ON UNIMODULAR ROWS OVER POLYNOMIAL RINGS

A.S. Sivatski

Department of Mathematics Sankt-Petersburg Electrotechnical University 194356, Sankt-Petersburg, Russia

Abstract. Let A be a commutative ring, n ≥ 3 positive number. In this note we consider unimodular rows (f_{1}(x), …, f_{n}(x)) over A[x]. We prove that if the row of the leading coefficients of f_{i}(x) is unimodular over A and a ∈ A, then there exists β ∈ E_{n}(A[x]) such that (f_{1}(x), …, f_{n}(x))β = (f_{1}(a), …, f_{n}(a)). Also if A is a Noetherian ring with finite Krull dimension and the row of leading coefficients satisfies the same condition, then we give a bound for the length of β in terms of elementary transvections.

In his work [Su] A. Suslin proved that if A is a commutative ring, a ∈ A, n ≥ 3, f(x) = (f_{1}(x), …, f_{n}(x)) is a unimodular row over A[x] and one of f_{i} is unitary, then there exists β ∈ E_{n}(A[x])SL_{2}(A[x]) such that f(x)β = f(a). (In fact this theorem is a crucial point in his proof of Serre’s conjecture). The natural question is: Is this true if f(x) = (f_{1}(x), …, f_{n}(x)) is an arbitrary unimodular row over A[x] and β is chosen from E_{n}(A[x])? We give a positive answer to this question in the case where the leading coefficients of f_{i} form a unimodular row. Notice that if A is a local ring and the answer to the question is negative (without any restriction on the leading coefficients of the polynomials), then it follows that srA[t] ≥ 3. As far as I know there is no such an example yet.

Also we give a bound for the length of β in terms of transvections in the case where A is a Noetherian ring of finite dimension and the row of leading coefficients of f_{i}(x) belongs to (0, …, 0, 1)GL_{n}(A). Besides we prove that in the case of a Laurent polynomials ring A[x, x^{−1}] the answer in general is negative even for a = 1.

In this note we use the following notation. All the rings considered below are supposed to be commutative, dim A means the Krull dimension of the ring A, SpecmA is the maximal spectrum of A, n is a positive integer not less than 3, E_{n}(A) is the elementary group of order n over the ring A. If 𝔪 is an ideal of A, then we put GL_{n}(A, 𝔪) = {α ∈ GL_{n}(A) : α ≡ 1_{n} mod 𝔪 }. The elementary transvection with the element a at the (i, j)-th position is denoted by t_{ij}(a). By

Typeset by AMS-TEX

1991 Mathematics Subject Classification. 19A13, 19B14, 13C10, (13B25, 13F20).

Key words and phrases. Polynomial ring, Elementary subgroup, Word length, Unimodular row.

The work under this publication was partially supported by INTAS 93-436EXT and DFG– RFBR grant No. 96-01-00092G

Um_{n}(A) we denote the set of all unimodular rows over A. If f, g ∈ A[x], then res (f, g) is the resultant of f and g. The leading coefficient of h ∈ A[x] is denoted by h(∞). If f(x) = (f_{1}(x), …, f_{n}(x)), then f(∞) = (f_{1}(∞), …, f_{n}(∞)) ∈ A^{n}. We set e_{i} = (0, …, 1, …, 0) ∈ A^{n}, where 1 stands at i-th position. The sign over a symbol means that this symbol must be omitted. If A is a ring and α ∈ E_{n}(A), then by L(α) we denote the word length of α, that is the smallest number k such that α can be written as a product of k elementary transvections from E_{n}(A). When giving a bound for the word length we are interested only in the principal part, and so we use in this case the sign ∼.

Theorem 1. Let A be a ring, f(x) = (f_{1}(x), …, f_{n}(x)) ∈ Um_{n}(A[x]), f(∞) ∈ Um_{n}(A), a ∈ A. Then there exists β ∈ E_{n}(A[x]) such that f(x)β = f(a).

Proof. We shall use the following lemmas:

Lemma 2. [SV] Let A be a ring, α ∈ GL_{n}(A), β ∈ E_{n}(A). Then αβα^{−1} ∈ E_{n}(A) and there exists a function l(n, β) ∼ 4n^{2}L(β) independent of α such that L(αβα^{−1}) ≤ l(n, β).

Lemma 3. Let A be a Noetherian ring, dim A = d < ∞, I be an infinite set of indeces, a_{i} ∈ A, ∑_{i∈I} Aa_{i} = A. Then there exist i_{1}, …, i_{d+1} ∈ I such that

∑_{j=1}^{d+1} Aa_{i_{j}} = A.

