2.1. $L^2(X,\mu )$ is a flat $L^\infty (X,\mu )$ -module

Proof. Let $r_1 m_1+\cdots +r_nm_n=0\in L^2(X,\mu )$ for some $n\in {\mathbb {N}}$ , and functions $r_1,\ldots , r_n\in L^\infty (X,\mu ) $ and $m_1,\ldots , m_n \in L^2(X,\mu )$ . Then, there exists a subset N of X with $\mu (N)=0$ such that for all $x\in X\setminus N$ ,

$$ \begin{align*}\textstyle r_1(x)m_1(x)+ \cdots + r_n(x)m_n(x)\!=\!0, \end{align*} $$

that is, ${\mathbf {m}}(x):=(m_1(x),\ldots , m_n(x))\in {\mathbb {C}}^n$ belongs to the orthogonal complement ${\mathbf {r}}(x)^\perp $ of ${\mathbf {r}}(x):=(r_1(x), \ldots , r_n(x))\in {\mathbb {C}}^n$ with respect to the usual Euclidean inner product $\langle \cdot ,\cdot \rangle $ (and corresponding norm denoted by $\|\cdot \|$ below) on ${\mathbb {C}}^n$ . If ${\mathbf {r}}(x)\!\neq \! 0$ , there exist $n-1$ orthonormal vectors ${\mathbf {e}}_1(x),\ldots , {\mathbf {e}}_{n-1}(x)$ forming a basis for ${\mathbf {r}}(x)^\perp $ , and we set ${\mathbf {e}}_n(x)\!=\!{\mathbf {0}}\!\in \! {\mathbb {C}}^n$ . If ${\mathbf {r}}(x)\!=\!0$ , then let $\{{\mathbf {e}}_1(x),\ldots , {\mathbf {e}}_{n}(x)\}$ be any orthonormal basis for ${\mathbb {C}}^n$ . For $x\in X\setminus N$ , define $\mu _1(x),\ldots , \mu _{n}(x)$ by $\mu _j(x)=\langle {\mathbf {m}}(x), {\mathbf {e}}_j(x)\rangle $ , $1\le j\le n$ . For $x\in X\setminus N$ , and $1\le i,j\le n$ , set $\rho _{ij}(x)=\langle {\mathbf {e}}_j(x), e_i\rangle $ , where $e_1,\ldots , e_n$ are the standard basis vectors for ${\mathbb {C}}^n$ . Then, by the Cauchy–Schwarz inequality, $|\rho _{ij}(x)|\le \|{\mathbf {e}}_j(x)\| \| e_i\|= 1$ . Thus, we have defined an element $\rho _{ij}$ of $ L^\infty (X,\mu ) $ for each $1\le i,j,\le n$ . Moreover, for all $x\in X\setminus N$ , we have for $1\le j\le n$ that

$$ \begin{align*}\textstyle \begin{array}{rcl} 0=\langle {\mathbf{r}}(x), {\mathbf{e}}_j(x)\rangle &=& \left[\begin{smallmatrix} r_1(x) &\cdots & r_n(x)\end{smallmatrix}\right] \left[\begin{smallmatrix} \langle {\mathbf{e}}_j(x),e_1\rangle \\ \vdots \\ \langle {\mathbf{e}}_j(x),e_n\rangle \end{smallmatrix}\right]\\[0.51cm] &=& r_1(x) \rho_{1j}(x)+\cdots +r_{n}(x) \rho_{nj}(x). \end{array} \end{align*} $$

Thus, in $L^\infty (X,\mu )$ , we have $r_1 \rho _{1j}+\cdots +r_n\rho _{nj}=0$ , $1\le j\le n$ . Moreover, $\mu _1,\ldots , \mu _n\in L^2(X, \mu ) $ because for $1\le i\le n$ , we have

$$ \begin{align*}\textstyle \begin{array}{rcl} \int_{X\setminus N} |\mu_i(x)|^2 \;d\mu(x) &=& \int_{X\setminus N} |\langle {\mathbf{m}}(x), {\mathbf{e}}_i(x)\rangle|^2 \;d\mu(x)\\[0.21cm] &\le& \int_{X\setminus N} \|{\mathbf{m}}(x)\|^2 1 \;d\mu(x)\\[0.21cm] &=& \|m_1\|_2^2+\cdots+\|m_n\|_2^2. \end{array} \end{align*} $$

Finally, for $x\in X\setminus N$ , we have

$$ \begin{align*}\textstyle \begin{array}{rcl} \sum\limits_{i=1}^{n} m_i(x) e_i &=& {\mathbf{m}}(x) =\sum\limits_{j=1}^{n} \langle {\mathbf{m}}(x),{\mathbf{e}}_j(x)\rangle\;\! {\mathbf{e}}_j(x) = \sum\limits_{j=1}^{n} \mu_j(x) {\mathbf{e}}_j(x) \\ &=& \sum\limits_{j=1}^{n} \mu_j(x) \sum\limits_{i=1}^{n} \langle {\mathbf{e}}_j(x) , e_i\rangle e_i = \sum\limits_{i=1}^{n} \sum\limits_{j=1}^{n} \rho_{ij}(x) \mu_j(x)e_i, \end{array} \end{align*} $$

and so, we get $ \textstyle m_i =\sum \limits _{j=1}^{n} \rho _{ij} \mu _j$ in $L^2(X,\mu )$ for all $1\le i\le n$ .

2.2. The case of finitely generated maximal ideals

We claim that for any finitely generated, nonzero, proper ideal $\mathfrak {n}$ of $L^\infty (X,\mu )$ , we have $\mathfrak {n} L^2(X,\mu )\subsetneq L^2(X,\mu )$ . To do this, we first show that $L^\infty (X,\mu )$ is a Bézout ring. Recall that a commutative ring is called Bézout if every finitely generated ideal is principal.

