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\fancyhead[LO]{\textbf{Finite Field Riesz-Frechet Representation}}
\fancyhead[RE]{\textbf{K. MAHESH KRISHNA}}


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{\bf{Finite Field Riesz-Frechet Representation}}\\
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\textbf{K. MAHESH KRISHNA}\\
School of Mathematics and Natural Sciences\\
Chanakya University Global Campus\\
NH-648, Haraluru Village\\
Devanahalli Taluk, 	Bengaluru  North District\\
Karnataka State 562 110 India\\
Email: kmaheshak@gmail.com\\

Date: \today
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\textbf{Abstract}: We derive a Riesz-Frechet representation for bounded linear functionals defined on the inner product spaces over sub-valued finite fields. 


\textbf{Keywords}: Hilbert Space, Riesz-Frechet representation, Finite Field.

\textbf{Mathematics Subject Classification (2020)}: 12E20, 46C20. \\

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\section{Introduction}
Derived over a century ago, the Riesz-Frechet representation (theorem) states that every bounded linear functional on a Hilbert space is uniquely determined by a single element in that same space.
\begin{theorem}\cite{KREYSZIG, DEBNATHMIKUSINSKI, DIEUDONNE, BIRKHOFFKREYSZIG, GRAY, FRECHET, RIESZ, RIESZ2, HALMOS, HALMOS2,  PAVONE,  DOUGLAS, RUDIN, KADISONRINGROSE, YOUNG, DUNFORDSCHWARTZ, RIESZNAGY, LIMAYE, PIETSCH, STONE, HANSEN, MUSCAT, AKHIEZERGLAZMAN, WEIDMANN, FUHRMANN, CONWAY, ALIPRANTISBORDER}
	(\textbf{Riesz-Frechet Representation}) Let $\mathcal{H}$ be a  Hilbert space over $\mathbb{K}$ ($\mathbb{R}$ or $\mathbb{C}$). Let $f:\mathcal{H} \to \mathbb{K}$ be a bounded linear functional. Then there is a unique element $\tau \in \mathcal{H}$ satisfying the following conditions. 
\begin{enumerate}[\upshape(i)]
	\item $f(h)=\langle h, \tau \rangle$ for all $h \in \mathcal{H}$.
	\item $\|f\|=\|\tau\|$.
\end{enumerate}
\end{theorem}
Over time, the Riesz–Frechet representation has been broadened to encompass Lebesgue spaces \cite{MCSHANE, SHIOJI, ROYDEN}, as well as continuous functionals on compact Hausdorff spaces (known as Riesz–Markov–Kakutani representation)  \cite{BACHMANNARICI, RIESZ2, KAKUTANI, MARKOV, GARLING, RUDIN2, YEH, COHN, CONWAY, ALIPRANTISBORDER}. It is known that the Riesz–Frechet representation extends naturally to quaternionic Hilbert spaces \cite{GHILONIMORETTIPEROTTI}.  Giles \cite{GILES} and James \cite{JAMES} derived Riesz-Frechet representation  for certain classes of Banach spaces. The traditional proof of the Riesz-Frechet representation relies on the orthogonal decomposition theorem, which fundamentally depends on the parallelogram law  \cite{HALMOS2, PAVONE}. If the Hilbert space is separable, one can also prove the Riesz-Frechet representation using orthonormal bases, a method that relies on the Gram-Schmidt orthonormalization process \cite{HALMOS}.
We are fundamentally motivated by the following question.
\begin{question}
What is the finite field version of Riesz-Frechet representation?
\end{question}
We begin by  the notion of sub-modulus field. 
\begin{definition}
	Let $\mathbb{F}$ be a  (finite) field. A map $|\cdot|: \mathbb{F}\to [0, \infty)$ is said to be a  \textbf{sub-modulus} or \textbf{sub-valued} if the following conditions hold.
	\begin{enumerate}[\upshape (i)]
		\item If $\lambda  \in \mathbb{F}$ is such that $|\lambda|=0$, then $\lambda=0$.
		\item $|\lambda \mu|\leq |\lambda||\mu|$ for all $\lambda, \mu \in \mathbb{F}$.
		\item $|\lambda +\mu|\leq |\lambda|+|\mu|$ for all $\lambda, \mu \in \mathbb{F}$.
