ON THE WORD LENGTH OF COMMUTATORS IN GL_{n}(R)

A.S. Sivatski, A.V. Stepanov

Chair of Mathematics II Sankt-Petersburg Electrotechnical University 194356, Sankt-Petersburg, Russia

Abstract. Van der Kallen proved that the elementary group E_{n}(ℂ[x]) does not have bounded word length with respect to the set of all the elementary transvections. Later, Dennis and Vaserstein showed that the same is true even with respect to the set of all commutators. The natural question is: Does the set of all commutators in E_{n}(R) have bounded word length with all the elementary transvections as generators. The article provides a positive answer over a finite dimensional commutative ring R. More precisely, it is proven that if dim Specm R = d and n ≥ 3 then there exists ℓ ∈ ℕ depending only on d and n such that any commutator [a, b] ( a ∈ E_{n}(R), b ∈ SL_{n}(R) ) can be written as a product of ℓ elementary transvections. When R is a localization of an affine algebra over a field, one can take b ∈ GL_{n}(R). Also, it is shown that for such a ring R the set of all commutators in the Steinberg group St_{n}(R) (n ≥ 5) has bounded word length with respect to the usual generators.

INTRODUCTION

Let G be a group with a fixed set of generators Γ. For g ∈ G we denote by L_{Γ}(g) the length of g with respect to Γ, i.e. the smallest number k such that g can be written as a product of k generators. If C ⊆ G, we set L_{Γ}(C) = sup_{g∈C} L_{Γ}(G). We say

that C has bounded word length (with respect to Γ) if L_{Γ}(C) is finite. In the case where G = E_{n}(R) is the elementary group we take Γ to be the set of all elementary transvections unless explicitly stated otherwise.

In his article [3] van der Kallen showed that there exists a commutative ring R such that the length of words in any elementary group E_{n}(R) is not bounded. In fact he proved a stronger theorem, namely that for any field k and n ≥ 3 boundedness of words in E_{n}(k[x]) in terms of transvections implies boundedness of words in K_{2}(k) in terms of Steinberg symbols. Besides he showed that from the latter condition it follows that for sufficiently large m, K_{m}^{M}(k) is torsion. On the other hand, it is well known that for any uncountable field k K_{m}^{M}(k) is not torsion (see for example [7]). In particular, it follows that E_{n}(ℂ[x]) does not have bounded word length.

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Key words and phrases. General linear group, Elementary subgroup, Word length, Dimension of maximal spectrum of a ring.

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

In the article [2] Dennis and Vaserstein showed that for uncountable fields E_{n}(k[x]) remains an unbounded word length group with respect to the set of all commutators [a, b] (a, b ∈ E_{n}(R)) as the generators. So, it is a rather natural problem to determine for an arbitrary ring R whether the set of all commutators in E_{n}(R) has bounded word length.

In this note we give a positive answer to this question for rings of finite dimension. The crucial point in the proof of the main result is a modification of the localization method of A.Bak [1].

The notation used in this paper is more or less standard. All the rings considered in the paper are supposed to be associative, commutative and with unit. A j-prime (Jacobson radical prime) of a ring is a prime which is an intersection of maximal ideals. Recall that the dimension of the maximal spectrum Specm R of a ring R (dim Specm(R)) can be defined as the length of a maximal proper chain of j-primes in R. The localization of a ring R with respect to a multiplicative system S is denoted by S^{−1}R. If the multiplicative system S coincides with {f^{n}, n ≥ 0} then we write S^{−1}R = R_{f}. By GL_{n}(R), E_{n}(R) and St_{n}(R) we denote respectively the general linear, elementary, and Steinberg groups of order n over a ring R. If 𝔭 is an ideal in R then GL_{n}(R, 𝔭) is the principal congruence subgroup, i.e. the kernel of the canonical homomorphism GL_{n}(R) → GL_{n}(R/𝔭). The relative elementary group E_{n}(R, 𝔭) is the minimal normal subgroup of E_{n}(R) containing all the elementary transvections from GL_{n}(R, 𝔭). The elementary transvection with the element ξ at the (i, j)-th position will be denoted by t_{ij}(ξ). If a is an invertible matrix then a_{ij} denotes the entries of a. The entries of a^{−1} are denoted by a′_{ij}. The symbol [a, b] stands for the commutator a^{−1}b^{−1}ab = a^{−1}a^{b}, where a^{b} = b^{−1}ab.