Before we give the proof, we collect some useful observations first. For $f\in L^\infty (X,\mu )$ , let $ |f|, \overline {f}\in L^\infty (X,\mu )$ be the functions obtained by taking pointwise complex absolute value, and pointwise complex conjugation, respectively. Then, for a representative function $f:X\to {\mathbb {C}}$ of an element in $L^\infty (X,\mu )$ , $f=|f|\cdot u_f$ , where $u_f\in L^\infty (X,\mu )$ is given by

$$ \begin{align*}\textstyle u_f(x)=\left\{\begin{array}{cl} \frac{f(x)}{|f(x)|} & \textrm{if }\; f(x)\neq 0,\\[0.21cm] 1 & \textrm{if }\;f(x)=0. \end{array}\right. \end{align*} $$

We have $f\in L^\infty (X,\mu )$ if and only if $|f|\in L^\infty (X,\mu )$ . Also, $f\in L^\infty (X,\mu )$ if and only if $\overline {f}\in L^\infty (X,\mu )$ . Also, $u_f u_{\overline {f}}={\mathbf {1}}$ (the constant function, taking value $1$ everywhere on X) and $|f|=f \overline {u_f}$ .

In a commutative ring R, the ideal in R generated by $r_1,\ldots , r_n$ , $n\in {\mathbb {N}}$ , is denoted by $\langle r_1,\ldots , r_n\rangle $ .

Proof. It suffices to show that an ideal $\langle f,g\rangle $ generated by $f,g\!\in \! L^\infty (X,\mu )$ is principal. We claim that $\langle f,g\rangle =\langle |f|+|g|\rangle. $ Since $\overline {u_f}, \overline {u_g}\in L^\infty (X,\mu )$ , we have $|f|+|g|=f \overline {u_f}+g \overline {u_g}\in \langle f,g\rangle. $ Thus, $\langle |f|+|g|\rangle \subset \langle f,g\rangle $ . To show the reverse inclusion, let us define F by

$$ \begin{align*}\textstyle F(x) =\left\{\begin{array}{cl} \frac{f(x)}{|f(x)|+|g(x)|} & \textrm{if }\; |f(x)|\!+\!|g(x)|\!\neq \!0,\\[0.21cm] 1 & \textrm{if }\; |f(x)|\!+\!|g(x)|\!=\!0, \end{array}\right. \end{align*} $$

$x\in X$ . Then, $|F(x)|\leq 1$ for all $x\in X$ , and so $F \in L^\infty (X,\mu )$ . Moreover, $f=F \cdot (|f|+|g|)$ , and so $f\in \langle |f|+|g|\rangle $ . Similarly, $g\in \langle |f|+|g|\rangle $ too. Hence, $\langle f,g\rangle \subset \langle |f|+|g|\rangle $ . Consequently, $\langle f,g\rangle =\langle |f|+|g|\rangle $ .

If the nonzero maximal ideal $\mathfrak {m}$ of $L^\infty (X,\mu )$ is finitely generated, then is $\mathfrak {m}L^2(X,\mu )\subsetneq L^2(X,\mu )$ ? We claim that for any finitely generated, nonzero, proper ideal $\mathfrak {n}$ of $L^\infty (X,\mu )$ , $\mathfrak {n} L^2(X,\mu )\subsetneq L^2(X,\mu )$ .

Proof. As $L^\infty (X,\mu )$ is a Bézout ring, $\mathfrak {n}=\langle g\rangle $ for some $g\in L^\infty (X,\mu )$ , and as $\mathfrak {n}$ is nonzero, so is g. Then,

$$ \begin{align*}\textstyle \mathfrak{n}L^2(X,\mu)=\langle g\rangle L^2(X,\mu) =gL^\infty(X,\mu) L^2(X,\mu) = gL^2(X,\mu). \end{align*} $$

We claim that $gL^2(X,\mu )\subsetneq L^2(X,\mu )$ . Suppose on the contrary that $gL^2(X,\mu )=L^2(X,\mu )$ .

We will first show that $gL^2(X,\mu )=L^2(X,\mu )$ implies that $g\neq 0$ almost everywhere in X. Indeed, if $\mu \{x\in X: g(x)=0\}>0$ , then since $\mu $ is a Radon measure, there exists a compact set ${K\subset \{x\in X\!:\! g(x)\!=\!0\}}$ with $0<\mu (K)<\infty $ . Let $f=\mathbf {1}_K$ be the indicator function of K (i.e., pointwise equal to $1$ on K and $0$ on $X\setminus K$ ). Then, clearly, $f\in L^2(X,\mu )$ , and so (thanks to our assumption that $gL^2(X,\mu )=L^2(X,\mu )$ ), there exists an $h\in L^2(X,\mu )$ such that $f=gh$ . But at any point $x\in K$ , we then get $1=\mathbf {1}_K(x)=f(x)=g(x)h(x)=0h(x)=0$ , a contradiction.

As $g(x)\neq 0$ almost everywhere in X, it follows that the multiplication map $M_g: L^2(X,\mu ) \to L^2(X,\mu )$ , $h\mapsto gh$ , is injective. But thanks to our assumption $gL^2(X,\mu )=L^2(X,\mu )$ , we also know that $M_g$ is surjective. Thus, $M_g$ is invertible. As $\mu $ is $\sigma $ -finite, we conclude that g is invertible as an element of $L^\infty (X,\mu )$ , that is, $g^{-1}\in L^\infty (X,\mu )$ (see, e.g., [Reference Halmos5, Problem 67]). But then $\mathfrak {n}=\langle g\rangle =L^\infty (X,\mu )$ , contradicting the properness of $\mathfrak {n}$ .

Ultrafilters and $M(\ell ^\infty )$ :

In order to show the non-faithfulness, we will use the maximal ideal space of $\ell ^\infty $ , which is known to be in correspondence with the set of ultrafilters on ${\mathbb {N}}$ . We begin by recalling some preliminaries about ultrafilters.

