Polynomial Graded Subalgebras of Polynomial Algebras

Polynomial graded subalgebrasof polynomial algebras

Piotr J ↪edrzejewicz, Andrzej NowickiFaculty of Mathematics and Computer Science

Nicolaus Copernicus University

Torun, Poland

Abstract

Let k[x1, . . . , xn] be the polynomial algebra over a field k. Wedescribe polynomial graded subalgebras of k[x1, . . . , xn], containingk[xp11 , . . . , xpnn ], where p1, . . . , pn 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 naturalZ-grading. By 〈v1, . . . , vn〉 we denote the k-linear space spanned by the ele-ments v1, . . . , vn and by 〈vi; i ∈ T 〉, where T is a set, we denote the k-linearspace spanned by the set {vi; 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 ofarbitrary field of characteristic p > 0 by Daigle in [2].

0Corresponding author: Piotr J ↪edrzejewicz, Nicolaus Copernicus University, Faculty ofMathematics and Computer Science, ul. Chopina 12/18, 87–100 Torun, Poland. E-mail:[email protected].

1

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

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

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

k[xp1, . . . , xpn] ⊆ k[f1, . . . , fn],

the following holds:

k[f1, . . . , fn] =

k[xl11 , . . . , xlnn ] if char k 6= p,

k[yl11 , . . . , ylnn ] if char k = p,

for some l1, . . . , ln ∈ {1, p} and some k-linear basis y1, . . . , yn of〈x1, . . . , xn〉.”

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 nand p. Moreover, we consider more general situation. Assume that

k[xp11 , . . . , xpnn ] ⊆ k[f1, . . . , fn]

for some homogeneous polynomials f1, . . . , fn ∈ k[x1, . . . , xn] and some primenumbers p1, . . . , pn. We show in Theorem 2.1 that:

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

k[f1, . . . , fn] = k[xl11 , . . . , xlnn ]

for some l1, . . . , ln such that li ∈ {1, pi};b) if char k = pi for some i, then

k[f1, . . . , fn] = k[yl11 , . . . , ylnn ]

2

for some l1, . . . , ln such that li ∈ {1, pi} and some k-linear basis y1, . . . , yn of〈x1, . . . , xn〉 with certain properties.

If char k = p > 0, then the ring of constants of a k-derivation of thepolynomial algebra k[X] = k[x1, . . . , xn] always contains the subalgebrak[xp1, . . . , x

pn]. If the derivation is homogeneous (in, particular, linear) then

its ring of constants is a graded subalgebra of k[X] (necessary definitionscan be found in [8], [9]). So, if we are interested in rings of constants,which are polynomial algebras, in the case of homogeneous derivations andchar k = p > 0 we ask about polynomial graded subalgebras of k[X], con-taining k[xp1, . . . , x

pn]. Note that every polynomial graded subalgebra of k[X],

containing k[xp1, . . . , xpn], is of the form k[f1, . . . , fn] for some homogeneous

polynomials f1, . . . , fn (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 trivialfields of constants, in the case of char k = 0. The first author in [4] describedlinear derivations, which rings of constants are polynomial algebras generatedby linear forms. Now we can deduce in Theorem 4.1 that in the case ofchar k = p > 0, if the ring of constants of a homogeneous derivation is apolynomial k-algebra, then it is generated over k[xp1, . . . , x

pn] by linear forms.

1 Preparatory facts

In this paper we are interested in polynomial graded subalgebras of thepolynomial algebra k[X] = k[x1, . . . , xn], containing k[xp11 , . . . , x

pnn ], where

p1, . . . , pn 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 thatk[xp11 , . . . , x

pnn ] ⊆ B, where p1, . . . , pn are prime numbers, then

B = k[f1, . . . , fn]

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

Hence, we can consider only subalgebras of the form k[f1, . . . , fn], wheref1, . . . , fn are homogeneous polynomials. Note that the transcendence degreeof a subalgebra of k[X], containing k[xp11 , . . . , x

pnn ], equals n, so in this case

the polynomials f1, . . . , fn are algebraically independent. Note also that theabove proposition is true even if p1, . . . , pn are arbitrary positive integers, butwe are interested only in the case when they are prime numbers.

