#  $\mathbb{F}$-valued trace of a finite-dimensional commutative
$\mathbb{F}$-algebra

A. K. Bhagat  and  R. Sarma  Anuj Kumar Bhagat: Department of Mathematics
Indian Institute of Technology, Delhi, Hauz Khas New Delhi-110016 India E-mail
address<EMAIL_ADDRESS>Ritumoni Sarma: Department of
Mathematics Indian Institute of Technology, Delhi, Hauz Khas New Delhi-110016
India E-mail address<EMAIL_ADDRESS>
###### Abstract.

A non-zero $\mathbb{F}$-valued $\mathbb{F}$-linear map on a finite dimensional
$\mathbb{F}$-algebra is called an $\mathbb{F}$-valued trace if its kernel does
not contain any non-zero ideals. However, given an $\mathbb{F}$-algebra such a
map may not always exist. We find an infinite class of finite-dimensional
commutative $\mathbb{F}$-algebras which admit an $\mathbb{F}$-valued trace. In
fact, in these cases, we explicitly construct a trace map. The existence of an
$\mathbb{F}$-valued trace on a finite dimensional commutative
$\mathbb{F}$-algebra induces a non-degenerate bilinear form on the
$\mathbb{F}$-algebra which may be helpful both theoretically and
computationally. In this article, we suggest a couple of applications of an
$\mathbb{F}$-valued trace map of an $\mathbb{F}$-algebra to algebraic coding
theory.

##  1\. Introduction

Throughout this manuscript, $\mathbb{F}$ denotes a field and $\mathbb{F}_{q}$
denotes the finite field of order $q,$ where $q$ is a prime power.

The $\mathbb{F}$-valued trace map of an extension of $\mathbb{F}$ plays an
important role in the theory of both finite and infinite fields. For instance,
every functional of a finite extension $\mathbb{K}$ over $\mathbb{F}$ can be
described as
$\alpha\mapsto\textnormal{Tr}_{\mathbb{K}/\mathbb{F}}(\alpha\beta),$ for a
unique $\beta\in\mathbb{K}.$ In fact,
$\textnormal{Tr}_{\mathbb{K}/\mathbb{F}}$ induces a non-degenerate bilinear
form, namely,
$(\alpha,\beta)\mapsto\textnormal{Tr}_{\mathbb{K}/\mathbb{F}}(\alpha\beta)$ on
$\mathbb{K}$ and consequently, given any basis
$\\{\alpha_{1}<\alpha_{2}<\dots<\alpha_{n}\\}$ for $\mathbb{K}$ over
$\mathbb{F},$ there is a unique basis
$\\{\beta_{1}<\beta_{2}<\dots<\beta_{n}\\}$ for $\mathbb{K}$ over $\mathbb{F}$
such that
$\textnormal{Tr}_{\mathbb{K}/\mathbb{F}}(\alpha_{i}\beta_{j})=\delta_{ij}.$
Because of these properties, the trace map turns out to be an important tool
in the theory of fields. As its application, in coding theory, the trace map
is used to construct “trace codes” and it also helps in the computation of
“subfield codes”[1].

Motivated by the wide range of applications of the trace map, we got curious
to find if there is an analogue to the trace map for a finite-dimensional
$\mathbb{F}$-algebra. In this manuscript, we study the trace maps of finite
dimensional commutative $\mathbb{F}$-algebras. For a commutative ring $R$ with
unity, and an $R$-algebra $A,$ a surjective $R$-linear map $\tau:A\to R$ is
called a $R$-valued trace if $\ker(\tau)$ contains no non-zero left ideals of
$A$ (see [3], [6], [7]). Not every finite-dimensional $\mathbb{F}$-algebra has
an $\mathbb{F}$-valued trace. For example, $\mathbb{F}_{2}[u,v]/\langle
u^{2},v^{2},uv\rangle$ does not admit any $\mathbb{F}_{2}$-valued trace. In
this article, we show by construction that a finite-dimensional commutative
$\mathbb{F}$-algebra of the form
$\mathcal{R}=\mathbb{F}[x_{1},x_{2},\dots,x_{n}]/\langle
g_{1}(x_{1}),g_{2}(x_{2}),\dots,g_{n}(x_{n})\rangle,$
$g_{i}(x_{i})\in\mathbb{F}[x_{i}],$ has an $\mathbb{F}$-valued trace.

Let $\mathbb{K}$ be a finite field extension of $\mathbb{F}$, then
$\mathbb{K}$ can be viewed as a vector space over $\mathbb{F}$. The
multiplication by $\alpha\in\mathbb{K},$
$\mathfrak{m}_{\alpha}:\mathbb{K}\to\mathbb{K}$ given by $x\mapsto\alpha x$ is
an $\mathbb{F}$-linear transformation. Then
$\textnormal{Tr}_{\mathbb{K}/\mathbb{F}}(\alpha)$ is defined as the trace of
the linear operator $\mathfrak{m}_{\alpha}.$ One may try to extend this
definition to finite-dimensional $\mathbb{F}$-algebras as they are
$\mathbb{F}$-vector spaces. But, sometimes this map turns out to be the zero
map. For instance, if $\mathcal{R}=\mathbb{F}_{2}[x]/\langle x^{2}\rangle,$
then $\textnormal{Tr}_{\mathcal{R}/\mathbb{F}_{2}}(a+bx+\langle
x^{2}\rangle)=0,$ for all $a,b\in\mathbb{F}_{2}.$ Thus, the obvious
generalization of trace does not lead to a map that possesses the expected
properties of the trace. However, in the case of
$\mathcal{R}=\mathbb{F}_{2}[x]/\langle x^{2}\rangle,$ $a+bx+\langle
x^{2}\rangle\mapsto a+b$ satisfies the desired properties of a trace map.

In recent years, linear codes over certain $\mathbb{F}_{q}$-algebras(see [8],
[11], [9], [2]) are studied and from them, by means of Gray maps, one produces
codes over $\mathbb{F}_{q}.$ Other means of obtaining codes over
$\mathbb{F}_{q}$ from codes over an $\mathbb{F}_{q}$-algebra could be using a
trace map, for instance, the trace code. Moreover, the subfield codes can be
determined with the help of an $\mathbb{F}_{q}$-valued trace map, if exists,
from codes over $\mathbb{F}_{q}$-algebras.

The remaining sections are arranged as follows. Preliminaries are provided in
Section 2. The main section is Section 3, where we show the existence and
construction of an $\mathbb{F}$-valued trace of
$\mathcal{R}=\mathbb{F}[x_{1},x_{2},\dots,x_{n}]/\langle
g_{1}(x_{1}),g_{2}(x_{2}),\dots,g_{n}(x_{n})\rangle,$
$g_{i}(x_{i})\in\mathbb{F}[x_{i}].$ Certain applications of
$\mathbb{F}$-valued trace are discussed in Section 4 before we conclude the
article in Section 5.

##  2\. Preliminaries

######  Definition 2.1.

A ring $\mathcal{R}$ with unity $1_{\mathcal{R}}$ is called an
$\mathbb{F}$-algebra if there is a ring homomorphism from $\mathbb{F}$ to
$\mathcal{R}$ such that $1_{\mathbb{F}}\mapsto 1_{\mathcal{R}}$ and the image
of $\mathbb{F}$ is contained in the center of $\mathcal{R}.$

If $\mathcal{R}$ is an $\mathbb{F}$-algebra, then $\mathcal{R}$ is a vector
space over $\mathbb{F}$ and if $\dim_{\mathbb{F}}(\mathcal{R})<\infty,$ then
$\mathcal{R}$ is called a _finite-dimensional_ $\mathbb{F}$-algebra.

######  Definition 2.2.

Let $\mathcal{R}$ be a commutative $\mathbb{F}$-algebra and let
$S=\\{\bm{r}_{1},\dots,\bm{r}_{n}\\}\subseteq\mathcal{R}.$ If there exists a
surjective ring homomorphism $\mathbb{F}[x_{1},\dots,x_{n}]\to\mathcal{R}$
such that $x_{i}\mapsto\bm{r}_{i},$ then $S$ is called a set of generators of
$\mathcal{R}$ over $\mathbb{F}$. In this case, we say that $\mathcal{R}$ is
finitely generated $\mathbb{F}$-algebra.

Obviously, finite-dimensional $\mathbb{F}$-algebras are finitely generated.

######  Proposition 2.3.

Let $\mathcal{R}$ be a finite-dimensional commutative $\mathbb{F}$-algebra and
let $\tau:\mathcal{R}\to\mathbb{F}$ be a non-zero $\mathbb{F}$-linear map.
Then the following statements are equivalent.