Let $\ell ^\infty $ be the Banach algebra of all complex bounded sequences on ${\mathbb {N}}=\{1,2,3,\ldots \}$ with pointwise operations and the supremum norm, which is given by

$$ \begin{align*}\textstyle \|f\|_\infty:=\sup_{n\in {\mathbb{N}}} |f(n)|\; \text{ for }\;f=(f(n))_{n\in {\mathbb{N}}} \in \ell^\infty. \end{align*} $$

Let $\ell ^2 $ be the Hilbert space of all square summable complex sequences with the inner product corresponding to the norm $\|\cdot \|_2$ , where

$$ \begin{align*}\textstyle \|h\|_2^2=\sum\limits_{n=1}^\infty |h(n)|^2 \;\text{ for all }\; h=(h(n))_{n\in {\mathbb{N}}} \in \ell^2. \end{align*} $$

Then, with the action of $\ell ^\infty $ on $\ell ^2 $ given by termwise multiplication, $ \textstyle f\cdot h=\pmb {(}\;\!f(n)h(n){\pmb {)}}_{n\in {\mathbb {N}}}$ for $ f\in \ell ^\infty $ , $h\in \ell ^2$ , $\ell ^2 $ is an $\ell ^\infty $ -module. As a consequence of our results in this section, it will follow that as an $\ell ^\infty $ -module, $\ell ^2$ is flat, but not faithfully flat (see Example 2.5).

For $n\in {\mathbb {N}}$ , define $\varphi _n: \ell ^\infty \to {\mathbb {C}}$ by $\varphi _n(f)=f(n)$ for all $f\in \ell ^\infty $ . Point evaluations $\varphi _n$ , $n\in {\mathbb {N}}$ , are nonzero complex homomorphisms. Thus, ${\mathbb {N}}$ can be identified with a subset of $M(\ell ^\infty )$ (and the topology on ${\mathbb {N}}$ , induced from the Gelfand topology on $M(\ell ^\infty )$ , coincides with the usual Euclidean topology on ${\mathbb {N}}$ as a subset of ${\mathbb {R}}$ ). For each $n\in {\mathbb {N}}$ , the maximal ideal $\mathfrak {m}_n:=\ker \varphi _n$ has the property that $\mathfrak {m}_n\ell ^2\subsetneq \ell ^2$ . Indeed, for example, if $f\in \ell ^2$ is the sequence with $1$ as the nth term, and all other terms $0$ , then $f\not \in \mathfrak {m}_n\ell ^2$ (otherwise $f\!=\!gh$ , for some $g\!\in \! \mathfrak {m}_n$ , $h\!\in \! \ell ^2$ , and so $0=\widehat {g}(\varphi _n)=g(n)$ , giving $1=f(n)=g(n)h(n)=0h(n)=0$ , a contradiction). Despite the fact that $\mathfrak {m}_n\ell ^2\subsetneq \ell ^2$ for each $n\in {\mathbb {N}}$ , there exist many maximal ideals $\mathfrak {m}$ of $\ell ^\infty $ for which $\mathfrak {m}\ell ^2=\ell ^2$ . We recall that $M(\ell ^\infty )$ can be identified with the set of all ultrafilters on ${\mathbb {N}}$ as described below. We refer to the article Komjáth and Totik [Reference Komjáth and Totik7] for background on ultrafilters.

An ultrafilter ${\mathcal {F}}$ on ${\mathbb {N}}$ is a collection of subsets of ${\mathbb {N}}$ with the properties listed below:

1. (F1) $\; {\mathbb {N}}\in {\mathcal {F}}$ , $\emptyset \not \in {\mathcal {F}}$ .

2. (F2) If $A\in {\mathcal {F}}$ and $A\subset B$ , then $B\in {\mathcal {F}}$ .

3. (F3) If $A,B\in {\mathcal {F}}$ , then $A\cap B\in {\mathcal {F}}$ .

4. (F4) $\; A\in {\mathcal {F}}$ if and only if ${\mathbb {N}}\setminus A \not \in {\mathcal {F}}$ .

A family of subsets of ${\mathbb {N}}$ with the properties (F1)–(F3) is called a filter on ${\mathbb {N}}$ . For a fixed $n\in {\mathbb {N}}$ , the ultrafilter

$$ \begin{align*}{\mathcal{F}}_n=\{A\subset {\mathbb{N}}: n\in A\} \end{align*} $$

is called a principal ultrafilter on ${\mathbb {N}}$ . As ${\mathbb {N}}$ is an infinite set, it can be shown (using Zorn’s lemma) that there exists a non-principal ultrafilter on ${\mathbb {N}}$ . If ${\mathcal {F}}$ is a non-principal ultrafilter on ${\mathbb {N}}$ , then it contains all cofinite sets in ${\mathbb {N}}$ (i.e., all subsets of ${\mathbb {N}}$ whose complements are finite sets). For a bounded complex sequence $(a_n)_{n\in {\mathbb {N}}}$ , a complex number L, and an ultrafilter ${\mathcal {F}}$ on ${\mathbb {N}}$ , we write

$$ \begin{align*}\textstyle \lim\limits_{\mathcal{F}} a_n=L \end{align*} $$

if for every $\epsilon \!>\!0$ , the set $\{n\in {\mathbb {N}}: |a_n-L|<\epsilon \}$ belongs to ${\mathcal {F}}$ . It is clear that if a complex sequence $ (a_n)_{n\in {\mathbb {N}}}$ is convergent in ${\mathbb {C}}$ with limit L, then for each non-principal ultrafilter ${\mathcal {F}}$ on ${\mathbb {N}}$ , we have

$$ \begin{align*}\textstyle \lim\limits_{\mathcal{F}} a_n\!=\!L \end{align*} $$

as well. For $m\in {\mathbb {N}}$ , and the principal ultrafilter ${\mathcal {F}}_m$ on ${\mathbb {N}}$ , we have

$$ \begin{align*}\textstyle \lim\limits_{{\mathcal{F}}_m}a_n =L \end{align*} $$

if and only if $a_m=L$ . The maximal ideal $\mathfrak {m}_{\mathcal {F}}$ of $\ell ^\infty $ corresponding to an ultrafilter ${\mathcal {F}}$ on ${\mathbb {N}}$ is

$$ \begin{align*}\textstyle \mathfrak{m}_{\mathcal{F}}=\{f\in \ell^\infty : \lim\limits_{\mathcal{F}} f(n)=0\}. \end{align*} $$

This exhausts $M(\ell ^\infty )$ . See, for example, [Reference Gillman and Jerison4, Theorem, §2.5] (where only the real-valued case is considered, but the proof for the complex-valued case is the same, mutatis mutandis). From the above, we know that if $m\in {\mathbb {N}}$ , then $\mathfrak {m}_{{\mathcal {F}}_m} \ell ^2 \subsetneq \ell ^2 $ . With these preliminaries about $M(\ell ^\infty )$ in hand, we show the non-faithfulness of the flat $L^\infty (X,\mu )$ -module $L^2(X,\mu )$ in the following section.

