Abstract
Let , and let denote the ring of all functions such that , where , and equipped with pointwise operations. (Here denotes the Laplace transform.) It is shown that the ring is not coherent, answering a question of Alban Quadrat. In fact, we present two principal ideals in the domain whose intersection is not finitely generated.
1. Introduction
The aim of this paper is to show that the ring (defined below) is not coherent.
We first recall the notion of a coherent ring.
Definition 1.1. Let be a commutative ring with identity element ,
and let ( times). Suppose that .
(1) An element is called a relation on if (2) Let denote the set of all relations on .
(Then is an -submodule of the -module .)(3) The ring is called coherent if for all and all , is finitely generated, that is, there exists a and there exist , ,
such that for all ,
there exist , such that .
An integral domain is coherent if and only if the intersection of any two finitely generated ideals of the ring is again finitely generated; see [1, Theorem 2.3.2, page 45].
The coherence of some rings of analytic functions has been investigated in earlier works. For example, McVoy and Rubel [2] showed that the Hardy algebra is coherent, while the disc algebra is not. Mortini and von Renteln proved that the Wiener algebra (of all absolutely convergent Taylor series in the open unit disc) is not coherent [3]. In this article, we will show that the ring (defined below, and which is useful in control theory) is not coherent.
Notation 1. Throughout the article, we will use the
following notation:
Definition 1.2. Let denote the Banach algebraequipped with pointwise operations and the normHere denotes the Laplace transform of , given by
The above algebra arises as a natural class of transfer functions of stable distributed parameter systems in control theory; see [4, 5].
Our main result is the following.
Theorem 1.3. The ring is not coherent.
The relevance of the coherence property in control theory can be found in [6, 7]. We will prove Theorem 1.3 following the same method as in the proof of the noncoherence of given by Mortini and von Renteln in [3].
In Section 3, we will give the proof of Theorem 1.3. But before doing that, in Section 2, we first prove a few technical results needed in the sequel.
2. Preliminaries
We first recall the definition of the Hardy algebra of the open right half plane.
Definition 2.1. Let denote the Hardy space of all bounded analytic functions in the open right half plane equipped with the norm
In order to prove our main result (Theorem 1.3), we will use the relation between the convergence in versus that in .
Lemma 2.2. If , then and .
Proof. LetFor , we haveand moreover,So the result follows.
The maximal ideal (defined below) of will play an important role in the remainder of this article.
Notation. Let denote the kernel of the complex algebra
homomorphism ,
that is,
Then is a maximal ideal of , and this maximal ideal plays an important role in the proof of our main result in the next section. We will prove a few technical results about in this section, which will be used in the sequel. The following result is analogous to [3, Lemma 1].
Lemma 2.3. Let be an ideal in contained in the maximal ideal . If , that is, if every function can be factorized in a product of two functions and , then cannot be finitely generated.
Proof. Suppose thatis a finitely generated ideal in contained in the maximal ideal . By our assumption, there are functions , withSince , there exist functions withFrom this it follows thatwhere is a constant chosen so that(Here denotes the supnorm over .) This implies together with the Cauchy-Schwarz inequality thatThis inequality holds for all . With , we obtain the inequalityfor all points , whereSince , is a dense subset of (for otherwise, if is such that it has a neighbourhood in where there is no point of , then each is identically zero in , and by the identity theorem for holomorphic functions, each is zero; consequently each is zero, and so , a contradiction). So by continuity, inequality (2.11) holds in . But this contradicts the fact that each vanishes at .
Remark 2.4. Lemma 2.3 can be proved purely algebraically using Nakayama's lemma. Indeed, it holds in the following more general algebraic situation: if is a nonzero ideal of a commutative domain contained in a maximal ideal and , then cannot be finitely generated. However, we have given an analytic proof in our special case above.
Since every maximal ideal is closed, is a commutative Banach subalgebra of , but obviously without identity element. But there is a substitute, namely the notion of the approximate identity, which turns out to be useful.
Definition 2.5. Let be a commutative Banach algebra (without identity element). One says that has an approximate identity if there exists a bounded sequence of elements in such that for any ,
We will now prove the following result, which shows that the maximal ideal in has an approximate identity.
Theorem 2.6. Let Then is an approximate identity for .
The existence of an approximate identity for the maximal ideal in is not obvious. In order to prove Theorem 2.6, we will need the following lemma.
Lemma 2.7. Suppose . Then, for all , there exists a such that has compact support in , and .
Proof. Let be given. Suppose thatwhere , , and . Since , we can choose an large enough such thatWith if , and otherwise, we have that is compactly supported andFurthermore, select such thatNow let be any number satisfying , and defineSetThen andSo . Clearly has compact support contained in . We haveThusThis completes the proof.
We are now ready to prove the existence of an approximate identity for the maximal ideal in .
