On a Banach algebra of entire functions with a weighted Hadamard multiplication

New algebraic-analytic properties of a previously studied Banach algebra $\mathcal{A}({\bf{p}})$ of entire functions are established. For a given fixed sequence $(\bf{p}(n))_{n\geq 0}$ of positive real numbers, such that $\lim_{n\rightarrow \infty} {\bf{p}}(n)^{\frac{1}{n}}=\infty$, the Banach algebra $\mathcal{A}({\bf{p}})$ is the set of all entire functions $f$ such that $f(z)=\sum_{n=0}^\infty \hat{f}(n) z^n $ ($z\in \mathbb{C}$), where the sequence $(\hat{f}(n))_{n\geq 0}$ of Taylor coefficients of $f$ satisfies $\hat{f}(n)=O({\bf{p}}(n)^{-1})$ for $n\rightarrow \infty$, with pointwise addition and scalar multiplication, a weighted Hadamard multiplication $\ast$ with weight given by ${\bf{p}}$ (i.e., $(f\ast g)(z)=\sum_{n=0}^\infty {\bf{p}}(n) \hat{f}(n)\hat{g}(n)z^n$ for all $z\in \mathbb{C}$), and the norm $\|f\|=\sup_{n\geq 0} {\bf{p}}(n)|\hat{f}(n)|$. The following results are shown: The Bass and the topological stable ranks of $\mathcal{A}({\bf{p}})$ are both $1$. $\mathcal{A}({\bf{p}})$ is a Hermite ring, but not a projective-free ring. Idempotents and exponentials in $\mathcal{A}({\bf{p}})$ are described, and it is shown that every invertible element of $\mathcal{A}({\bf{p}})$ has a logarithm. A generalised necessary and sufficient `corona-type condition' on the matricial data $(A,b)$ with entries from $\mathcal{A}({\bf{p}})$ is given for the solvability of $Ax = b$ with $x$ also having entries from $\mathcal{A}({\bf{p}})$. The Krull dimension of $\mathcal{A}({\bf{p}})$ is infinite. $\mathcal{A}({\bf{p}})$ is neither Artinian nor Noetherian, but is coherent. The special linear group over $\mathcal{A}({\bf{p}})$ is generated by elementary matrices.

1 n = ∞, the Banach algebra A(p) is the set of all entire functions f such that where the sequence ( f (n)) n≥0 of Taylor coefficients of f satisfies f (n) = O(p(n) −1 ) for n ∞, with pointwise addition and scalar multiplication, a weighted Hadamard multiplication * with weight given by p (i.e., (f * g)(z) = ∞ n=0 p(n) f (n) g(n)z n for all z ∈ C), and the norm f = sup n≥0 p(n)| f (n)|.The following results are shown: • The Topological stable rank of A(p) is 1.
• The Bass stable rank of A(p) is 1.
• A(p) is a Hermite ring.
• A(p) is not a projective-free ring.
• Idempotents in A(p) are described.
• Exponentials in A(p) are described, and it is shown that every invertible element of A(p) has a logarithm, so that the first Čech cohomology group H 1 (M (A(p)), Z) with integer coefficients of the maximal ideal space M (A(p)) is trivial.• A generalised necessary and sufficient 'corona-type condition' on the matricial data (A, b) with entries from A(p) is given for the solvability of Ax = b with x also having entries from A(p).• The Krull dimension of A(p) is infinite.
• The special linear group over A(p) is generated by elementary matrices.
The O-notation here means as usual that there exists a constant C > 0 such that p(n)|a n | < C for all n ∈ N 0 .For f ∈ A(p), we set . With pointwise addition and scalar multiplication, A(p) is a complex vector space.We equip A(p) with the weighted Hadamard multiplication * , given by (f * g)(z) = ∞ n=0 p(n) f (n) g(n)z n (z ∈ C), for all f, g ∈ A(p), and the norm • , defined by Then A(p) is a complex commutative unital Banach algebra with the unit element ε given by For example, if p * (n) = n! (n ∈ N 0 ), then lim n ∞ (n!) 1 n = ∞, and the corresponding Banach algebra A(p * ) has the identity element exp z.This Banach algebra A(p * ) was introduced and studied in [26].In [28], the more general Banach algebras A(p) were introduced, their ideal structure was studied, and the following results were shown.
(R1) g ∈ A(p) is a divisor of f ∈ A(p) if and only if there exists a constant C > 0 such that | f (n)| ≤ C| g(n)| for all n ∈ N 0 .
In particular, f is invertible in A(p) if and only if there exists a δ > 0 such that | f (n)| ≥ δ p(n) for all n ∈ N 0 .(R2) Every finite collection of functions has a greatest common divisor d ∈ A(p).Up to invertible elements, d is given by d In this article, we will show the following: (1) The topological stable rank of A(p) is equal to 1.
(2) The Bass stable rank of A(p) is equal to 1.
(3) A(p) is a Hermite ring.(4) A(p) is not a projective-free ring.
( The outline of this article is as follows: In each section, we will first give the background of the property, by recalling key definitions, and then prove the property, possibly with additional commentary.