2.3. $L^2(X,\mu )$ is not a faithfully flat $L^\infty (X,\mu )$ -module

Finally, we show the main result of this section. We will make a mild assumption, which will be satisfied in all the examples of our interest.

Theorem 2.4. Let X be a locally compact Hausdorff topological space $, \mu $ be a positive Radon measure on $X,$ such that there is a family of Borel sets $U_n$ of $X, n\in {\mathbb {N}},$ such that:

• • $U_1\subset U_2\subset U_3\subset \dots ,$ and $X=\textstyle \bigcup \limits _{n\in {\mathbb {N}}} U_n;$

• • for all $n\!\in \! {\mathbb {N}}, \overline {U_n}$ is compact $,$ and $\mu (U_{n}\setminus U_{n-1})>0,$ where $U_0:=\emptyset $ .

Let ${\mathcal {F}}$ be any non-principal ultrafilter on ${\mathbb {N}},$ and for each $n\in {\mathbb {N}},$ let $\varphi _n\in M\pmb {(}L^\infty (U_n\setminus U_{n-1}, \mu )\pmb {)}$ . Define $\varphi \in M\pmb {(} L^\infty (X,\mu )\pmb {)}$ by

$$ \begin{align*}\textstyle \varphi(f)=\lim\limits_{\mathcal{F}} \varphi_n(f|_{U_n\setminus U_{n-1}}) \;\text{ for all }\; f\in L^\infty(X,\mu). \end{align*} $$

If $\mathfrak {m}:=\ker \varphi ,$ then $\mathfrak {m}L^2(X,\mu )=L^2(X,\mu )$ .

Note that under the above assumptions, X is necessarily an infinite set. As each $U_n\setminus U_{n-1}$ is a Borel set, the restriction of the Radon measure $\mu $ to $U_n\setminus U_{n-1}$ is again a Radon measure (see, e.g., [Reference Folland2, Chapter 7, Example 7]). For each $n\in {\mathbb {N}}$ , $L^\infty (U_n\setminus U_{n-1}, \mu )$ is a Banach algebra with pointwise operations and the (essential) supremum norm. If the complex homomorphisms $\varphi _n\in M\pmb {(}L^\infty (U_n\setminus U_{n-1}, \mu )\pmb {)}$ , $n\in {\mathbb {N}}$ , then for any $g\in L^\infty (X,\mu )$ , the complex sequence $\pmb {(}\varphi _n(g|_{U_n\setminus U_{n-1}}){\pmb {)}}_{n\in {\mathbb {N}}}\in \ell ^\infty $ because for all $n\in {\mathbb {N}}$ , $ |\varphi _n(g|_{U_n\setminus U_{n-1}})| \le \|\varphi _n\|\|g|_{U_n\setminus U_{n-1}}\|_\infty \le 1\|g\|_\infty. $ Here, $\|\varphi _n\|$ denotes the operator norm of $\varphi _n:L^\infty (U_n\setminus U_{n-1}, \mu )\to {\mathbb {C}}$ which is equal to $1$ (see, e.g., [Reference Douglas1, Proposition 2.22]). If ${\mathcal {F}}$ is any non-principal ultrafilter on ${\mathbb {N}}$ and $\varphi _n\in M\pmb {(}L^\infty (U_n\setminus U_{n-1}, \mu )\pmb {)}$ , $n\in {\mathbb {N}}$ , then $\varphi : L^\infty (X,\mu )\to {\mathbb {C}}$ , as defined in the theorem statement above, is a continuous linear transformation, and it respects multiplication. If $\mathbf {1}$ denotes the constant function on X taking value $1$ everywhere, then $\varphi (\mathbf {1})=1$ . So $\varphi \in M\pmb {(}L^\infty (X,\mu )\pmb {)}$ .

Proof. Let $f\in L^2(X,\mu )$ . We will construct a function $w:X\to (0,+\infty )$ such that $w(x)\to +\infty $ as $x\to \boldsymbol {\infty }$ , and $fw\in \ell ^2 $ . Here, $X\cup \{\boldsymbol {\infty }\}$ is the one-point/Alexandroff compactification of X. Below, by $f\in L^2(X,\mu )$ having a compact support, we mean that there exists a representative of f that is identically $0$ outside a compact set.

1. 1^° If f has compact support, then we simply set $w|_{U_1}\!=\!1$ and for $n\ge 2$ , $w|_{U_n\setminus U_{n-1}}=n$ .

2. 2^° Suppose f does not have compact support. We set $w|_{U_1}\!=\!1$ and for all $n\ge 2$ , $w|_{U_n\setminus U_{n-1}}=\frac {1}{\sqrt [4]{r_{n-1}}}$ , where for $n\in {\mathbb {N}}$ ,

   $$ \begin{align*}\textstyle r_n\!:=\!\sum\limits_{k=n+1}^\infty a_k^2, \;\text{ and }\; a_k^2\!:=\!\int_{x\in U_k\setminus U_{k-1}} |f(x)|^2 d\mu(x) \;\text{ for }\;k\in {\mathbb{N}}. \end{align*} $$
   Note that $r_n>0$ since f does not have compact support. We have $r_n\to 0$ as $n\to +\infty $ because $f\in L^2(X,\mu )$ . Thus, $w(x)\to +\infty $ as $x\to {\boldsymbol {\infty }}$ . Moreover, using Proposition 1.1, we now show that $fw\in L^2 (X,\mu )$ . We have
   $$ \begin{align*}\begin{array}{rcl} \quad\quad&& \int_X |f(x)|^2 w(x)^2d\mu(x)\\[0.18cm] &=& \int_{U_1} |f(x)|^2d\mu(x)+ \sum\limits_{n=2}^\infty \int_{x\in U_{n}\setminus U_{n-1}} |f(x)|^2 \frac{1}{\sqrt{r_{n-1}}} d\mu(x) \quad\quad \quad (\star)\\ &\le & \|f\|_2^2 + \sum\limits_{n=2}^\infty \frac{a_n^2}{\sqrt{r_{n-1}}}<\infty. \end{array} \end{align*} $$

