GEORGIAN MATHEMATICAL JOURNAL: Vol. 3, No. 3, 1996, 263-274

ON KRULL DIMENSION OF ORE EXTENSIONS

E. RTVELIASHVILI

Abstract. The Krull dimension of rings of skew polynomials is studied. Earlier the problem of Krull dimension was investigated only for some particular cases, namely, for Weyl algebras [2], a ring of differential operators [7,8], as well as for rings of Laurent skew polynomials [9–10].

Let R be a ring with unity and let R[x] be a ring of left polynomials (i.e., polynomials with coefficients from the left to the powers of x) over R. Suppose that α is an endomorphism of R and δ is an α-differentiation of R, i.e., δ(a+b) = δ(a)+δ(b) and δ(ab) = δ(a)b+α(a)δ(b) for any a, b ∈ R. Let R[x; α; δ] denote the ring of left skew polynomials over R [1] (the additive group of this ring coincides with the one of R[x] and the multiplication in it is defined by means of operators in R[x] and the following commutation formula:

x · a = α(a)x + δ(a), a ∈ R).  (1)

If δ is the zero mapping of R, we use the notation R[x; α] for R[x; α; δ]. Denote by K. dim(A) the Krull dimension of a ring A in the sense of

Gabriel and Rentschler (i.e., the deviation of the set of left ideals of A) [2].

Theorem 1. Let R be a ring with unity, let α be its automorphism, and let δ be a nilpotent (δ^{d} = 0) α-differentiation of R. Suppose that δ^{−i}(1_{R}) ≠ ∅ for i = 1, 2, …, d − 1. Then

K. dim(R[x; α; δ]) = K. dim(R[x; α]) = K. dim(R[x]).

Theorem 1 is a trivial consequence of the following propositions: 263

1991 Mathematics Subject Classification. 16S36.

Key words and phrases. Skew polynomial ring, α-differentiation.

1072-947X/96/0500-0263$09.50/0 \begin{matrix}c \\ ◯ \end{matrix} 1996 Plenum Publishing Corporation

Proposition 1. Let R be a ring with unity, let α be its automorphism, and let δ be a nilpotent (δ^{d} = 0) α-differentiation of R. If δ^{−i}(1_{R}) ≠ ∅ for i = 1, 2, …, d − 1, then

K. dim(R[x; α]) ≤ K. dim(R[x; α; δ]).

Proposition 2. Let α be an injective endomorphism of a ring R with unity satisfying

α(𝔞) < Rα(𝔪) ⇒ 𝔞 < 𝔪

for any left ideals 𝔞 and 𝔪 of R, where

Rα(𝔪) = { ∑_{p=1}^{n} λ_{p}α(m_{p}); m_{p} ∈ 𝔪; λ_{p} ∈ R}.

Then

K. dim(R[x]) ≤ K. dim(R[x; α]).

Proposition 3. Let α be an automorphism of a ring R with unity. Then

K. dim(R[x; α; δ]) ≤ K. dim(R[x]).

The proofs of these propositions as well as of the other ones given in this paper are based on

Lemma 1 [2]. Let E and F be partially ordered sets. If there exists a strictly isotonic mapping Φ : E → F, then dev E ≤ dev F.

To prove Proposition 1, we shall also need

Lemma 2. Let α be an automorphism of a ring R with unity, and let δ be an α-differentiation of R. Then the condition

f_{1}c_{1} + f_{2}c_{2} + ⋯ + f_{n}c_{n} = c,  (2)

where f_{1}, f_{2}, …, f_{n} ∈ R[x; α; δ] and c_{1}, c_{2}, …, c_{n}, c ∈ R, implies

c = b_{1}c_{1} + b_{2}c_{2} + ⋯ + b_{n}c_{n}

with b_{1}, b_{2}, …, b_{n} ∈ R.

Proof. Suppose that k is the maximum of degrees of polynomials f_{1}, f_{2}, …, f_{n}. We can write these polynomials as

f_{1} = a_{1k}x^{k} + a_{1,k−1}x^{k−1} + ⋯ + a_{10},
f_{2} = a_{2k}x^{k} + a_{2,k−1}x^{k−1} + ⋯ + a_{20},
. . . . . . . . . . . . . . . . . . . . . . . . .
f_{n} = a_{nk}x^{k} + a_{n,k−1}x^{k−1} + ⋯ + a_{n0},
a_{ip} ∈ R, i = 1, 2, …, n, p = 0, 1, …, k.