  1. (a)

$\ker(\tau)$ does not contain any non-zero ideals of $\mathcal{R}.$

  2. (b)

$f:\mathcal{R}\times\mathcal{R}\to\mathbb{F}$ defined by
$f(\bm{x},\bm{y})=\tau(\bm{xy})$ is a non-degenerate bilinear form, that is,
$\tau(\bm{xy})=0,\forall\,\bm{x}\in\mathcal{R}\implies\bm{y}=\bm{0}.$

######  Definition 2.4.

Let $\mathcal{R}$ be a finite-dimensional commutative $\mathbb{F}$-algebra. A
non-zero $\mathbb{F}$-linear map $\tau:\mathcal{R}\to\mathbb{F}$ is called an
$\mathbb{F}$-valued trace of $\mathcal{R},$ if it satisfies any one of the
statements of Proposition 2.3.

######  Remark 2.5.

In [3], the notion of trace (or Generalized Frobenius trace) is defined for a
general ring. Let $S$ be a ring (need not be commutative) with unity and $R$
be a subring of $S$ sharing the same unity. Then a homomorphism of left
$R$-modules $\textnormal{Tr}_{R}^{S}:S\to R$ is called a trace from $S$ to $R$
if it is surjective and $\ker(\textnormal{Tr}_{R}^{S})$ does not contain any
non-zero left ideals of $S.$ We note that if $R=\mathbb{F}$ and $S$ is an
$\mathbb{F}$-algebra, then Definition 2.4 is a special case of the definition
in [3].

Let $V$ be a finite-dimensional vector space over $\mathbb{F}$ and let $T$ be
a linear operator on $V.$

######  Definition 2.6.

[4] If $\bm{\alpha}$ is a vector in $V$, then the subspace
$Z(\bm{\alpha};T):=\\{g(T)\bm{\alpha}:g(x)\in\mathbb{F}[x]\\}$ is called the
$T$-cyclic subspace generated by $\bm{\alpha}.$ If $Z(\bm{\alpha};T)=V,$ then
$\bm{\alpha}$ is called a cyclic vector for $T.$

######  Definition 2.7.

[4] If $\bm{\alpha}$ is a vector in $V$, then then ideal
$M(\bm{\alpha};T):=\\{g(x)\in\mathbb{F}[x]:g(T)\bm{\alpha}=0\\}$ of
$\mathbb{F}[x]$ is called the $T$-annihilator of $\bm{\alpha}.$

Since $\mathbb{F}[x]$ is an Euclidean domain, there is a unique monic
polynomial $p_{\bm{\alpha}}(x)$ which generates $M(\bm{\alpha};T)$; the
polynomial $p_{\bm{\alpha}}(x)$ is also called the $T$-annihilator of
$\bm{\alpha}$. With these notations we note the following result.

######  Theorem 2.8 ([4]).

Let $\bm{\alpha}\in V\setminus\\{\bm{0}\\}.$ Then

  1. (1)

$\dim_{\mathbb{F}}Z(\bm{\alpha};T)=\deg p_{\bm{\alpha}}(x)$

  2. (2)

If $\deg p_{\bm{\alpha}}(x)=k,$ then $\\{\bm{\alpha},T\bm{\alpha},\dots
T^{k-1}\bm{\alpha}\\}$ is a basis for $Z(\bm{\alpha};T).$

The following result that describes a basis of a tensor product of vector
spaces is helpful for computation.

######  Proposition 2.9.

Let $V$ and $W$ be finite-dimensional vector spaces over $\mathbb{F}$ with
bases $\\{\bm{\alpha}_{1}\dots,\bm{\alpha}_{m}\\}$ and
$\\{\bm{\beta}_{1}\dots,\bm{\beta}_{n}\\}$ respectively. Then
$\\{\bm{\alpha}_{i}\otimes\bm{\beta}_{j}:1\leq i\leq m,1\leq j\leq n\\}$ is a
basis for $V\otimes_{\mathbb{F}}W.$

######  Theorem 2.10.

Let, for $1\leq i\leq n,$ $g_{i}(x)\in\mathbb{F}[x]$. Then

$\frac{\mathbb{F}[x_{1},x_{2},\dots,x_{n}]}{\langle
g_{1}(x_{1}),g_{2}(x_{2}),\dots,g_{n}(x_{n})\rangle}\cong\frac{\mathbb{F}[x_{1}]}{\langle
g_{1}(x_{1})\rangle}\otimes_{\mathbb{F}}\frac{\mathbb{F}[x_{2}]}{\langle
g_{2}(x_{2})\rangle}\otimes_{\mathbb{F}}\cdots\otimes_{\mathbb{F}}\frac{\mathbb{F}[x_{n}]}{\langle
g_{n}(x_{n})\rangle}.$

##  3\. Trace on finite dimensional commutative $\mathbb{F}$-algebras

######  Theorem 3.1.

Let $g(x)\in\mathbb{F}[x]$ be an irreducible polynomial over $\mathbb{F}$ of
degree $n$, and let $T:\mathbb{F}[x]/\langle g(x)\rangle\to\mathbb{F}$ be a
non-zero functional. Set $\mathcal{R}=\mathbb{F}[x]/\langle g(x)^{r}\rangle,$
for $r\in\mathbb{N}.$ Then the map $\tau:\mathcal{R}\to\mathbb{F}$ given by

$p_{0}(x)+p_{1}(x)g(x)+\dots+p_{r-1}(x)g(x)^{r-1}+\langle
g(x)^{r}\rangle\mapsto\sum_{i=0}^{r-1}T(p_{i}(x)+\langle g(x)\rangle)$

is an $\mathbb{F}$-valued trace of $\mathcal{R}.$

###### Proof.

Note that $\mathcal{B}:=\\{\bm{e}_{i,j}=x^{i}g(x)^{j}:0\leq i\leq n-1,0\leq
j\leq r-1\\}$ is a basis for $\mathcal{R}$ and hence, every element
$p(x)+\langle g(x)^{r}\rangle$ of $\mathcal{R}$ can be uniquely expressed in
the form $p_{0}(x)+p_{1}(x)g(x)+\cdots+p_{r-1}(x)g(x)^{r-1}+\langle
g(x)^{r}\rangle,$ where $p_{i}(x)\in\mathbb{F}[x]$ with $\deg p_{i}(x)\leq
n-1.$ For convenience, identify $\mathbb{F}[x]/\langle g(x)\rangle$ with
$\mathbb{F}^{n}$ via the map $f(x)+\langle
g(x)\rangle\mapsto\bm{f}=(a_{0},a_{1},\dots,a_{n-1})\in\mathbb{F}^{n},$ if
$f(x)=a_{0}+a_{1}x+\cdots+a_{n-1}x^{n-1}.$ In turn, get the identification of
$\mathcal{R}$ with $\mathbb{F}^{nr}$ as $\mathbb{F}$-vector spaces via the
vector space isomorphism

$p(x)+\langle
g(x)^{r}\rangle\mapsto\overline{\bm{p}}=(\bm{p}_{0},\bm{p}_{1},\dots,\bm{p}_{r-1})$
(3.1)

where $\bm{p}_{i}=(p_{i,0},p_{i,1},\dots,p_{i,(n-1)})\in\mathbb{F}^{n}.$

Since $T$ is a non-zero functional on $\mathbb{F}[x]/\langle g(x)\rangle,$
there exists
$\bm{s}=(s_{0},\dots,s_{n-1})\in\mathbb{F}^{n}\setminus\\{\bm{0}\\}$ such that
$T(\bm{a})=\bm{s}\cdot\bm{a},$ the usual dot product of $\bm{s}$ and $\bm{a}.$
With the above identification, $\tau$ can be expressed as

$\tau(\overline{\bm{p}})=\sum_{i=0}^{r-1}\bm{s}\cdot\bm{p}_{i}.$ (3.2)

Next consider the operator

$\displaystyle\mathfrak{m}_{x}:\mathcal{R}$ $\displaystyle\to\mathcal{R}$
$\displaystyle p(x)+\langle g(x)^{r}\rangle$ $\displaystyle\mapsto
xp(x)+\langle g(x)^{r}\rangle.$

Order $\mathcal{B}$ such that $\bm{e}_{i,j}<\bm{e}_{k,l}$ if either $j=l$ and
$i<k$ or $j<l.$ Suppose $A=[\mathfrak{m}_{x}]_{\mathcal{B}}$ is the matrix of
$\mathfrak{m}_{x}$ relative to the ordered basis $\mathcal{B}.$ Now, for
$1\leq j\leq r-1,$ compute $g(x)^{j}\left(p(x)+\langle
g(x)^{r}\rangle\right)$. Observe,