Define $g\in L^\infty (X,\mu )$ by

$$ \begin{align*}\textstyle g(x)=\frac{1}{w(x)}\;\text{ for all }\;x\in X. \end{align*} $$

Then, $g(x)\to 0$ as $x\to \boldsymbol {\infty }$ . Thus, as $w|_{U_n\setminus U_{n-1}}$ is a constant function, taking value which we denote by $\omega _n$ , and since by construction, $(\omega _n)_{n\in {\mathbb {N}}}$ diverges to $+\infty $ as $n\to +\infty $ , we get

$$ \begin{align*}\textstyle \varphi(g)=\lim\limits_{\mathcal{F}} \varphi_n(g|_{U_n\setminus U_{n-1}})= \lim\limits_{\mathcal{F}} \frac{1}{\omega_n}=0. \end{align*} $$

So $g\in \ker \varphi =\mathfrak {m}$ . Next, define h by

$$ \begin{align*}\textstyle h(x)=\frac{f(x)}{g(x)}\;\text{ for all }\;x\in X. \end{align*} $$

By ( $\star $ ) above, h belongs to $L^2(X,\mu )$ . So $f=gh$ , where $g\in \mathfrak {m}$ and $h\in L^2(X,\mu ) $ . This completes the proof.

An immediate consequence of Theorem 2.4 is that, under the same assumptions on X and $\mu $ , the $L^\infty (X,\mu )$ -module $L^2(X,\mu )$ is not faithfully flat.

2.4. Applications/examples

The following list of examples is motivated by the consideration of signal spaces arising in control theory (see, e.g., Quadrat [Reference Quadrat10]).

3.1. $H^2$ is a flat $H^\infty $ -module

The vector-valued Beurling theorem (the Lax–Halmos theorem, see, e.g., [Reference Nikolskiĭ9, Corollary 6, pp. 17–18]), characterizing shift-invariant subspaces of $H^2$ , implies that $H^2$ is a flat $H^\infty $ -module. This result was also shown in [Reference Quadrat10, Proposition 8] (in the half-plane setting), but we include a short proof for completeness. Below, $(H^2)^n$ is the Hilbert space which is the direct sum of n ( $\in {\mathbb {N}}$ ) copies of $H^2$ .

Proof. Let $n\in {\mathbb {N}}$ and $r_1,\ldots , r_n\in H^\infty $ . Consider the ${\mathbb {C}}$ -linear map ${\mathbf {r}}:(H^2)^n\to H^2$ given by ${\mathbf {r}}(h_1,\ldots , h_n) =r_1 h_1+\cdots +r_n h_n$ for all $ (h_1,\ldots , h_n)\in (H^2)^n$ . It is clear that $\ker {\mathbf {r}}$ is a shift-invariant subspace (i.e., $(zh_1,\ldots , zh_n)\in \ker {\mathbf {r}}$ whenever $(h_1,\ldots , h_n)\in \ker {\mathbf {r}}$ ). By the Beurling–Lax–Halmos theorem, there exists a $k\in {\mathbb {N}}$ and a matrix $[\rho _{ij}]$ of size $n\times k$ with $H^\infty $ entries, such that

$$ \begin{align*}\textstyle \ker {\mathbf{r}}=\Big\{\sum\limits_{j=1}^k \rho_{ij} \varphi_j: \varphi_1, \ldots, \varphi_k\in H^2\Big\}. \end{align*} $$

For a $j\!\in \! \{1,\ldots , k\}$ , taking $\varphi _{i}\!=\!\delta _{ij}$ (equal to the constant function ${\mathbf {0}}$ if $i \neq j$ and the constant function ${\mathbf {1}}$ if $i\!=\!j$ ), $1\!\le \! i\!\le \! k$ , we get $(\rho _{1j},\ldots , \rho _{nj})\!\in \! \ker {\mathbf {r}}$ . So $H^2$ is a flat $H^\infty $ -module.

3.2. The case of finitely generated maximal ideals

We claim that for any finitely generated, nonzero, closed, proper ideal $\mathfrak {n}$ of $H^\infty $ , we have $\mathfrak {n} H^2\subsetneq H^2$ . Using a result from Mortini [Reference Mortini8] (saying that any finitely generated, nonzero ideal of $H^\infty $ is closed if and only if it is a principal ideal generated by an inner function), $\mathfrak {n}=\langle g\rangle $ , where $g\in H^\infty $ is an inner function. Hence, $\mathfrak {n} H^2=\langle g\rangle H^2=gH^\infty H^2= gH^2 \subsetneq H^2$ . (That the last inclusion is strict: Otherwise, in particular, ${\mathbf {1}}\in H^2=gH^2$ , and so by looking at boundary values on ${\mathbb {T}}$ , the pointwise complex conjugate $\overline {g}$ of g satisfies $\overline {g}=\frac {1}{g} \in H^2$ . Thus, $g,\frac {1}{g}\in H^2$ , implying that the Fourier coefficients $\hat {g}_m$ of g are zero for all $m\neq 0$ , i.e., $g=\alpha $ with $|\alpha |=1$ . But then $\mathfrak {n}=\langle g\rangle =H^\infty $ , contradicting the properness of $\mathfrak {n}$ .)

3.3. Some background on Hardy spaces

We recall a few facts we need for the proof of the non-faithfulness of the flat $H^\infty $ -module $H^2$ . Background on Hardy spaces can be found, for example, in Garnett [Reference Garnett3] and Hoffman [Reference Hoffman6].

