The maximal order property for quantum determinantal rings

T H Lenagan and L Rigal^{∗}

Abstract

We develop a method of reducing the size of quantum minors in the algebra of quantum matrices O_{q}(M_{n}). We use the method to show that the quantum determinantal factor rings of O_{q}(M_{n}(ℂ)) are maximal orders, for q an element of ℂ transcendental over ℚ.

2000 Mathematics subject classification: 16P40, 16W35, 20G42. Keywords: quantum matrices, quantum minors, quantum determinantal rings, maximal orders.

Introduction

Throughout, 𝕂 will denote a base field, q a non-zero element of 𝕂, and m, n positive integers. We denote by O_{q}(M_{m,n}) the quantization of the ring of regular functions on m × n matrices with entries in 𝕂; it is the 𝕂-algebra generated by mn indeterminates X_{ij}, 1 ≤ i ≤ m and 1 ≤ j ≤ n, subject to the following relations:

\begin{aligned}X_{ij}X_{il} & = & qX_{il}X_{ij}, \\ X_{ij}X_{kj} & = & qX_{kj}X_{ij}, \\ X_{il}X_{kj} & = & X_{kj}X_{il}, \\ X_{ij}X_{kl} − X_{kl}X_{ij} & = & (q − q^{−1})X_{il}X_{kj}, \end{aligned}  (1)

for 1 ≤ i < k ≤ m and 1 ≤ j < l ≤ n.

It will be convenient to use the following notation: setting X = (X_{ij})_{1≤i≤m,1≤j≤n} we will denote O_{q}(M_{m,n}) by 𝕂_{q}[X].

Let A be any 𝕂-algebra, we say that an m × n matrix (a_{ij}) with entries in A is a qquantum matrix if the map X_{ij} ↦ a_{ij} induces a homomorphism of algebras 𝕂_{q}[X] ⟶ A. If, in addition, this homomorphism is a monomorphism, then we say that the matrix (a_{ij}) is a generic q-quantum matrix; clearly, X is a generic q-quantum matrix, by definition.

∗This research was partially supported by grants from the Edinburgh and London Mathematical Societies and the European Science Foundation programme ‘Noncommutative Geometry’.

In the case where m = n, the quantum determinant of X is defined by

det_{q} = det_{q}X := ∑_{σ∈S_{n}}(−q)^{l(σ)}X_{1,σ(1)} … X_{n,σ(n)};

from [8, Theorem 4.6.1], we know that det_{q} is in the centre of O_{q}(M_{n}) := O_{q}(M_{n,n}). Clearly, if I ⊆ ℕ_{m} := {1, …, m} and J ⊆ ℕ_{n} := {1, …, n} with |I| = |J| = t, that is, I and J both have t elements, then the submatrix obtained from X by keeping the rows indexed by elements of I and the columns indexed by elements of J is a generic q-quantum matrix; and so, we can speak of its quantum determinant. Such an element is called a t × t quantum minor of X and is denoted by [I|J]. In order to simplify the notation, if I and J are given by the explicit list of their elements: I = {i_{1} < … < i_{t}} and J = {j_{1} < … < j_{t}} we will use the notation [i_{t}, …, i_{1}|j_{1}, …, j_{t}]. It follows at once from the above that,

for I ⊆ ℕ_{m}, J ⊆ ℕ_{n}, with |I| = |J| = t, i ∈ I and j ∈ J, [I|J]X_{ij} = X_{ij}[I|J]. (2)

We need to use several identities involving quantum minors. Many of these are obtained from [8]. However, it is worth noting that the conventions used in [8] are slightly different from ours. In order to make the notation fit we must replace q by q^{−1} each time we use a relation from [8]. For convenience, some of the identities we use are collected in an Appendix.

For any positive integer t such that t ≤ min{m, n}, we denote by I_{t}(X) the twosided ideal of 𝕂_{q}[X] generated by the t × t quantum minors of X; such ideals will be referred to as quantum determinantal ideals. For each such t, we define the quantum determinantal ring R_{t}(X) := 𝕂_{q}[X]/I_{t}(X). It has been proved in [1] that R_{t}(X) is a domain. Here, we are interested in the question as to whether or not R_{t}(X) is a maximal order in its division ring of fractions. This has already been established for R_{2}(X) in [9]. We will show that the localised ring R_{t}(X)[x_{1n}^{−1}], where x_{1n} := X_{1n} + I_{t}(X), is a maximal order for each t, and are then able to deduce that R_{t}(X) is a maximal order, in the case that 𝕂 = ℂ and q is an element of ℂ transcendental over ℚ. The question for general 𝕂 and any q remains open: we conjecture that all quantum determinantal rings are maximal orders. The problem is a technical one: it is necessary to show that a certain ideal is a prime ideal. The restriction to q an element of ℂ transcendental over ℚ is because the relevant ideal is shown to be prime in [5] for this case. We can answer the question for general 0 ≠ q ∈ 𝕂 for the case of R_{n}(X) (= O_{q}(M_{n})/⟨det_{q}⟩), when X is n × n, and also for the case R_{3}(X) for general m × n. A proof of the former case is included in this paper since it is relatively short, and of independent interest. The proof of the latter is not given here, since it is somewhat ad-hoc and rather long.

1 A new set of generators for 𝕂_{q}[X]_{X_{1n}}

It follows at once from (1) that X_{1n} is a normal element of 𝕂_{q}[X]. So, {X_{1n}^{i}, i ∈ ℕ} is a right denominator set of 𝕂_{q}[X]; the corresponding localisation of 𝕂_{q}[X] will be denoted 𝕂_{q}[X]_{X_{1n}}.

For the remainder of this section we assume that min{m, n} ≥ 2. In 𝕂_{q}[X]_{X_{1n}}, we consider the following elements, for 2 ≤ i ≤ m and 1 ≤ j ≤ n − 1:

Xij^{′} = X_{ij} − q^{−1}X_{1j}X_{in}X_{1n}^{−1} = −q^{−1}[i, 1|j, n]X_{1n}^{−1}.  (3)

Set

X̃ = ( \begin{matrix}X_{11} & … & X_{1,n−1} & X_{1n} \\ X′_{21} & … & X′_{2,n−1} & X_{2n} \\ ⋮ & & ⋮ & ⋮ \\ X′_{m1} & … & X′_{m,n−1} & X_{mn} \end{matrix} ) and X′ = ( \begin{matrix}X′_{21} & … & X′_{2,n−1} \\ ⋮ & & ⋮ \\ X′_{m1} & … & X′_{m,n−1} \end{matrix} )

with the convention that the index of a row in X′ is actually the index of the corresponding row in X̃. So, for instance X′_{21} … X′_{2,n−1} is referred to as the row of index 2 in X′.