1. THE MAIN RESULT

1.1. Theorem. Let n ≥ 3 and let R be a Noetherian ring with dim Specm R = d < ∞. Then there exists a natural number ℓ depending only on n and d such that each commutator [a, b] of matrices a ∈ E_{n}(R) and b ∈ SL_{n}(R) is a product of at most ℓ elementary transvections.

We start with some lemmas. The first of them is a consequence of the Gauss decomposition.

1.2. Lemma. Let R be a ring of stable rank 1. Then any matrix from SL_{n}(R) is a product of at most \frac{(n−1)(3n+2)}{2} transvections.

Proof. The statement is obvious for n = 1. Let n > 1 and b ∈ SL_{n}(R). There exists λ ∈ R such that the row (b_{n1}, …, b_{n,n−2}, b_{n,n−1} + b_{nn}λ) is unimodular (see [12, theorem 2.3]). Multiplying the matrix bt_{n,n−1}(λ) by n − 1 transvections on the right we can get 1 at the position (n, n). Using another n − 1 right and n − 1 left elementary transformations one gets a matrix with the last row and the last column equal to those of the identity matrix. Thus, by the induction hypothesis, we can represent b as a product of 3n − 2 + \frac{(n−2)(3n−1)}{2} = \frac{(n−1)(3n+2)}{2} transvections.

1.3. Lemma. Let R be a ring such that dim Specm R = d, and b ∈ SL_{n}(R). Then there exist s_{0}, …, s_{d} ∈ R generating R as an ideal such that for each k = 0, …, d the image of b in E_{n}(R_{s_{k}}) is a product of \frac{(n−1)(3n+2)}{2} transvections.

Proof. Let 𝔭_{1}^{(0)}, 𝔭_{2}^{(0)}, …, 𝔭_{h_{0}}^{(0)} be all the minimal j-primes of R (the number of them is finite because they correspond to the irreducible components of Specm R). Consider the localization S_{0}^{−1}R of R at the multiplicative subset S_{0} = R \ ⋃_{i=1}^{h_{0}} 𝔭_{i}^{(0)}. Since this

ring is semilocal, its stable rank equals 1. By the previous lemma the image of b in SL_{n}(S_{0}^{−1}R) is a product of \frac{(n−1)(3n+2)}{2} transvections, and hence the same is true over the ring R_{s_{0}} for some s_{0} ∈ S_{0}.

Let further 𝔭_{1}^{(1)}, 𝔭_{2}^{(1)}, …, 𝔭_{h_{1}}^{(1)} be all the minimal j-primes in R containing s_{0} (i.e. the preimages of the minimal j-primes of R/s_{0}R). Note that each of the 𝔭_{i}^{(1)}’s contains at least one of the 𝔭_{j}^{(0)}’s. As in the first paragraph of the proof, there

exists an element s_{1} ∈ S_{1} = R \ ( ⋃_{i=1}^{h_{1}} 𝔭_{i}^{(1)}) such that the image of b in SL_{n}(R_{s_{1}}) is

a product of at most \frac{(n−1)(3n+2)}{2} transvections.

Continuing the process we obtain elements s_{0}, …, s_{k} such that inverting each of them we can represent b as a product of at most \frac{(n−1)(3n+2)}{2} transvections. Note that the height of any ideal 𝔭_{i}^{(k)} is at least k. Thus for some k ≤ d there is no prime containing all the elements s_{0}, …, s_{k}. That means that the ideal generated by these elements coincides with R.