3

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. Thenthe height of P is not greater than n.

Note the following useful lemma.

Lemma 1.3 Let h1, . . . , hs 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]rk be a homogeneous polynomialof degree m such that w ∈ k[h1, . . . , hs], and let

A ={

(α1, . . . , αs) ∈ Ns; α1r1 + · · ·+ αsrs = m}.

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

w =∑α∈A

aαhα,

where if α = (α1, . . . , αs) then hα means hα11 · · ·hαs

s . Moreover, if j ∈{1, . . . , s}, and αj = 0 for all α ∈ A, then w ∈ k[h1, . . . , hj−1, hj+1, . . . , hs].

Proof. Every element hα = hα11 · · ·hαs

s , with α = (α1, . . . , αs) ∈ A, is ahomogeneous polynomial of degree m. Since w ∈ k[h1, . . . , hs] and w ishomogeneous, we have an equality of the form w =

∑α∈A aαh

α, where eachaα belongs to k. The uniqueness follows from the assumption that h1, . . . , hsare algebraically independent over k. The remain part of this lemma isobvious. �

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

Proposition 1.4 Let f1, . . . , fs ∈ k[x1, . . . , xn] be homogeneous polynomials,algebraically independent over k, of degrees r1 6 . . . 6 rs, respectively. Letg1, . . . , gs ∈ k[x1, . . . , xn] be homogeneous polynomials of degrees t1 6 . . . 6ts, respectively. Assume that

k[f1, . . . , fs] ⊆ k[g1, . . . , gs].

Then k[f1, . . . , fs] = k[g1, . . . , gs] if and only if ri = ti for all i = 1, . . . , s.

4

Proof. Note first that the polynomials g1, . . . , gs are algebraically indepen-dent over k, because tr degk k[g1, . . . , gs] = s. For each i = 1, . . . , s, considerthe set Ai defined by

Ai ={α(i) =

(α(i)1 , . . . , α

(i)s

)∈ Ns; α

(i)1 t1 + · · ·+ α(i)

s ts = ri

}.

Since every fi belongs to k[g1, . . . , gs], we have (by Lemma 1.3) the equalitiesof the form

(a) fi =∑

α(i)∈Ai

aα(i)gα(i)

=∑

α(i)∈Ai

aα(i)gα(i)1

1 · · · gα(i)s

s ,

with unique elements aα(i) belonging to k.

Part 1. Let us assume that k[f1, . . . , fs] = k[g1, . . . , gs]. For each i =1, . . . , s, consider the set Bi defined by

Bi ={β(i) =

(β(i)1 , . . . , β(i)

s

)∈ Ns; β

(i)1 r1 + · · ·+ β(i)

s rs = ti

}.

Since every gi belongs to k[f1, . . . , fs], we have (by Lemma 1.3) the equalitiesof the form

(b) gi =∑

β(i)∈Bi

bβ(i)fβ(i)

=∑

β(i)∈Bi

bβ(i)fβ(i)1

1 · · · fβ(i)s

s ,

with unique elements bβ(i) belonging to k.

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

β(i)1 r1 + · · ·+ β(i)

s rs = ti,

for every β(i) ∈ Bi, the element β(i)s is equal to zero. This means, by Lemma

1.3, that all the polynomials g1, . . . , gs belong to the ring k[f1, . . . , fs−1]. Butit is a contradiction with transcendence degrees. Hence, rs 6 ts. By thesame way we show that ts 6 rs. Therefore, rs = ts.

We will show that r1 = t1. Suppose that r1 > t1. Then all the numbersr1, . . . , rs are strictly greater than t1, and so, in each equality of the form

β(i)1 r1 + · · ·+ β(i)

s rs = t1,

for all β(i) ∈ B1, the numbers β(i)1 , . . . , β

(i)s are equal to zero. This means

that t1 = 0, that is, g1 ∈ k. But the polynomials g1, . . . , gs are algebraically

5

independent over k, so we have a contradiction. Hence, r1 6 t1. By the sameway 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 rl = tl. Suppose that rl > tl. Then we have

t1 6 t2 6 · · · 6 tl < rl 6 rl+1 6 · · · 6 rs.