ON KRULL DIMENSION OF ORE EXTENSIONS 265

Here some of the a_{ip}’s may be equal to zero.

Calculate the left side of (2) (using a^{α} and a^{δ} instead of α(a) and δ(a)).

f_{1}c_{1} + f_{2}c_{2} + ⋯ + f_{n}c_{n} = ∑_{i=1}^{n} a_{ik}c_{i}^{α^{k}}x^{k} +

+ ∑_{i=1}^{n} (a_{ik}(c_{i}^{δα^{k−1}} + c_{i}^{αδα^{k−2}} + ⋯ + c_{i}^{α^{k−1}δ}) + a_{i,k−1}c_{i}^{α^{k−1}})x^{k−1} + ⋯ +

+ ∑_{i=1}^{n} (a_{ik}(c_{i}^{δ^{d}α^{k−d}} + c_{i}^{δ^{d−1}αδα^{k−d−1}} + ⋯ + c_{i}^{δ^{d−1}α^{k−d}δ} + c_{i}^{δ^{d−2}αδ^{2}α^{k−d−1}} +

+ ⋯ + c_{i}^{α^{k−d}δ^{d}}) + a_{i,k−1}(c_{i}^{δ^{d−1}α^{k−d}} + ⋯ + c_{i}^{α^{k−d}δ^{d−1}}) + ⋯ +

+a_{id}c_{i}^{α^{d}})x^{d} + ⋯ + ∑_{i=1}^{n}(a_{ik}c_{i}^{δ^{k}} + a_{i,k−1}c_{i}^{δ^{k−1}} + ⋯ + a_{i0}c_{i}).

Taking into consideration (2), we have

∑_{i=1}^{n} a_{ik}c_{i}^{α^{k}} = 0,

∑_{i=1}^{n} (a_{ik}(c_{i}^{δα^{k−1}} + c_{i}^{αδα^{k−2}} + ⋯ + c_{i}^{α^{k−1}δ} + a_{i,k−1}c_{i}^{α^{k−1}}) = 0,

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∑_{i=1}^{n} (a_{ik}(c_{i}^{δ^{k−1}α} + c_{i}^{δ^{k−2}αδ} + ⋯ + c_{i}^{αδ^{k−1}}) +

+ a_{i,k−1}(c_{i}^{δ^{k−2}α} + ⋯ + c_{i}^{αδ^{k−2}}) + ⋯ + a_{i1}c_{i}^{α}) = 0,

∑_{i=1}^{n}(a_{ik}c_{i}^{δ^{k}} + a_{i,k−1}c_{i}^{δ^{k−2}} + ⋯ + a_{i0}c_{i}) = c.

Using these equalities and calculating ( ∑_{i=1}^{n} a_{ik}c_{i}^{α^{k}})^{(α^{−1}δ)^{k}}, we obtain

0=( ∑_{i=1}^{n} a_{ik}^{α^{−1}}c_{i}^{α^{k−1}})^{δ(α^{−1}δ)^{k−1}} =( ∑_{i=1}^{n}(a_{ik}^{α^{−1}δ}c_{i}^{α^{k−1}} +a_{ik}c_{i}^{α^{k−1}δ}))^{(α^{−1}δ)^{k−1}} =

=( ∑_{i=1}^{n} (a_{ik}^{α^{−1}δ}c_{i}^{α^{k−1}} −a_{ik}(c_{i}^{δα^{k−1}} +⋯+c_{i}^{α^{k−2}δα})−a_{i,k−1}c_{i}^{α^{k−1}}))^{(α^{−1}δ)^{k−1}} =

=( ∑_{i=1}^{n} (a_{ik}^{α^{−1}δα^{−1}δ}c_{i}^{α^{k−2}} +a_{ik}^{α^{−1}δ}c_{i}^{α^{k−2}δ} −a_{ik}^{α^{−1}δ}(c_{i}^{δα^{k−2}} +⋯+c_{i}^{α^{k−2}δ})−

−a_{ik}(c_{i}^{δα^{k−2}δ} +⋯+c_{i}^{α^{k−2}δ^{2}})−a_{i,k−1}^{α^{−1}δ}c_{i}^{α^{k−2}} −a_{i,k−1}c_{i}^{α^{k−2}δ}))^{(α^{−1}δ)^{k−2}} =