Topological stable rank of A(p) is 1
The notion of the topological stable rank for topological algebras was introduced by Rieffel in [23] analogous to the K-theoretic concept of (Bass) stable rank, as well as several related numerical invariants.
Definition 2.1.Let R be a commutative unital ring with identity element 1.We assume that 1 = 0, that is, R is not the trivial ring {0}.
Now suppose that A is a commutative unital Banach algebra.The topological stable rank of A is the minimum n ∈ N such that U n (A) is dense in A n , and it is infinite if no such n exists.
Here, A n is the normed space with the 'Euclidean norm' • 2 given by for all v in A n , where v has the components v 1 , . . ., v n ∈ A.
Theorem 2.2.The topological stable rank of A(p) is 1.
Proof.Let f ∈ A(p), and ǫ > 0. Define g by Consequently, g is invertible in A(p) by (R1) on page 2.

Bass stable rank of A(p) is 1
The notion of the Bass stable rank was introduced in algebraic Ktheory by Bass [4], but it has since proved useful in other branches of mathematics as well, for example, in analysis (see e.g.[20]).
Definition 3.1.Let R be a commutative unital ring with identity 1.Let n ∈ N.
as n ∞, and so u ∈ A(p).Moreover, we have We have F ≤ 1 because the above yields The element H 1 'approximates' G 1 , since for all n ∈ N 0 , we have [24,Thm. 10.7]).Setting h = H −1 1 * u * g 2 , we have , which is invertible in A(p), as it is the product of invertible elements from A(p).

A(p) is a Hermite ring, but not a projective free ring
The study of Serre's conjecture and algebraic K-theory naturally led to the notion of Hermite rings; see e.g., [18].Definition 4.1.Let R be a commutative unital ring.The ring R is called Hermite if every finitely generated stably free R-module is free.The ring R is called projective-free if every finitely generated projective R-module is free.
If R-modules M, N are isomorphic, then we write M ∼ = N. Recall that if M is a finitely generated R-module, then: It is clear that every projective free ring is Hermite.
For m, n ∈ N, R m×n denotes the set of matrices with m rows and n columns having entries from R. The identity element in R k×k having diagonal elements 1, and zeroes elsewhere will be denoted by I k .
In terms of matrices, R is Hermite if and only if left-invertible tall matrices over R can be completed to invertible ones (see e.g.[18, p.VIII], [27, p.1029]):For all k, K ∈ N such that k < K, and for all f ∈ R K×k such that there exists a g ∈ R k×K so that gf = I k , there exists an f c ∈ R K×(K−k) and there exists a In terms of matrices (see e.g.[8,Proposition 2.6] or [2, Lemma 2.2]), the ring R is projective-free if and only if every idempotent matrix P is conjugate (by an invertible matrix S) to a diagonal matrix with elements 1 and 0 on the diagonal, that is, for every m ∈ N and every P ∈ R m×m satisfying P 2 = P , there exists an S ∈ R m×m such that S is invertible as an element of R m×m , and for some integer r ≥ 0, S −1 P S = [ Ir 0 0 0 ].In 1976, it was shown independently by Quillen and Suslin, that if F is a field, then the polynomial ring F[x 1 , • • • , x n ] is projective-free, settling Serre's conjecture from 1955 (see [18]).