It is clear that the entries of X̃ together with X_{1n}^{−1} form a new set of algebra generators for 𝕂_{q}[X]_{X_{1n}}. Our next aim is to give a description of 𝕂_{q}[X]_{X_{1n}} based on this set of generators.

In order to do this, we need to know the relations between the entries of the matrix X̃. This problem is dealt with in Lemma 1.1 and Proposition 1.2 where we obtain a set of relations. We will show later that the relations obtained in these results give the complete set of relations between entries of X̃.

Lemma 1.1 The matrix X′ is a q-quantum matrix, and all its entries commute with X_{1n}.

Proof: Let 2 ≤ i, k ≤ m and 1 ≤ j, l ≤ n − 1. From (2) we see that X_{1n} commutes with the 2×2 quantum minor [i, 1|j, n]. So X′_{ij}X_{1n} = X_{1n}X′_{ij} follows from the second equality of (3). Moreover, from (3) again, we have X′_{ij} = −q^{−1}[i, 1|j, n]X_{1n}^{−1} and X′_{kl} = −q^{−1}[k, 1|l, n]X_{1n}^{−1}. The desired relations between X′_{ij} and X′_{kl} are deduced in an obvious manner from [8, Theorem 5.2.1] applied to the submatrix of X defined by rows 1, i, k and columns j, l, n.

The following proposition is a list of relations between entries of X̃. Relations between two elements of the first row or of the last column of X̃ are deduced from the fact that X is a q-quantum matrix.

Proposition 1.2 The following relations hold between entries of the matrix X̃.

1) If 1 ≤ j ≤ n − 1 and 2 ≤ i ≤ m then:

X_{1j}X_{in} − q^{2}X_{in}X_{1j} = q(q^{2} − 1)X′_{ij}X_{1n}.

2.1) For 1 ≤ j ≤ n − 1 and 2 ≤ k ≤ m,

X_{1j}X′_{kl} − X′_{kl}X_{1j} = (q^{−1} − q)X_{1l}X′_{kj}, for 1 ≤ l ≤ j − 1,
X_{1j}X′_{kj} = q^{−1}X′_{kj}X_{1j},
X_{1j}X′_{kl} = X′_{kl}X_{1j}, for j + 1 ≤ l ≤ n − 1.

2.2) For 2 ≤ i ≤ m and 1 ≤ l ≤ n − 1,

X_{in}X′_{kl} = X′_{kl}X_{in}, for 2 ≤ k ≤ i − 1,
X_{in}X′_{il} = qX′_{il}X_{in},
X_{in}X′_{kl} − X′_{kl}X_{in} = (q − q^{−1})X_{kn}X′_{il}, for i + 1 ≤ k ≤ m.

3) If 1 ≤ k < l ≤ n and 1 ≤ i < j ≤ m then

X_{1k}X_{1l} = qX_{1l}X_{1k} and X_{in}X_{jn} = qX_{jn}X_{in}.

Proof: Part 3) is obvious. The other relations are all obtained in the following way. First, use the definition (3) to translate the desired relation into a new one that uses only the entries of X, by multiplying by X_{1n}. The relation so obtained involves entries coming from a certain submatrix of X of size at most 3 × 3. Use Relation 5.1 and Relations 5.2 to check that the relation holds. Since these are easy but tedious computations we omit the details.

The next thing we want to do is to show that X′ is a generic q-quantum matrix. We use a computation with Gelfand-Kirillov dimension to do this. The following lemma is probably well-known, but we have not located the exact statement that we need; and so we include a proof.

Lemma 1.3 Let B be a 𝕂-algebra. Suppose A is a subalgebra of B and x an element of B such that B is generated by A and x as an algebra. Further, suppose there exists a finite dimensional subspace V of A that generates A as an algebra and such that xV ⊆ V x + A. Then GKdim (B) ≤ GKdim (A) + 1.

Proof: Without loss of generality, we may assume that 1 ∈ V. We denote by W the subspace of B spanned by x and V. Thus, W is a finite dimensional subspace of B that generates B as an algebra. Since xV ⊆ V x + A and V is finite dimensional, there exists m ∈ ℕ^{∗} such that xV ⊆ V x + V^{m}. Then, an easy induction shows that, for n ∈ ℕ, xV^{n} ⊆ V^{n}x + V^{m+n}. We claim that,

W^{n} ⊆ V^{nm} + V^{nm}x + ... + V^{nm}x^{n},  (4)

for all n ∈ ℕ. The inclusion (4) is trivial for n = 0 (recall the standard convention that V^{0} = W^{0} = 𝕂). Now, assume (4) holds for p ∈ ℕ, that is W^{p} ⊆ V^{pm} +V^{pm}x+...+V^{pm}x^{p}. It follows that

xW^{p} ⊆ xV^{pm} + xV^{pm}x + ... + xV^{pm}x^{p}
⊆ V^{pm}x + V^{pm+m} + V^{pm}x^{2} + V^{pm+m}x + ... + V^{pm}x^{p+1} + V^{pm+m}x^{p}.

So xW^{p} ⊆ V^{m(p+1)} + ... + V^{m(p+1)}x^{p+1}. On the other hand, V W^{p} ⊆ V^{mp+1} + ... + V^{mp+1}x^{p}. We finally obtain W^{p+1} = (V + 𝕂x)W^{p} ⊆ V^{m(p+1)} + ... + V^{m(p+1)}x^{p+1}. This establishes (4) by induction on n.

From (4), it follows that dim W^{n} ≤ (n + 1) dim V^{mn}, for all n ∈ ℕ, and thus

\overline{lim} log_{n} dim W^{n} ≤ \overline{lim} log_{n} dim V^{mn} + 1.

This establishes that GKdim B ≤ GKdim A + 1.

Proposition 1.4 The matrix X′ is a generic q-quantum matrix.

Proof: In this proof, we denote by A the subalgebra of 𝕂_{q}[X]_{X_{1n}} generated by the entries of X′. By Lemma 1.1, there is a surjective morphism ϕ : O_{q}(M_{m−1,n−1}) ⟶ A. We want to prove that ker ϕ = (0). Assume that this is not the case. Then, since O_{q}(M_{m−1,n−1}) is a domain, ker ϕ must contain a regular element of O_{q}(M_{m−1,n−1}) and thus we have GKdim A < GKdim O_{q}(M_{m−1,n−1}) = (m − 1)(n − 1).