=( ∑_{i=1}^{n} (a_{ik}^{(α^{−1}δ)^{2}}c_{i}^{α^{k−2}} −a_{ik}^{α^{−1}δ}(c_{i}^{δα^{k−2}} +⋯+c_{i}^{α^{k−3}δα})+a_{ik}(c_{i}^{δ^{2}α^{k−2}} +

+c_{i}^{δαδα^{k−3}} + ⋯ + c_{i}^{α^{k−3}δ^{2}α}) + a_{i,k−1}(c_{i}^{δα^{k−2}} + ⋯ + c_{i}^{α^{k−3}δα}) −

−a_{i,k−1}^{α^{−1}δ}c_{i}^{α^{k−2}} + a_{i,k−2}c_{i}^{α^{k−2}}))^{(α^{−1}δ)^{k−2}} = ⋯ = ( ∑_{i=1}^{n} (a_{ik}^{(α^{−1}δ)^{k−1}}c_{i}^{α} −

−a_{ik}^{(α^{−1}δ)^{k−2}}c_{i}^{δα} + ⋯ + (−1)^{k−1}a_{ik}c_{i}^{δ^{k−2}α} − a_{i,k−1}^{(α^{−1}δ)^{k−2}}c_{i}^{α} +

+a_{i,k−1}^{(α^{−1}δ)^{k−3}}c_{i}^{δα} + ⋯ + (−1)^{k−1}a_{i,k−1}c_{i}^{δ^{k−3}α} + ⋯ + (−1)^{k−1}a_{i1}c_{i}^{α})^{α^{−1}δ} =

= ∑_{i=1}^{n} (a_{ik}^{(α^{−1}δ)^{k}}c_{i} − a_{i,k−1}^{(α^{−1}δ)^{k−1}}c_{i} + ⋯ +

+(−1)^{k−1}a_{i1}^{α^{−1}δ}c_{i} + (−1)^{k−1}(a_{ik}c_{i}^{δ^{k}} + a_{i,k−1}c_{i}^{δ^{k−1}} + ⋯ + a_{i1}c_{i}^{δ})),

and therefore

c = ∑_{i=1}^{n} ((−1)^{k}(a_{ik}^{(α^{−1}δ)^{k}} − a_{i,k−1}^{(α^{−1}δ)^{k−1}} + ⋯ + (−1)^{k−1}a_{i1}^{α^{−1}δ}) + a_{i0})c_{i}.

Proof of Proposition 1. Taking into account that the Krull dimension of the ring A is equal to dev(Id A), where Id A is the set of left ideals of A, by Lemma 1 it suffices to construct a strictly isotonic mapping F from the set of left ideals of R[x; α] into the set of left ideals of R[x; α; δ].

First of all, let us show by induction on k that for any a, b ∈ R we can write δ^{k}(a · b) as

δ^{k}(ab) = a_{k}δ^{k}(b) + a_{k−1}δ^{k−1}(b) + ⋯ + a_{1}δ(b) + a_{0}b,  (3)

where a_{0}, a_{1}, …, a_{k} ∈ R are the coefficients found from the representation

x^{k}a = a_{k}x^{k} + a_{k−1}x^{k−1} + ⋯ + a_{0}.

Indeed, if k = 1, then

δ(ab) = α(a)δ(b) + δ(a)b and xa = α(a)x + δ(a).

Suppose the validity of (3) for some natural k. Then

δ^{k+1}(ab) = δ(δ^{k}(ab)) = δ(a_{k}δ^{k}(b) + a_{k−1}δ^{k−1}(b) + ⋯ + a_{0}b) =
= a′_{k+1}δ^{k+1}(b) + a′_{k}δ^{k}(b) + ⋯ + a′_{0}b,

where a′_{k+1} = α(a_{k}), a′_{0} = δ(a_{0}), and a′_{n} = δ(a_{n}) + α(a_{n−1}) for n = 1, 2, …, k.

On the other hand,

x^{k+1}a = x(x^{k}a) = x(a_{k}x^{k} + a_{k−1}x^{k−1} + ⋯ + a_{0}) =

= α(a_{k})x^{k+1} + (δ(a_{k}) + α(a_{k−1}))x^{k} + ⋯ + (δ(a_{1}) + α(a_{0}))x + δ(a_{0}) =

= a′_{k+1}x^{k+1} + a′_{k}x^{k} + ⋯ + a′_{0}.