A(p) is a Hermite ring.
It is known that a commutative unital ring having Bass stable rank ≤ 2 is Hermite (see e.g., [20,Corollary 36.17]).In light of this result and Theorem 3.2, we have:

A(p)
is not a projective-free ring.While every projective free ring is Hermite, the converse may not hold.In fact A(p) is such an example, by using the matricial characterisation of projective free rings.
Then P ∈ A(p), and So P * P = P .As we have assumed A(p) is projective free, there is an f ∈ {0, ǫ}, D = [ f ], and S, S −1 ∈ A(p) such that P = S −1 * D * S.But then P = 0 or P = ε, and either case is false.Consequently, A(p) is not projective free.
For a Banach space A, we denote the set of all continuous linear maps from A to C by A * .Recall that for a commutative unital complex Banach algebra A, the maximal ideal space M(A) ⊂ A * is the set of nonzero homomorphisms A C , endowed with the Gelfand topology, the weak- * topology of A * .It is a compact Hausdorff space contained in the unit sphere of A * .Contractibility of the maximal ideal space M(A) in the Gelfand topology suffices for A to be projective-free (see, e.g., [6,Corollary 1.4]).Thus the maximal ideal space M(A(p)) is not contractible.

Idempotents in A(p)
The following result characterises idempotents in A(p).

Theorem 5.1. f ∈ A(p) is an idempotent if and only if for all
Let A be a commutative unital complex Banach algebra.The Gelfand transform of a ∈ A, defined by â(ϕ ϕ ϕ) := ϕ ϕ ϕ(a) for ϕ ϕ ϕ ∈ M(A), is a nonincreasing-norm morphism from A into C(M(A)), the Banach algebra of complex-valued continuous functions on M(A) equipped with the supremum norm • ∞ (given by f From the special case of (R3) on page 3 when f = ε (the identity element of A(p)), we have the following corona theorem:
We will now show that {ϕ k : k ∈ N 0 }, with the topology induced from M(A(p)), is homeomorphic to N 0 , with the topology induced from R. In the proof below, we will use the notation δ m,n := 0 if m = n and δ m,m := 1 for all m, n ∈ N 0 .
Proof.'If' part: Let (k i ) i∈I be eventually constant, equal to k 0 .There exists an i * ∈ I such that for all i ≥ i * (where ≥ denotes the order on the directed set I), k i = k 0 , and so for all f ∈ A(p), we have that 'Only if' part: Suppose that (ϕ k i ) i∈I converges to ϕ k 0 in M(A(p)).Then with f = z k 0 ∈ A(p), by the definition of the Gelfand topology, the net ( As M(A(p)) is compact, while N 0 with its usual topology is not compact, it follows that not all elements of M(A(p)) are of the form ϕ k for some k ∈ N 0 .Explicit examples are known (see below), and these were mentioned as non-fixed maximal ideals in [28,Remark,.
Example 5.4.Let k = (k n ) n∈N be any subsequence of the sequence of natural numbers.Define and so We note that for each m ∈ N 0 , M = ker ϕ m (since for any m ∈ N 0 , f := z m ∈ I k ⊂ M, and then ϕ m (f ) = 1 = 0).
Recall the Shilov idempotent theorem (see, e.g., [11,Corollary 6.5]):If E is an open-closed subset of M(A), then there is a unique element f of A such that f 2 = f and f = 1 E (the characteristic function of E, which is identically equal to 1 on E, and 0 elsewhere on M(A) \ E).From Theorem 5.1, we get the following.
is closed and open, then there exists an idempotent

Exponentials in A(p)
We will show that every invertible element in A(p) has a logarithm.
So exchange of summations below is allowed.We have Let A(p) −1 denote the multiplicative group of invertible elements of A(p), and e A(p) the subgroup of A(p) consisting of all exponentials e f , f ∈ A(p).Proof.Let g ∈ A(p) −1 .By (R1) on page 2, there exists a δ > 0 such that | g(k)| ≥ δ p(k) for all k ∈ N 0 .In particular, g(k) = 0, and we define denotes the principal branch of the logarithm.We have 0 < δ ≤ p(k)| g(k)| ≤ g for all k ∈ N 0 , and so showing that f ∈ A(p).It follows from Lemma 6.2 that e f = g.