We recall that an inner function is a $g\in {\mathcal {O}}({\mathbb {D}})$ such that $|g(z)|\le 1$ for all $z\in {\mathbb {D}}$ and such that $|g(e^{i\theta })|=1$ for almost all $\theta \in (-\pi ,\pi ]$ . An outer function is an analytic function $F\in {\mathcal {O}}({\mathbb {D}})$ having the form

$$ \begin{align*}\textstyle F(z)=\alpha \exp\big( \frac{1}{2\pi} \int_{-\pi}^\pi \frac{e^{i\theta}+z}{e^{i\theta}-z} k(\theta) d\theta\big), \end{align*} $$

where $k:{\mathbb {T}}\to {\mathbb {R}}$ belongs to $L^1({\mathbb {T}})$ and $\alpha \in {\mathbb {T}}$ . Then, $k(\theta )=\log |F(e^{i\theta })|$ for almost all $\theta \in (-\pi ,\pi ]$ .

For the results claimed below, we refer to [Reference Hoffman6, pp. 160–162]. Consider the identity function ${\mathbf {z}} \in H^\infty $ , given by ${\mathbb {D}}\owns z\mapsto z$ . Then, the map $\pi : M( H^\infty )\to {\mathbb {C}}$ , $ \pi (\varphi )=\varphi ({\mathbf {z}})\text { for all }\varphi \in M( H^\infty ), $ is a continuous map onto the closed unit disk $\overline {{\mathbb {D}}}$ in ${\mathbb {C}}$ . Over the open unit disk ${\mathbb {D}}$ , $\pi $ is one-to-one, and maps ${\mathbb {D}}$ homeomorphically onto an open subset of $M( H^\infty )$ , be sending $\lambda \in {\mathbb {D}}$ to $\varphi _\lambda \in M(H^\infty )$ , where $\varphi _\lambda $ is given by $\varphi _\lambda (f)=f(\lambda )$ for all $f\in H^\infty $ . The remainder of $M( H^\infty )$ is mapped by $\pi $ onto the unit circle. If $|\alpha |=1$ , then $\pi ^{-1}\{\alpha \}$ is the fiber of $M( H^\infty )$ over $\alpha $ . If $g\in H^\infty $ and $\alpha \in {\mathbb {T}}$ , then the range of $\widehat {g}$ on the fiber $\pi ^{-1}\{\alpha \}$ is the set of all $\zeta \in {\mathbb {C}}$ such that there exists a sequence $(\lambda _n)_{n\in {\mathbb {N}}}$ in ${\mathbb {D}}$ with the properties that $ \lim \limits _{n\to \infty }\lambda _n=\alpha \text { and } \lim \limits _{n\to \infty } g(\lambda _n)=\zeta. $

It is clear that for each point evaluation $\varphi _\lambda $ , where $\lambda \in {\mathbb {D}}$ , the corresponding maximal ideal ${\mathfrak {m}_\lambda :=\ker \varphi _\lambda }$ of $H^\infty $ has the property that $\mathfrak {m}_\lambda H^2\subsetneq H^2$ (because any $g\in \mathfrak {m}_\lambda $ has a zero at $\lambda $ , and so each element of $\mathfrak {m}_\lambda H^2$ will also have a zero at $\lambda $ , but ${\mathbf {1}}\in H^2$ does not vanish at $\lambda $ ).

However, $H^2$ is not faithfully flat, since we will show below that for any $\mathfrak {m}:=\ker \varphi $ , where ${\varphi \in M( H^\infty )\setminus \pi ^{-1}({\mathbb {D}})}$ , we have $\mathfrak {m} H^2= H^2$ . More explicitly, if $\alpha \in {\mathbb {T}}$ , $\varphi \in \pi ^{-1} \{\alpha \}$ and $\mathfrak {m}:=\ker \varphi $ , then $\mathfrak {m} H^2= H^2$ . We prove this just when $\alpha =1$ , but rotational symmetry yields the result for arbitrary $\alpha \in {\mathbb {T}}$ .

The proof of Theorem 3.2 is based on the following email reply (paraphrased on disc instead of the half-plane) by Donald Sarason to the question (of faithfulness of the flat $H^\infty $ -module $H^2$ ) put to him by Alban Quadrat [Reference Quadrat and Sarason11]:

We give a complete proof by making the outline above explicit below, since the construction of such a w with the required properties is not at all trivial.

Proof. Let $f\!\in \! H^2$ be nonzero. Then, by the F. and M. Riesz theorem (see, e.g., [Reference Douglas1, Theorem 6.13]), it follows that the boundary function $f\in L^2({\mathbb {T}})$ cannot be identically zero on any arc of ${\mathbb {T}}$ with a positive measure. Without loss of generality, we may also assume that $\|f\|_2\le 1$ . Let $a_n>0$ be defined by

$$ \begin{align*}\textstyle a_n^2=\int_{\frac{1}{n+1}\le |\theta| <\frac{1}{n}} |f(e^{i\theta})|^2 d\theta \text{ for all } n\in {\mathbb{N}}. \end{align*} $$

Then, $a_n\in (0,1)$ , and also $(a_n)_{n\in {\mathbb {N}}}\in \ell ^2$ . For $n\in {\mathbb {N}}$ , set

$$ \begin{align*}\textstyle r_n=\sum\limits_{k=n+1}^\infty a_k^2. \end{align*} $$

Then, $r_n\in (0,1]$ (thanks to $\|f\|_2\le 1$ and as each $a_n>0$ ) for all $n\in {\mathbb {N}}$ . Define w as follows:

$$ \begin{align*}\textstyle w(e^{i\theta}) = \left\{ \begin{array}{ll} 1 & \text{if }\theta\in (-\pi,\pi]\setminus (-\frac{1}{2},\frac{1}{2}),\\[0.1cm] \min\{\frac{1}{\sqrt[4]{r_{n-1}}},n\} & \text{if } \frac{1}{n+1}\le |\theta|<\frac{1}{n}, \;n\ge 2. \end{array}\right. \end{align*} $$

We note that w is pointwise $\ge 1$ (as $ \textstyle r_{n-1}\le \|f\|_2^2\le 1$ ). Also, we have that $w(e^{i\theta })\to +\infty $ as $\theta \to 0$ .