The induction is thus completed.

Now calculate x^{d}a. Suppose that

x^{d}a = a_{d}x^{d} + a_{d−1}x^{d−1} + ⋯ + a_{0}, a_{0}, a_{1}, …, a_{d} ∈ R.

We have

δ^{d}(ab) = a_{d}δ^{d}(b) + a_{d−1}δ^{d−1}(b) + ⋯ + a_{0}b, b ∈ R,

where a_{0} = δ^{d}(a). Since δ^{d} = 0, we obtain

a_{d−1}δ^{d−1}(b) + ⋯ + a_{2}δ^{2}(b) + a_{1}δ(b) = 0.  (4)

Since b is an arbitrary element of R and δ^{−1}(1_{R}) ≠ ∅, we can assume in (4) that b ∈ δ^{−1}(R). Taking into account that δ(0) = 0 and δ(1_{R}) = δ(1_{R} · 1_{R}) = δ(1_{R}) + δ(1_{R}) = 0, we have a_{1} = 0. Further, assuming that b ∈ δ^{−2}(1_{R}), we obtain a_{2} = 0, and so on. Finally, we have

a_{0} = a_{1} = ⋯ = a_{d−1} = 0.

Therefore x^{d}a = a_{d}x^{d}. Using the commutation formula (1), we easily obtain that a_{d} = α^{d}(a), and hence

x^{d}a = α^{d}(a)x^{d}.

Moreover, if p = md + q, we can write

x^{p}a = x^{q}α^{md}(a)x^{md}, p, q, m ∈ ℕ ∪ {0}.  (5)

Now we begin the construction of the mapping F.

Let I be an arbitrary left ideal of R[x; α]. Let n be the minimum of degrees of nonzero polynomials from I. For any k ≥ n denote by 𝔞_{k} the set of highest-degree coefficients of kth polynomials from I. Thus we obtain the sequence of left ideals of R:

𝔞_{n}, 𝔞_{n+1}, …

satisfying

α(𝔞_{n+i}) ⊆ 𝔞_{n+i+1} for i ∈ ℕ ∪ {0}.  (6)

(This means that this sequence is α-nondecreasing.)

Consider all monomials of the form

α^{(n+i)(d−1)}(a_{n+i})x^{(n+i)d}, a_{n+i} ∈ 𝔞_{n+i}, i ∈ ℕ ∪ {0},  (7)

and let J be the left ideal of R[x; α; δ] generated by them. Define the mapping F by

F (I) = J,

and show that it is strictly isotonic.

Let us first study the structure of J. Taking into account (5) and the set of generators of J, we easily check that any polynomial from J with degree divisible by d is a monomial. Let

g = α^{r(d−1)}(a_{r})x^{dr}, a_{r} ∈ 𝔞_{r},

be any of them, and suppose that

g = ∑_{λ=1}^{m} f_{λ}g_{λ}, f_{λ} ∈ R[x; α; δ],

where g_{λ} is of the form (7).

Take any λ ∈ {1, 2, …, m} and let the degree of g_{λ} be d · k. By (5) we can assume: if k > r, then f_{λ} = 0; if k = r, then the degree of f_{λ} is less than d; if k < r, then f_{λ} contains only terms of the degree from the interval [dr − dk; d(r + 1) − dk) of natural numbers.

Suppose that

g_{λ} = α^{k(d−1)}(a_{k})x^{dk}, k < r, a_{k} ∈ 𝔞_{k}.

We can assume that

f_{λ} = c_{1}x^{d(r+1)−1−dk} + ⋯ + c_{d}x^{dr−dk}; c_{1}, c_{2}, …, c_{d} ∈ R.

Then using (5), we obtain

f_{λ} · g_{λ} = (c_{1}x^{d−1} + c_{2}x^{d−2} + ⋯ + c_{d})α^{dr−dk}(α^{k(d−1)}(a_{k}))x^{dr}.

But

α^{dr−dk}(α^{k(d−1)}(a_{k})) = α^{rd−k}(a_{k}) = α^{r(d−1)}(α^{r−k}(a_{k})).