$\displaystyle
g(x)^{j}\left(\underset{i=0}{\overset{r-1}{\sum}}p_{i}(x)g(x)^{i}\right)$
$\displaystyle=\underset{i=j}{\overset{r-1}{\sum}}p_{i-j}(x)g(x)^{i}$

and

$\displaystyle
g(x)^{r}\left(\underset{i=0}{\overset{r-1}{\sum}}p_{i}(x)g(x)^{i}\right)$
$\displaystyle=\bm{0}.$

Hence,

$g(A)=\left({\begin{array}[]{ccccc}O&O&\cdots&O&O\\\ I&O&\cdots&O&O\\\
O&I&\cdots&O&O\\\ \vdots&\vdots&\ddots&\vdots&\vdots\\\ O&O&\cdots&I&O\\\
\end{array}}\right)_{nr\times nr}$

where $O$ is the null matrix of order $n\times n$ and $I$ is the identity
matrix of order $n\times n$ over $\mathbb{F}.$ Observe that $r$ is the least
positive integer such that $g(A)^{r}=O.$ Hence, the minimal polynomial of
$\mathfrak{m}_{x}$ divides $g(x)^{r}.$ Since $g(x)$ is irreducible and
$g(x)^{i}$ does not annihilate $\mathfrak{m}_{x}$ for $0\leq i\leq r-1,$ the
minimal polynomial of $\mathfrak{m}_{x}$ is $g(x)^{r}.$

Consider $\overline{\bm{s}}:=(\bm{s},\bm{s},\dots,\bm{s})\in\mathbb{F}^{nr},$
where $\bm{s}$ is identified with
$s_{0}+s_{1}x+\dots+s_{n-1}x^{n-1}\in\mathbb{F}[x]/\langle g(x)\rangle.$ Since
$\mathfrak{m}_{x}$-annihilator of $\overline{\bm{s}}$ is $g(x)^{r}$ and $\deg
g(x)^{r}=nr=\dim_{\mathbb{F}}\mathcal{R},$ the vector $\overline{\bm{s}}$ is a
cyclic vector for $\mathfrak{m}_{x}$ and
$\mathcal{B}^{{}^{\prime}}=\\{\overline{\bm{s}}^{t},A\overline{\bm{s}}^{t},\dots,A^{nr-1}\overline{\bm{s}}^{t}\\}$
is a basis for $\mathcal{R}$ via the identification given in (3.1).

Now,

$\tau\left(x^{j}(p(x)+\langle
g(x)^{r}\rangle)\right)=\tau(A^{j}\overline{\bm{p}})=\overline{\bm{s}}\cdot\left(A^{j}\overline{\bm{p}}\right)=\left(\overline{\bm{s}}A^{j}\right)\cdot\overline{\bm{p}}=({A^{j}}^{t}\overline{\bm{s}}^{t})^{t}\cdot\overline{\bm{p}}.$

If $\tau\left(x^{j}(p(x)+\langle g(x)^{r}\rangle)\right)=0,\forall\,0\leq
j\leq nr-1,$ then
$({A^{j}}^{t}\overline{\bm{s}}^{t})^{t}\cdot\overline{\bm{p}}=0,\forall\,0\leq
j\leq nr-1,\,\text{we have}\,\,\overline{\bm{p}}=\overline{\bm{0}}$ as
$\mathcal{B}^{{}^{\prime}}$ is a basis for $\mathcal{R}$ and
$\overline{\bm{s}}$ is also a cyclic vector for $A^{t}.$ ∎

######  Remark 3.2.

The usual trace map $T=\textnormal{Tr}_{\mathbb{F}[\alpha]/\mathbb{F}}$, where
$\alpha=x+\langle g(x)\rangle$ is always a choice of $T$ in Theorem 3.1.

######  Lemma 3.3.

Let $\mathcal{R}_{1},\mathcal{R}_{2},\dots,\mathcal{R}_{s}$ be finite-
dimensional commutative $\mathbb{F}$-algebras. If each $\mathcal{R}_{i}$
admits an $\mathbb{F}$-valued trace, then so does
$\prod_{i=1}^{s}\mathcal{R}_{i}.$

###### Proof.

Suppose for each $i,$ $\tau_{i}:\mathcal{R}_{i}\to\mathbb{F}$ is an
$\mathbb{F}$-valued trace of $\mathcal{R}_{i}$. Set
$\mathcal{R}=\prod_{i=1}^{s}\mathcal{R}_{i}$ and define

$\displaystyle\tau:\mathcal{R}$ $\displaystyle\to\mathbb{F}$
$\displaystyle\bm{x}=(x_{1},\dots,x_{s})$
$\displaystyle\mapsto\sum_{i=1}^{s}\tau_{i}(x_{i}).$

Then $\tau$ is clearly a non-zero $\mathbb{F}$-linear map from $\mathcal{R}$
to $\mathbb{F}.$ Suppose $\tau(\bm{x})=0,$ where
$\bm{x}=(x_{1},\dots,x_{s})\in\mathcal{R}\setminus\\{\bm{0}\\}.$ Without loss
of generality, assume that $x_{1}\neq 0.$ Then there is
$y_{x_{1}}\in\mathcal{R}_{1}\setminus\\{0\\}$ such that
$\tau_{1}(x_{1}y_{x_{1}})\neq 0.$ If
$\bm{y}=(y_{x_{1}},0,\dots,0)\in\mathcal{R},$ then

$\tau(\bm{xy})=\tau(x_{1}y_{x_{1}},0,\dots,0)=\tau_{1}(x_{1}y_{x_{1}})\neq 0.$

∎

######  Theorem 3.4.

Let $\mathcal{R}$ be a finite-dimensional commutative $\mathbb{F}$-algebra
generated by only one element. Then there is an $\mathbb{F}$-valued trace of
$\mathcal{R}.$

###### Proof.

Since $\mathcal{R}$ is a finite-dimensional commutative $\mathbb{F}$-algebra
generated by one element, $\mathcal{R}\cong\mathbb{F}[x]/\langle h(x)\rangle$
for some $h(x)\in\mathbb{F}[x]\setminus\\{\bm{0}\\}.$ Let
$h(x)=\prod_{i=1}^{s}h_{i}(x)^{r_{i}}$ be the irreducible factorization of
$h(x)$ in $\mathbb{F}[x].$ Then by Chinese Remainder Theorem,

$\mathcal{R}\cong\prod_{i=1}^{s}\mathcal{R}_{i}$

where $\mathcal{R}_{i}=\mathbb{F}[x]/\langle h_{i}(x)^{r_{i}}\rangle.$ By
Theorem 3.1, for each $i$, there exists an $\mathbb{F}$-valued trace of
$\mathcal{R}_{i}$ and consequently, by Lemma 3.3 there is an
$\mathbb{F}$-valued trace of $\mathcal{R}$. ∎

However, if $\mathcal{R}$ is a finite-dimensional commutative
$\mathbb{F}$-algebra generated by two elements, then there may not exist any
$\mathbb{F}$-valued trace of $\mathcal{R}.$

######  Example 3.5.

Consider the $\mathbb{F}_{2}$-algebra $\mathcal{R}=\mathbb{F}_{2}[x,y]/\langle
x^{2},y^{2},xy\rangle.$ Let $u$ denote $x+\langle x^{2},y^{2},xy\rangle$ and
$v$ denote $y+\langle x^{2},y^{2},xy\rangle$. Then any $\mathbb{F}_{2}$-linear
map $\tau:\mathcal{R}\to\mathbb{F}_{2}$ is of the form:

$\tau(a+bu+cv)=\alpha a+\beta b+\gamma c,$

for some $\alpha,\beta,\gamma\in\mathbb{F}_{2}$. If one of $\beta$ or $\gamma$
is non-zero, set $\bm{s}=\gamma u+\beta v\in\mathcal{R}.$ Then

$\tau\left(\bm{rs}\right)=\tau\left(a\gamma u+a\beta
v\right)=a\beta\gamma+a\beta\gamma=0,\forall\,\bm{r}=a+bu+cv\in\mathcal{R}.$

If both $\gamma$ and $\beta$ are zero, then set $\bm{s}=u$ and note that
$\tau(\bm{rs})=0$ for all $\bm{r}\in\mathcal{R}.$ Hence, for any non-zero
$\tau$, there is $\bm{s}\in\mathcal{R}\setminus\\{\bm{0}\\}$ such that
$\tau(\bm{rs})=0,\,\forall\,\bm{r}\in\mathcal{R},$ proving that there is no
$\mathbb{F}_{2}$-valued trace of $\mathcal{R}.$

We note from the following example that writing elements of
$\mathcal{R}=\mathbb{F}[x]/\langle g(x)^{r}\rangle$ in a special form as
described in Theorem 3.1 is crucial to determine an $\mathbb{F}$-valued trace
of $\mathcal{R}.$

######  Example 3.6.

Let $\mathcal{R}=\mathbb{F}_{2}[x]/\langle(1+x)^{2}\rangle$ and let
$u=x+\langle(1+x)^{2}\rangle.$ Then $\sigma:\mathcal{R}\to\mathbb{F}_{2}$
given by $\sigma(a+bu)=a+b$ for $a,b\in\mathbb{F}_{2}$ is not an
$\mathbb{F}_{2}$-valued trace of $\mathcal{R}$ as $\ker(\sigma)=\langle
1+u\rangle.$ But by Theorem 3.1, for $a,b\in\mathbb{F}_{2},$ the map
$\tau:\mathcal{R}\to\mathbb{F}_{2}$ given by by
$\tau(a+bu)=\tau(a+b+b(1+u))=a+b+b=a$ is an $\mathbb{F}_{2}$-valued trace of
$\mathcal{R}$.