We wish to construct an outer function g with modulus $\frac {1}{w}$ on ${\mathbb {T}}$ , that is, with $k:=\log |g|_{{\mathbb {T}}}|=\log \frac {1}{w}$ . So we must check that the w constructed above has the property that $k=\log \frac {1}{w}\in L^1({\mathbb {T}})$ . We have

$$ \begin{align*}\textstyle \begin{array}{rcl} \int_{-\pi}^\pi |\log \frac{1}{w(e^{i\theta})}| d\theta &=& \int_{\pi>|\theta|\ge \frac{1}{2}} |\log 1| d\theta + \sum\limits_{n=2}^\infty \int_{\frac{1}{n+1}\le |\theta| <\frac{1}{n}} \log w(e^{i\theta}) d\theta \\ &\le & 0+ \sum\limits_{n=2}^\infty \frac{2}{n(n+1)} \log n \le 2 \sum\limits_{n=2}^\infty \frac{\log n}{n^2}. \end{array} \end{align*} $$

But there exists an $N\in {\mathbb {N}}$ such that for all $n>N$ , $\log n<\sqrt {n}$ because

$$ \begin{align*}\textstyle \lim\limits_{x\to \infty} \frac{\log x}{\sqrt{x}}=0. \end{align*} $$

For $n>N$ , $\frac {\log n}{n^2}<\frac {1}{n^{3/2}}$ . As $ \sum \limits _{n=1}^\infty \frac {1}{n^{3/2}}<\infty $ , we get by comparison that

$$ \begin{align*}\textstyle \sum\limits_{n=2}^\infty \frac{\log n}{n^2}<\infty. \end{align*} $$

So from the above, we conclude that $\log \frac {1}{w}\in L^1({\mathbb {T}})$ .

Let g be the outer function with modulus $\frac {1}{w}$ on ${\mathbb {T}}$ , that is, with $k=\log \frac {1}{w}$ ,

$$ \begin{align*}\textstyle g(z)=\exp \big( \frac{1}{2\pi} \int_{-\pi}^\pi \frac{e^{i\theta}+z}{e^{i\theta}-z} k(\theta) \;\!d\theta\big). \end{align*} $$

Then, $g\in {\mathcal {O}}({\mathbb {D}})$ , $\frac {1}{w}\in L^\infty ({\mathbb {T}})$ , and $g\in H^\infty $ . By the construction of w, we have $w(e^{i\theta })\to \infty $ as $\theta \to 0$ , that is, $g(e^{i\theta })=\frac {1}{w(e^{i\theta })}\to 0$ as $\theta \to 0$ . By Lindelöf’s theorem (see, e.g., [Reference Garnett3, Example 7(c), pp. 88–89]), it follows that

$$ \begin{align*}\quad \quad \quad\quad \quad \quad\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \lim\limits_{{\mathbb{D}}\owns z\to 1} g(z)=0. \quad \quad \quad\quad \quad \quad\quad \quad \quad\quad \quad \quad \quad \quad \quad \quad (\star\star) \end{align*} $$

We claim that $g\in \mathfrak {m}$ , that is, $\varphi (g)=0$ . This follows from the fact that the range of $\widehat {g}$ on $ \pi ^{-1}\{1\}$ is the set of all $\zeta \in {\mathbb {C}}$ such that there exists a sequence $(\lambda _n)_{n\in {\mathbb {N}}}$ in ${\mathbb {D}}$ satisfying

$$ \begin{align*}\lim\limits_{n\to \infty}\lambda_n=1 \; \text{ and } \; \lim\limits_{n\to \infty} g(\lambda_n)=\zeta. \end{align*} $$

But for each such sequence $(\lambda _n)_{n\in {\mathbb {N}}}$ in ${\mathbb {D}}$ converging to $1$ , we have by ( $\star \star $ ) that $(g(\lambda _n))_{n\in {\mathbb {N}}}$ converges to $0$ . Hence, the range of $\widehat {g}$ on $ \pi ^{-1}\{1\}$ is just $\{0\}$ . In particular, $\widehat {g}(\varphi )=0$ , that is, $\varphi (g)=0$ . Hence, $g\in \mathfrak {m}$ .

As g is given by an exponential, g is never $0$ , and so $\frac {1}{g}\in {\mathcal {O}}({\mathbb {D}})$ . Define $h\in {\mathcal {O}}({\mathbb {D}})$ by $h=\frac {1}{g} f$ . We claim that $h\in H^2$ by showing that $wf\in L^2({\mathbb {T}})$ . We have

$$ \begin{align*}\textstyle \! \begin{array}{rcl} \int_{-\pi}^\pi \!|f(e^{i\theta})|^2 (w(e^{i\theta}))^2 d\theta &\le & \int_{\pi>|\theta|\ge \frac{1}{2}} \! |f(e^{i\theta})|^2d\theta + \sum\limits_{n=2}^\infty \int_{\frac{1}{n+1}\le |\theta| \le \frac{1}{n}}\! \frac{|f(e^{i\theta})|^2}{\sqrt{r_{n-1}}} d\theta \\ &\le& \|f\|_2^2+ \sum\limits_{n=2}^\infty \frac{a_n^2}{\sqrt{r_{n-1}}}<\infty, \end{array} \end{align*} $$

where we used Proposition 1.1 to obtain the last inequality. Thus, $f=gh$ with $g\in \mathfrak {m}$ and $h\in H^2$ , showing that $\mathfrak {m} H^2=H^2$ .

3.4. The half plane case

In control theory, besides the Hardy spaces on the disc (corresponding to “discrete-time systems” arising from difference equations), one also encounters Hardy spaces on the right half plane ${\mathbb {C}}_+=\{s\in {\mathbb {C}}: \text {Re}\;s>0\}$ (corresponding to “continuous-time systems” arising from differential equations). We observe that the results above also hold in this case, as explained below.