By (6), α^{r−k}(a_{k}) ∈ 𝔞_{r}. Therefore any product f_{λ} · g_{λ} such that the degree of g_{λ} is less than the degree of g can be replaced by the product f′_{λ} · g′_{λ}, where the degree of f′_{λ} is less than d, the degree of g′_{λ} is equal to the degree of g, and g′_{λ} ∈ J. Hence by Lemma 2 the coefficient of g can be rewritten as

∑_{λ=1}^{m} b_{λ}α^{r(d−1)}(a_{r_{λ}}), b_{λ} ∈ R.

Since 𝔞_{r} is an ideal, we conclude that

a = α^{r(d−1)}(a) with a ∈ 𝔞_{r}.  (8)

Now we can show that the mapping F constructed above is strictly isotonic.

Let I_{1} be any left ideal of R[x; α] such that I ⊂ I_{1}, and let F (I_{1}) = J_{1}. We have to show that J ⊂ J_{1}. By the construction of F it is clear that J ⊆ J_{1}.

Since I_{1} ⊃ I, there exists a polynomial

h = b_{k}x^{k} + b_{k−1}x^{k−1} + ⋯ + b_{0}, b_{k}, b_{k−1}, …, b_{0} ∈ R,

in I_{1} such that it is not an element of I. More than that we can assume that b_{k} cannot be obtained as a left linear combination of the highest coefficients of polynomials of kth degree from I. Indeed, if it is not the case, we can take the appropriate difference h − ∑_{μ} c_{μ}f_{μ} which will be an element of I_{1}\I whose degree will be less than that of h, and so on.

Obviously, α^{k(d−1)}(b_{k})x^{dk} ∈ J_{1}. Let us show that this monomial does not belong to J. Suppose the contrary. Then by (8),

b_{k}x^{k} ∈ I

which contradicts the choice of h.

Before proving Proposition 2 note that for the first time an endomorphism of the type considered in this proposition has been studied by Lesieur [3]. He proved that the condition

α(a) ⊂ Rα(𝔪) ⇒ 𝔞 ⊂ 𝔪

is equivalent to

α(a) = ∑_{j=1}^{n} λ_{j}α(b_{j}) ⇒ a = ∑_{j=1}^{n} μ_{j}b_{j}, a, λ_{j}, μ_{j}, b_{j} ∈ R.  (9)

It was also shown by him that elements of the left ideal

\underbrace{\underbrace{Rα}(Rα(⋯ (Rα(𝔞)) ⋯))}_{n times}

of R generally have the form

∑_{i=1}^{d} μ_{i}α^{n}(a_{i}), μ_{i} ∈ R; a_{i} ∈ 𝔞.

Proof of Proposition 2. By Lemma 1 it suffices to construct a strictly isotonic mapping F from the set of left ideals of R[x] into the set of left ideals of R[x; α].

Let I be any left ideal of R[x]. Let n be the minimum of the degrees of nonzero polynomials from I. Consider the nondecreasing sequence

𝔞_{n} ⊆ 𝔞_{n+1} ⊆ ⋯, n ∈ ℕ ∪ {0},  (10)

of left ideals of R, where 𝔞_{n+i} (i = 0, 1, 2, …) is the set (in fact, the left ideal) of highest coefficients of all polynomials of degree n + i from I. With the help of this sequence we can construct the α-nondecreasing sequence of left ideals of R:

𝔞′_{n} ↪^{α} Rα(𝔞′_{n+1}) ↪^{α} Rα(Rα(𝔞′_{n+2})) ↪^{α} ⋯,  (11)

where 𝔞′_{n+i} = \underbrace{\underbrace{Rα}(Rα(⋯ (Rα(𝔞_{n+i})) ⋯))}_{n times}, i = 0, 1, 2, . . ..

Consider all monomials of the form

a_{n+i}x^{n+i}, a_{n+i} ∈ \underbrace{Rα(Rα(⋯ (Rα(𝔞′_{n+i})) ⋯))}_{i times}, i = 0, 1, …, (12)

and let J be the left ideal of R[x; α] generated by them.

Define the mapping F by

F (I) = J.

Let I_{1} ⊃ I be any left ideal, and let J_{1} = F (I_{1}). It follows from the construction of F that J ⊆ J_{1}.

As in proving Proposition 1, choose a polynomial

g = a_{d}x^{d} + ⋯ + a_{0}, a_{0}, ⋯, a_{d} ∈ R, d ∈ ℕ ∪ {0},

from I_{1} which does not belong to I. Here we can also assume that the highest term of g cannot be obtained as a left linear combination of highest terms of the polynomials from g having the same degree. Obviously, α^{d}(a_{d})x^{d} ∈ J_{1}. Show that this monomial is not in J.