1.4. The following formulae follows easily from the Steinberg relations. We will use them in the proof of the next lemma to put a transvection to an appropriate place of a product. If i, j, k are distinct indexes then

\begin{aligned}t_{ij}(λ)t_{jk}(ξ) & = t_{jk}(ξ) · t_{ij}(λ)t_{ik}(λξ) \\ t_{ij}(λμ)t_{ji}(ξ) & = t_{ji}(ξ) · t_{jk}(ξλ)t_{ik}(−λ)t_{ki}(−μξ)t_{kj}(−μ)t_{jk}(−ξλ)t_{ik}(λ)t_{ki}(μξ)t_{kj}(μ) \end{aligned}

Let s ∈ R and m ∈ ℕ. We will say that t_{ij}(ξ) ∈ E_{n}(R_{s}) is a transvection of level s^{m} if ξ is in the image of s^{m}R in R_{s}.

1.5. Lemma. Let R be a Noetherian ring, s ∈ R, and let b ∈ SL_{n}(R) be such that its image in SL_{n}(R_{s}) is a product of h transvections. Then there exists a positive integer m such that given a ∈ GL_{n}(R, s^{m}R) the commutator [a, b] is a product of at most f(n, h) ∼ 8n^{2}h transvections.

Proof. Note that given s ∈ R there is a positive integer m′ such that GL_{n}(R, s^{m′}R) injects into GL_{n}(R_{s}) [1, lemma 4.10]. Clearly, if m ≥ m′ then [a, b] ∈ GL_{n}(R, s^{m′}R). We shall show by induction on h that given m′ there is m ≥ m′ such that the image of [a, b] in SL_{n}(R_{s}) can be written as a product of f(n, h) transvections of level s^{m′}. Obviously, this will imply the result.

Let t_{ij}(ξ) ∈ E_{n}(R_{s}) and let c be the image in GL_{n}(R_{s}) of an arbitrary matrix from GL_{n}(R, s^{m}R). For the case h = 1 we have to show that [t_{ij}(ξ), c] is a product of f(n, 1) transvections of level s^{m′} providing m is large enough. This fact will be used also for the induction step. Choose an index p distinct from i and j and

decompose t_{ij}(ξ) into the product ∏_{r=1}^{n} t_{ij}(ξc′_{rp}c_{pr})t_{ip}(−ξc′_{rp}c_{jr}) (cf. “the Theme” of

[8]). The conjugate of each factor of the product above by c is equal to

\begin{aligned}b^{(r)} & = c^{−1}t_{ij}(ξc′_{rp}c_{pr})t_{ip}(−ξc′_{rp}c_{jr})c = e + c′_{∗i}ξc′_{rp}(c_{pr}c_{j∗} − c_{jr}c_{p∗}) = \\ & = [∏_{q≠r} t_{qr}(c′_{qi}ξc′_{rp}), ∏_{q≠r} t_{rq}(c_{pr}c_{jq} − c_{jr}c_{pq})] ∏_{q≠r} t_{rq}(c′_{ri}ξc′_{rp}(c_{pr}c_{jq} − c_{jr}c_{pq})) \end{aligned}

(the latter equality is a version of Whitehead’s lemma). Here c′_{∗i} and c′_{j∗} stand for the i-th column and j-th row of c^{−1} respectively. Note that if m is sufficiently large then all the transvection in the product above are of level s^{m′′} (where m′′ can be chosen arbitrarily) except t_{rp}(c_{pr}c_{jp} − c_{jr}c_{pp}) if r = j and t_{rj}(c_{pr}c_{jj} − c_{jr}c_{pj}) and t_{ir}(c′_{ii}ξc′_{rp}) if r = p. Thus b^{(r)} is a product of 4(n − 1) transvections of level s^{m′′} providing r ≠ j, p. Using formulae 1.4 with m′′ large enough one shows that b^{(j)} (resp. t_{ij}(−ξ)b^{(p)}) is a product of 5n + 21 (resp. 6n + 22) transvections of level s^{m′}. Since b^{(p)} commutes with each of the b^{(r)}’s, the commutator t_{ij}(−ξ)c^{−1}t_{ij}(ξ)c is a product of 4(n − 1)(n − 2) + 11n + 43 transvections of level s^{m′}. Thus, we have shown that f(n, 1) ∼ 4n^{2}.