	\end{enumerate} 
	In this case, we say $\mathbb{F}$ as sub-modulus or sub-valued field. 
\end{definition}
\begin{example}
	Let $p$ be a prime, let $\mathbb{Z}_p\coloneqq \{0, 1, \dots, p-1\}$ be the standard field of integers modulo $p$. Given $n \in \mathbb{Z}_p$, we define 
	\begin{align*}
		|n|\coloneqq \text{unique } a \text{ such that } 0\leq a \leq p-1 \text{ and }n \equiv a (\text{mod }p).
	\end{align*}
\end{example}
We next introduce the notion of sub-normed linear spaces and inner product spaces.
\begin{definition}
	Let $\mathcal{X}$ be a vector space over a sub-modulus field $\mathbb{F}$. A map $\|\cdot\|: \mathcal{X}\to [0, \infty)$ is said to be a \textbf{sub-norm} if the following conditions hold.	
	\begin{enumerate}[\upshape (i)]
		\item If $x \in \mathcal{X}$ is such that $\|x\| =0$, then $x=0$.
		\item $\|\lambda x\|\leq |\lambda|\|x\|$ for all $ \lambda \in \mathbb{F}$, for all $x \in \mathcal{X}$.
		\item $\|x+y\|\leq \|x\|+\|y\|$ for all $x,y \in \mathcal{X}$.
	\end{enumerate} 
	In this case, we say $\mathcal{X}$ is a \textbf{sub-normed linear space}. 
\end{definition}
\begin{definition}
	Let $\mathbb{K}$ be a sub-modulus field (with sub-modulus $|\cdot|$) and $\mathcal{X}$ be a sub-normed space (with sub-norm $\|\cdot\|$) over $\mathbb{F}$. We say that $\mathcal{X}$ is an \textbf{inner product space} if there is a map (called as \textbf{inner product}) $\langle \cdot, \cdot \rangle: \mathcal{X} \times \mathcal{X} \to \mathbb{F}$ satisfying the following conditions:
	\begin{enumerate}[\upshape (i)]
		\item If $x \in \mathcal{X}$ is such that $\langle x,y \rangle =0$ for all $y \in \mathcal{X}$, then $x=0$.
		\item $\langle x, y \rangle =\langle y, x \rangle$ for all $x,y \in \mathcal{X}$.
		\item $\langle \alpha x, y \rangle =\alpha \langle x, y \rangle$ for all $\alpha  \in \mathbb{F}$, for all $x,y \in \mathcal{X}$.
		\item $\langle x+y, z \rangle = \langle x,  z \rangle+\langle y,z\rangle$ for all $x,y,z \in \mathcal{X}$.
		\item $|\langle x, y \rangle |\leq \|x\|\|y\|$ for all $x,y \in \mathcal{X}$.
	\end{enumerate}
\end{definition}
\begin{example}
	  For  $d \in \mathbb{N}$, let $\mathbb{Z}_p^d$ be the standard vector space over  $\mathbb{Z}_p$. We define  the inner product 
	\begin{align*}
		\langle (a_j)_{j=1}^d,(b_j)_{j=1}^d\rangle \coloneqq \sum_{j=1}^da_jb_j \in \mathbb{Z}_p,  \quad \forall (a_j)_{j=1}^d,(b_j)_{j=1}^d \in \mathbb{Z}_p^d
	\end{align*}
	and the sub-norm 
	\begin{align*}
		\|(c_j)_{j=1}^d\|^2\coloneqq \sum_{j=1}^d|c_j|^2 \in \{0, 1, \dots, d(p-1)^2\}, \quad \forall (c_j)_{j=1}^d\in 	\mathbb{Z}_p^d.
	\end{align*}
\end{example}
\begin{example}
	Let $\mathbb{F}$ be a sub-modulus field. Define 
\begin{align*}
	c_{00}(\mathbb{N}, \mathbb{Z}_p)	\coloneqq\left\{\{a_n\}_{n=1}^\infty: a_n \in \mathbb{Z}_p, \forall n \in \mathbb{N}, a_n \neq 0 \text{ only for finitely many } n\right\}. 
\end{align*}
We define the inner product 
\begin{align*}
	\langle \{a_n\}_{n=1}^\infty, \{b_n\}_{n=1}^\infty\rangle \coloneqq \sum_{n=1}^{\infty}a_nb_n, \quad \forall \{a_n\}_{n=1}^\infty, \{b_n\}_{n=1}^\infty \in c_{00}(\mathbb{N}, \mathbb{Z}_p)
\end{align*}
and the sub-norm 
\begin{align*}
	\|\{a_n\}_{n=1}^\infty\|\coloneqq\left(\sum_{n=1}^{\infty}|a_n|^2\right)^\frac{1}{2}, \quad \forall \{a_n\}_{n=1}^\infty\in c_{00}(\mathbb{N}, \mathbb{Z}_p).
\end{align*}
Then $\|\cdot\|_{00}$ is a sub-norm on  $c_{00}(\mathbb{N}, \mathbb{F}).$		
\end{example}
We establish two distinct formulations of the Riesz-Frechet representation within the context of finite fields, each employing unique assumptions and yielding different conclusions.