The Hardy Hilbert space $H^2({\mathbb {C}}_+)$ is the set of all $F\in {\mathcal {O}}({\mathbb {C}}_+)$ with

$$ \begin{align*}\textstyle \|F\|_2^2:=\sup\limits_{x>0} \int_{-\infty}^\infty |F(x+iy)|^2\;\! dy<\infty. \end{align*} $$

Then, $H^2({\mathbb {C}}_+)$ is a Hilbert space with pointwise operations and the norm $\|\cdot \|_2$ defined above (which arises from an inner product). The Hardy algebra $H^\infty ({\mathbb {C}}_+)$ is the set of all $F\in {\mathcal {O}}({\mathbb {C}}_+)$ such that

$$ \begin{align*}\textstyle \|F\|_\infty:=\sup\limits_{x>0} |F(x+iy)|<\infty. \end{align*} $$

With pointwise operations and the $\|\cdot \|_\infty $ norm, $H^\infty ({\mathbb {C}}_+)$ is a Banach algebra. With the action of the ring $H^\infty ({\mathbb {C}}_+)$ on $H^2({\mathbb {C}}_+)$ given by pointwise multiplication, $H^2({\mathbb {C}}_+)$ is an $H^\infty ({\mathbb {C}}_+)$ -module. It was shown in [Reference Quadrat10, Proposition 8] that $H^2({\mathbb {C}}_+)$ is a flat $H^\infty ({\mathbb {C}}_+)$ -module.

Define $\varphi : {\mathbb {C}}_+\to {\mathbb {D}}$ by

(3.1) $$ \begin{align} \textstyle \varphi(s)=\frac{s-1}{s+1}\;\text{ for all }\;s\in {\mathbb{C}}_+. \end{align} $$

Then, $\varphi $ is biholomorphic, and

$$ \begin{align*}\textstyle \varphi^{-1}(z)=\frac{1+z}{1-z}\;\text{ for all }\;z\in {\mathbb{D}}. \end{align*} $$

It is clear that $f\in H^\infty $ if and only if $f\circ \varphi \in H^\infty ({\mathbb {C}}_+)$ . Equivalently, $F\in H^\infty ({\mathbb {C}}_+)$ if and only if $F\circ \varphi ^{-1} \in H^\infty $ . By [Reference Hoffman6, Theorem, p. 130], $f\in H^2$ if and only if the function

$$ \begin{align*}\textstyle {\mathbb{C}}_+\owns s\mapsto \frac{(f\circ \varphi)(s)}{1+s} \end{align*} $$

belongs to $ H^2({\mathbb {C}}_+)$ . Equivalently, $F\in H^2({\mathbb {C}}_+)$ if and only if

$$ \begin{align*}\textstyle {\mathbb{D}}\owns z\mapsto \frac{(F\circ \varphi^{-1})(z)}{1-z} \end{align*} $$

belongs to $H^2$ .

Proof. Let $F\in H^2({\mathbb {C}}_+)$ . Then, $f\in H^2$ , where

$$ \begin{align*}\textstyle f(z):=2 \frac{(F\circ \varphi^{-1})(z)}{1-z} \;\text{ for all }\;z\in {\mathbb{D}}. \end{align*} $$

Thus, there exist $g\in \mathfrak {m}$ and $h\in H^2$ such that $f=gh$ . But then $G:=g\circ \varphi \in \mathfrak {M}$ , and $H\in H^2({\mathbb {C}}_+)$ , where

$$ \begin{align*}\textstyle H(s)=\frac{(h\circ \varphi)(s)}{1+s} \;\text{ for all }\;s\in {\mathbb{C}}_+. \end{align*} $$

Also, from $f=gh$ , we get $f\circ \varphi = (g\circ \varphi )(h\circ \varphi )$ , and so for all $s\in {\mathbb {C}}_+$ ,

$$ \begin{align*}\textstyle \frac{(f\circ \varphi)(s)}{1+s} = (g\circ \varphi)(s) \frac{(h\circ \varphi^{-1})(s)}{1+s}=G(s)H(s). \end{align*} $$

Thus, $ \textstyle G(s)H(s) \!=\! \frac {(f\circ \varphi )(s)}{1+s}\!=\! \frac {1}{1+s} f(\varphi (s))\!=\! \frac {1}{1+s}2\frac {(F\circ \varphi ^{-1})(\varphi (s))}{1-\varphi (s)}\!=\! F(s)$ .

It can also be shown that if $\mathfrak {N}$ is any finitely generated, nonzero, closed, proper ideal of $H^\infty ({\mathbb {C}}_+)$ , then $\mathfrak {n}H^2({\mathbb {C}}_+)\subsetneq H^2({\mathbb {C}}_+)$ . Indeed, if we set $\mathfrak {n}=\{G\circ \varphi ^{-1}: G\in \mathfrak {N}\}$ , then $\mathfrak {n}$ is a finitely generated, nonzero, closed, proper ideal of $H^\infty $ , and so there exists an $f\in H^2 \setminus (\mathfrak {n}H^2)$ . We claim that $F\in H^2({\mathbb {C}}_+)$ defined by

$$ \begin{align*}\textstyle F(s)=2\frac{(f\circ \varphi)(s)}{1+s}\;\text{ for all }\; s\in {\mathbb{C}}_+, \end{align*} $$

does not belong to $\mathfrak {N} H^2({\mathbb {C}}_+)$ . Otherwise, there exist elements $G\in \mathfrak {N} $ and $H\in H^2({\mathbb {C}}_+)$ such that $F=GH$ . Define $h\in H^2$ by

$$ \begin{align*}\textstyle h(z)=\frac{(H\circ \varphi^{-1})(z)}{1-z}\;\text{ for all }\;z\in {\mathbb{D}}. \end{align*} $$

Thus, $F\circ \varphi ^{-1}=(G\circ \varphi ^{-1})(H\circ \varphi ^{-1})$ , and so for all $z\in {\mathbb {D}}$ , we have

$$ \begin{align*}\textstyle \frac{(F\circ \varphi^{-1})(z)}{1-z}=(G\circ \varphi^{-1})(z) \frac{(H\circ \varphi^{-1})(z)}{1-z}= g(z)h(z). \end{align*} $$

Thus, $\textstyle g(z)h(z)\!=\!\frac {(F\circ \varphi ^{-1})(z)}{1-z}\!=\!\frac {1}{1-z}F(\varphi ^{-1}(z)) \!=\!\frac {1}{1-z}2\frac {(f\circ \varphi )(\varphi ^{-1}(z))}{1+\varphi ^{-1}(z)} \!=\!f(z)$ . So $f=gh\in \mathfrak {n}H^2$ , a contradiction.