Suppose the contrary. Then the monomial α^{d}(a_{d})x^{d} can be represented as the sum of the products of monomials of the form

c_{d−p}x^{d−p} · a_{p}x^{p}, c_{d−p} ∈ R, a_{p} ∈ \underbrace{Rα(Rα(⋯ (Rα(𝔞_{p})) ⋯))}_{p times}, (13)

where p ≤ d and a_{p} = ∑_{i=1}^{k} μ_{i}α^{p}(b_{i}), b_{i} ∈ 𝔞_{p}.

If we carry out the multiplication in (13), we obtain

c_{d−p}x^{d−p}a_{p}x^{p} = ∑_{i=1}^{k} λ_{i}α^{d}(b_{i})x^{d},

where λ_{i} = c_{d−p}α^{d−p}(μ_{i}), λ_{i} ∈ R. Moreover, by (11) we can assume that b_{i} ∈ 𝔞_{d} (i = 1, 2, …, k).

Thus we have

α^{d}(a_{d}) = ∑_{j=1}^{n} ∑_{i=1}^{k} λ_{ij}α^{d}(b_{ij}), λ_{ij} ∈ R; b_{ij} ∈ 𝔞_{d}.

This equality can be rewritten as

α^{d}(a_{d}) = ∑_{l=1}^{q} ν_{l}α^{d}(b_{l}), ν_{l} ∈ R, b_{l} ∈ 𝔞_{d}, q ≤ nk.

Using here equality (9) d times, we get

a_{d} = ∑_{l=1}^{q} β_{l}b_{l}, β_{l} ∈ R, b_{l} ∈ 𝔞_{d}.

But this contradicts the choice of g.

Thus we have proved that F is strictly isotonic.

Proof of Proposition 3. Let I be any left ideal of R[x; α; δ]. Consider all the monomials of the form

α^{−n}(a_{n})x^{n}, a_{n} ∈ 𝔞_{n},  (14)

where 𝔞_{n} is the left ideal of all polynomials of degree n from I.

Let J be the left ideal of R[x] generated by monomials (14). Define the mapping F from the set of left ideals of R[x; α; δ] into the set of left ideals of R[x] by

F (I) = J.

Since α is the automorphism of R and α^{−n}(𝔞_{n}) ⊆ α^{−(n+1)}(𝔞_{n+1}), ∀n ∈ ℕ, it can be proved quite analogously to the proofs of Propositions 1 and 2 that F is strictly isotonic.

Remark 1. As an example let us show that the equality given in Proposition 3 can be strict.

Let R = K[y], where K is a field of characteristic zero. Let α be the identical automorphism of R, and let δ be the partial differentiation by the variable y. Then δ is the α-differentiation of R, and thus we obtain the ring of skew polynomials R[x; α; δ] which can be considered as the Weyl algebra over K. Its Krull dimension is equal to 1 [2]. On the other hand, R[x] = K[y][x] = K[y; x]. Therefore, the Krull dimension of R[x] is 2.

Remark 2. Let R be any division ring and let α be its automorphism. Then the definition of the Krull dimension and Proposition 3 imply that

K. dim(R[x; α; δ]) = 1.

Taking into consideration the already known results concerning the Krull dimension of polynomial rings, from Theorem 1 we obtain

Corollary 1. Let R be a left Noetherian ring with the Krull dimension n. Let α be an automorphism of R and let δ be a nilpotent (δ^{d} = 0) αdifferentiation of R. Moreover, let δ^{−i}(1) ≠ ∅, i = 1, 2, …, d − 1. Then

K. dim(R[x; α; δ]) = K. dim(R[x; α]) = n + 1.

Let R be a ring and let α and δ satisfy the conditions of Theorem 1. Denote by R[[x; α; Δ]] the ring of left skew formal power series over R[s] for which

xa = a^{δ_{0}}x + a^{δ_{1}}x^{2} + ⋯ + a^{δ_{d−1}}x^{d−1}, a ∈ R,

where δ_{0} = α, δ_{1} = δα, …, δ_{d−1} = δ^{d−1}α.

Proposition 4. K. dim(R[x; α; δ]) ≤ K. dim(R[[x; α; Δ]]).