By Lemma 1.1, we see that the subalgebra B of 𝕂_{q}[X]_{X_{1n}} generated by A and X_{1n} is an extension of A of the type investigated in Lemma 1.3. Thus, we must have GKdim B < (m − 1)(n − 1) + 1. Moreover, since X_{1n} is central in B, [4, Proposition 4.2] shows that GKdim B_{X_{1n}} < (m − 1)(n − 1) + 1. Now, it is clear that 𝕂_{q}[X]_{X_{1n}} can be obtained by successive algebra extensions starting from B_{X_{1n}} and adding (in this order) X_{11}, …, X_{1,n−1} and then X_{mn}, …, X_{2n} (this is because 𝕂_{q}[X]_{X_{1n}} is generated by the entries of X̃ and X_{1n}^{−1}). Moreover, Proposition 1.2 shows that at each step, the extension is of the type investigated in Lemma 1.3. Thus, (m − 1) + (n − 1) applications of this Lemma show that we must have GKdim 𝕂_{q}[X]_{X_{1n}} < (m − 1)(n − 1) + 1 + (m − 1) + (n − 1) = mn.

However, we know that GKdim 𝕂_{q}[X]_{X_{1n}} ≥ GKdim 𝕂_{q}[X] = mn. This is a contradiction and thus we have proved that ker ϕ = (0).

Proposition 1.4 states that the subalgebra of 𝕂_{q}[X]_{X_{1n}} generated by X′ is isomorphic to O_{q}(M_{m−1,n−1}); for this reason, we denote it by 𝕂_{q}[X′].

The following remark will be useful in what follows.

Remark 1.5 We denote by F⟨X_{1}, …, X_{p}⟩ the free 𝕂-algebra on p generators X_{1}, …, X_{p}. Let I be the ideal generated by elements f_{1}, …, f_{s} ∈ F⟨X_{1}, …, X_{p}⟩, and set A := F⟨X_{1}, …, X_{p}⟩/I. Finally, let σ be an automorphism of F⟨X_{1}, …, X_{p}⟩ and δ be a left σ-derivation of F⟨X_{1}, …, X_{p}⟩ such that σ(I) = I and δ(I) ⊆ I. We denote by \overline{σ} the automorphism of A induced by σ and by \overline{δ} the left \overline{σ}-derivation of A induced by δ. Then, A[x; \overline{σ}, \overline{δ}] is isomorphic to the algebra F ⟨X_{1}, …, X_{n}, X⟩/J where J is the ideal of F ⟨X_{1}, …, X_{n}, X⟩ generated by f_{1}, …, f_{s} and the p elements XX_{i} − σ(X_{i})X − δ(X_{i}) (here we identify F⟨X_{1}, …, X_{p}⟩ and the subalgebra of F⟨X_{1}, …, X_{p}, X⟩ generated by X_{1}, …, X_{p}).

We now proceed to show that 𝕂_{q}[X]_{X_{1n}} can be described as an iterated Ore extension of its subalgebra 𝕂_{q}[X′] (recall the notation we fixed immediately after Proposition 1.4. For this, we first construct an algebra B which is an iterated Ore extension of O_{q}(M_{m−1,n−1}). We then show that B is isomorphic to 𝕂_{q}[X]_{X_{1n}}.

Let us start with the algebra O_{q}(M_{m−1,n−1}). For convenience of notation, we denote its canonical generators by Y′_{ij} for 2 ≤ i ≤ m and 1 ≤ j ≤ n − 1. Thus, Y′ = (Y′_{ij})_{2≤i≤m,1≤j≤n−1} is a generic q-quantum matrix and, following our previous notation, we have O_{q}(M_{m−1,n−1}) = 𝕂_{q}[Y′]. Now, we consider the Laurent polynomial extension A = 𝕂_{q}[Y′][Y_{1n}^{±1}] obtained from 𝕂_{q}[Y′] adding a central indeterminate denoted Y_{1n} as well as its inverse.

By Remark 1.5, we know how to describe A as a quotient of a free algebra. Then, it is easy to see that we can define a (unique) left skew derivation (σ_{11}, δ_{11}) of A such that δ_{11} = 0 and such that, σ_{11}(Y_{1n}) = qY_{1n} and, for 2 ≤ k ≤ m and 2 ≤ l ≤ n − 1, σ_{11}(Y′_{k1}) = q^{−1}Y′_{k1} and σ_{11}(Y′_{kl}) = Y′_{kl}. We put A_{11} = A[Y_{11}; σ_{11}, δ_{11}].

Now, for 2 ≤ p ≤ n − 1, we construct an algebra A_{1p} by induction on p. For 1 ≤ p ≤ n − 2, by Remark 1.5, we know how to describe A_{1p} as a quotient of a free algebra. It is then easy to check that we can define a (unique) left skew derivation (σ_{1,p+1}, δ_{1,p+1}) of A_{1p} such that, for 2 ≤ k ≤ m :

σ_{1,p+1}(Y′_{kl}) = Y′_{kl}  and δ_{1,p+1}(Y′_{kl}) = (q^{−1} − q)Y_{1l}Y′_{k,p+1} for 1 ≤ l ≤ p,
σ_{1,p+1}(Y′_{k,p+1}) = q^{−1}Y′_{k,p+1} and δ_{1,p+1}(Y′_{k,p+1}) = 0
σ_{1,p+1}(Y′_{kl}) = Y′_{kl} and δ_{1,p+1}(Y′_{kl}) = 0 for p + 2 ≤ l ≤ n − 1
σ_{1,p+1}(Y_{1n}) = qY_{1n}  and δ_{1,p+1}(Y_{1n}) = 0
σ_{1,p+1}(Y_{1l}) = q^{−1}Y_{1l}  and δ_{1,p+1}(Y_{1l}) = 0  for 1 ≤ l ≤ p.

Then, for 1 ≤ p ≤ n − 2, we put A_{1,p+1} = A_{1,p}[Y_{1,p+1}; σ_{1,p+1}, δ_{1,p+1}].

Thus, we now have a first sequence of Ore extensions: A, A_{11},..., A_{1,n−1}. To finish the construction of B, we need a second such sequence which we now define.

By Remark 1.5, we know how to descibe A_{1,n−1} as a quotient of a free algebra. Then, it is easy to see that we can define a (unique) left skew derivation (σ_{mn}, δ_{mn}) of A_{1,n−1} such that, for 1 ≤ l ≤ n − 1:

σ_{mn}(Y′_{kl}) = Y′_{kl}  and δ_{mn}(Y′_{kl}) = 0  for 2 ≤ k ≤ m − 1,
σ_{mn}(Y′_{ml}) = qY′_{ml}  and δ_{mn}(Y′_{ml}) = 0
σ_{mn}(Y_{1n}) = q^{−1}Y_{1n} and δ_{mn}(Y_{1n}) = 0
σ_{mn}(Y_{1l}) = q^{−2}Y_{1l} and δ_{mn}(Y_{1l}) = (q^{−1} − q)Y′_{ml}Y_{1n}.

We put A_{mn} = A_{1,n−1}[Y_{mn}; σ_{mn}, δ_{mn}].