In each equality of the form

β(i)1 r1 + · · ·+ β

(i)l rl + β

(i)l+1rl+1 + · · ·+ β(i)

s = ti,

for i = 1, . . . , l, the numbers β(i)l , β

(i)l+1, . . . , β

(i)s are equal to zero. This means,

that in the equalities (b) for i = 1, . . . , l, there are not the polynomialsfl, fl+1, . . . , fs. Thus, in this case k[g1, . . . , gl] ⊆ k[f1, . . . , fl−1], but it isa contradiction with transcendence degrees. Hence, rl 6 tl. By the sameway we show that tl 6 rl. Therefore, rl = tl. Therefore, we proved that ifk[f1, . . . , fs] = k[g1, . . . , gs], then ri = ti for all i = 1, . . . , s.

Part 2. Assume that ri = ti for all i = 1, . . . , s. We will show thatk[f1, . . . , fs] = k[g1, . . . , gs]. We arrange the proof in two steps.

Step 1. Let m be the maximal number belonging to {1, . . . , s} such thatr1 = r2 = · · · = rm. We will show that k[f1, . . . , fm] = k[g1, . . . , gm].

Look at the equalities (a) for i = 1, . . . ,m. Observe that in these equali-ties there are not polynomials gj with j > m. Moreover, each equality of the

form α(i)1 t1 + · · · + α

(i)s ts = ri, for every α(i) ∈ Ai with i = 1, . . . ,m, reduces

to the equality α(i)1 r1 + · · ·+ α

(i)m r1 = r1, that is, α

(i)1 + · · ·+ α

(i)m = 1. Hence,

the equalities (a), for i = 1, . . . ,m, are of the formsf1 = a11g1 + a12g2 + · · ·+ a1mgm,f2 = a21g1 + a22g2 + · · ·+ a2mgm,

...fm = am1g1 + am2g2 + · · ·+ ammgm,

where each aij belongs to k. Since f1, . . . , fm are algebraically independentover k, the determinant of the m ×m matrix [aij] is nonzero. This impliesthat k[f1, . . . , fm] = k[g1, . . . , gm].

Ifm = s, then we are done; we have the equality k[f1, . . . , fs] = k[g1, . . . , gs],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 ru < ru+1.

6

Step 2. Assume that u is a number belonging to {1, . . . , s−1} such thatru < ru+1 and k[f1, . . . , fu] = k[g1, . . . , gu]. Let v be the maximal naturalnumber satisfying the equalities ru+1 = ru+2 = · · · = ru+v. We will showthat k[f1, . . . , fu+v] = k[g1, . . . , gu+v].

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

α(i)1 t1 + · · ·+ α(i)

s ts = ri,

for every α(i) ∈ Ai with i ∈ {u+ 1, . . . , u+ v}, reduces to the equality

α(i)1 r1 + · · ·+ α(i)

u ru +(α(i)u+1 + · · ·+ α

(i)u+v

)ru+1 = ru+1

Hence, the equalities (a), for i = u+ 1, . . . , u+ v, are of the formsfu+1 = a11gu+1 + a12gu+2 + · · ·+ a1vgu+v + H1(g1, . . . , gu),fu+2 = a21gu+1 + a22gu+2 + · · ·+ a2vgu+v + H2(g1, . . . , gu),

...fu+v = av1gu+1 + av2gu+2 + · · ·+ avvgu+v + Hv(g1, . . . , gu),

where each aij belongs to k, and H1, . . . , Hv are polynomials in u variablesover k. Put

hj = aj1gu+1 + aj2gu+2 + · · ·+ ajvgu+v

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

k[f1, . . . , fu+v] ⊆ k[g1, . . . , gu, h1, . . . , hv],

because k[f1, . . . , fu] = k[g1, . . . , gu]. But the polynomials f1, . . . , fu+v arealgebraically independent over k, so it follows from the above inclusion, thenthe polynomials

g1, . . . , gu, h1, . . . , hv

are also algebraically independent over k. In particular, the polynomialsh1, . . . , hv are linearly independent over k. This means that the determinantof the v× v matrix [aij] is nonzero, and we have the equality k[h1, . . . , hv] =k[gu+1, . . . , gu+v]. Now we have:

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

= k[g1, . . . , gu] [h1 +H1(g1, . . . , gu), . . . , hv +Hv(g1, . . . , gu)]

= k[g1, . . . , gu][h1, . . . , hv] = k[g1, . . . , gu][gu+1, . . . , gu+v]

= k[g1, . . . , gu+v].

