∈ {1, . . . , t}, so q_i0 = char k = p. Put z_i0 = v_i0 + w_i0, v_i0 = c₁f₁+ . . . + c_mf_m, w_i0 = c_m+1z_m+1+ . . . + c_nz_n for some c₁, . . . , c_n ∈ k.

Then

z_i^p₀ = c^p₁f₁^p+ . . . + c^p_mf_m^p + c^p_m+1z_m+1^p + . . . + c^p_nz_n^p,

because char k = p. By the assumption of the theorem and the result of Step 2 we have

z_i^p

0 ∈ k[f₁, . . . , f_m, z_m+1^qm+1, . . . , z_n^qn].

However, every element of k[f₁, . . . , f_m, z^q_m+1^m+1, . . . , z^q_nⁿ] as a polynomial in variables f₁, . . . , f_m, z_m+1, . . . , z_n, is a sum of monomials, which degrees with respect to z_i are divisible by q_i for i > m. So, if q_i 6= char k for some i > m, then c_i = 0.

We may assume that there exists u ∈ {m, . . . , n} such that for i ∈ {m + 1, . . . , n} we have







q_i = p if i 6 u, q_i 6= p if i > u.

Denote

V = hv₁, . . . , v_t, z_t+1, . . . , z_mi, W = hz_m+1, . . . , z_ui.

Then w1, . . . , wt belong to W , so z1, . . . , zt belong to V + W . Note that z₁, . . . , z_m, z_m+1, . . . , z_u form a k-linear basis of V + W . Thus dim V = m, so v₁, . . . , v_t, z_t+1, . . . , z_m are linearly independent and form a basis of hf1, . . . , fmi. We obtain that

k[f₁, . . . , f_n] = k[v₁, . . . , v_t, z_t+1, . . . , z_m, z_m+1^qm+1, . . . , z_n^qn],

so it is enough to put y_ji = v_i for i = 1, . . . , t, y_ji = x_ji for i = t + 1, . . . , n, l_ji = 1 for i = 1, . . . , m and l_ji = q_i = p_ji for i = m + 1, . . . , n. 

Remark. Note that actually we have proved a stronger result than the the- sis of Theorem 2.1 b). We obtain a subalgebra B of the form k[y₁^l1, . . . , y^l_nⁿ], where we can additionally put y_i = x_i for all i such that l_i > 1. More pre- cisely, if char k = p and we assume that p₁ = · · · = p_s = p and p_s+1, . . . , p_n6=

p for some s, then the subalgebra in case b) is of the form B = k[y₁^l1, . . . , y^l_s^s, x^l_s+1^s+1, . . . , x^l_nⁿ],

where y1, . . . , ys form a basis of hx1, . . . , xsi and li ∈ {1, p} for i = 1, . . . , s, l_i ∈ {1, p_i} for i = s + 1, . . . , n. We can choose the basis y₁, . . . , y_s in such

a way that l₁ = · · · = l_t = 1 and l_t+1 = · · · = l_s = p for some t, so we can assume that

(∗) B = k[y₁, . . . , y_t, y_t+1^p , . . . , y_s^p, x^l_s+1^s+1, . . . , x^l_nⁿ].

If we take any other elements y⁰_t+1, . . . , y_s⁰ ∈ hx₁, . . . , x_si such that the elements y₁, . . . , y_t, y_t+1⁰ , . . . , y_s⁰ form a basis of hx₁, . . . , x_si, then

hy^p₁, . . . , y_t^p, y^0p_t+1, . . . , y_s^0pi = hy₁^p, . . . , y_t^p, y_t+1^p , . . . , y^p_si, so

k[y1, . . . , yt, y^0p_t+1, . . . , y_s^0p] = k[y1, . . . , yt, y^p_t+1, . . . , y_s^p].

The elements y_t+1⁰ , . . . , y⁰_scan be chosen from the set {x₁, . . . , x_s}. Then, like in the proof of Theorem 2.1, we obtain

(∗∗) B = k[y₁, . . . , y_t, x^p_j

t+1, . . . , x^p_j

s, x^l_s+1^s+1, . . . , x^l_nⁿ], where 1 6 jt+1 < · · · < j_s 6 s.

Each algebra of the form (∗) can be presented in the form (∗∗). The form (∗) is simpler, but the form (∗∗) is more precise. We see that the subalgebra B satisfying the assumptions of Theorem 2.1 is uniquely determined by a k-linear subspace

hy_i; i ∈ T, l_i = 1i ⊆ hx_i; i ∈ T i

(that is, hy1, . . . , yti in our example above) and by li ∈ {1, pi} for i ∈ {1, . . . , n} \ T , where T = {i ∈ {1, . . . , n}; char k = p_i}.