Now, for 2 ≤ s ≤ m − 1, we construct an algebra A_{sn}, by decreasing induction on s. For 3 ≤ s ≤ m, by Remark 1.5, we know how to describe A_{sn} as a quotient of a

free algebra. It is then easy to check that we can define a (unique) left skew derivation (σ_{s−1,n}, δ_{s−1,n}) of A_{sn} such that, for 1 ≤ l ≤ n − 1 :

σ_{s−1,n}(Y′_{kl}) = Y′_{kl}  and δ_{s−1,n}(Y′_{kl}) = 0  for 2 ≤ k ≤ s − 2
σ_{s−1,n}(Y′_{s−1,l}) = qY′_{s−1,l} and δ_{s−1,n}(Y′_{s−1,l}) = 0
σ_{s−1,n}(Y′_{kl}) = Y′_{kl}  and δ_{s−1,n}(Y′_{kl}) = (q − q^{−1})Y_{kn}Y′_{s−1,l} for s ≤ k ≤ m
σ_{s−1,n}(Y_{1n}) = q^{−1}Y_{1n}, and δ_{s−1,n}(Y_{1n}) = 0
σ_{s−1,n}(Y_{1l}) = q^{−2}Y_{1l},  and δ_{s−1,n}(Y_{1l}) = (q^{−1} − q)Y′_{s−1,l}Y_{1n}
σ_{s−1,n}(Y_{kn}) = qY_{kn},  and δ_{s−1,n}(Y_{kn}) = 0  for s ≤ k ≤ m.

Finally, we get a sequence of Ore extensions: A, A_{11},..., A_{1,n−1},A_{mn},...,A_{2n}, and we put B = A_{2n}. Iterative applications of Remark 1.5 show that B can be easily described as the quotient of a free algebra in mn generators. Clearly, we have constructed B in such a way that the relations between the mn generators of this algebra are exactly the same as those holding between the generators of X̃ as noted in Proposition 1.1.2. It follows at once that we can define a morphism of algebras

φ : B ⟶ 𝕂_{q}[X]_{X_{1n}}
Y′_{kl} ↦ X′_{kl} for 2 ≤ k ≤ m and 1 ≤ l ≤ n − 1
Y_{1n} ↦ X_{1n}
Y_{1l} ↦ X_{1l}  for 1 ≤ l ≤ n − 1
Y_{kn} ↦ X_{kn}  for 2 ≤ k ≤ m

Proposition 1.6 The morphism φ is an isomorphism.

Proof: The surjectivity of φ is obvious since the entries of X̃ together with X_{1n}^{−1} form a set of algebra generators for 𝕂_{q}[X]_{X_{1n}}. It remains to prove that ker φ = (0). Recall that B is obtained from A = O_{q}(M_{m−1,n−1})[X_{1n}^{−1}] by (m − 1) + (n − 1) succesive Ore extensions that are all extensions of algebras of the type investigated in Lemma 1.3. Thus, we have GKdim B ≤ GKdim A + (m − 1) + (n − 1). On the other hand, A is a Laurent polynomial extension of O_{q}(M_{m−1,n−1}) thus GKdim A = GKdim O_{q}(M_{m−1,n−1})+1 = (m−1)(n−1)+1, by [4, Proposition 3.5 and Proposition 4.2]. All this together gives

GKdim B ≤ (m − 1)(n − 1) + 1 + (m − 1) + (n − 1) = mn.

On the other hand, we have GKdim 𝕂_{q}[X]_{X_{1n}} ≥ GKdim 𝕂_{q}[X] = mn, since 𝕂_{q}[X] is a subalgebra of 𝕂_{q}[X]_{X_{1n}}.

Now, B is clearly a domain; and so if we assume that ker φ ≠ 0 then GKdim B/ ker φ < GKdim B = mn. But this is a contradiction since B/ ker φ ≅ 𝕂_{q}[X]_{X_{1n}}. Thus, φ is injective.

2 Reduction of the size of quantum minors

Using the previous section we are now able to link k × k quantum minors of X, for k ≥ 2, with (k − 1) × (k − 1) quantum minors of X′ (provided X_{1n} is invertible, of course).

Recall from the introduction that the expression [i_{k}, …, i_{1}|j_{1}, …, j_{k}] stands for a k × k quantum minor of X, namely the quantum determinant of the submatrix of X obtained from X using rows i_{1}, …, i_{k} and columns j_{1}, …, j_{k}. This notation is extended to X′ adding a “′ ” to avoid confusion. Thus, a quantum k × k minor of X′ will be expressed by a symbol [i_{k}, …, i_{1}|j_{1}, …, j_{k}]′. The convention on the index of rows of X′ (see introduction of Section 1) is in order; so, in such an expression, we shall always have i_{1} ≥ 2 and j_{k} ≤ n−1.

The results we need will follow from the special case where m = n. In this context, the role played by (n − 1) × (n − 1) minors is of special importance; thus we use a more convenient notation for them (coming from [8]). For 1 ≤ i, j ≤ n, the (n − 1) × (n − 1) quantum minor of X obtained by deleting the i-th row and j-th column is denoted A(ij). Moreover, for 2 ≤ i ≤ n and 1 ≤ j ≤ n − 1 we also define A′(ij) to be the (n − 2) × (n − 2) quantum minor of X′ obtained from X′ by deleting the i-th row and j-th column.

Theorem 2.1 Assume that m = n. Then, with the above notation:

(det_{q}X′)X_{1n} = X_{1n}(det_{q}X′) = (−q)^{1−n}det_{q}X.

Proof: Note that (det_{q}X′)X_{1n} = X_{1n}(det_{q}X′) is clear from 1.1. The proof is by induction on n. The case where n = 2 is an obvious consequence of (3). We suppose now that the result is true for any integer less than or equal to n−1. Because of Lemma 1.1 the relations of [8, Corollary 4.4.4] give us the expansion:

det_{q}X′ = ∑_{j=1}^{n−1}(−q)^{j−1}X′_{2j}A′(2j).  (5)

Now, X_{1n}X′_{2j} = X′_{2j}X_{1n} = X_{2j}X_{1n} − q^{−1}X_{1j}X_{2n} for all j ∈ {1, …, n − 1}; and so

X_{1n}det_{q}X′ = ∑_{j=1}^{n−1}(−q)^{j−1}X_{1n}X′_{2j}A′(2j)
= ∑_{j=1}^{n−1}(−q)^{j−1}(X_{2j}X_{1n} − q^{−1}X_{1j}X_{2n})A′(2j)
= ∑_{j=1}^{n−1}(−q)^{j−1}X_{2j}X_{1n}A′(2j) − q^{−1} ∑_{j=1}^{n−1}(−q)^{j−1}X_{1j}X_{2n}A′(2j).