Proof. It suffices to choose a_{i_{0}} = 0, a_{i_{1}}, …, a_{i_{d+1}} ∈ A such that ht (a_{i_{0}}, …, a_{i_{k}}) ≥ k for 0 ≤ k ≤ d+1. Let we have a_{i_{0}}, …, a_{i_{k}} satisfying this condition, (a_{i_{0}}, …, a_{i_{k}}) ≠ (1) and let p_{1}, … p_{m} be all the minimal primes containing (a_{i_{0}}, …, a_{i_{k}}) (i.e. the

preimages of minimal primes in A/(a_{i_{0}}, …, a_{i_{k}})). Choose a_{i_{k+1}} ∉ ⋃_{j=1}^{m} p_{j}. Since

ht (p_{j}) ≥ k + 1 we get ht (a_{i_{0}}, a_{i_{1}}, …, a_{i_{k+1}}) ≥ k + 1 and the proof is concluded by induction.

Also we need the following lemma which is a slight modification of Lemma 2.3 in [Su].

Lemma 4. Let A be the ring, f(x) = (f_{1}(x), …, f_{n}(x)) ∈ Um_{n}(A[x]), f(∞) ∈ Um_{n}(A), 𝔭 be the ideal of A generated by (res(f_{i}, (f_{1}, …, f̂_{i}, …, f_{n})γe_{1}^{T}) where γ runs over all matrices from E_{n−1}(A[x]) such that L(γ) ≤ n − 2.

i) If f(∞) ∈ Um_{n}(A), then 𝔭 = A.

ii) If f(∞) ∈ Um_{n}(A), A is Noetherian and dim A = d < ∞, then the number of resultants in i) can be bounded by d + 1.

iii) If 𝔭 = A, then for any u ∈ A there exists α ∈ SL_{n}(A[x]) such that f(x)α = f(u).

Proof. i) Let 𝔪 be an arbitrary maximal ideal of A and k = A/𝔪. Since f(∞) is unimodular, f_{i}(∞) ∉ 𝔪 for some i. Let, for instance i = 1, \overline{q} be a prime divisor of \overline{f_{1}} ∈ k[x]. Obviously, there exist \overline{h_{3}}, … \overline{h_{n}} ∈ k[x] such that \overline{q} does not divide

\overline{h} = (\overline{f_{2}} + ∑_{i=3}^{n} \overline{f_{i}}\overline{h_{i}}). Hence by the Chinese remainder theorem we can suppose that

(\overline{f_{1}}, \overline{h}) = (\overline{1}), that is res (\overline{f_{1}}, \overline{h}) ≠ 0. So 𝔭 ⊄ 𝔪. Since \overline{h} = \overline{(f_{2}, …, f_{n})γe_{1}^{T}}, where

γ = ∏_{i=3}^{n} t_{i1}(h_{i}) ∈ E_{n−1}(A[x]) and L(γ) ≤ n − 2, we are done. ii) This follows immediately from part i) and Lemma 3.

iii) This is a slight modification of [Su, Cor. 2.5]. The proof will be left to the reader.

We return to the proof of the theorem. Note that it suffices to prove the theorem only for a = 0 because if f(x)β = f(0), then f(x)ββ(a)^{−1} = f(a).

Let a_{ip} be the coefficients of f_{i}. By the previous lemma there exists a finite

number of γ_{ij} = ∏_{kl} t_{kl}(b_{ijkl}) ∈ E_{n−1}(A[x]) and c_{ij} ∈ A such that

∑_{ij} res(f_{i}, (f_{1}, …, f̂_{i}, …, f_{n})γ_{ij}e_{1}^{T})c_{ij} = 1.

Let further ã_{ip}, b̃_{ijkl}, c̃_{ij}, be independent variables over ℤ corresponding to a_{ip},

b_{ijkl}, c_{ij}. Put F_{i}(x) = ∑_{p} ã_{ip}x^{p} and let γ̃_{ij} ∈ E_{n−1}(ℤ[ã_{ip}, b̃_{ijkl}, c̃_{ij}]) be the matrices

corresponding to γ_{ij}. Consider the ring

Ã = ℤ[ã_{ip}, b̃_{ijkl}, c̃_{ij}]/(∑_{ij} res(F_{i}, (F_{1}, …, F̂_{i}, … F_{n})γ̃_{ij}e_{1}^{T})c̃_{ij} − 1).