3 Some special cases

In this section we present explicitly some particular cases of Theorem 2.1.

For p₁ = . . . = p_n we obtain the following theorem.

Theorem 3.1 Let n be a positive integer and p a prime number. Let k be a field of arbitrary characteristic and let f₁, . . . , f_n ∈ k[x₁, . . . , x_n] be homoge- neous polynomials such that

k[x^p₁, . . . , x^p_n] ⊆ k[f₁, . . . , f_n].

a) If char k 6= p, then

k[f₁, . . . , f_n] = k[x^l₁¹, . . . , x^l_nⁿ]

for some l₁, . . . , l_n ∈ {1, p}.

b) If char k = p, then

k[f₁, . . . , f_n] = k[y₁, . . . , y_m, y_m+1^p , . . . , y_n^p]

for some m ∈ {0, 1, . . . , n} and some k-linear basis y1, . . . , yn of hx1, . . . , xni.

Observe that in Theorem 3.1 b), if m = 0, then k[f1, . . . , fn] = k[x^p₁, . . . , x^p_n], and if m = n, then k[f₁, . . . , f_n] = k[x₁, . . . , x_n].

Note also an important special case of char k = 0, when the hypothesis of Theorem 2.1 a) is automatically satisfied.

Theorem 3.2 Let k be a field of characteristic 0 and let f₁, . . . , f_n ∈ k[x₁, . . . , x_n] be homogeneous polynomials such that

k[x^p₁¹, . . . , x^p_nⁿ] ⊆ k[f₁, . . . , f_n] for some prime numbers p1, . . . , pn. Then

k[f₁, . . . , f_n] = k[x^l₁¹, . . . , x^l_nⁿ] for some l₁ ∈ {1, p₁}, . . . , l_n ∈ {1, p_n}.

In the case of pairwise different primes we have the same thesis for arbi- trary characteristic of k, since the k-linear space hxi; pi = char ki is at most one-dimensional.

Theorem 3.3 Let k be a field and let f₁, . . . , f_n ∈ k[x₁, . . . , x_n] be homoge- neous polynomials such that

k[x^p₁¹, . . . , x^p_nⁿ] ⊆ k[f1, . . . , fn]

for some pairwise different prime numbers p₁, . . . , p_n. Then k[f₁, . . . , f_n] = k[x^l₁¹, . . . , x^l_nⁿ]

for some l₁ ∈ {1, p₁}, . . . , l_n ∈ {1, p_n}.

4 Conclusions for homogeneous derivations

Let k[X] = k[x₁, . . . , x_n].

Recall that a k-linear map d : k[X] → k[X] such that d(f g) = d(f )g + f d(g) for every f, g ∈ k[X], is called a k-derivation. The kernel of a k- derivation d is called the ring of constants and is denoted by k[X]^d. If char k = p > 0, then

k[x^p₁, . . . , x^p_n] ⊆ k[X]^d.

If a k-derivation is homogeneous with respect to natural grading of k[X], then its ring of constants is a graded subalgebra. In this case Theorem 2.1 yields the following.

Theorem 4.1 Let d be a homogeneous k-derivation of the polynomial algebra k[X] = k[x₁, . . . , x_n], where k is a field of characteristic p > 0. Then k[X]^d is a polynomial k-algebra if and only if

k[X]^d= k[y₁, . . . , y_m, y_m+1^p , . . . , y^p_n]

for some m ∈ {0, 1, . . . , n} and some k-linear basis y₁, . . . , y_n of hx₁, . . . , x_ni.

A homogeneous k-derivation of degree 0 is called linear. The restriction of a linear derivation to the k-linear space hx₁, . . . , x_ni is a k-linear endomor- phism. Every endomorphism of hx₁, . . . , x_ni determines the unique linear k-derivation. We have the following corollary from the above theorem and Theorem 3.2 from [4].

Corollary 4.2 Let d be a linear derivation of the polynomial algebra k[X] = k[x₁, . . . , x_n], where k is a field of characteristic p > 0. Then k[X]^d is a polynomial k-algebra if and only if the Jordan matrix of the endomorphism d|hx1,...,xni satisfies the following conditions.

(1) Nonzero eigenvalues of different Jordan blocks are pairwise different and linearly independent over Fp.

(2) At most one Jordan block has dimension greater than 1 and, if such a block exists, then:

(a) its dimension is equal to 2 in the case of p > 3, (b) its dimension is equal to 2 or 3 in the case of p = 2.

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