Suppose now that the image of b in SL_{n}(R_{s}) is written in the form b̃t_{ij}(ξ) where b̃ is a product of h − 1 transvections, and ã is the image of a in GL_{n}(R_{s}). Then, by a well-known commutator formula,

[b̃t_{ij}(ξ), ã] = [b̃, ã][[b̃, ã], t_{ij}(ξ)][t_{ij}(ξ), ã]

By the induction hypothesis one can choose m large enough to guarantee that [b̃, ã] is a product of f(n, h − 1) transvections of level m′′ for any given m′′. Thus, by the previous paragraph of the proof, f(n, h) = f(n, h−1)+2f(n, 1) ∼ 8n^{2}(h−1)+8n^{2} = 8n^{2}h.

1.6. Remark. Instead of decomposition of a transvection one can use Suslin’s lemma ([9, lemma 1.3]). But our bound on f(n, h) is a bit better.

Proof of Theorem 1.1. By Lemma 1.3 we can choose elements s_{0}, …, s_{d}, generating the ring R as an ideal, such that the image of b in E_{n}(R_{s_{k}}) is a product of at most \frac{(n−1)(3n+2)}{2} transvections for each k = 0, …, d. Clearly, for any m the elements s_{0}^{m}, …, s_{d}^{m} generate R. Choose m large enough and decompose a in a product of a^{(k)} ∈ E_{n}(R, s_{k}^{m}R) (k = 0, …, d). By Lemma 1.5, each commutator [a^{(k)}, b] is a product of at most f(n, \frac{(n−1)(3n+2)}{2}) ∼ 12n^{4} transvections. The following lemma completes the proof.

1.7. Lemma. Let b, a^{(0)}, …, a^{(d)} ∈ GL_{n}(R) be such that each commutator [a^{(k)}, b] is a product of at most h transvections. Then the commutator [a^{(0)} ⋯ a^{(d)}, b] is a product of at most g(n, h, d) ∼ 4n^{2}h(d + 1) transvections.

Proof. We proceed by induction on d. The case d = 1 is obvious. Let now a = a^{(1)} ⋯ a^{(d)}. Then [a^{(0)} ⋯ a^{(d)}, b] = [a^{(0)}a, b] = [a^{(0)}, b]^{a}[a, b]. By hypothesis the length of [a^{(0)}, b] is at most h. The same arguments as in the proof of Lemma 1.5 show that the length of a^{−1}ua is equivalent to at most 4n^{2} for any transvection u. Thus, the first factor of the product above has length equivalent to at most 4n^{2}h whenever the second one has length at most g(n, h, d − 1).

This completes the proof of the lemma and of Theorem 1.1.

1.8. Remark. The final bound of the length of a commutator is given by the formula ℓ = ℓ(n, d) ∼ 48n^{6}d (by the equivalence sign we always mean the leading monomial of a polynomial). This bound seems to be reasonable but of course not the best possible. For instance, if n > 2(d + 1) ≥ 2 sr R then using surjective stability of K_{1} and Whitehead’s lemma one can show that ℓ(n, d) = O(n^{4}).

2. IMPROVEMENTS FOR RINGS OF GEOMETRIC ORIGIN

Now we consider the Steinberg group with the usual generators. As an easy consequence of the previous result we get the following result.

2.1. Theorem. Let R be a ring of geometric origin, i.e. a localization of an affine algebra A over a field k, n ≥ 5. Then there exists a constant ℓ which depends only on R and n such that L([x, y]) ≤ ℓ for any x, y ∈ St_{n}(R).

Proof. Since any finite number of elements of R can be regarded as elements of an affine algebra A_{f}, where f ∈ A, and any affine algebra is a quotient of a polynomial ring, we may suppose that x, y ∈ GL_{n}(R) = GL_{n}(F [t_{1}, …, t_{m}]), where m is independent on x and y. First, suppose that m = 0 and so R = F. Let π : St_{n}(F) → E_{n}(F) be the canonical projection. It is evident that there exist elements s_{x}, s_{y} ∈ St_{n}(R) such that xs_{x}^{−1}, ys_{y}^{−1} ∈ K_{2,n}(R) and L(s_{x}), L(s_{y}) ≤ k = \frac{(n−1)(3n+2)}{2}. Since K_{2,n}(R) is central in St_{n}(R) (see for example [7]), we have [x, y] = [s_{x}, s_{y}] and so L([x, y]) ≤ 4k.