\section{Finite Field Riesz-Frechet Representation}
We begin by introducing following notion of orthonormal basis.
\begin{definition}\label{PONB}
	Let $\mathcal{X}$  be an inner product space over sub-modulus field $\mathbb{F}$.   A (Hamel) basis  $\{\tau_n\}_{n}$ for  $\mathcal{X}$ is said to be an  \textbf{orthonormal basis} for    $\mathcal{X}$   if  the following conditions hold:
	\begin{enumerate}[\upshape(i)]
		\item $\|\tau_n\|\leq 1$ for all $n \in \mathbb{N}$.
		\item $	\langle \tau_n, \tau_m\rangle =\delta_{n ,m}$ for all $ n, m \in \mathbb{N}$.
	\end{enumerate}
\end{definition}
Note that given an orthonormal basis $\{\tau_n\}_{n}$ for  $\mathcal{X}$,  condition (ii) in Definition \ref{PONB} says that 
\begin{align*}
	1=|\langle \tau_n, \tau_n \rangle |\leq \|\tau_n\|^2, \quad \forall n \in \mathbb{N}.
\end{align*}
By using condition (i) in Definition \ref{PONB} we then get $\|\tau_n\|= 1$ for all $n \in \mathbb{N}$. Thus all elements of orthonormal basis lie on the unit sphere. 

Gram-Schmidt orthonormalization says that every separable Hilbert space has an orthonormal basis \cite{LEONBJORCKGANDER}. It is clear that an inner product space over a sub-modulus field does not necessarily have an orthonormal basis. 
Let $\mathcal{X}$ be an inner product space over a sub-modulus field $\mathbb{F}$. A linear map $f: \mathcal{X} \to \mathbb{F}$ is said to be bounded if 
\begin{align*}
	\sup\left\{\frac{|f(x)|}{\|x\|}: x \in \mathcal{X} \setminus \{0\}\right\}<\infty.
\end{align*}
In this case, we define the norm of $f$ as 
\begin{align*}
	\|f\|\coloneqq 	\sup\left\{\frac{|f(x)|}{\|x\|}: x \in \mathcal{X} \setminus \{0\}\right\}. 
\end{align*}
Following proposition is obvious. 
\begin{proposition}
	Let $\mathcal{X}$ be an inner product space over a sub-modulus field $\mathbb{F}$. Let $f, g: \mathcal{X} \to \mathbb{F}$ be bounded linear maps. Then 
	\begin{enumerate}[\upshape(i)]
		\item If $\|f\| =0$, then $f=0$.
		\item $\|\lambda f\|\leq |\lambda|\|f\|$ for all $ \lambda \in \mathbb{F}$.
		\item $\|f+g\|\leq \|f\|+\|g\|$. 
	\end{enumerate}
In particular, the space $\mathcal{X}^*\coloneqq\{f: \mathcal{X} \to \mathbb{F} \text{ is bounded and linear}\}$ is a sub-normed space over $\mathbb{F}$ w.r.t. sub-norm $\|\cdot\|$. 
\end{proposition}
It is  clear that every bounded linear functional on an inner product space over a sub-modulus field cannot be represented using a vector in the space. 