######  Example 3.7.

Consider the $\mathbb{F}_{2}$-algebra $\mathcal{R}=\mathbb{F}_{2}[x]/\langle
x^{3}-x\rangle\cong\mathbb{F}_{2}\times\mathbb{F}_{2}[x]/\langle(1+x)^{2}\rangle,$
where the isomorphism is given by:

$a+bx+cx^{2}+\langle
x^{3}-x\rangle\mapsto\left(a,a+b+c+b(1+x)+\langle(1+x)^{2}\rangle\right).$

If $u$ denotes $x+\langle x^{3}-x\rangle,$ then

$\displaystyle\tau:\mathcal{R}$ $\displaystyle\to\mathbb{F}_{2}$
$\displaystyle a+bu+cu^{2}$ $\displaystyle\mapsto a+(a+b+c+b)=c$

is an $\mathbb{F}_{2}$-valued trace of $\mathcal{R}$ by Theorem 3.1 and
Theorem 3.4.

######  Theorem 3.8.

Let $\mathcal{R}$ and $\mathcal{S}$ be finite-dimensional commutative
$\mathbb{F}$-algebras. If both $\mathcal{R}$ and $\mathcal{S}$ admit
$\mathbb{F}$-valued trace, then so does
$\mathcal{R}\otimes_{\mathbb{F}}\mathcal{S}.$

###### Proof.

Let $\tau_{\mathcal{R}}$ and $\tau_{\mathcal{S}}$ be $\mathbb{F}$-valued trace
maps of $\mathcal{R}$ and $\mathcal{S}$ respectively. Suppose
$\mathcal{B}=\\{\bm{\alpha}_{1}<\bm{\alpha}_{2}<\dots<\bm{\alpha}_{m}\\}$ and
$\mathcal{B}^{{}^{\prime}}=\\{\bm{\beta}_{1}<\bm{\beta}_{2}\dots<\bm{\beta}_{n}\\}$
are ordered bases for $\mathcal{R}$ and $\mathcal{S}$ respectively over
$\mathbb{F}$ so that $\\{\bm{\alpha}_{i}\otimes\bm{\beta}_{j}:1\leq i\leq
m,1\leq j\leq n\\}$ is a basis for
$\mathcal{R}\otimes_{\mathbb{F}}\mathcal{S}$ over $\mathbb{F}.$
Let $\bm{x}=x_{1}\bm{\alpha}_{1}+\cdots+x_{m}\bm{\alpha}_{m}\in\mathcal{R}$
and let $[\bm{x}]_{\mathcal{B}}$ be the coordinate vector of $\bm{x}$ relative
to $\mathcal{B}.$ We often write $[\bm{x}]_{\mathcal{B}}$ as
$\mathbf{x}=\begin{bmatrix}x_{1}&x_{2}&\cdots&x_{m}\end{bmatrix}^{t}.$ Define
$\mathfrak{m}_{\bm{\alpha}_{i}}:\mathcal{R}\to\mathcal{R}$ by
$\mathfrak{m}_{\bm{\alpha}_{i}}(\bm{x})=\bm{\alpha}_{i}\bm{x}.$ Then
$\mathfrak{m}_{\bm{\alpha}_{i}}$ is an $\mathbb{F}$-linear map. Denote the
matrix of $\mathfrak{m}_{\bm{\alpha}_{i}}$ relative to $\mathcal{B}$ by
$A_{i}.$ Then $[\bm{\alpha}_{i}\bm{x}]_{\mathcal{B}}=A_{i}\mathbf{x}.$ Thus,

$\tau_{\mathcal{R}}(\bm{x})=\sum_{i=1}^{m}x_{i}\tau_{\mathcal{R}}(\bm{\alpha}_{i})=\mathbf{x}^{t}\bm{\tau}_{\mathcal{R}},\text{where}\hskip
2.84544pt\bm{\tau}_{\mathcal{R}}=\begin{bmatrix}\tau_{\mathcal{R}}(\bm{\alpha}_{1})&\tau_{\mathcal{R}}(\bm{\alpha}_{2})&\cdots&\tau_{\mathcal{R}}(\bm{\alpha}_{m})\end{bmatrix}^{t}$

and consequently,
$\tau_{\mathcal{R}}(\bm{\alpha}_{i}\bm{x})=\mathbf{x}^{t}A_{i}^{t}\bm{\tau}_{\mathcal{R}}.$
Since $\tau_{\mathcal{R}}$ is an $\mathbb{F}$-valued trace of $\mathcal{R},$

$\tau_{\mathcal{R}}(\bm{\alpha}_{i}\bm{x})=0,\forall\,1\leq i\leq
m\implies\mathbf{x}=\bm{0}.$ (3.3)

Let $\bm{y}=y_{1}\bm{\beta}_{1}+\cdots+y_{n}\bm{\beta}_{n}\in\mathcal{S}$ and
$[\bm{y}]_{\mathcal{B}^{{}^{\prime}}}$ be the coordinate vector of $\bm{y}$
relative to $\mathcal{B}^{{}^{\prime}}.$ If $B_{j}$ is the matrix of the
$\mathbb{F}$-linear map
$\mathfrak{M}_{\bm{\beta}_{j}}:\mathcal{S}\to\mathcal{S}$ defined by
$\mathfrak{M}_{\bm{\beta}_{j}}(\bm{y})=\bm{\beta}_{j}\bm{y}$ relative to
$\mathcal{B}^{{}^{\prime}}$ and writing $[\bm{y}]_{\mathcal{B}^{{}^{\prime}}}$
as $\mathbf{y}=\begin{bmatrix}y_{1}&y_{2}&\cdots&y_{n}\end{bmatrix}^{t},$ then
by a similar argument as above, we obtain
$\tau_{\mathcal{S}}(\bm{\beta}_{j}\bm{y})=\mathbf{y}^{t}B_{j}^{t}\bm{\tau}_{\mathcal{S}},$
where
$\bm{\tau}_{\mathcal{S}}=\begin{bmatrix}\tau_{\mathcal{S}}(\bm{\beta}_{1})&\tau_{\mathcal{S}}(\bm{\beta}_{2})&\cdots&\tau_{\mathcal{S}}(\bm{\beta}_{n})\end{bmatrix}^{t},$
and,

$\tau_{\mathcal{S}}(\bm{\beta}_{j}\bm{y})=0,\forall\,1\leq j\leq
n\implies\mathbf{y}=\bm{0}.$ (3.4)

Define

$\displaystyle T:\mathcal{R}\otimes_{\mathbb{F}}\mathcal{S}$
$\displaystyle\to\mathbb{F}$
$\displaystyle\bm{z}:=\sum_{i=1}^{m}\sum_{j=1}^{n}z_{i,j}\bm{\alpha}_{i}\otimes\bm{\beta}_{j}$
$\displaystyle\mapsto\sum_{i=1}^{m}\sum_{j=1}^{n}z_{i,j}\tau_{\mathcal{R}}(\bm{\alpha}_{i})\tau_{\mathcal{S}}(\bm{\beta}_{j}).$

Then,

$\displaystyle T(\bm{z})$
$\displaystyle=\sum_{j=1}^{n}\left(\sum_{i=1}^{m}z_{i,j}\tau_{\mathcal{R}}(\bm{\alpha}_{i})\right)\tau_{\mathcal{S}}(\bm{\beta}_{j})$
$\displaystyle=\sum_{j=1}^{n}\left(\mathbf{z}_{\bm{\cdot}j}^{t}\bm{\tau}_{\mathcal{R}}\right)\tau_{\mathcal{S}}(\bm{\beta}_{j})$
$\displaystyle\text{where}\hskip
2.84544pt\mathbf{z}_{\bm{\cdot}j}=\begin{bmatrix}z_{1,j}&z_{2,j}&\cdots&z_{m,j}\end{bmatrix}^{t}$
$\displaystyle=\sum_{j=1}^{n}a_{j}\tau_{\mathcal{S}}(\bm{\beta}_{j})$
$\displaystyle\text{where}\hskip
2.84544pta_{j}=\mathbf{z}_{\bm{\cdot}j}^{t}\bm{\tau}_{\mathcal{R}}$
$\displaystyle=\bm{a}^{t}\bm{\tau}_{\mathcal{S}}$
$\displaystyle\text{where}\hskip
2.84544pt\bm{a}=\begin{bmatrix}a_{1}&a_{2}&\cdots&a_{n}\end{bmatrix}^{t}$