Solvability of Ax = b
We will show the following: Then the following are equivalent: (1) There exists an x ∈ A(p) n×1 such that A * x = b.
(2) There exists a δ > 0 such that for all k ∈ N 0 and all y ∈ C m×1 , Here •, • 2 denotes the usual Euclidean inner product on C ℓ×1 for ℓ ∈ N, and • 2 is the corresponding induced norm.Also, if a ij ∈ A(p) denotes the entry in the i th row and j th column of A, then A(k) ∈ C m×n is the matrix whose entry in the i th row and For a matrix M ∈ C m×n , M * denotes its Hermitian adjoint (obtained by taking the entrywise complex conjugate of M and then taking the transpose of the resulting matrix).We will use the following elementary linear algebraic result; see [25,Lemma 8.2].We include the short proof for the sake of completeness.= 0, then we can take x = 0, and the estimate on x 2 is obvious.So we assume that b = 0 and then A * y 0 = 0 (since we know that AA * y 0 = b).We have, using the given inequality (⋆), Since A * y 0 = 0, we obtain x 2 = A * y 0 2 ≤ 1 δ .
Proof of Theorem 7.1.(1)⇒(2): As x ∈ A(p) n×1 , there exists a C > 0 such that for all k ∈ N 0 , x(k) 2 ≤ C p(k) .So for all y ∈ C m×1 and k Setting δ := 1 C > 0 and rearranging gives (2).( 2)⇒(1): Fix k ∈ N 0 .The condition in statement (2) and Lemma 7.2 implies the existence of an In this way, we obtain a sequence (x k ) k≥0 in C n×1 .Let the components of x k be denoted by x for all k ∈ N 0 ), and so the column vector x having these components belongs to A(p) n×1 .Also,

Krull dimension of A(p) is infinite
Definition 8.1.The Krull dimension of a commutative ring R is the supremum of the lengths of chains of distinct proper prime ideals of R.
Recall that the Hardy algebra H ∞ is the Banach algebra of bounded and holomorphic functions on the unit disc D := {z ∈ C : |z| < 1}, with pointwise operations and the supremum norm • ∞ .In [29], von Renteln showed that the Krull dimension of H ∞ is infinite.We adapt the idea given in [29], to show that the Krull dimension of A(p) is infinite too.A key ingredient of the proof in [29] was the use of a canonical factorisation of H ∞ elements used to create ideals with zeroes at prescribed locations with prescribed orders.Instead of zeroes of our entire functions, we will look at the indices for the vanishing coefficients in the Taylor expansion centred at 0, and instead of orders of zeroes, we will use the notion of 'index-order' introduced below.
If f ∈ A(p) and k ∈ N 0 is an index such that f (k) = 0, then we define the index-order m(f, k) of the index k for f by If f (k+ℓ) = 0 for all ℓ ∈ N 0 , then we set m(f, k) = ∞.If f (k) = 0, then we set m(f, k) = 0. Analogous to the order of a zero of a holomorphic function, the index-order satisfies the following property.
The order of a zero ζ of the pointwise product of two holomorphic functions is the sum of the orders of ζ as a zero of each of of the two holomorphic functions.For the index order, and the weighted Hadamard product, we have the following instead: We will use the following known result; see [12,Theorem,§0.16,p.6].Proposition 8.2.If J is an ideal in a ring R, and M is a set that is closed under multiplication and M ∩ J = ∅, then there exists an ideal P such that J ⊂ P and P ∩ M = ∅, and P maximal with respect to these properties.Moreover, such an ideal P is necessarily prime.