Now we can treat the general case. By theorem 1.1 there exists z ∈ St_{n}(R) such that [x, y]z^{−1} ∈ K_{2,n}(R) and L(z) ≤ ℓ_{m}. Since n ≥ 5, by [11], [x, y]z^{−1} = [x(0), y(0)]z(0)^{−1} ∈ K_{2,n}(F), where x(0), y(0), z(0) are the specializations of x, y, z by the projection K_{2,n}(R) → K_{2,n}(R/(t_{1}, …, t_{m})) = K_{2,n}(F). It follows from the previous paragraph that L([x, y]) ≤ 2ℓ_{m} + 2(n − 1)(3n + 2).

2.2. Theorem. Let R be a ring of geometric origin and n ≥ 3. Then there exists ℓ ∈ ℕ depending only on R and n such that L([a, b]) ≤ ℓ for any a ∈ E_{n}(R) and b ∈ GL_{n}(R).

Proof. Let d = ( \begin{matrix}e & 0 \\ 0 & det(b) \end{matrix} ). Then bd^{−1} ∈ SL_{n}(R) and [b, a] = [bd^{−1}d, a] =

[bd^{−1}, a]^{d}[d, a]; so by theorem 1.1 it suffices to prove the claim for matrices a and d. Moreover, similarly to the previous theorem we may suppose that R =

k[t_{1}, …, t_{m}, u, u^{−1}], b = ( \begin{matrix}e & 0 \\ 0 & u \end{matrix} ), and a ∈ E_{n}(k[t_{1}, …, t_{m}]), where t_{1}, …, t_{m}, u are

independent variables. Now we shall argue by induction on m. The case m = 0 is evident. Let A = k[t_{1}, …, t_{m−1}] and B = A[u, u^{−1}]. Without loss of generality we may assume that a_{n2} ≠ 0.

By a suitable change of the variables we may also suppose that a_{n2} ∈ A[t_{m}] is a nonzero polynomial in t_{m} with the leading coefficient in k (see for example [5, Ch.10, Theorem 6]). Let t_{m} = t. By the induction hypothesis applied to the ring B there exists v ∈ E_{n}(B) such that α = [a, b]v^{−1} ∈ GL_{n}(B[t], tB[t]) and L(v) is bounded by some constant c depending only on m and n. To finish the proof we need a few lemmas.

2.3. Lemma. [6, Ch.1, Lemma 4.1] Let C be a semilocal ring, g_{1}, …, g_{p} ∈ C[t], and let \overline{g_{1}}, …, \overline{g_{p}} be their images in C/ Rad(C)[t]. Suppose further that the leading coefficient of at least one of g_{i} is invertible and (\overline{g_{1}}, …, \overline{g_{p}}) = C/ Rad(C)[t]. Then (g_{1}, …, g_{p}) = C[t].

Proof. Let M = C[t]/(g_{1}, g_{2}, …, g_{p}). Since some g_{i} has an invertible leading coefficient, M is finite over C. Thus, because M/ Rad(C)M = 0, Nakayama’s lemma can be applied to get M = 0.

2.4. Lemma. For any finite set of primes 𝔭_{1}, …, 𝔭_{r} ∈ Spec B let S = B \ ⋃_{i=1}^{r} 𝔭_{i}.

Consider a and b as elements of GL_{n}(S^{−1}B[t]). Then L([a, b]) ≤ ρ(n) ∼ 16n^{5}.

Proof. Obviously, l = S^{−1}B/ Rad(S^{−1}B) is a direct product of a finite number of fields. By [7, lemma 13.6], there is a v ∈ S^{−1}B[t] such that (\overline{a_{n1} + va_{nn}}, \overline{a_{n2}}, …, \overline{a_{n,n−1}}) = l[t]. By the previous lemma, we get (a_{n1} + va_{nn}, a_{n2}, …, a_{n,n−1}) = S^{−1}B[t]. Carrying out 2(n − 1) and n − 1 elementary operations on the columns and the rows of the matrix a, respectively, we can transform it to some matrix c ∈ GL(n − 1, S^{−1}B[t]). So, we have a = xcy. Evidently cb = bc, and therefore L([a, b]) = L([xcy, b]) = L([x, b]^{cy}[cy, b]) = L([x, b]^{cy}[y, b]) ≤ ρ(n) ∼ 4n^{2} · 4n^{2} · (n − 1) ∼ 16n^{5}.