It is well known that the resultant of any two polynomials is a linear combination of them. Since (F_{1}, …, F̂_{i}, … F_{n})γ̃_{ij}e_{1}^{T} is in turn a linear combination of F_{1}, …, F̂_{i}, … F_{n} with coefficients in ℤ[ã_{ip}, b̃_{ijkl}, c̃_{ij}]) we conclude that (F_{1}(x), …, F_{n}(x)) is a unimodular row over Ã[x]. By Lemma 4, iii) there exists α ∈ SL_{n}(Ã[x]) such that

(F_{1}(x), …, F_{n}(x))α = (F_{1}(0), …, F_{n}(0))

whence

(F_{1}(0), …, F_{n}(0))α(0) = (F_{1}(0), …, F_{n}(0)).

It follows that

(F_{1}(x), …, F_{n}(x))αα(0)^{−1} = (F_{1}(0), …, F_{n}(0)).

Evidently, α_{1} = αα(0)^{−1} ∈ GL_{n}(Ã[x], xÃ[x]). We want to show that α_{1} ∈ E_{n}(Ã[x]). To do this it suffices to check that α_{1}_{𝔪} ∈ E_{n}(Ã_{𝔪}[x]) ([Su, Theorem 3.1]) for each 𝔪 ∈ Specm(Ã). Let 𝔪 ∈ Specm(Ã). By the definition of Ã u = res(F_{p}, (F_{1}, …, F̂_{p}, … F_{n})γ̃_{pq}e_{1}^{T}) ∉ 𝔪 for some p, q, whence u is invertible in Ã_{𝔪}. Evidently, Ã_{u} = ℤ̃_{u}, where

ℤ̃ = ℤ[ã_{ij}, b̃_{ijkl}, c̃_{ij}, ˆ c̃_{pq}].

Therefore,

Ã_{𝔪}[x] = (Ã_{u})_{𝔪}[x] = ℤ̃_{𝔪}[x].

(We abuse notation to some extent denoting 𝔪 and its preimage in ℤ̃ identically).

By [Su, Cor. 6.5] we have

GL_{n}(ℤ̃[x]) = GL_{n}(ℤ̃)E_{n}(ℤ̃[x]).

From this and [V, Lemma 2.1] we get

GL_{n}(ℤ̃_{𝔪}[x]) = GL_{n}(ℤ̃_{m})E_{n}(ℤ̃_{𝔪}[x]),

and therefore,

α_{1}_{𝔪} ∈ SL_{n}(ℤ̃_{𝔪}[x]) = SL_{n}(ℤ̃_{𝔪})E_{n}(ℤ̃_{𝔪}[x]) = E_{n}(ℤ̃_{𝔪})E_{n}(ℤ̃_{𝔪}[x]) = E_{n}(ℤ̃_{𝔪}[x]).

Thus α_{1} ∈ E_{n}(Ã[x]). Finally, let π : Ã → A be the obvious map. It is evident that

(f_{1}(x), …, f_{n}(x))π(α_{1}) = (f_{1}(0), …, f_{n}(0)),

so the theorem is proved.

Lemma 5. Let A be a ring, f(x) = (f_{1}(x), … f_{n}(x)) ∈ Um_{n}(A[x]), f(∞) ∈

Um_{n}(A). { \atop f_{i}(x)} = x^{deg f_{i}}f_{i}(\frac{1}{x}). Then f̃ = (f̃_{1}, …, f̃_{n}) ∈ Um_{n}(A[x]).

Proof. Let 𝔭 = (f̃_{1}, …, f̃_{n}). Let g_{i} ∈ A[x] such that ∑_{i=1}^{n} f_{i}(x)g_{i}(x) = 1. Then

∑_{i=1}^{n} f_{i}(\frac{1}{x})g_{i}(\frac{1}{x}) = 1. Hence x^{k} ∈ 𝔭 for some k. On the other hand

f̃_{i} ≡ f_{i}(∞) (mod xA[x]),

hence, since (f_{1}(∞), …, f_{n}(∞)) = (1), there exists p ∈ A[x] such that 1 − xp ∈ 𝔭. Therefore, 1 − x^{k}p^{k} ∈ 𝔭, and so 1 ∈ 𝔭.

Corollary 6. Let f_{1}, …, f_{n} be as in Theorem 1. Then there exists β ∈ E_{n}(A[x]) such that f(x)β = f(∞).