Theorem 8.3. The Krull dimension of
Note that m(f n , a k ) ≥ k n+1 , but for each fixed n ∈ N, there exists a K n ∈ N 0 such that the gap between the indices, The set I is nonempty since 0 ∈ I. Clearly I is closed under addition, and f * g ∈ I whenever f ∈ I and g ∈ A(p).So I is an ideal of A(p).
For n ∈ N, we define Clearly f n ∈ I n , and so I n is not empty.Using (P1), we see that if It is easy to check that for all n ∈ N, I n+1 ⊂ I n and M n ⊂ M n+1 .We now prove that the inclusions are strict for each n ∈ N. From (3), it follows that We will now show that the Krull dimension of A(p) is infinite by showing that for all N ∈ N, we can construct a chain of strictly decreasing prime ideals Fix an N ∈ N. Applying Proposition 8.2, taking J = I N +1 and M = M N +1 , we obtain the existence of a prime ideal P = P N +1 in A(p), which satisfies I N +1 ⊂ P N +1 and P N +1 ∩ M N +1 = ∅.

We claim the ideal
, where f ∈ I N and g ∈ P N +1 .Since g ∈ P N +1 , by the construction of P N +1 it follows that g ∈ M N +1 .But M N ⊂ M N +1 , and so g ∈ M N as well.Thus there exists a subsequence Thus h ∈ M N .Consequently, (I Clearly I N ⊂ I N +P N +1 .Applying Proposition 8.2 again, now taking J = I N + P N +1 and M = M N , we obtain the existence of a prime ideal P = P N in A(p) such that I N + P N +1 ⊂ P N and P N ∩ M N = ∅.Thus P N +1 ⊂ I N + P N +1 ⊂ P N .The first inclusion is strict because Proceeding in this manner, we obtain the chain of distinct prime ideals P N +1 As N ∈ N was arbitrary, it follows that the Krull dimension of A(p) is infinite.

A(p) is neither Artinian nor Noetherian
Even Noetherian rings can have an infinite Krull dimension (see e.g.[21, Appendix, Example E1] or [10, Exercise 9.6]).However, in our case, A(p) is not Noetherian.Definition 9.1.A commutative ring R is called Noetherian if there is no infinite increasing sequence of ideals, that is, for every increasing sequence I 1 ⊂ I 2 ⊂ I 3 ⊂ • • • of ideals of R, there exists an N ∈ N such that I n = I N for all n > N. A commutative ring R is called Artinian if there is no infinite descending sequence of ideals, that is, for every decreasing sequence I 1 ⊃ I 2 ⊃ I 3 ⊃ • • • of ideals of R, there exists an N ∈ N such that I n = I N for all n > N.
1 Let h = f + g ∈ I N −1 + P N , where f ∈ I N −1 and g ∈ P N .Since g ∈ P N , by the construction of P N , g ∈ M N ⊃ M N −1 , and so g ∈ M N −1 .Thus there exists a subsequence (k ℓ ) ℓ∈N0 of (k) k∈N0 such that lim In absence of the Noetherian 'finiteness' property, a natural finiteness question is that of coherence.We refer the reader to the article [14] and the monograph [13] for the relevance of the property of coherence in commutative holomogical algebra.Definition 10.1.A commutative unital ring R is called coherent if every finitely generated ideal I is finitely presentable, that is, there exists an exact sequence 0 K F I 0 of R-modules, where F is a finitely generated free R-module and K is a finitely generated R-module.
All Noetherian rings are coherent, but not all coherent rings are Noetherian.(For example, the polynomial ring C[x 1 , x 2 , x 3 , • • • ] is not Noetherian (because the sequence of ideals A commutative ring in which every finitely generated ideal is principal is said to be Bézout.By property (R4) (p.3), A(p) is a Bézout ring.It is known that Bézout domains are coherent, but we cannot use this to conclude that A(p) is coherent, since A(p) is not a domain (as A(p) has nontrivial zero divisors: e.g.z * z 2 = 0).Theorem 10.2.A(p) is coherent.
Proof.Let I be a finitely generated ideal in A(p).Then I is principal by the property (R4) (p.3).So there exists an f I ∈ A(p) such that I = f I .Define χ ∈ A(p) by setting Define K = χ .Then K is a finitely generated A(p)-module.Let F := A(p) = ε .Then F is a finitely generated free module.Consider the A(p)-module morphism ϕ : F I given by ϕ(h) = f I * h for all h ∈ A(p).We will show that the sequence 0 K ֒ F I 0 is exact, where K ֒ F denotes the inclusion map.The exactness at K and at I is clear.It remains to show {h ∈ A(p) : If k ∈ N 0 is such that f I (k) = 0, then by the definition of χ, we have χ(k) = 0, and moreover, then (4) above implies that h(k) = 0, so that If k ∈ N 0 is such that f I (k) = 0, then we have χ(k) = 1 p(k) , and so (5) and ( 6) together imply that | h(k)| ≤ C| χ(k)| for all k ∈ N 0 , and so by the criterion (R1) (p. 2), χ divides h in A(p), that is, there exists some f ∈ A(p) such that h = χ * f , i.e., h ∈ χ = K, as wanted.