2.5. Lemma. [9, Lemma 3.5] Let α ∈ GL_{n}(B[t], tB[t]), f_{1}, …, f_{h} ∈ B, and let α_{f_{i}} be the image of α in GL_{n}(B_{f_{i}}[t], tB_{f_{i}}[t]). Suppose that (f_{1}, …, f_{h}) = B and that for every i α_{f_{i}} ∈ E_{n}(B_{f_{i}}[t]) and L(α_{f_{i}}) ≤ c. Then α ∈ E_{n}(B[t]) and L(α) ≤ g(n, h, c) ∼ (12n^{3})^{h−1}c.

(In the original lemma there is no bound for L(α), but one can extract it easily from the proof. This will be left to the reader).

Now we can go on as in Lemma 1.3 and take some f_{1}, …, f_{m+1} ∈ B such that (f_{1}, …, f_{m+1}) = B and L([a, b]_{f_{i}}) ≤ ρ(n) for any i. But then L(α_{f_{i}}) = L(([a, b]v^{−1})_{f_{i}}) ≤ ρ(n) + r(m − 1, n) and therefore by Lemma 2.5 L(α) ≤ g(n, m + 1, ρ(n) + r(m − 1, n)). Finally L([a, b]) = L(αv) ≤ r(m, n) = g(n, m + 1, ρ(n) + r(m − 1, n)) + r(m − 1, n) Q.E.D.

(One can easily establish from the proof that r(m, n) ∼ 32n^{5}(12n^{3})^{m(m+1)/2}) It is not known yet if the main result of the article holds for rings of non finite

dimension. If this is not the case then it will be possible to give a simple purely algebraic proof of the following result of van der Kallen [4] and Bak [1].

Let R be a ring, n ≥ 3. Then K_{1,n}(R) = GL_{n}(R)/ E_{n}(R) is not abelian in general.

Indeed, let u, v, x_{ij}, y_{ij} (1 ≤ i, j ≤ n) be independent variables, and R = ℤ[u, v, x_{ij}, y_{ij}]/(det(x_{ij})u = 1, det(y_{ij})v = 1). Let further a = (x_{ij}), b = (y_{ij}). Suppose that [a, b] ∈ E_{n}(R), [a, b] = ∏ t_{kh}(ξ_{kh}), and L([a, b]) = ℓ. Then, since in any ring any commutator of n-square invertible matrices a′ and b′ is a specialization of [a, b], we have that L([a′, b′]) ≤ ℓ, contradiction.

We provide now a result on the length of commutators in the stable range. Apparently, the result has been known by specialists for at least 25 years, but has never been published. It follows from the injective stability theorem for K_{1} (see [10], [13]) which says that if R is a ring (not necessarily commutative) and n ≥ sr(R)+2 then E_{n}(R)∩GL_{n−1}(R) = E_{n−1}(R). From Suslin–Tulenbaev’s proof of this theorem one can easily obtain the following statement.

2.7. Proposition. Let R be a ring, k ≥ sr(R) + 2, and a ∈ GL_{k−1}(R) ∩ E_{k}(R) = E_{k−1}(R). Denote by L_{k}(a) and L_{k−1}(a) the length of a in E_{k}(R) and E_{k−1}(R), respectively. Then

L_{k−1}(a) ≤ (32n − 54)L_{k}(a) + (2n + 2)(n − 2).

From this and from Whitehead’s lemma we get the following result which can be considered as a stable version of the theorems of the current article:

2.8. Proposition. Let R be a ring, n ≥ sr(R) + 1, and a, b ∈ GL_{n}(R). Then c = [a, b] ∈ E_{n}(R) and

L_{n}(c) ≤ f(n) ∼ \frac{18n^{2} · 32^{n} · (2n − 2)!}{(n − 2)!}

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e-mail: alexei@stepanov.spb.su