By the induction hypotheses we have X_{1n}A′(2j) = (−q)^{2−n}A(2j) for all j ∈ {1, …, n − 1}. Hence,

\begin{aligned}X_{1n}det_{q}X′ & = & ∑_{j=1}^{n−1}(−q)^{j−1}X_{2j}(−q)^{2−n}A(2j) − q^{−1} ∑_{j=1}^{n−1}(−q)^{j−1}X_{1j}X_{2n}X_{1n}^{−1}(−q)^{2−n}A(2j) \\ & = & ∑_{j=1}^{n−1}(−q)^{j+1−n}X_{2j}A(2j) − q^{−1} ∑_{j=1}^{n−1}(−q)^{j+1−n}X_{1j}X_{2n}X_{1n}^{−1}A(2j) \end{aligned}

Setting R = −q^{−1} ∑_{j=1}^{n−1}(−q)^{j+1−n}X_{1j}X_{2n}X_{1n}^{−1}A(2j), we obtain:

X_{1n}R = ∑_{j=1}^{n−1}(−q)^{j−n}X_{1j}X_{2n}A(2j)
= ∑_{j=1}^{n−1}(−q)^{j−n}(X_{2n}X_{1j} + (q − q^{−1})X_{1n}X_{2j})A(2j)
= ∑_{j=1}^{n−1}(−q)^{j−n}X_{2n}X_{1j}A(2j) + (q − q^{−1}) ∑_{j=1}^{n−1}(−q)^{j−n}X_{1n}X_{2j}A(2j).

On the other hand, from [8, Corollary 4.4.4], we have the relation ∑_{j=1}^{n}(−q)^{j−2}X_{1j}A(2j) = 0.

Thus,

\begin{aligned}X_{1n}R & = & (−q)^{2−n}X_{2n} ∑_{j=1}^{n−1}(−q)^{j−2}X_{1j}A(2j) + (q − q^{−1})(−q)^{2−n}X_{1n} ∑_{j=1}^{n−1}(−q)^{j−2}X_{2j}A(2j). \\ & = & (−q)^{2−n}X_{2n}(0 − (−q)^{n−2}X_{1n}A(2n)) + (q − q^{−1})(−q)^{2−n}X_{1n} ∑_{j=1}^{n−1}(−q)^{j−2}X_{2j}A(2j). \end{aligned}

It follows that:

R = (−q)^{−1}X_{2n}A(2n) + (q − q^{−1})(−q)^{2−n} ∑_{j=1}^{n−1}(−q)^{j−2}X_{2j}A(2j).

Thus,

X_{1n}det_{q}X′ = ∑_{j=1}^{n−1}(−q)^{j+1−n}X_{2j}A(2j) + (−q)^{−1}X_{2n}A(2n)

+(q − q^{−1})(−q)^{2−n} ∑_{j=1}^{n−1}(−q)^{j−2}X_{2j}A(2j)

= (−q)^{2−n} ∑_{j=1}^{n−1}((−q)^{j−1} + (q − q^{−1})(−q)^{j−2})X_{2j}A(2j) + (−q)^{−1}X_{2n}A(2n)
= (−q)^{2−n} ∑_{j=1}^{n−1}((−q)^{j−1} − (−q)^{j−1} − q^{−1}(−q)^{j−2})X_{2j}A(2j) + (−q)^{−1}X_{2n}A(2n)
= (−q)^{1−n} ∑_{j=1}^{n−1}(−q)^{j−2}X_{2j}A(2j) + (−q)^{1−n}(−q)^{n−2}X_{2n}A(2n)
= (−q)^{1−n} ∑_{j=1}^{n}(−q)^{j−2}X_{2j}A(2j).

Again, by using [8, Corollary 4.4.4], it follows that X_{1n}det_{q}X′ = (−q)^{1−n}det_{q}X.

By using Theorem 2.1, we can establish Corollary 2.2 which links (p − 1) × (p − 1) minors of X′ with p × p minors of X, for p ≥ 2, that involve the first row and the last column of X.

Corollary 2.2 Let p ≥ 2. Suppose that I = {1 = i_{1} < … < i_{p}} ⊆ ℕ_{m} and J = {j_{1} < … < j_{p} = n} ⊆ ℕ_{n} and set I′ = {i_{2} < … < i_{p}} and J′ = {j_{1} < … < j_{p−1}}. Then

[I′|J′]′ = (−q)^{1−p}[I|J]X_{1n}^{−1} = (−q)^{1−p}X_{1n}^{−1}[I|J].

Proof: This is an immediate consequence of Theorem 2.1 applied to the square submatrix obtained from X by using rows i_{1}, …, i_{p} and columns j_{1}, …, j_{p} of X.

Recall from the introduction that, for 1 ≤ t ≤ min{m, n}, we denote by I_{t}(X) the ideal generated in 𝕂_{q}[X] by the t × t minors of X. Clearly, the ideal generated by the t × t minors of X in 𝕂_{q}[X]_{X_{1n}} is just Ĩ_{t}(X) := I_{t}(X)𝕂_{q}[X]_{X_{1n}}.

Lemma 2.3 Let 1 ≤ t ≤ min{m, n}.

(i) I_{t}(X) coincides with the right ideal of 𝕂_{q}[X] generated by the t × t minors of X. (ii) Ĩ_{t}(X) coincides with the right ideal of 𝕂_{q}[X]_{X_{1n}} generated by the t × t minors [I|J] of X such that 1 ∈ I and n ∈ J.

Proof: (i) The case where m = n follows at once from [1, Corollary A.2]. Now, set s := max{m, n}. There is a surjective algebra morphism

π : O_{q}(M_{s}) ⟶ O_{q}(M_{m,n})
X_{ij} ↦ \{ \begin{matrix}X_{ij} & if i ≤ m and j ≤ n \\ 0 & otherwise \end{matrix}

and, for I, J ⊆ ℕ_{s} such that |I| = |J| = t, a t × t minor [I|J] of O_{q}(M_{s}) is sent to [I|J] if I ⊆ ℕ_{m} and J ⊆ ℕ_{n} and is sent to 0 otherwise. It follows that the ideal of O_{q}(M_{m,n}) generated by the t × t minors is the image under π of the ideal of O_{q}(M_{s}) generated by the t × t minors. From this, we see that point (i) for arbitrary positive integers m and n follows from the special case m = n.

(ii) By part (i), Ĩ_{t}(X) coincides with the right ideal of 𝕂_{q}[X]_{X_{1n}} generated by the t × t minors. For the purpose of this proof, denote by S the set of t × t minors [I|J] of X such that 1 ∈ I and n ∈ J.