Following is our first version of Riesz-Frechet representation.
\begin{theorem} (\textbf{Finite Field Riesz-Frechet Representation})\label{p1} Let $\mathcal{X}$ be an inner product space over a sub-valued field $\mathbb{K}$. Let $f:\mathcal{X} \to \mathbb{F}$ be a bounded linear functional. Assume that there is an  orthonormal basis $\{\omega_n\}_{n}$ for  $\mathcal{X}$ such that 
	\begin{align*}
		f(\omega_n)\neq 0 ~  \text{ only for finitely many } n.
	\end{align*}
	Then there is a unique element $\tau \in \mathcal{X}$ satisfying the following conditions:
	\begin{enumerate}[\upshape(i)]
		\item $f(x)=\langle x, \tau \rangle$ for all $x \in \mathcal{X}$.
		\item 
		\begin{align*}
			\frac{\left|\sum_{n=1}^{\infty}f(\omega_n)^2\right|}{\left\|\sum_{n=1}^{\infty}f(\omega_n)\omega_n\right\|}=	\frac{|\langle \tau, \tau \rangle|}{\|\tau\|}\leq \|f\|\leq \|\tau\|=\left\|\sum_{n=1}^{\infty}f(\omega_n)\omega_n\right\|.
		\end{align*}
	\end{enumerate}
\end{theorem}
	\begin{proof}
	 Define 
	\begin{align*}
		\tau\coloneqq \sum_{n=1}^{\infty}f(\omega_n)\omega_n.
	\end{align*}
We then have 
\begin{align*}
	f(x)&=f\left(\sum_{n=1}^{\infty}\langle x, \omega_n \rangle \omega_n\right)=\sum_{n=1}^{\infty}\langle x, \omega_n \rangle f(\omega_n)\\
	&=\left\langle x, \sum_{n=1}^{\infty}f(\omega_n)\omega_n\right \rangle=\langle x, \tau \rangle, \quad \forall x \in \mathcal{X}.
\end{align*}
Now 
\begin{align*}
	\|f\|=\sup\left\{\frac{|f(x)|}{\|x\|}: x \in \mathcal{X}, x \neq 0\right\}=\sup\left\{\frac{|\langle x, \tau \rangle|}{\|x\|}: x \in \mathcal{X}, x \neq 0\right\}\leq \|\tau\|.
\end{align*}
 We now observe that 
 \begin{align*}
 		\|\tau\|=\left\|\sum_{n=1}^{\infty}f(\omega_n)\omega_n\right\|, 
 \end{align*}
 \begin{align*}
 	\langle \tau, \tau \rangle=\left\langle \sum_{n=1}^{\infty}f(\omega_n)\omega_n, \sum_{m=1}^{\infty}f(\omega_m)\omega_m\right \rangle =\sum_{n=1}^{\infty}f(\omega_n)^2
 \end{align*}
 and 
 \begin{align*}
 	|\langle \tau, \tau \rangle|=|f(\tau)|\leq \|f\|\|\tau\|.
 \end{align*}
Uniqueness easily follows using the definition of  inner product.
	\end{proof}
Let $\mathcal{X}$ be an inner product space over sub-modulus field $\mathbb{F}$. A  subspace $\mathcal{Y}$ of $ \mathcal{X}$ is said to be orthogonally complementable if there exists a unique subspace $\mathcal{Z}$ of $ \mathcal{X}$ satisfying the following: for every $ x \in \mathcal{X}$, there exist unique $ y \in \mathcal{Y}$ and $ z \in \mathcal{Z}$ such that 
\begin{enumerate}[\upshape(i)]
	\item $x=y+z$.
	\item $\langle y, z \rangle =0$.
	\item $\|x\|^2=\|y\|^2+ \|z\|^2$.
\end{enumerate}
In this case, we write $x=y\oplus z$, $\mathcal{Z}=\mathcal{Y}^\perp$ and $\mathcal{X}=\mathcal{Y}\oplus \mathcal{Y}^\perp$.