Thus, if $\mathbf{z}=[z_{i,j}]_{m\times n},$ then

$T(\bm{z})=\bm{\tau}_{\mathcal{R}}^{t}\mathbf{z}\bm{\tau}_{\mathcal{S}}\,.$
(3.5)

Let $\mathbf{e}_{k}^{(r)}=\begin{bmatrix}0&0&\cdots&1&\cdots&0\end{bmatrix}$
denote the row vector of size $r$ whose $k$-th component is $1$ and all other
components are $0$. Then $\mathbf{e}_{k}^{(m)}A_{i}$ is the $k$th row of
$A_{i}$ and $B_{j}{\mathbf{e}_{k}^{(n)}}^{t}$ is the $k$th column of $B_{j}.$
Now, for $1\leq r\leq m,$

$\displaystyle\bm{\alpha}_{r}\bm{z}$
$\displaystyle=\sum_{j=1}^{n}\left(\sum_{i=1}^{m}z_{i,j}\bm{\alpha}_{i}\bm{\alpha}_{r}\right)\otimes\bm{\beta}_{j}$
$\displaystyle=\sum_{j=1}^{n}\sum_{i=1}^{m}\mathbf{e}_{i}^{(m)}A_{r}\mathbf{z}_{\bm{\cdot}j}(\bm{\alpha}_{i}\otimes\bm{\beta}_{j})$

Similarly, for $1\leq s\leq n,$ we have

$\displaystyle\bm{\beta}_{s}\bm{z}=\sum_{j=1}^{n}\sum_{i=1}^{m}\mathbf{e}_{j}^{(n)}B_{s}\mathbf{z}_{i\bm{\cdot}}^{t}(\bm{\alpha}_{i}\otimes\bm{\beta}_{j})$

where
$\mathbf{z}_{i\bm{\cdot}}=\begin{bmatrix}z_{i,1}&z_{i,2}&\cdots&z_{i,n}\end{bmatrix}$
and consequently

$\bm{\beta}_{s}\bm{\alpha}_{r}\bm{z}=\sum_{j=1}^{n}\sum_{i=1}^{m}\mathbf{e}_{j}^{(n)}B_{s}\mathbf{w}_{i\bm{\cdot}}^{t}(\bm{\alpha}_{i}\otimes\bm{\beta}_{j})$

where

$\displaystyle\mathbf{w}_{i\bm{\cdot}}^{t}$
$\displaystyle=\begin{bmatrix}\mathbf{e}_{i}^{(m)}A_{r}\mathbf{z}_{\bm{\cdot}1}&\mathbf{e}_{i}^{(m)}A_{r}\mathbf{z}_{\bm{\cdot}2}&\cdots&\mathbf{e}_{i}^{(m)}A_{r}\mathbf{z}_{\bm{\cdot}n}\end{bmatrix}^{t}$
$\displaystyle=\left(\mathbf{e}_{i}^{(m)}A_{r}\mathbf{z}\right)^{t}$
$\displaystyle=\mathbf{z}^{t}A_{r}^{t}{\mathbf{e}_{i}^{(m)}}^{t}$

Hence,

$\bm{\beta}_{s}\bm{\alpha}_{r}\bm{z}=\sum_{j=1}^{n}\sum_{i=1}^{m}u_{i,j}(\bm{\alpha}_{i}\otimes\bm{\beta}_{j})$

where
$u_{i,j}=\mathbf{e}_{j}^{(n)}B_{s}\mathbf{z}^{t}A_{r}^{t}{\mathbf{e}_{i}^{(m)}}^{t}.$
Then by (3.5),

$T(\bm{\beta}_{s}\bm{\alpha}_{r}\bm{z})=\bm{\tau}_{\mathcal{R}}^{t}\mathbf{u}\bm{\tau}_{\mathcal{S}},\,\text{where}\,\,\bm{u}=[u_{i,j}]_{m\times
n}.$ (3.6)

Observe, $\mathbf{u}=A_{r}\mathbf{z}B_{s}^{t}.$ Hence, by (3.6)

$T(\bm{\beta}_{s}\bm{\alpha}_{r}\bm{z})=\left(\mathbf{z}^{t}A_{r}^{t}\bm{\tau}_{\mathcal{R}}\right)^{t}B_{s}^{t}\bm{\tau}_{\mathcal{S}}.$
(3.7)

Suppose $T(\bm{\beta}_{s}\bm{\alpha}_{r}\bm{z})=0,\forall\,1\leq r\leq m$ and
$\forall\,1\leq s\leq n.$ Then

$\displaystyle\left(\mathbf{z}^{t}A_{r}^{t}\bm{\tau}_{\mathcal{R}}\right)^{t}B_{s}^{t}\bm{\tau}_{\mathcal{S}}=0,\forall\,1\leq
s\leq n.$
$\displaystyle\implies\mathbf{z}^{t}A_{r}^{t}\bm{\tau}_{\mathcal{R}}=\mathbf{0},\forall\,1\leq
r\leq m$ (using (3.4))
$\displaystyle\implies{\mathbf{z}_{i\bm{\cdot}}}^{t}A_{r}^{t}\bm{\tau}_{\mathcal{R}}=0,\forall\,1\leq
r\leq m,\forall\,1\leq i\leq m$
$\displaystyle\implies\mathbf{z}_{i\bm{\cdot}}=\bm{0},1\leq i\leq m$ (using
(3.3)) $\displaystyle\implies\mathbf{z}=\bm{0}$

∎

######  Remark 3.9.

The map $T$ is independent of the choice of the bases of $\mathcal{R}$ and
$\mathcal{S}.$ For instance, if
$\\{\bm{\alpha}_{1}^{{}^{\prime}}<\dots<\bm{\alpha}_{m}^{{}^{\prime}}\\}$ is
another ordered basis for $\mathcal{R},$ then $T^{{}^{\prime}}=T,$ where

$T^{{}^{\prime}}\left(\sum_{i=1}^{m}\sum_{j=1}^{n}z_{i,j}\bm{\alpha}_{i}^{{}^{\prime}}\otimes\bm{\beta}_{j}\right)=\sum_{i=1}^{m}\sum_{j=1}^{n}z_{i,j}\tau_{\mathcal{R}}(\bm{\alpha}_{i}^{{}^{\prime}})\tau_{\mathcal{S}}(\bm{\beta}_{j}).$

Write
$\bm{\alpha}_{i}^{{}^{\prime}}=\sum_{k=1}^{m}p_{k}^{(i)}\bm{\alpha}_{k}.$ Then
for $1\leq i\leq m,1\leq j\leq n,$

$\displaystyle
T^{{}^{\prime}}(\bm{\alpha}_{i}^{{}^{\prime}}\otimes\bm{\beta}_{j})$
$\displaystyle=\tau_{\mathcal{R}}(\bm{\alpha}_{i}^{{}^{\prime}})\tau_{\mathcal{S}}(\bm{\beta}_{j})$
$\displaystyle=\tau_{\mathcal{R}}\left(\sum_{k=1}^{m}p_{k}^{(i)}\bm{\alpha}_{k}\right)\tau_{\mathcal{S}}(\bm{\beta}_{j})$
$\displaystyle=\sum_{k=1}^{m}p_{k}^{(i)}\tau_{\mathcal{R}}(\bm{\alpha}_{k})\tau_{\mathcal{S}}(\bm{\beta}_{j})$
$\displaystyle=T\left(\sum_{k=1}^{m}p_{k}^{(i)}\bm{\alpha}_{k}\otimes\bm{\beta}_{j}\right)$
$\displaystyle=T(\bm{\alpha}_{i}^{{}^{\prime}}\otimes\bm{\beta}_{j})$

so that $T^{{}^{\prime}}=T.$

From the proof of Theorem 3.8, we record the explicit form of an
$\mathbb{F}$-valued trace of $\mathcal{R}\otimes_{\mathbb{F}}\mathcal{S}$ in
the following corollary.