Proof. By Theorem 1 and Lemma 5 there exists β_{1}, β_{2} ∈ E_{n}(A[x]) such that f(x)β_{1} = f(1) and f̃(x)β_{2} = f(∞). Putting x = 1 in the last equality we get f(1)β_{2}(1) = f(∞). Hence we can set γ = β_{1}β_{2}(1).

Proposition 7. Theorem 1 does not generalize to the case of a ring of Laurent polynomials over a Noetherian ring for a = 1.

Proof. To give a counterexample we need to use some topological techniques.

For any ring R there is a natural action of E_{n}(R) on Um_{n}(R). Let X be a topological space, C(X) be the ring of real continious functions on X. Evidently every f ∈ Um_{n}(C(X)) determines a map X → ℝ^{n} \ {0}. Since ℝ^{n} \ {0} is homotopically equivalent to 𝕊^{n−1} there is a map

Um_{n}(C(X))/E_{n}(C(X)) → π^{n−1}(X).

In particular, if X = 𝕊^{n−1} we get a map

h : Um_{n}(C(𝕊^{n−1}))/E_{n}(C(𝕊^{n−1})) → ℤ.

Let t_{1}, …, t_{n} be the coordinates on 𝕊^{n−1} so that ∑_{i=1}^{n} t_{i}^{2} = 1. It is clear that

h((t_{1}, …, t_{n−1}, t_{n})) = 1 and h((t_{1}, …, t_{n−1}, −t_{n})) = −1,

hence

(t_{1}, …, t_{n−1}, t_{n}) and (t_{1}, …, t_{n−1}, −t_{n})

are not on the same orbit. Let A = C(𝕊^{n−1}), f = (t_{1}, …, t_{n−1}, t_{n}x) ∈ Um_{n}(A[x, x^{−1}]). Suppose that f(1) = fα with α ∈ E_{n}(A[x, x^{−1}]). Then f(1) = f(−1)α(−1), a contradiction.

Now put A_{1} = ℤ[u_{1}, …, u_{n}]/(∑_{i=1}^{n} u_{i}^{2} − 1). There exists an evident homomor-

phism π : A_{1} ⟶ A that takes u_{i} to t_{i}. Consider the unimodular row f_{1} = (u_{1}, …, u_{n−1}, u_{n}x) over A_{1}[x, x^{−1}]. If f_{1}(1) = f_{1}α_{1} with α ∈ E_{n}(A_{1}[x, x^{−1}]), then f(1) = fπ(α_{1}) and π(α_{1}) ∈ E_{n}(A[x, x^{−1}]) what, as we have just seen, does not hold. On the other hand the ring A_{1} is Noetherian.

The proofs of the following two lemmas are very like the ones of Lemma 2.2 and Proposition 2.4 in [Su].

Lemma 8. Let f(x) = (f_{1}(x), …, f_{n}(x)) ∈ Um_{n}(A[x]), γ ∈ E_{n−1}(A[x]), (f_{1}, …, f̂_{k}, …, f_{n})γ = (g_{1}, …, ĝ_{k}, …, g_{n}), c ∈ (f_{k}A[x]+g_{1}A[x])∩A, (c ∈ (f_{1}A[x]+ g_{2}A[x]) ∩ A if k = 1), b_{1}, b_{2} ∈ A[x], b_{1} ≡ b_{2} (mod cA[x]). Then there exist β ∈ E_{n}(A[x]), δ ∈ SL_{2}(A[x]) and a function r(n, γ) ∼ 4n^{2}L(γ) such that f(b_{2}) = f(b_{1})βδ and L(β) ≤ r(n, γ).

Proof. We give only a sketch of proof. For each 2 ≤ k ≤ n put

SL_{2}^{(k)}(A[x]) = {α = (a_{ij}) ∈ SL_{n}(A[x]) : a_{11}a_{kk} − a_{k1}a_{1k} = 1},

the other elements of α coincide with the corresponding ones of E_{n}. Also put SL_{2}^{(1)}(A[x]) = SL_{2}(A[x]). An evident modification of Lemma 2.2 from [Su] together with Lemma 2 show that there exist β_{1} ∈ E_{n}(A[x]) and δ_{1} ∈ SL_{2}^{(k)}(A[x]) such that f(b_{2}) = f(b_{1})β_{1}δ_{1} and L(β_{1}) ≤ r(n, γ) ∼ 4n^{2}L(γ). Note that SL_{2}^{(k)}(A[x]) = α_{k}^{−1}SL_{2}(A[x])α_{k} for some α_{k} ∈ E_{n}(A[x]) with L(α_{k}) = 2. Therefore,

f(b_{2}) = f(b_{1})β_{1}δ_{1} = f(b_{1})β_{1}α_{k}^{−1}δα_{k} = f(b_{1})β_{1}α_{k}^{−1}(δα_{k}δ^{−1})δ,

for some δ ∈ SL_{2}(A[x]) and we can put β = β_{1}α_{k}^{−1}(δα_{k}δ^{−1}).