Let [I|J] be a t × t minor of X such that 1 ∈ I but n ∉ J. We may apply [8, Corollary 4.4.4] to the subalgebra of 𝕂_{q}[X] generated by those X_{ij} such that i ∈ I and j ∈ J ∪ {n}. This leads to the equation:

∑_{j∈J∪{n}}(−q)^{•}[I|J_{j}]X_{1j} = 0,

where, for j ∈ J ∪ {n}, we put J_{j} = J ∪ {n} \ {j}, and occurences of (−q)^{•} denote integer powers of −q which it is not necessary to specify exactly. Since J_{n} = J, it follows that the equation

[I|J] = − ∑_{j∈J}(−q)^{•}[I|J_{j}]X_{1j}X_{1n}^{−1}

holds in 𝕂_{q}[X]_{X_{1n}}. Now, n ∈ J_{j} for each j ∈ J; and so we have shown that [I|J] is in the right ideal generated by S in 𝕂_{q}[X]_{X_{1n}}. By a similar argument, we prove that a t×t minor [I|J] such that 1 ∉ I but n ∈ J is in the right ideal generated by S in 𝕂_{q}[X]_{X_{1n}}.

It remains to deal with a t × t minor [I|J] such that 1 ∉ I and n ∉ J. In this case, we apply [8, Corollary 4.4.4] in the subalgebra of 𝕂_{q}[X] generated by those X_{ij} such that i ∈ I ∪ {1} and j ∈ J ∪ {n}. This gives us the relation

[I ∪ {1}|J ∪ {n}] = ∑_{j∈J∪{n}}(−q)^{•}[I|J_{j}]X_{1j} = ∑_{j∈J}(−q)^{•}[I|J_{j}]X_{1j} + (−q)^{•}[I|J]X_{1n} (6)

where, for j ∈ J ∪ {n}, we put J_{j} = J ∪ {n} \ {j}. Another application of [8, Corollary 4.4.4] gives

[I ∪ {1}|J ∪ {n}] = ∑_{j∈J∪{n}}(−q)^{•}[I ∪ {1} \ {s}|J_{j}]X_{sj},  (7)

where, s = max I. Equation (7) and the results we established above show that the (t + 1) × (t + 1) minor [I ∪ {1}|J ∪ {n}] is in the right ideal generated in 𝕂_{q}[X]_{X_{1n}} by S. Thus, using (6), it follows that [I|J] is also in the right ideal generated in 𝕂_{q}[X]_{X_{1n}} by S. The proof is now complete.

Recall that, by Lemma 1.1 and Proposition 1.4 we know that the subalgebra of 𝕂_{q}[X]_{X_{1n}} generated by the X′_{ij} for 2 ≤ i ≤ m and 1 ≤ j ≤ n − 1 is isomorphic to O_{q}(M_{m−1,n−1}) and that we denote it by 𝕂_{q}[X′]. Following our conventions, if 2 ≤ t ≤ min{m, n}, we denote by I_{t−1}(X′) the ideal of 𝕂_{q}[X′] generated by the (t − 1) × (t − 1) minors of X′. In this notation, we have the following important result.

Proposition 2.4 For 2 ≤ t ≤ min{m, n}, the following equality holds:

Ĩ_{t}(X) = I_{t−1}(X′)𝕂_{q}[X]_{X_{1n}}.

Proof: By Lemma 2.3 (ii), we know that Ĩ_{t}(X) coincides with the right ideal of 𝕂_{q}[X]_{X_{1n}} generated by the t × t minors [I|J] of X such that 1 ∈ I and n ∈ J. On the other hand, let [I|J] be a t × t minor of X such that 1 ∈ I and n ∈ J. Applying Corollary 2.2, we have (−q)^{1−t}[I|J] = [I \ {1}|J \ {n}]′X_{1n}. Thus, Ĩ_{t}(X) coincides with the right ideal of 𝕂_{q}[X]_{X_{1n}} generated by the (t − 1) × (t − 1) minors [I|J]′ of X′ (such that I ⊆ {2, …, n} and J ⊆ {1, …, m − 1}). On the other hand, Lemma 2.3 (i) shows that I_{t−1}(X′) is the right ideal of 𝕂_{q}[X′] generated by the (t−1)×(t−1) minors X′; so, the proof is complete.

Recall from the introduction that, for any positive integer t such that t ≤ min{m, n}, we define the quantum determinantal ring R_{t}(X) := 𝕂_{q}[X]/I_{t}(X). This is a domain by [1, Corollary 2.6]. If we put x_{ij} := X_{ij} + I_{t}(X), for 1 ≤ i ≤ m and 1 ≤ j ≤ n, then there is a canonical isomorphism

R_{t}(X)_{x_{1n}} ≅ 𝕂_{q}[X]_{X_{1n}}/Ĩ_{t}(X).  (8)

We finish this subsection by showing that, for 2 ≤ t ≤ min{m, n}, the ring R_{t}(X)_{x_{1n}} can be described as an iterated Ore extension of R_{t−1}(X′). To achieve this aim, we will have to make use of Proposition 1.6 which shows that 𝕂_{q}[X]_{X_{1n}} can be obtained from its subalgebra 𝕂_{q}[X′] by iterated Ore extensions adding successively X_{1n}^{±1}, X_{11},...,X_{1,n−1},X_{mn},...,X_{2n}.

Theorem 2.5 For 2 ≤ t ≤ min{m, n}, the ring R_{t}(X)_{x_{1n}} is a localisation of an iterated Ore extension of R_{t−1}(X′).

Proof: We start by giving a list of relations between a (t − 1) × (t − 1) minor of X′ and the generators X_{1n}^{±1}, X_{11},...,X_{1,n−1},X_{mn},...,X_{2n}. (Note that, by Lemma 1.1, we already know that a (t − 1) × (t − 1) minor of X′ commutes with X_{1n}^{±1}.)

Let I′ ⊆ {2, …, m} and J′ ⊆ {1, …, n−1} be sets of indices such that |I′| = |J′| = t−1. Setting I = I′ ∪{1} and J = J′ ∪{n}, Corollary 2.2, shows that (−q)^{1−t}[I|J] = [I′|J′]′X_{1n}. On the other hand, [8, Lemma 4.5.1 and Theorem 4.6.1] give the following relations, for k ∈ {2, …, m} and l ∈ {1, …, n − 1}:

1) X_{1l}[I|J] = [I|J]X_{1l} if l ∈ J′;

2) X_{1l}[I|J] − q[I|J]X_{1l} = q(q − q^{−1}) ∑_{j<l,j∈J′}(−q)^{•}X_{1j}[I|J ∪ {l} \ {j}] if l ∉ J′;

3) X_{kn}[I|J] = [I|J]X_{kn} if k ∈ I′;

4) X_{kn}[I|J] − q^{−1}[I|J]X_{kn} = q^{−1}(q^{−1} − q) ∑_{j>k,j∈I′}(−q)^{•}X_{jn}[I ∪ {k} \ {j}|J] if k ∉ I′.