7

Therefore, k[f1, . . . , fu+v] = k[g1, . . . , gu+v].

Now, starting from Step 1, and using one or several times Step 2, weobtain 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, . . . , fn ∈ k[x1, . . . , xn] be homoge-neous polynomials such that

k[xp11 , . . . , xpnn ] ⊆ k[f1, . . . , fn]

for some prime numbers p1, . . . , pn.

a) If char k 6= pi for every i ∈ {1, . . . , n}, then

k[f1, . . . , fn] = k[xl11 , . . . , xlnn ]

for some l1 ∈ {1, p1}, . . . , ln ∈ {1, pn}.b) If char k belongs to the set {p1, . . . , pn}, then

k[f1, . . . , fn] = k[yl11 , . . . , ylnn ]

for some l1 ∈ {1, p1}, . . . , ln ∈ {1, pn} and some k-linear basis y1, . . . , yn of〈x1, . . . , xn〉 such that 〈yi; i ∈ T 〉 = 〈xi; i ∈ T 〉,

yi = xi for i ∈ {1, . . . , n} \ T,

where T = {i ∈ {1, . . . , n}; char k = pi}.

Proof. Let r1, . . . , rn be the degrees of f1, . . . , fn, respectively. We arrangethe proof in three steps.

Step 1. Assume that r1, . . . , rn > 2. We will show that there exists apermutation σ of the set {1, . . . , n} such that ri = pσ(i) for i = 1, . . . , n.

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

Aq = {i ∈ {1, . . . , n}; pi = q}, Bq = {i ∈ {1, . . . , n}; ri = q}.

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

8

Let Q = {p1, . . . , pn} ∪ {r1, . . . , rn}, and let q ∈ Q. Put s = |Aq| andt = |Bq|; s, t ∈ {0, . . . , n}. For i ∈ {1, . . . , n} we may assume that pi = q if i 6 s,

pi 6= q if i > s,

and we may also assume that ri = q if i 6 t,

ri 6= q if i > t.

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

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

xpii =∑

α1,...,αn>0α1r1+...+αnrn=pi

a(i)α fα11 . . . fαn

n ,

where a(i)α ∈ k, α = (α1, . . . , αn). Note that if t = n, then each equality

α1r1 + . . .+αnrn = pi yields that q divides pi, what gives a contradiction fori > s. Therefore t < n.

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

xpii = g(i)n fn + . . .+ g(i)t+2ft+2 + h(i)

(if t = n − 1, we just put xpii = h(i)), where g(i)j ∈ k[f1, . . . , fj] for j =

t+ 2, . . . , n and h(i) ∈ k[f1, . . . , ft+1],

h(i) =∑

α1,...,αt+1>0α1r1+...+αt+1rt+1=pi

b(i)α fα11 . . . f

αt+1

t+1 ,

where b(i)α ∈ k, α = (α1, . . . , αt+1). Since r1, . . . , rt = q, q > 1 and q 6= pi,

the equality α1r1 + . . . + αt+1rt+1 = pi implies that αt+1 6= 0. This meansthat h(i) belongs to the principal ideal generated by ft+1, so xpii belongs tothe 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 isnot greater than n − t. On the other side, we have xpii ∈ P , so xi ∈ P for

9

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

Now observe that the sets from the family {Aq}q∈Q are pairwise disjointand their union is {1, . . . , n}. The same we can tell about the family {Bq}q∈Q.We have proved that |Aq| > |Bq| for every q ∈ Q, so in fact |Aq| = |Bq| forevery q ∈ Q. Bijections from Bq to Aq for every q taken together form arequired permutation σ of the set {1, . . . , n} such that ri = pσ(i) for i =1, . . . , n.

Step 2. We may assume that there exists m ∈ {0, 1, . . . , n} such that fori ∈ {1, . . . , n} we have ri = 1 if i 6 m,

ri > 1 if i > m.

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

k[f1, . . . , fn] = k[f1, . . . , fm, xpjm+1

jm+1, . . . , x

pjnjn

]

for some jm+1 < . . . < jn.