Lemma 9. Let dim A = d, f(x) = (f_{1}(x), …, f_{n}(x)) ∈ Um_{n}(A[x]), f(∞) ∈ Um_{n}(A), b, b′ ∈ A[x]. Then there exist β ∈ E_{n}(A[x]), δ ∈ SL_{2}(A[x]) and a function u(n, d) ∼ 16n^{5}d such that f(b)βδ = f(b′) and L(β) ≤ u(n, d).

Proof. By Lemma 4 there exist b = b_{0}, b_{1}, …, b_{d+1} = b′ ∈ A and γ_{1}, …, γ_{d+1} ∈ E_{n−1}(A[x]) such that b_{i} ≡ b_{i−1} mod (c_{i}A[x]) and L(γ_{i}) ≤ (n − 2), where c_{i} = res(f_{i}, (f_{1}, …, f̂_{i}, … f_{n})γ_{i}e_{1}^{T}). By the previous lemma we have f(b_{i}) = f(b_{i−1})β_{i}δ_{i} where β_{i} ∈ E_{n}(A[x]), δ_{i} ∈ SL_{2}(A[x]) and L(β_{i}) ≤ r(n, γ_{i}) ∼ 4n^{2}(n − 2). Hence f(b)β_{1}δ_{1} … β_{d+1}δ_{d+1} = f(b′) or, in other words,

f(b)β_{1}(δ_{1}β_{2}δ_{1}^{−1})(δ_{1}δ_{2}β_{3}δ_{2}^{−1}δ_{1}^{−1}) … (δ_{1}δ_{2} … δ_{d+1}) = f(b′).

Thus we can set

β = β_{1}(δ_{1}β_{2}δ_{1}^{−1}) … (δ_{1} … δ_{d}β_{d+1}(δ_{1} … δ_{d})^{−1}), δ = δ_{1} … δ_{d+1}.

Applying Lemma 2 we get the required statement.

In the sequel if f(x) = (f_{1}(x), …, f_{n}(x)) is a row of polynomials and deg f = max deg f_{i}, we put f̃(x) = (x^{deg f}f_{1}(x), …, x^{deg f}f_{n}(x)). Evidently Lemma 5 remains valid in this interpretation of f̃. By the leading coefficient of f_{i} we mean now the coefficient of f_{i} at x^{deg f} and by f(∞) we denote the row of the leading coefficients of f_{i}.

Proposition 10. Let A be a Noetherian ring, dim A = d < ∞, a_{1}, a_{2} ∈ A ∪ {∞}, f(x) = (f_{1}(x), …, f_{n}(x)) ∈ Um_{n}(A[x]), f(∞) ∈ Um_{n}(A). Then the following holds:

i) There exist β ∈ E_{n}(A[x]), δ ∈ SL_{2}(A[x]) and a function F (n, d) ∼ 64n^{7}d such that f(x)βδ = f(a_{1}) and L(β) ≤ F (n, d).

ii) If f_{i} is unitary for some i, then there exist β ∈ E_{n}(A[x]) and a function G(n, d) ∼ 8n^{2}F (n, d) such that f(x)β = f(a_{1}) and L(β) ≤ G(n, d).

iii) If f(a_{2}) ∈ e_{1}GL_{n}(A), then there exist β ∈ E_{n}(A[x]) and a function H(n, d) ∼ 64n^{4}F (n, d) such that f(x)β = f(a_{1}), and L(β) ≤ H(n, d).

Proof. i) If a_{1} ≠ ∞, then the assertion follows from Lemma 9. Let a_{1} = ∞. By Lemma 5 f̃(x) ∈ Um_{n}(A[x]). Since f̃(∞) = f(0) ∈ Um_{n}(A) we obtain by Lemma 4 that there exist β_{1}, β_{2} ∈ E_{n}(A[x]) and δ_{1}, δ_{2} ∈ SL_{2}(A[x]) such that f̃(x)β_{1}δ_{1} = f̃(1) = f(1) = f(x)β_{2}δ_{2}. Putting here x = 0 we get f(∞)β_{1}(0)δ_{1}(0) = f̃(0)β_{1}(0)δ_{1}(0) = f(1). Therefore,

f(∞) = f(x)β_{2}δ_{2}δ_{1}^{−1}(0)β_{1}^{−1}(0) = f(x)(δ_{2}δ_{1}^{−1}(0)(δ_{1}(0)δ_{2}^{−1}β_{2}δ_{2}δ_{1}^{−1}(0))β_{1}^{−1}(0),

and so we can choose β ∈ E_{n}(A[x]) such that L(β) ≤ F (n, d) ∼ 4n^{2}·16n^{5}d = 64n^{7}d.