(Points 1) and 3) follow from [8, Theorem 4.6.1], point 2) is [8, Lemma 4.5.1(1) first relation], point 4) is [8, Lemma 4.5.1(3) second relation].) Hence, for k ∈ {2, …, n} and l ∈ {1, …, n − 1}, Corollary 2.2 gives:

1’) X_{1l}[I′|J′]′ = q^{−1}[I′|J′]′X_{1l} if l ∈ J′; 2’) X_{1l}[I′|J′]′ − [I′|J′]′X_{1l} = q(q − q^{−1}) ∑_{j<l,j∈J′}(−q)^{•}X_{1j}[I′|J′ ∪ {l} \ {j}]′ if l ∉ J′; 3’) X_{kn}[I′|J′]′ = q[I′|J′]′X_{kn} if k ∈ I′; 4’) X_{kn}[I′|J′]′ − [I′|J′]′X_{kn} = q^{−1}(q^{−1} − q) ∑_{j>k,j∈I′}(−q)^{•}X_{jn}[I′ ∪ {k} \ {j}|J′]′ if k ∉ I′.

By Proposition 1.6, 𝕂_{q}[X]_{X_{1n}} can be obtained from its subalgebra 𝕂_{q}[X′] by iterated Ore extensions adding successively X_{1n}^{±1}, X_{11},...,X_{1,n−1},X_{mn},...,X_{2n}. The relations 1’) to 4’) together with the fact that X_{1n} commutes with any element of the subalgebra 𝕂_{q}[X′] show that, at each step of this Ore extension, the ideal of the base algebra generated by the (t − 1) × (t − 1) minors of X′ is invariant under the corresponding skew derivation. It follows that 𝕂_{q}[X]_{X_{1n}}/I_{t−1}(X′)𝕂_{q}[X]_{X_{1n}} is isomorphic to an iterated Ore extension of 𝕂_{q}[X′]/I_{t−1}(X′). But Proposition 2.4 shows that Ĩ_{t}(X) = I_{t−1}(X′)𝕂_{q}[X]_{X_{1n}}. The proof is thus complete.

3 Quantum determinantal rings are maximal orders

Let R be a noetherian domain with division ring of fractions Q. Then R is said to be a maximal order in Q if the following condition is satisfied: if T is a ring such that R ⊆ T ⊆ Q and such that there exist nonzero elements a, b ∈ R with aT b ⊆ R, then T = R. This condition is the natural noncommutative analogue of normality for commutative domains, see, for example, [7, Section 5.1]. In this section, we investigate the maximal order condition for determinantal rings.

Recall that the quantum determinantal ring R_{t}(X) := 𝕂_{q}[X]/I_{t}(X) is a domain by [1, Corollary 2.6]. Also, recall that R_{2}(X) has been shown to be a maximal order in [9]. In this section, we prove that R_{t}(X) is a maximal order in its division ring of fractions, when 𝕂 = ℂ and for q an element of ℂ transcendental over ℚ.

The following Lemma from [9] (see [9, Lemma 1.1]) is recalled here for the convenience of the reader.

Lemma 3.1 Let R be a noetherian domain and Q = Frac R its division ring of fractions. Assume there exists a nonzero normal element x in R such that

1) xR = ∩_{i=1}^{r}𝔭_{i} where, for i ∈ {1, …, r}, 𝔭_{i} is a completely prime ideal of R, and

2) the localisation R_{x} of R with respect to the set {x^{i}, i ∈ ℕ} is a maximal order in Q. Let τ be the automorphism of R associated with x; that is ax = xτ(a) for all a ∈ R. Suppose that τ(𝔭_{i}) ⊆ 𝔭_{i} for i ∈ {1, …, r}. Then R is a maximal order in Q.

One case where the above result immediately applies is the case that the ideal xR itself is a completely prime ideal. It is this case that we want to use. The applicability of the above lemma to determinantal rings is a consequence of the results of the previous section, since we can deduce the following result.

Theorem 3.2 Assume that 𝕂 = ℂ and let q be an element of ℂ transcendental over ℚ. Let t be an integer such that 0 < t ≤ min{m, n}. Then R_{t}(X) is a maximal order.

Proof: For t = 1, this is trivial and for t = 2 it is [9, Théorème 2.3.11]. We proceed by induction on t. Assume that the result is true for an integer s ≥ 2 and set t = s + 1. By Theorem 2.5, R_{t}(X)_{x_{1n}} is a localisation of an iterated Ore extension of R_{t−1}(X′). The induction hypothesis shows that R_{t−1}(X′) is a maximal order in its quotient ring and so R_{t}(X)_{x_{1n}} is a maximal order by [6, V.2.5 and IV.2.1]. However, [5, Corollaire 11.7] shows that ⟨x_{1n}⟩ is a completely prime ideal of R_{t}(X). Thus, Lemma 3.1 shows that R_{t}(X) is a maximal order.

We conjecture that this result holds for arbitrary non-zero q in any field 𝕂 and for all t. We conclude by proving that the factor by the quantum determinantal ideal ⟨det_{q}⟩ in O_{q}(M_{n}) is a maximal order. All that remains to be proved after the above discussion is that the ideal ⟨det_{q}, X_{1n}⟩ is a completely prime ideal. This is what we do next.

4 O_{q}(M_{n})/⟨det_{q}⟩ is a maximal order

In this section we need to use the preferred bases in O_{q}(M_{u,v}) developed in [1], and we follow the notation of that paper. See, in particular, [1, Corollary 1.11]. We recall the notation [T |T′] for the product of quantum minors corresponding to an allowable bitableau (T, T′). We recall also that it is sometimes convenient to label rows of (T, T′) in the form (I, J) where I and J are sets of row and column indices, respectively (of course, I ⊆ {1, …, u} and J ⊆ {1, …, v}); such a pair is called an index pair (see [1, Section 1]). Many of the results in [1] are stated for the square case O_{q}(M_{u,u}), and there are easy extensions to the rectangular case O_{q}(M_{u,v}), see, for example, [1, 1.11] and [2, Section 2] for more details of this standard procedure.

We aim to prove that the ideal ⟨det_{q}, X_{1n}⟩ of O_{q}(M_{n}) is a completely prime ideal for each n ≥ 3, by using the following result of David Jordan, [3].

Proposition 4.1 Let σ be an automorphism and δ be a σ-derivation of a domain A. Let R = A[x; σ, δ]. Let c be a normal element of R of the form dx+e, where d, e ∈ A and d ≠ 0. Let β be the automorphism of R such that cr = β(r)c for all r ∈ R. Then β(A) = A, the element d is normal in A and β(a)d = dσ(a) for all a ∈ A. Furthermore, if e is regular modulo the ideal Ad = dA then R/Rc is a domain.