Generation of SL n (A(p)) by elementary matrices
Let R be a commutative unital ring with multiplicative identity 1 and additive identity element 0. Let m ∈ N. The general linear group of invertible matrices in R m×m is denoted by GL m (R).The special linear group SL m (R) is the subgroup of GL m (R) of all matrices M whose determinant det M = 1.An elementary matrix E ij (α) is a matrix having form E ij = I m + αe ij , where i = j, α ∈ R, and e ij is the m × m matrix whose entry in the i th row and j th column is 1, and all the other entries of e ij are zeros.E m (R) is the subgroup of SL m (R) generated by elementary matrices.A classical question in algebra is: , having only one nonzero entry, which is the j th component, and is equal to ε.Then we obtain Proof.For z ∈ C, we have To apply Proposition 11.1, we will need the following result.Recall that A ∈ GL n (C) possesses a logarithm, which can be obtained as follows (see e.g.[17,Example 5.20]).We will be a bit more particular about the construction of the logarithm, since we will apply this to each A(k), k ∈ N 0 , for our A ∈ SL n (A(p)), and we will then need a uniform estimate on log(p(k) A(k)), k ∈ N 0 .

Denote the spectrum (the set of eigenvalues) of A by σ(A).
There exists an open sector Ω θ = {z ∈ C \ {0} : |arg z − θ| < π n } of angular width 2π n that does not intersect the spectrum of A. Moreover, we have 0 < r := min λ∈σ(A) |λ| ≤ R := max λ∈σ(A) |λ|.Let γ be the path C R+1 + S 1 + C r/2 + S 2 (see the following picture), where C r/2 is a circular arc of radius r/2 centred at 0 traversed in the clockwise direction, C R+1 is a circular arc of radius R+1 centred at 0 traversed in the anticlockwise direction, S 1 is a radial straight line segment joining the arc C R+1 to the arc C r/2 with the fixed argument θ − π 2n , and S 2 is a radial straight line segment joining the arc C r/2 to the arc C R+1 with the fixed argument θ + π 2n .
PSfrag replacements The spectrum σ(A) is contained in the shaded region.
If log denotes the logarithm branch with a cut along the radial ray with fixed argument θ, then we have 3 This lower bound is obtained by dropping a perpendicular from the corner of the shaded region onto S 1 , which has a length r sin π 2n , and the distance between C r/2 and the shaded region is clearly r/2. 4 The exchange of the two summations is justified, since:

)
Idempotents in A(p) are described.(6) Exponentials in A(p) are described, and it is shown that every invertible element of A(p) has a logarithm, so that the first Čech cohomology group H 1 (M(A(p)), Z) with integer coefficients of the maximal ideal space M(A(p)) is trivial.(7) A generalised necessary and sufficient 'corona-type condition' on the matricial data (A, b) with entries from A(p) is given for the solvability of Ax = b with x also having entries from A(p).(8) The Krull dimension of A(p) is infinite.(9) A(p) is neither Artinian nor Noetherian.(10) A(p) is coherent.(11) The special linear group over A(p) is generated by elementary matrices.
The Bass stable rank of R is the smallest integer n such that every element in U n+1 (R) is reducible.It is infinite if no such n exists.