######  Corollary 3.10.

Let $\mathcal{R}$ and $\mathcal{S}$ be finite-dimensional commutative
$\mathbb{F}$-algebras. If $\\{\bm{\alpha}_{1},\dots,\bm{\alpha}_{m}\\}$ and
$\\{\bm{\beta}_{1},\dots,\bm{\beta}_{n}\\}$ are bases for $\mathcal{R}$ and
$\mathcal{S}$ respectively over $\mathbb{F},$ then
$T\left(\sum_{i=1}^{m}\sum_{j=1}^{n}z_{i,j}\bm{\alpha}_{i}\otimes\bm{\beta}_{j}\right)=\sum_{i=1}^{m}\sum_{j=1}^{n}z_{i,j}\tau_{\mathcal{R}}(\bm{\alpha}_{i})\tau_{\mathcal{S}}(\bm{\beta}_{j})$
defines an $\mathbb{F}$-valued trace of
$\mathcal{R}\otimes_{\mathbb{F}}\mathcal{S},$ where $\tau_{\mathcal{R}}$ and
$\tau_{\mathcal{S}}$ are $\mathbb{F}$-valued traces of $\mathcal{R}$ and
$\mathcal{S}$ respectively.

By induction we have:

######  Corollary 3.11.

Let $\mathcal{R}_{1},\mathcal{R}_{2},\dots,\mathcal{R}_{n}$ be finite-
dimensional commutative $\mathbb{F}$-algebras. If there is an
$\mathbb{F}$-valued trace of each $\mathcal{R}_{i},$ then there is an
$\mathbb{F}$-valued trace of
$\mathcal{R}_{1}\otimes_{\mathbb{F}}\mathcal{R}_{2}\otimes_{\mathbb{F}}\cdots\otimes_{\mathbb{F}}\mathcal{R}_{n}.$

######  Corollary 3.12.

If $\mathcal{R}=\mathbb{F}[x_{1},x_{2},\dots,x_{n}]/\langle
g_{1}(x_{1}),g_{2}(x_{2}),\dots,g_{n}(x_{n})\rangle,$ where
$g_{i}(x_{i})\in\mathbb{F}[x_{i}]\setminus\\{\bm{0}\\},$ then $\mathcal{R}$
admits an $\mathbb{F}$-valued trace.

###### Proof.

By Theorem 2.10, $\mathcal{R}\cong\mathbb{F}[x_{1}]/\langle
g_{1}(x_{1})\rangle\otimes_{\mathbb{F}}\mathbb{F}[x_{2}]/\langle
g_{2}(x_{2})\rangle\otimes_{\mathbb{F}}\cdots\otimes_{\mathbb{F}}\mathbb{F}[x_{n}]/\langle
g_{n}(x_{n})\rangle.$ By Theorem 3.4, there is an $\mathbb{F}$-valued trace of
each $\mathbb{F}[x_{i}]/\langle g_{i}(x_{i})\rangle,$ hence, by Corollary
3.11, there is an $\mathbb{F}$-valued trace of $\mathcal{R}.$ ∎

######  Example 3.13.

Let $\mathcal{R}=\mathbb{F}[x,y]/\langle x^{2},y^{2}\rangle$ and set
$w=x+\langle x^{2}\rangle.$ Then $\tau:\mathbb{F}[x]/\langle
x^{2}\rangle\to\mathbb{F}$ defined by $\tau(a+bw)=a+b$ is an
$\mathbb{F}$-valued trace on $\mathbb{F}[x]/\langle x^{2}\rangle$ by Theorem
3.1 and hence $T:\mathcal{R}\to\mathbb{F}$ defined by $T(a+bu+cv+duv)=a+b+c+d$
is an $\mathbb{F}$-valued trace on $\mathcal{R}$ by Corollary 3.10, where
$u=x+\langle x^{2},y^{2}\rangle$ and $v=y+\langle x^{2},y^{2}\rangle.$

##  4\. Applications

###  4.1. Bases and dual basis

Let $\mathcal{R}$ be a finite-dimensional commutative $\mathbb{F}$-algebra
that admits an $\mathbb{F}$-valued trace $\tau.$ For
$\bm{\beta}\in\mathcal{R},$ define

$\displaystyle f_{\bm{\beta}}:\mathcal{R}$ $\displaystyle\to\mathbb{F}$
$\displaystyle f_{\bm{\beta}}(\bm{\alpha})$
$\displaystyle=\tau(\bm{\beta\alpha}),\,\forall\,\bm{\alpha}\in\mathcal{R}.$

By $\mathbb{F}$-linearity of $\tau,$ $f_{\bm{\beta}}$ is a linear functional
on $\mathcal{R}.$ Moreover, we have

######  Theorem 4.1.

If $\mathcal{R}$ is a finite-dimensional commutative $\mathbb{F}$-algebra that
admits an $\mathbb{F}$-valued trace $\tau,$ then for every functional
$f:\mathcal{R}\to\mathbb{F},$ there exists $\bm{\beta}\in\mathcal{R}$ such
that $f=f_{\bm{\beta}}.$

###### Proof.

Let $\bm{\beta},\bm{\gamma}\in\mathcal{R},$ with $\bm{\beta}\neq\bm{\gamma}.$
Since $\bm{\beta}-\bm{\gamma}\neq\bm{0},$ it follows from the definition of
$\tau$ that there is some $\bm{\alpha}\in\mathcal{R}$ such that
$\tau((\bm{\beta}-\bm{\gamma})\bm{\alpha})\neq 0.$ Hence,
$f_{\bm{\beta}}(\bm{\alpha})-f_{\bm{\gamma}}(\bm{\alpha})=\tau((\bm{\beta}-\bm{\gamma})\bm{\alpha})\neq
0,$ and so the maps $f_{\bm{\beta}}$ and $f_{\bm{\gamma}}$ are different. ∎

Let $\mathcal{R}$ be a commutative $\mathbb{F}$-algebra of dimension $n$ that
admits an $\mathbb{F}$-valued trace $\tau.$ If
$\\{\bm{\alpha}_{1},\dots,\bm{\alpha}_{n}\\}$ is a basis for $\mathcal{R}$
over $\mathbb{F},$ then the projection map
$\pi_{j}:\mathcal{R}\to\mathbb{F},\sum_{i=1}^{n}x_{i}\bm{\alpha}_{j}\mapsto
x_{j}$ is an $\mathbb{F}$-linear map from $\mathcal{R}$ to $\mathbb{F},$ hence
by Theorem 4.1, there is $\bm{\beta}_{j}\in\mathcal{R}$ such that
$\pi_{j}(\bm{\alpha})=\tau(\bm{\beta}_{j}\bm{\alpha})$ for all
$\bm{\alpha}\in\mathcal{R}.$ Putting $\bm{\alpha}=\bm{\alpha}_{i},1\leq i\leq
n,$ we have

$\tau(\bm{\beta}_{j}\bm{\alpha}_{i})=\begin{cases}0&\text{if}\,\,i\neq j\\\
1&\text{if}\,\,i=j\end{cases}$

Thus, $\\{\beta_{1},\dots,\beta_{n}\\}$ is also a basis for $\mathcal{R}$ over
$\mathbb{F}.$

######  Definition 4.2.

Let $\mathcal{R}$ be a finite-dimensional commutative $\mathbb{F}$-algebra
that admits an $\mathbb{F}$-valued trace $\tau.$ Then the two bases
$\\{\bm{\alpha}_{1},\dots,\bm{\alpha}_{n}\\}$ and
$\\{\bm{\beta}_{1},\dots,\bm{\beta}_{n}\\}$ for $\mathcal{R}$ over
$\mathbb{F}$ are said to be $\tau$-dual (or $\tau$\- complementary) bases if
$\tau(\bm{\alpha}_{i}\bm{\beta}_{j})=\delta_{ij},$ where $\delta_{ij}$ denotes
the Kronecker symbol.

From the above discussion, we have the following proposition.

######  Proposition 4.3.

Let $\mathcal{R}$ be a finite-dimensional commutative $\mathbb{F}$-algebra. If
there is an $\mathbb{F}$-valued trace $\tau$ of $\mathcal{R},$ then for any
basis for $\mathcal{R}$ over $\mathbb{F},$ there exists a unique $\tau$-dual
basis.

######  Example 4.4.

Let $\mathcal{R}=\mathbb{F}[x_{1},x_{2},\dots,x_{n}]/\langle
g_{1}(x_{1}),g_{2}(x_{2}),\dots,g_{n}(x_{n})\rangle,$ where
$g_{i}(x_{i})\in\mathbb{F}[x_{i}].$ Then by Corollary 3.12, there is an
$\mathbb{F}$-valued trace $\tau$ of $\mathcal{R}.$ Hence, for any basis for
$\mathcal{R},$ there is a $\tau$-dual basis.