Following is our second version of Riesz-Frechet representation.
\begin{theorem}(\textbf{Finite Field Riesz-Frechet Representation}) \label{p2} Let $\mathcal{X}$ be an inner product space over a sub-valued field $\mathbb{F}$. Let $f:\mathcal{X} \to \mathbb{F}$ be a non-zero bounded linear functional. Assume that $\operatorname{Ker}(f)$ is orthogonally complementable in $\mathcal{X}$ and there exists a (nonzero) vector $\omega \in \operatorname{Ker}(f)^\perp$ such that $\langle \omega, \omega \rangle\neq 0$. 
Then there is a unique element $\tau \in \mathcal{X}$ satisfying the following conditions:
\begin{enumerate}[\upshape(i)]
	\item $f(x)=\langle x, \tau \rangle$ for all $x \in \mathcal{X}$.
	\item 
	\begin{align*}
		\frac{|\langle \tau, \tau \rangle|}{\|\tau\|}\leq \|f\|\leq \|\tau\|.
	\end{align*}
\end{enumerate}
Moreover, 
\begin{align*}
	\|\tau\|=\frac{|f(\omega)|}{|\langle \omega, \omega \rangle|} \|\omega\|\leq \frac{\|f\|\|\omega\|^2}{|\langle \omega, \omega \rangle|}.
\end{align*}
\end{theorem}	
\begin{proof}
	Define 
	\begin{align*}
		\tau\coloneqq \frac{f(\omega)}{\langle \omega, \omega \rangle} \omega.
	\end{align*}
Let $x \in \mathcal{X}$. Then 
\begin{align*}
	f\left(x-\frac{f(x)}{f(\omega)}\omega\right)=f(x)-f(x)=0.
\end{align*}
Hence 
\begin{align*}
	x-\frac{f(x)}{f(\omega)}\omega \in \operatorname{Ker}(f).
\end{align*}
We then have 
\begin{align*}
	0&=\left\langle x-\frac{f(x)}{f(\omega)}\omega, \tau \right\rangle =\left\langle x, \tau \right\rangle-\left\langle \frac{f(x)}{f(\omega)}\omega, \tau \right\rangle\\
	&=\left\langle x, \tau \right\rangle- \frac{f(x)}{f(\omega)}\langle \omega, \tau \rangle=\left\langle x, \tau \right\rangle- \frac{f(x)}{f(\omega)}\left\langle \omega, \frac{f(\omega)}{\langle \omega, \omega \rangle} \omega\right\rangle \\
	&=\langle x, \tau \rangle -f(x).
\end{align*}
Note that
\begin{align*}
	|\langle \tau, \tau \rangle|=|f(\tau)|\leq \|f\|\|\tau\|.
	\end{align*}
\end{proof}
We next extend the finite field Riesz-Frechet representation to finite  fields  admitting conjugation (involution). We use the following definition of conjugation.
\begin{definition}
	Let $\mathbb{F}$ be a finite sub-valued field (with valuation $|\cdot|$). A map $\sigma: \mathbb{F} \to \mathbb{F}$ is said to be a \textbf{conjugation} or \textbf{involution} if the following conditions hold:
	\begin{enumerate}[\upshape(i)]
		\item $\sigma(\alpha+\beta)=\sigma(\alpha)+\sigma(\beta)$ for all $\alpha, \beta \in \mathbb{F}$.
		\item  $\sigma(\alpha\beta)=\sigma(\alpha)\sigma(\beta)$ for all $\alpha, \beta \in \mathbb{F}$.
		\item $\sigma(\sigma(\alpha))=\alpha$ for all $\alpha  \in \mathbb{F}$.
		\item $|\sigma(\alpha)|=|\alpha|$ for all $\alpha  \in \mathbb{F}$.
	\end{enumerate}
	In this case, we say that $\mathbb{F}$ is  $\sigma$-conjugated. 
\end{definition}
\begin{definition}
	Let $\mathbb{F}$ be a $\sigma$-conjugated sub-valued field  and $\mathcal{X}$ be a sub-normed space  over $\mathbb{F}$. We say that $\mathcal{X}$ is a \textbf{ $\sigma$-inner product space} if there is a map (called as \textbf{$\sigma$-inner product}) $\langle \cdot, \cdot \rangle: \mathcal{X} \times \mathcal{X} \to \mathbb{K}$ satisfying the following:
	\begin{enumerate}[\upshape (i)]
		\item If $x \in \mathcal{X}$ is such that $\langle x,y \rangle =0$ for all $y \in \mathcal{X}$, then $x=0$.
	\item $\langle x, y \rangle =\sigma(\langle y, x \rangle)$ for all $x,y \in \mathcal{X}$.
	\item $\langle \alpha x, y \rangle =\alpha \langle x, y \rangle$ for all $\alpha  \in \mathbb{F}$, for all $x,y \in \mathcal{X}$.