Proposition 4.2 The ideal ⟨det_{q}, X_{1n}⟩ of O_{q}(M_{n}) is a completely prime ideal for each n ≥ 3.

Proof: The idea is first to factor out X_{1n} obtaining a domain and then to factor out D_{n} := det_{q} and see that we still have a domain by invoking Jordan’s Domain Theorem. We use the notation from that theorem. Let T be the subalgebra of O_{q}(M_{n}) generated by all X_{ij} except X_{nn}. Set A := T/⟨X_{1n}⟩. Then A is a domain. Let R := A[X_{nn}; σ, δ] ≅ O_{q}(M_{n})/⟨X_{1n}⟩. Let D_{n} denote the quantum determinant of O_{q}(M_{n}) and let D_{n−1} denote the quantum determinant of the copy of O_{q}(M_{n−1}) generated by the X_{ij} with i, j < n. Note that D_{n−1} = A(nn) in the notation introduced before Theorem 2.1. Set x := X_{nn} ∈ R and d := \overline{D_{n−1}} ∈ A ⊆ R. Note that d ≠ 0 in R. Finally, set c := \overline{D_{n}} ∈ R. The quantum Laplace expansion of D_{n} by the nth column gives D_{n} = ∑_{i=1}^{n} ±q^{•}A(in)X_{in} so that c = dx + e, where e is the image in A of ∑_{i=1}^{n−1} ±q^{•}A(in)X_{in} ∈ T.

Note that R/⟨c⟩ ≅ O_{q}(M_{n})/⟨D_{n}, X_{1n}⟩. We show that this is a domain by showing that Jordan’s Theorem applies. Note that c is normal (in fact, central) in R. All we need to do is to observe that e is regular modulo the ideal dA = Ad of the ring A. However, A/dA is isomorphic to an iterated Ore extension of O_{q}(M_{n−1})/⟨D_{n−1}⟩. Now O_{q}(M_{n−1})/⟨D_{n−1}⟩ is a domain by [1, Theorem 2.5] or by [3, Example 2]; and so A/dA is a domain. Thus, all we have to do is to show that e ∉ dA.

Suppose that e ∈ dA. Then e = \overline{D_{n−1}α} for some α ∈ T. Taking pre-images, we obtain

∑_{i=1}^{n−1} ±q^{•}A(in)X_{in} = A(nn)α + βX_{1n}

for some β ∈ T. Now, each term on the left hand side is an element of O_{q}(M_{n}) of bidegree (1, …, 1; 1, …, 1) in the ℤ^{n} × ℤ^{n} grading of O_{q}(M_{n}) described in [1, 1.5]. Hence, we may assume that each term on the right hand side also has this bidegree. This implies that α = λX_{nn} for some scalar λ ∈ 𝕂. Since X_{nn} ∉ T, this implies that α = 0. In the resulting equation

∑_{i=1}^{n−1} ±q^{•}A(in)X_{in} = βX_{1n}

each of the terms on the left hand side is a preferred product. However, if we write β = ∑ λ_{i}[T_{i}|T′_{i}] in terms of the preferred basis then βX_{1n} = ∑ λ_{i}[T_{i}|T′_{i}]X_{1n} is again in preferred form. The equation

∑_{i=1}^{n−1} ±q^{•}A(in)X_{in} = ∑ λ_{i}[T_{i}|T′_{i}]X_{1n}

then contradicts the independence of the preferred basis.

Thus e ∉ dA and we have all of the hypotheses of Jordan’s Theorem and conclude that R/⟨c⟩ ≅ O_{q}(M_{n})/⟨D_{n}, X_{1n}⟩ is a domain, as required.

It is interesting to note that when n = 2 the above result fails. In this case, the ideal ⟨det_{q}, X_{12}⟩ is semiprime; in fact, it is the intersection of two completely prime ideals, each of which is fixed by the automorphism determined by X_{1n}, so Lemma 3.1 is applicable. However, this case has already been dealt with in [9].

Theorem 4.3 O_{q}(M_{n})/⟨det_{q}⟩ is a maximal order in its division ring of quotients for each n ≥ 2.

Proof: When n = 2 this is proved in [9, Théorème 2.3.11] (and can be proved directly from Theorem 3.2 by the reasoning in the previous paragraph). An inductive argument similar to that used in Theorem 3.2 finishes the proof.

5 Appendix: some useful relations

In this section, we collect some useful relations. They are essentially derived from results of [8].

Relation 5.1 If ( \begin{matrix}a & b \\ c & d \end{matrix} ) is a 2 × 2 quantum matrix, then the following relation holds:

ad − q^{2}da = (1 − q^{2})(ad − qbc).

Relations 5.2 If ( \begin{matrix}X_{11} & X_{12} & X_{13} \\ X_{21} & X_{22} & X_{23} \\ X_{31} & X_{32} & X_{33} \end{matrix} ) is a 3 × 3 quantum matrix, then the following re-

lations hold.

1) For i = 2, 3: 1.1) X_{11}[i, 1|1, 3] = [i, 1|1, 3]X_{11}, (see [PW] (4.6.1)) 1.2) X_{11}[i, 1|2, 3] = q[i, 1|2, 3]X_{11}, (see [PW] (4.5.1)(1)) 1.3) X_{12}[i, 1|1, 3] − q[i, 1|1, 3]X_{12} = (q^{−1} − q)X_{11}[i, 1|2, 3], (see [PW] (4.5.1)(1)) 1.4) X_{12}[i, 1|2, 3] = [i, 1|2, 3]X_{12}. (see [PW] (4.6.1))

2) For j = 1, 2: 2.1) X_{33}[3, 1|j, 3] = [3, 1|j, 3]X_{33}, (see [PW] (4.6.1)) 2.2) X_{33}[2, 1|j, 3] = q^{−1}[2, 1|j, 3]X_{33}, (see [PW] (4.5.1)(3)) 2.3) X_{23}[3, 1|j, 3] − q^{−1}[3, 1|j, 3]X_{23} = (q − q^{−1})X_{33}[2, 1|j, 3], (see [PW] (4.5.1)(3)) 2.4) X_{23}[2, 1|j, 3] = [2, 1|j, 3]X_{23}. (see [PW] (4.6.1))

References

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T. H. Lenagan: School of Mathematics, University of Edinburgh, James Clerk Maxwell Building, King’s Buildings, Mayfield Road, Edinburgh EH9 3JZ, Scotland E-mail: tom@maths.ed.ac.uk

L. Rigal: Université Jean Monnet (Saint-Étienne), Faculté des Sciences et Techniques, Département de Mathématiques 23 rue du Docteur Paul Michelon 42023 Saint-Étienne Cédex 2, France E-mail: Laurent.Rigal@univ-st-etienne.fr