######  Definition 4.5.

Let $\mathcal{R}$ be a commutative $\mathbb{F}$-algebra of dimension $n$ that
admits an $\mathbb{F}$-valued trace $\tau.$ Then the discriminant
$\Delta_{\tau}(\bm{\alpha}_{1},\dots,\bm{\alpha}_{n})$ of the elements
$\bm{\alpha}_{1},\dots,\bm{\alpha}_{n}\in\mathcal{R}$ is defined by the
determinant of order $n$ given by

$\Delta_{\tau}(\bm{\alpha}_{1},\dots,\bm{\alpha}_{n})=\begin{vmatrix}\tau(\bm{\alpha}_{1}\bm{\alpha}_{1})&\tau(\bm{\alpha}_{1}\bm{\alpha}_{2})&\cdots&\tau(\bm{\alpha}_{1}\bm{\alpha}_{n})\\\
\tau(\bm{\alpha}_{2}\bm{\alpha}_{1})&\tau(\bm{\alpha}_{2}\bm{\alpha}_{2})&\cdots&\tau(\bm{\alpha}_{2}\bm{\alpha}_{n})\\\
\vdots&\vdots&\vdots&\vdots\\\
\tau(\bm{\alpha}_{n}\bm{\alpha}_{1})&\tau(\bm{\alpha}_{n}\bm{\alpha}_{2})&\cdots&\tau(\bm{\alpha}_{n}\bm{\alpha}_{n})\\\
\end{vmatrix}$

Note that $\Delta_{\tau}(\bm{\alpha}_{1},\dots,\bm{\alpha}_{n})\in\mathbb{F}.$

A simple characterization of bases for $\mathcal{R}$ which admits an
$\mathbb{F}$-valued trace can be given as follows:

######  Theorem 4.6.

Let $\mathcal{R}$ be a commutative $\mathbb{F}$-algebra of dimension $n$ that
admits an $\mathbb{F}$-valued trace $\tau.$ Then
$\\{\bm{\alpha}_{1},\dots,\bm{\alpha}_{n}\\}$ is a basis for $\mathcal{R}$
over $\mathbb{F}$ iff
$\Delta_{\tau}(\bm{\alpha}_{1},\dots,\bm{\alpha}_{n})\neq 0.$

###### Proof.

Let $\\{\bm{\alpha}_{1},\dots,\bm{\alpha}_{n}\\}$ be a basis for $\mathcal{R}$
over $\mathbb{F}.$ We show that the rows of the determinant
$\Delta_{\tau}(\bm{\alpha}_{1},\dots,\bm{\alpha}_{n})$ are linearly
independent. Suppose that
$c_{1}\tau(\bm{\alpha}_{1}\bm{\alpha}_{j})+\cdots+c_{n}\tau(\bm{\alpha}_{n}\bm{\alpha}_{j})=0$
for $1\leq j\leq n,$ where $c_{i}\in\mathbb{F}.$ If
$\bm{\beta}=c_{1}\bm{\alpha}_{1}+\cdots+c_{n}\bm{\alpha}_{n},$ then
$\tau(\bm{\beta\alpha}_{j})=0,$ for $1\leq j\leq n.$ However, this implies
$\bm{\beta}=\bm{0}$ as $\\{\bm{\alpha}_{1},\dots,\bm{\alpha}_{n}\\}$ forms a
basis for $\mathcal{R}.$ Hence, $c_{1}=\cdots=c_{n}=0,$ which ensures that
$\Delta_{\tau}(\bm{\alpha}_{1},\dots,\bm{\alpha}_{n})\neq 0.$

Conversely, suppose that
$\Delta_{\tau}(\bm{\alpha}_{1},\dots,\bm{\alpha}_{n})\neq 0.$ Let
$c_{1}\bm{\alpha}_{1}+\cdots+c_{n}\bm{\alpha}_{n}=\bm{0},$ for some
$c_{i}\in\mathbb{F}.$ Multiplying by $\bm{\alpha}_{j}$ and applying $\tau,$ we
obtain

$c_{1}\tau(\bm{\alpha}_{1}\bm{\alpha}_{j})+\cdots+c_{n}\tau(\bm{\alpha}_{n}\bm{\alpha}_{j})=0\,\,\text{for}\,1\leq
j\leq n.$

Since $\Delta_{\tau}(\bm{\alpha}_{1},\dots,\bm{\alpha}_{n})\neq 0,$ it follows
that $c_{1}=\cdots=c_{n}=0.$ ∎

###  4.2. Applications to Algebraic coding theory

Let $R$ be a finite commutative ring. Then $R^{n}$ is a free $R$-module of
rank $n.$ Any non-empty subset $\mathcal{C}$ of $R^{n}$ is called a code of
length $n$ over $R$. If, in addition, $\mathcal{C}$ is a $R$-submodule of
$R^{n},$ then the code $\mathcal{C}$ is called a linear code. One may refer to
[5] and [10] for more on codes over rings and fields.

Let $\mathcal{R}_{q}$ be a finite-dimensional commutative
$\mathbb{F}_{q}$-algebra that admits an $\mathbb{F}_{q}$-valued trace $\tau.$
Then the trace map $\tau$ can be used to go down from a code defined over
$\mathcal{R}_{q}$ to a code over $\mathbb{F}_{q}$ as follows:

######  Definition 4.7 ($\tau$-trace code).

For a linear code $\mathcal{C}$ of length $n$ over $\mathcal{R}_{q},$ define
the $\tau$-trace code of $\mathcal{C}$ by

$\tau\left(\mathcal{C}\right):=\left\\{\left(\tau(c_{1}),\dots,\tau(c_{n})\right):(c_{1},\dots,c_{n})\in\mathcal{C}\right\\},$

which is a linear code of length $n$ over $\mathbb{F}_{q}.$

Suppose $\varphi:\mathbb{F}_{q}\hookrightarrow\mathcal{R}_{q}$ defines the
$\mathbb{F}_{q}$-algebra structure of $\mathcal{R}_{q},$ then we identify
$\mathbb{F}_{q}$ with $\varphi(\mathbb{F}_{q}).$

Another way to go down from a code defined over $\mathcal{R}_{q}$ to a code
over $\mathbb{F}_{q}$ is the following:

######  Definition 4.8 (Subfield subcode).

Let $\mathcal{C}$ be a linear code of length $n$ over $\mathcal{R}_{q}.$ The
code $\mathcal{C}|_{\mathbb{F}_{q}}:=\mathcal{C}\cap\mathbb{F}_{q}^{n},$ is
called the subfield subcode.

If Tr is the usual trace map from $\mathbb{F}_{q^{m}}$ to $\mathbb{F}_{q},$
then there is a nice relationship between the trace and subfield subcodes.
Interestingly, the same relationship holds between $\tau$-trace code and
subfield subcodes.

######  Theorem 4.9.

Let $\tau$ be an $\mathbb{F}_{q}$-valued trace of $\mathcal{R}_{q}.$ Suppose
$\mathcal{C}$ is a linear code of length $n$ over $\mathcal{R}_{q},$ then

$\tau\left(\mathcal{C}^{\bot}\right)=\left(\mathcal{C}|_{\mathbb{F}_{q}}\right)^{\bot}$

where
$\mathcal{C}^{\bot}=\\{\bm{x}=(x_{1},\dots,x_{n})\in\mathcal{R}_{q}^{n}:\bm{x}\cdot\bm{y}=\sum_{i=1}^{n}x_{i}y_{i}=0,\,\forall\,\bm{y}=(y_{1},\dots,y_{n})\in\mathcal{C}\\}.$

###### Proof.

Let $\bm{c}=\left(c_{1},\dots,c_{n}\right)\in\mathcal{C}|_{\mathbb{F}_{q}}$
and $\bm{a}=\left(a_{1},\dots,a_{n}\right)\in\mathcal{C}^{\bot}$. Then

$\bm{c}\cdot\tau(\bm{a})=\sum_{i=1}^{n}c_{i}\tau(a_{i})=\tau\left(\sum_{i=1}^{n}c_{i}a_{i}\right)=\tau(\bm{c}\cdot\bm{a})=0,$

showing that
$\tau\left(\mathcal{C}^{\bot}\right)\subseteq\left(\mathcal{C}|_{\mathbb{F}_{q}}\right)^{\bot}.$
Next, we show that
$\tau\left(\mathcal{C}^{\bot}\right)\supseteq\left(\mathcal{C}|_{\mathbb{F}_{q}}\right)^{\bot}.$
This assertion is equivalent to

$\left(\tau\left(\mathcal{C}^{\perp}\right)\right)^{\perp}\subseteq\mathcal{C}|_{\mathbb{F}_{q}}.$