	\item $\langle x+y, z \rangle = \langle x,  z \rangle+\langle y,z\rangle$ for all $x,y,z \in \mathcal{X}$.
	\item $|\langle x, y \rangle |\leq \|x\|\|y\|$ for all $x,y \in \mathcal{X}$.
	\end{enumerate}
\end{definition}
We now have  generalized versions of Theorem \ref{p1} and Theorem \ref{p2}.
\begin{theorem} (\textbf{Finite Field Riesz-Frechet Representation}) Let $\mathcal{X}$ be a $\sigma$-inner product space over a $\sigma$-conjugated sub-valued field $\mathbb{F}$. Let $f:\mathcal{X} \to \mathbb{F}$ be a bounded linear functional. Assume that there is an orthonormal basis $\{\omega_n\}_{n}$ for  $\mathcal{X}$ such that 
	\begin{align*}
	f(\omega_n)\neq 0 ~  \text{only for finitely many } n.
\end{align*}
	Then there is a unique element $\tau \in \mathcal{X}$ satisfying the following conditions:
	\begin{enumerate}[\upshape(i)]
		\item $f(x)=\langle x, \tau \rangle$ for all $x \in \mathcal{X}$.
		\item 
			\begin{align*}
			\frac{\left|\sum_{n=1}^{\infty}f(\omega_n)^2\right|}{\left\|\sum_{n=1}^{\infty}f(\omega_n)\omega_n\right\|}=	\frac{|\langle \tau, \tau \rangle|}{\|\tau\|}\leq \|f\|\leq \|\tau\|=\left\|\sum_{n=1}^{\infty}f(\omega_n)\omega_n\right\|.
		\end{align*}
	\end{enumerate}
\end{theorem}
\begin{proof}
	Define 
	\begin{align*}
		\tau\coloneqq \sum_{n=1}^{\infty}\sigma(f(\omega_n))\omega_n.
	\end{align*}
The other arguments are similar to those used in the proof of Theorem \ref{p1}.
\end{proof}
\begin{theorem}(\textbf{Finite Field Riesz-Frechet Representation}) Let $\mathcal{X}$ be a p-adic $\sigma$-inner product space over a $\sigma$-conjugated  sub-valued field $\mathbb{F}$. Let $f:\mathcal{X} \to \mathbb{K}$ be a non-zero bounded linear functional. Assume that $\operatorname{Ker}(f)$ is orthogonally complementable in $\mathcal{X}$ and there exists a (nonzero) vector $\omega \in \operatorname{Ker}(f)^\perp$ such that $\langle \omega, \omega \rangle\neq 0$. 
	Then there is a unique element $\tau \in \mathcal{X}$ satisfying the following conditions:
	\begin{enumerate}[\upshape(i)]
		\item $f(x)=\langle x, \tau \rangle$ for all $x \in \mathcal{X}$.
		\item 
		\begin{align*}
			\frac{|\langle \tau, \tau \rangle|}{\|\tau\|}\leq \|f\|\leq \|\tau\|.
		\end{align*}
	\end{enumerate}
	Moreover, 
	\begin{align*}
		\|\tau\|=\frac{|f(\omega)|}{|\langle \omega, \omega \rangle|} \|\omega\|\leq \frac{\|f\|\|\omega\|^2}{|\langle \omega, \omega \rangle|}.
	\end{align*}
\end{theorem}	
\begin{proof}
		Define 
	\begin{align*}
		\tau\coloneqq \frac{\sigma(f(\omega))}{\langle \omega, \omega \rangle} \omega.
	\end{align*}
	The other arguments are similar to those used in the proof of Theorem \ref{p2}.
\end{proof}
\section{Conclusions}
\begin{enumerate}
	\item In 1907, Riesz \cite{RIESZ} and Frechet \cite{FRECHET} independently showed that every continuous linear functional on a Hilbert space is completely determined by a single element in that Hilbert space \cite{BIRKHOFFKREYSZIG}.
	\item In 1938, Markov \cite{MARKOV} and in 1941, Kakutani \cite{KAKUTANI} extended the Riesz-Frechet representation to function spaces. 
	\item In 1947, James \cite{JAMES} and  1967, Giles \cite{GILES} extended the Riesz-Frechet representation to certain classes of Banach spaces. 
	\item In this article, we derived two finite field Riesz representation theorems under certain conditions for inner product spaces over sub-modulus fields.
\end{enumerate}





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\end{thebibliography}

 


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