ii) By part i) there exists β_{1} ∈ E_{n}(A[x]) and δ ∈ SL_{2}(A[x]) such that f(x)β_{1}δ = f(∞) and L(β_{1}) ≤ F (n, d). It is easy to see that there exists β_{2} ∈ E_{n}(A[x]) such that f(∞)β_{2} = e_{n} and L(β_{2}) ≤ 2(n − 1). So f(x)β_{1}δβ_{2} = e_{n}, and, since e_{n}SL_{2}(A[x]) = e_{n}, we get f(x)β_{1}δβ_{2}δ^{−1} = e_{n}. Hence f(a_{1})β_{1}δβ_{2}δ^{−1}(a_{1}) = e_{n} for any a_{1} ∈ A. Thus f(x)β = f(a_{1}), where β = β_{1}(δβ_{2}δ^{−1})(δβ_{2}^{−1}δ^{−1})(a_{1})β_{1}^{−1}(a_{1}). Obviously L(β) ≤ γ(n, d) ∼ 8n^{2}F (n, d).

iii) Let at first a_{2} = 0, a_{1} = ∞, α ∈ GL_{n}(A) such that (f(x)α)(0) = e_{1}. Then by i) there exists β_{1} ∈ E_{n}(A[x]) and δ_{1} ∈ SL_{2}(A[x]) such that L(β_{1}) ≤ F (n, d) and (f(x)α)β_{1}δ_{1} = f(0)α = e_{n}. Since e_{n}SL_{2}(A[x]) = e_{n} we have f(x)αβ_{1}α^{−1} = f(0). On the other hand, since f̃α(∞) = f̃(∞)α = f(0)α, we have by ii) f(0)α = f̃(x)αβ_{2} for some β_{2} ∈ E_{n}(A[x]) such that L(β_{2}) ≤ G(n, d). Putting x = 0 in the last equality we get f(0)α = f̃(0)αβ_{2}(0), that is f(0) = f(∞)αβ_{2}(0)α^{−1}. Therefore, f(x)β = f(∞), where β = (αβ_{1}α^{−1})(αβ_{2}^{−1}(0)α^{−1}) and, evidently L(β) ≤ H(n, d) ∼ 4n^{2}G(n, d) ∼ 32n^{4}F (n, d). The case where a_{2} is an arbitrary element of A and a_{1} = ∞ is treated by substituting f(x) for g(x) = f(x + a_{2}). Hence if a_{1}, a_{2} ∈ A, then there exists β_{1} ∈ E_{n}(A[x]) such that f(x)β_{1} = f(∞), and so f(a_{1})β_{1}(a_{1}) = f(∞). Thus f(x)(β_{1}β_{1}(a_{1})^{−1}) = f(a_{1}) and so we can choose β satisfying iii) with L(β) ≤ 64n^{4}F (n, d). The case where a_{2} = ∞ is treated in an analogous way.

Finally we pose several open questions.

Question 1. Is Theorem 1 true in general if one does not suppose that f(∞) ∈ Um_{n}(A) (at least for a regular ring A) ?

Question 2. Is there any bound for L(β) in Theorem 1 independent of dim A ? Question 3. Is it possible to generalize Theorem 1 to the ring of Laurent

polynomials A[x, x^{−1}] and a = 1 if dim A is not ”large” in comparison with n?

Question 4. Is it possible anyhow to generalize Theorem 1 to the polynomial ring in many variables ?

Question 5. The bounds in Proposition 10 seem to be very rough. Is it possible to improve them ?

REFERENCES

[1] Suslin A.A., On the structure of the general linear group over polynomial rings, Soviet Math. Izv. 41 (1977), no. 2, 503–516.

[2] Stepanov A.V., Vavilov N.A., Decomposition of transvections. A theme with variations, to appear in K-theory.

[3] Vorst T., The general linear group of polynomial rings over regular rings, Comm. Alg 9 (1981), no. 5, 499–509.

e-mail: sivatsky@AS3476.spb.edu