Assume that 0 < m < n. Choose jm+1, . . . , jn ∈ {1, . . . , n}, jm+1 < . . . <jn, such that the elements

f1, . . . , fm, xjm+1 , . . . , xjn

form a basis of 〈x1, . . . , xn〉. For a simplicity denote zi = xji and qi = pji fori = m+ 1, . . . , n.

Consider a homomorphism of k-algebras

ϕ : k[x1, . . . , xn]→ k[zm+1, . . . , zn]

such that ϕ(fi) = 0 for i = 1, . . . ,m and ϕ(zi) = zi for i = m+ 1, . . . , n. Bythe assumption of the theorem we have the inclusion

k[zqm+1

m+1 , . . . , zqnn ] ⊆ k[f1, . . . , fn].

Applying the homomorphism ϕ we obtain

k[zqm+1

m+1 , . . . , zqnn ] ⊆ k[gm+1, . . . , gn],

where gm+1 = ϕ(fm+1), . . . , gn = ϕ(fn). Note that the polynomials gm+1,. . . , gn are nonzero, because tr degk k[gm+1, . . . , gn] = n − m. Moreover,gm+1, . . . , gn are homogeneous polynomials of degrees rm+1, . . . , rn > 2, re-spectively, because ϕ is a homogeneous homomorphism. Then, if we assume,

10

for example, p1 6 . . . 6 pn and r1 6 . . . 6 rn, by Step 1 we obtain ri = qi fori = m+ 1, . . . , n. Therefore, since

k[f1, . . . , fm, zqm+1

m+1 , . . . , zqnn ] ⊆ k[f1, . . . , fn],

by Proposition 1.4 we have

k[f1, . . . , fn] = k[f1, . . . , fm, zqm+1

m+1 , . . . , zqnn ].

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, . . . , fn] = k[xp11 , . . . , xpnn ].

Note also that if m = n, then obviously 〈f1, . . . , fn〉 = 〈x1, . . . , xn〉, so

k[f1, . . . , fn] = k[x1, . . . , xn].

Now, assume that 0 < m < n. Recall that f1, . . . , fm, xjm+1 , . . . , xjn is abasis of 〈x1, . . . , xn〉 and jm+1 < . . . < jn. Arrange all elements of the set{1, . . . , n} \ {jm+1, . . . , jn} in the increasing order: j1 < . . . < jm. For asimplicity denote zi = xji and qi = pji for 1, . . . ,m.

For i = 1, . . . ,m we present zi in the form zi = vi + wi, where vi ∈〈f1, . . . , fm〉 and wi ∈ 〈zm+1, . . . , zn〉. Observe that vi 6= 0 for every i ∈{1, . . . ,m}, because zi 6∈ 〈zm+1, . . . , zn〉. We may assume that there existst ∈ {0, . . . ,m} such that for i ∈ {1, . . . ,m} we have qi = char k if i 6 t,

qi 6= char k if i > t.

Let i ∈ {t + 1, . . . ,m}. Put K = k[f1, . . . , fm]. Then k[x1, . . . , xn] =K[zm+1, . . . , zn] is a polynomial algebra over K. By Step 2 we have k[f1,. . . , fn] = K[z

qm+1

m+1 , . . . , zqnn ]. By the assumption of the theorem, zqii ∈

k[f1, . . . , fn], so zqii as a polynomial in zm+1, . . . , zn over K has zero compo-nent of degree 1. On the other hand, zi = vi + wi, where vi ∈ K and wi, ifnonzero, is a homogeneous polynomial of degree 1 in K[zm+1, . . . , zn]. Thusqiv

qi−1i wi is the component of degree 1 of zqii = (vi +wi)

qi . Since char k 6= qi,we have wi = 0, so zi = vi and zi ∈ 〈f1, . . . , fm〉. We obtain that

〈zt+1, . . . , zm〉 ⊆ 〈f1, . . . , fm〉.