Proof of Theorem 2.6. We haveThus for
an ,Given ,
and arbitrarily small, in view of Lemma 2.7, we
can find a such that has compact support and .
ThenSo it is enough to prove thatfor all such that has compact support in .
We do this below.
We haveLet denote the convolution :We note that ,
since is an ideal in .
Let be such thatWe haveWe estimate () as follows:Since the integral () does not depend on ,
we obtain thatFurthermore,Since has compact support in , is an entire function by the Payley-Wiener
theorem (see, e.g., [8, Theorem 7.2.3, page 122]). Consequently,This completes the proof.
We will also need the following lemma, which is basically a repetition of key steps from Browder's proof of Cohen's factorization theorem; see [9, Theorem 1.6.5, page 74]. We will need this version since in our application in the proof of Theorem 1.3, we are not able to use Cohen's factorization theorem directly.
Lemma 2.8. Let ,
and .
Let denote the set of all invertible elements in .
Then there exists a sequence in such that
(1) for all , ;(2) is convergent in to a limit ;(3) for all , , .
Proof. We will first prove two general results in steps (A) and (B), which
we will use in the rest of the proof.
(A) Let and ,
where .
Then ,
where is a number chosen such thatIndeed,and so
(B) Furthermore, under the same assumptions and
notation as in (A) above, we now show that if is small for some ,
then so is ,
where .
Sincewe haveButHenceThis estimate will be used in
constructing the sequence of 's.
Let denote the approximate identity for from Theorem 2.6. Let be such that for all .
Choose such that
We will inductively define a sequence with terms from the approximate identity for such that ifthen we have , ,
and
(P1) for all , ,(P2)for all , , .
Since is an approximate identity for ,
we can choose such thatDefine .
So by (A), and using the calculation in (B), we see thatSuppose that have been constructed, so that defined by (2.44) satisfies (P1) and (P2). We
assert that if we choose such thatare sufficiently small, then defined by (2.44) satisfies (P1) and (P2),
completing the induction step.
Indeed, if ,
we haveLet be defined byThen we haveHence and moreover is small, provided only that is small for .
(Indeed, this is because is an open set in .)
Since ,
we have then , ,
and so for ,Moreover, recall that by (B), we
know thatThus if () and () are sufficiently small, we will have as small as we please. This completes the
induction step.
Since , ,
and is a Banach algebra, it follows thatand the proof is completed.
3. Noncoherence of
Proof of Theorem 1.3. We will use the characterization that an integral domain is
coherent if and only if the intersection of any two finitely generated ideals
of the ring is again finitely generated; see [1, Theorem 2.3.2, page 45]. In fact, we present two
finitely generated ideals and such that is not finitely generated.
Let be given byClearly we have .
It is known (see, e.g., [3, Remark after Theorem 1,
page 224]) thatHere .
So if 's are defined viathen we haveIf ,
then ,
and so from (3.3), we haveSince ,
the right-hand side in (3.5) belongs to .
So .
We define the ideals and of .
LetWe claim that .
Trivially .
To prove the reverse inclusion, let .
Then there exist two functions and in such that .
Since and is an integral domain, we obtain that .
So .
Let denote the idealThen .
Since has a singularity at ,
it follows that .
We will show that .
Let .
We would like to factor with and .
Applying Lemma 2.8 with and ,
for any ,
there exists a sequence in such that
(1) for all , ;(2) is convergent in to a limit ;(3) for all ,
PutThen .
Also ,
since is bounded by on and .
The estimates above imply that and are Cauchy sequences in .
Since is closed, they converge to elements and ,
respectively, in ,
that is, and .
Since convergence in implies convergence in (Lemma 2.2), it follows thatand so by the uniqueness of the
limit of the sequence in ,
we have .
Also, in -norm we havesince multiplication is
continuous in the Banach algebra .
Since and belong to ,
we see that .
Moreover, as ,
we have got the desired factorization and .
But ,
since .
By Lemma 2.3, it follows that cannot be finitely generated. Therefore, is not finitely generated.
Remark 3.1. The ideal in the above proof can be interpreted as an ideal of denominators; see [10, page 396]. Using the fact
that ,
we have ,
where denotes the field of fractions of .
We can then consider the fractional
ideal of (see [11, page 19]) and the ideal of denominators of ,
namely .
Based on the results in [12, Theorem 3, Example 3], it
follows that does not admit a weak coprime factorization,
since is not a principal ideal of .
In particular, does not admit a coprime factorization, that
is, there do not exist such that , and .
Moreover, is not
internally stabilizable, since otherwise would be generated by two elements. Finally,
the fact that is not finitely generated implies that is not a greatest common divisor domain: indeed,
were it the case that is a greatest common divisor domain, then by
[12, Corollary 3],
every element in would admit a weak coprime factorization.
Acknowledgment
The author thanks all the referees for their careful review, and in particular, two of the referees for the Remarks 2.4 and 3.1.