Proof. In order to construct a strictly isotonic mapping from the set of left ideals of R[x; α; δ] into the set of left ideals of R[[x; α; Δ]], it suffices to associate with any left ideal I of R[x; α; δ] the left ideal J of R[[x; α; Δ]] generated by the monomials of the form

α^{d^{r}}(a_{d}) · x^{dr}; a_{d} ∈ R,

where a_{d}x^{d} is the highest-degree term of the dth degree polynomial from I. The fact that this mapping is strictly isotonic, can be proved as above.

Consider now the ring R[x_{1}, …, x_{n}; α_{1}, …, α_{n}; δ_{1}, …, δ_{n}] of left skew polynomials in n variables over R [6], where

\begin{matrix}a^{α_{i}α_{j}} = a^{α_{j}α_{i}}, a^{δ_{i}δ_{j}} = a^{δ_{j}δ_{i}}, i, j = 1, 2, …, n; \\ a^{α_{i}δ_{j}} = a^{δ_{j}α_{i}}, i ≠ j; \\ x_{i}x_{j} = x_{j}x_{i}, x_{i}a = α_{i}(a)x_{i} + δ_{i}(a), a ∈ R. \end{matrix}  (15)

It is easy to show that if the endomorphisms α_{i} (i = 1, 2, …, n) of R and the corresponding α_{i}-differentiations δ_{i} (i = 1, 2, …, n) of R satisfy (15), then R[x_{1}, …, x_{n}; α_{1}, …, α_{n}; δ_{1}, …, δ_{n}] can be represented as the ring of left skew polynomials in one variable A_{n−1}[x_{n}; \overline{α}_{n}; \overline{δ}_{n}] over A_{n−1} = R[x_{1}, …, x_{n−1}; α_{1}, …, α_{n−1}; δ_{1}, …, δ_{n−1}], where the mappings \overline{α}_{n} and \overline{δ}_{n} are defined as follows: if

f = ∑_{ν} a_{ν}x_{1}^{ν_{1}} ⋯ x_{n−1}^{ν_{n−1}} ∈ A_{n−1}, a_{ν} ∈ R,

then

\overline{α}_{n}(f) = ∑_{ν} α_{n}(a_{ν})x_{1}^{ν_{1}} ⋯ x_{n−1}^{ν_{n−1}}

and

\overline{δ}_{n}(f) = ∑_{ν} δ_{n}(a_{ν})x_{1}^{ν_{1}} ⋯ x_{n−1}^{ν_{n−1}}

(the fact that \overline{δ}_{n} is an \overline{α}_{n}-differentiation of A_{n−1} can be checked by direct calculation). If α_{n} is an automorphism of R, then \overline{α}_{n} is an automorphism of A_{n−1}. This enables us to generalize Proposition 3.

Proposition 5. Let α_{i} (i = 1, 2, …, n) be automorphisms of R. Then

K. dim(R[x_{1}, …, x_{n}; α_{1}, …, α_{n}; δ_{1}, …, δ_{n}]) ≤ K. dim(R[x_{1}, …, x_{n}]).

Taking into account that if δ_{n} is a nilpotent α_{n}-differentiation of R, then \overline{δ}_{n} is a nilpotent \overline{α}_{n}-differentiation of A_{n−1}m from Theorem 1 we obtain

Theorem 2. Let α_{i} (i = 1, 2, …, n) be automorphisms of R, and let δ_{i} (i = 1, 2, …, n) be nilpotent (δ_{i}^{d_{i}} = 0) α_{i}-differentiations of R such that δ_{i}^{−k_{i}}(1) ≠ ∅ for k_{i} = 1, 2, …, d_{i} − 1. Then

K. dim(R[x_{1}, …, x_{n}; α_{1}, …, α_{n}; δ_{1}, …, δ_{n}]) =
K. dim(R[x_{1}, …, x_{n}; α_{1}, …, α_{n}]) = K. dim(R[x_{1}, …, x_{n}]).

If, in addition, R is a left Noetherian ring with finite Krull dimension, then

K. dim(R[x_{1}, …, x_{n}; α_{1}, …, α_{n}; δ_{1}, …, δ_{n}]) = K. dim R + n.

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(Received 02.09.1994)

Author’s address: Department of Mathematics (3) Georgian Technical University 77, M. Kostava St., Tbilisi 380075 Republic of Georgia