11

Now, let i0 ∈ {1, . . . , t}, so qi0 = char k = p. Put zi0 = vi0 + wi0 ,vi0 = c1f1 + . . .+ cmfm, wi0 = cm+1zm+1 + . . .+ cnzn for some c1, . . . , cn ∈ k.Then

zpi0 = cp1fp1 + . . .+ cpmf

pm + cpm+1z

pm+1 + . . .+ cpnz

pn,

because char k = p. By the assumption of the theorem and the result of Step2 we have

zpi0 ∈ k[f1, . . . , fm, zqm+1

m+1 , . . . , zqnn ].

However, every element of k[f1, . . . , fm, zqm+1

m+1 , . . . , zqnn ] as a polynomial in

variables f1, . . . , fm, zm+1, . . . , zn, is a sum of monomials, which degreeswith respect to zi are divisible by qi for i > m. So, if qi 6= char k for somei > m, then ci = 0.

We may assume that there exists u ∈ {m, . . . , n} such that for i ∈ {m+1, . . . , n} we have qi = p if i 6 u,

qi 6= p if i > u.

DenoteV = 〈v1, . . . , vt, zt+1, . . . , zm〉, W = 〈zm+1, . . . , zu〉.

Then w1, . . . , wt belong to W , so z1, . . . , zt belong to V + W . Note thatz1, . . . , zm, zm+1, . . . , zu form a k-linear basis of V + W . Thus dimV =m, so v1, . . . , vt, zt+1, . . . , zm are linearly independent and form a basis of〈f1, . . . , fm〉. We obtain that

k[f1, . . . , fn] = k[v1, . . . , vt, zt+1, . . . , zm, zqm+1

m+1 , . . . , zqnn ],

so it is enough to put yji = vi for i = 1, . . . , t, yji = xji for i = t + 1, . . . , n,lji = 1 for i = 1, . . . ,m and lji = qi = pji 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[yl11 , . . . , y

lnn ],

where we can additionally put yi = xi for all i such that li > 1. More pre-cisely, if char k = p and we assume that p1 = · · · = ps = p and ps+1, . . . , pn 6=p for some s, then the subalgebra in case b) is of the form

B = k[yl11 , . . . , ylss , x

ls+1

s+1 , . . . , xlnn ],

where y1, . . . , ys form a basis of 〈x1, . . . , xs〉 and li ∈ {1, p} for i = 1, . . . , s,li ∈ {1, pi} for i = s + 1, . . . , n. We can choose the basis y1, . . . , ys in such

12

a way that l1 = · · · = lt = 1 and lt+1 = · · · = ls = p for some t, so we canassume that

(∗) B = k[y1, . . . , yt, ypt+1, . . . , y

ps , x

ls+1

s+1 , . . . , xlnn ].

If we take any other elements y′t+1, . . . , y′s ∈ 〈x1, . . . , xs〉 such that the

elements y1, . . . , yt, y′t+1, . . . , y

′s form a basis of 〈x1, . . . , xs〉, then

〈yp1, . . . , ypt , y′pt+1, . . . , y

′ps 〉 = 〈yp1, . . . , y

pt , y

pt+1, . . . , y

ps〉,

sok[y1, . . . , yt, y

′pt+1, . . . , y

′ps ] = k[y1, . . . , yt, y

pt+1, . . . , y

ps ].

The elements y′t+1, . . . , y′s can be chosen from the set {x1, . . . , xs}. Then, like

in the proof of Theorem 2.1, we obtain

(∗∗) B = k[y1, . . . , yt, xpjt+1

, . . . , xpjs , xls+1

s+1 , . . . , xlnn ],

where 1 6 jt+1 < · · · < js 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 subalgebraB satisfying the assumptions of Theorem 2.1 is uniquely determined by ak-linear subspace

〈yi; i ∈ T, li = 1〉 ⊆ 〈xi; i ∈ T 〉

(that is, 〈y1, . . . , yt〉 in our example above) and by li ∈ {1, pi} for i ∈{1, . . . , n} \ T , where T = {i ∈ {1, . . . , n}; char k = pi}.

3 Some special cases

In this section we present explicitly some particular cases of Theorem 2.1.For p1 = . . . = pn we obtain the following theorem.

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

k[xp1, . . . , xpn] ⊆ k[f1, . . . , fn].

a) If char k 6= p, then

k[f1, . . . , fn] = k[xl11 , . . . , xlnn ]

13

for some l1, . . . , ln ∈ {1, p}.

b) If char k = p, then

k[f1, . . . , fn] = k[y1, . . . , ym, ypm+1, . . . , y

pn]

for some m ∈ {0, 1, . . . , n} and some k-linear basis y1, . . . , yn of 〈x1, . . . , xn〉.

Observe that in Theorem 3.1 b), ifm = 0, then k[f1, . . . , fn] = k[xp1, . . . , xpn],

and if m = n, then k[f1, . . . , fn] = k[x1, . . . , xn].

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

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

k[xp11 , . . . , xpnn ] ⊆ k[f1, . . . , fn]

for some prime numbers p1, . . . , pn. Then

k[f1, . . . , fn] = k[xl11 , . . . , xlnn ]

for some l1 ∈ {1, p1}, . . . , ln ∈ {1, pn}.

In the case of pairwise different primes we have the same thesis for arbi-trary characteristic of k, since the k-linear space 〈xi; pi = char k〉 is at mostone-dimensional.

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

k[xp11 , . . . , xpnn ] ⊆ k[f1, . . . , fn]

for some pairwise different prime numbers p1, . . . , pn. Then

k[f1, . . . , fn] = k[xl11 , . . . , xlnn ]

for some l1 ∈ {1, p1}, . . . , ln ∈ {1, pn}.

14

4 Conclusions for homogeneous derivations

Let k[X] = k[x1, . . . , xn].

Recall that a k-linear map d : k[X] → k[X] such that d(fg) = d(f)g +fd(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[xp1, . . . , xpn] ⊆ 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 2.1yields the following.

Theorem 4.1 Let d be a homogeneous k-derivation of the polynomial algebrak[X] = k[x1, . . . , xn], 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, . . . , ym, ypm+1, . . . , y

pn]

for some m ∈ {0, 1, . . . , n} and some k-linear basis y1, . . . , yn of 〈x1, . . . , xn〉.

A homogeneous k-derivation of degree 0 is called linear. The restrictionof a linear derivation to the k-linear space 〈x1, . . . , xn〉 is a k-linear endomor-phism. Every endomorphism of 〈x1, . . . , xn〉 determines the unique lineark-derivation. We have the following corollary from the above theorem andTheorem 3.2 from [4].

Corollary 4.2 Let d be a linear derivation of the polynomial algebra k[X] =k[x1, . . . , xn], where k is a field of characteristic p > 0. Then k[X]d is apolynomial k-algebra if and only if the Jordan matrix of the endomorphismd|〈x1,...,xn〉 satisfies the following conditions.

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

(2) At most one Jordan block has dimension greater than 1 and, if such ablock 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.

15

References

[1] S. Balcerzyk, T. Jozefiak, Commutative Rings. Dimension, Multiplicityand Homological Methods, PWN - Polish Scientific Publishers - Warszawa,Ellis Horwood Limited Publishers - Chichester, 1989.

[2] D. Daigle, Plane Frobenius sandwiches of degree p, Proc. Amer. Math.Soc. 117 (1993), 885–889.

[3] R. Ganong, Plane Frobenius sandwiches, Proc. Amer. Math. Soc. 84(1982), 474–478.

[4] P. J ↪edrzejewicz, Linear derivations with rings of constants generated bylinear forms, Coll. Math. 113 (2008), 279–286.

[5] P. J ↪edrzejewicz, On polynomial graded subalgebras of a polynomial alge-bra, to appear in Algebra Coll.

[6] H. Kraft, Geometrische Methoden in der Invariantentheorie, Vieweg &Sohn, Braunschweig, 1985.

[7] A. Nowicki, On the nonexistence of rational first integrals for systemsof linear differential equations, Linear Algebra And Its Applications 235(1996), 107–120.

[8] A. Nowicki, Polynomial derivations and their rings of constants, NicolausCopernicus University, Torun, 1994, www.mat.umk.pl/~anow/.

[9] A. Nowicki and M. Nagata, Rings of constants for k-derivations ink[x1, . . . , xn], J. Math. Kyoto Univ. 28 (1988), 111–118.

[10] O. Zariski, P. Samuel, Commutative Algebra, vol. I, D. Van Nostrand,New York, 1958.

16