Suppose the above relationship does not hold, then there exist some
$\bm{u}\in\left(\tau\left(\mathcal{C}^{\bot}\right)\right)^{\bot}\setminus\mathcal{C}|_{\mathbb{F}_{q}}$
and $\bm{v}\in\mathcal{C}^{\bot}$ with $\bm{u}\cdot\bm{v}\neq 0$. As $\tau$ is
an $\mathbb{F}_{q}$-valued trace of $\mathcal{R}_{q}$, there is an element
$\alpha\in\mathcal{R}_{q}$ such that $\tau(\alpha(\bm{u}\cdot\bm{v}))\neq 0$.
Hence,

$\bm{u}\cdot\tau(\alpha\bm{v})=\tau(\bm{u}\cdot\alpha\bm{v})=\tau(\alpha(\bm{u}\cdot\bm{v}))\neq
0.$

But, on the other hand, we have $\bm{u}\cdot\tau(\alpha\bm{v})=0$ because
$\bm{u}\in$ $\left(\tau\left(\mathcal{C}^{\bot}\right)\right)^{\bot}$ and
$\alpha\bm{v}\in C^{\bot}$. The desired result follows from this
contradiction. ∎

An $\mathbb{F}_{q}$-valued trace $\tau$ of $\mathcal{R}_{q}$ can also be used
as a tool to construct linear codes over $\mathbb{F}_{q}$ as follows:
Let $D=\\{\\{\bm{d}_{1}<\bm{d}_{2}<...<\bm{d}_{n}\\}\\}$ be an ordered
multiset, where each $\bm{d}_{i}\in\mathcal{R}_{q}.$ Define

$\mathcal{C}_{D}:=\\{\left(\tau(\bm{xd}_{1}),\tau(\bm{xd}_{2}),\dots,\tau(\bm{xd}_{n})\right):\bm{x}\in\mathcal{R}_{q}\\}$

Then $\mathcal{C}_{D}$ is a linear code of length $n$ over $\mathbb{F}_{q}$
and we call $D$ the _defining sequence_ of the code $\mathcal{C}_{D}.$

######  Example 4.10.

Consider $\mathcal{R}_{2}=\mathbb{F}_{2}[x]/\langle x^{3}-x\rangle$ and let
$u=x+\langle x^{3}-x\rangle.$ Then by Example 3.7, $\tau(a+bu+cu^{2})=c$ is an
$\mathbb{F}_{2}$-valued trace of $\mathcal{R}_{2}.$ Let
$D=\\{\\{1<u<1+u<1+u^{2}<u+u^{2}<1+u+u^{2}\\}\\}.$ Then $\mathcal{C}_{D}$ is a
binary $[6,3,3]$-quasicyclic linear code of degree $2.$

####  4.2.1. Construction of subfield codes

An $\mathbb{F}_{q}$-valued trace $\tau$ of $\mathcal{R}_{q}$ is useful in the
computation of subfield codes.
Suppose that $\mathcal{C}$ is a code of length $n$ over $\mathcal{R}_{q}$
generated by the (full rank) matrix $G$ and $\mathcal{B}$ be a basis for
$\mathcal{R}_{q}$ over $\mathbb{F}_{q}.$ The code $\mathcal{C}^{(q)}$ over
$\mathbb{F}_{q}$ generated by the matrix which is obtained by replacing each
entry of $G$ by its column representation relative to $\mathcal{B},$ is called
the subfield code of $\mathcal{C}$[1]. In fact, it is independent of the
choice of $\mathcal{B}.$

Let $\\{\bm{\alpha}_{1},\dots,\bm{\alpha}_{m}\\}$ be a basis for
$\mathcal{R}_{q}$ and let $\\{\bm{\beta}_{1},\dots,\bm{\beta}_{m}\\}$ be its
$\tau$-dual basis for $\mathcal{R}_{q}$ over $\mathbb{F}_{q}$. Then if
$\bm{r}=\sum_{i=1}^{m}r_{i}\bm{\alpha}_{i}\in\mathcal{R}_{q},$ for
$r_{i}\in\mathbb{F}_{q},$ then

$r_{i}=\tau(\bm{r}\bm{\beta}_{i}).$

With the above discussion, Theorem 2.4 of [1] gets generalized to
$\mathcal{R}_{q}.$

######  Theorem 4.11.

Let $\tau:\mathcal{R}_{q}\to\mathbb{F}_{q}$ be an $\mathbb{F}_{q}$-valued
trace of $\mathcal{R}_{q}$ and let
$\mathcal{B}=\\{\bm{\alpha}_{1},\dots,\bm{\alpha}_{m}\\}$ be a basis for
$\mathcal{R}_{q}$ over $\mathbb{F}_{q}.$ Suppose that $\mathcal{C}$ is a
linear code of length $n$ over $\mathcal{R}_{q}$ generated by:

$G=\begin{bmatrix}\bm{g}_{11}&\bm{g}_{12}&\cdots&\bm{g}_{1n}\\\
\bm{g}_{21}&\bm{g}_{22}&\cdots&\bm{g}_{2n}\\\ \vdots&\vdots&\vdots&\vdots\\\
\bm{g}_{k1}&\bm{g}_{k2}&\cdots&\bm{g}_{kn}\\\ \end{bmatrix}$

Then the subfield code $\mathcal{C}^{(q)}$ of $\mathcal{C}$ is generated by

$G^{(q)}=\begin{bmatrix}G_{1}^{(q)}\\\ G_{2}^{(q)}\\\ \vdots\\\
G_{k}^{(q)}\end{bmatrix}$

where for $1\leq i\leq k,$

$G_{i}^{(q)}=\begin{bmatrix}\tau(\bm{g}_{i1}\bm{\alpha}_{1})&\tau(\bm{g}_{i2}\bm{\alpha}_{1})&\dots&\tau(\bm{g}_{in}\bm{\alpha}_{1})\\\
\tau(\bm{g}_{i1}\bm{\alpha}_{2})&\tau(\bm{g}_{i2}\bm{\alpha}_{2})&\dots&\tau(\bm{g}_{in}\bm{\alpha}_{2})\\\
\vdots&\vdots&\vdots&\vdots\\\
\tau(\bm{g}_{i1}\bm{\alpha}_{m})&\tau(\bm{g}_{i2}\bm{\alpha}_{m})&\dots&\tau(\bm{g}_{in}\bm{\alpha}_{m})\\\
\end{bmatrix}.$

##  5\. Conclusion and Discussion

In this manuscript, we studied the $\mathbb{F}$-valued trace of a finite-
dimensional commutative $\mathbb{F}$-algebra. Given a finite-dimensional
commutative $\mathbb{F}$-algebra, an $\mathbb{F}$-valued trace may not exist;
however, we proved that a finite-dimensional commutative $\mathbb{F}$-algebra
of the form $\mathcal{R}=\mathbb{F}[x_{1},x_{2},\dots,x_{n}]/\langle
g_{1}(x_{1}),g_{2}(x_{2}),\dots,g_{n}(x_{n})\rangle,$
$g_{i}(x_{i})\in\mathbb{F}[x_{i}],$ possesses an $\mathbb{F}$-valued trace,
and we constructed such a map on $\mathcal{R}$. We showed that an
$\mathbb{F}$-valued trace on a finite-dimensional commutative
$\mathbb{F}$-algebra induces a non-degenerate bilinear form on $\mathcal{R}$
and hence determines all linear transformations from $\mathcal{R}$ to
$\mathbb{F},$ and it is helpful in the characterisation of bases for
$\mathcal{R}.$ In the field of algebraic coding theory, we presented how an
$\mathbb{F}_{q}$-valued trace can be used as a tool to descend from codes over
an $\mathbb{F}_{q}$-algebra to codes over $\mathbb{F}_{q}.$

A non-zero $\mathbb{F}$-valued $\mathbb{F}$-linear map on a non-commutative
$\mathbb{F}$-algebra is called an $\mathbb{F}$-valued trace if its kernel
contains no non-zero left ideals. Let $\mathcal{R}=M_{n}(\mathbb{F}),$ the
ring of all $n\times n$ matrices over $\mathbb{F}.$ Then $\mathcal{R}$ is a
non-commutative $\mathbb{F}$-algebra for $n\geq 2$. Define
$\tau:\mathcal{R}\to\mathbb{F}$ by $A=[a_{ij}]\mapsto\sum_{i=1}^{n}a_{ii},$
the usual trace of $A.$ Let $\mathcal{I}$ be any non-zero left ideal of
$\mathcal{R}$ and suppose that $B=(b_{ij})\in\mathcal{I}$ be such $b_{ij}\neq
0$ for some $1\leq i,j\leq n$. If $E_{ij}\in\mathcal{R}$ be such that whose
$(i,j)$th entry is $1$ and whose other entries are all $0,$ then
$\textnormal{Tr}(E_{jj}\sum_{k=1}^{n}E_{ki}B)=b_{ij}\neq 0,$ and consequently,
$\mathcal{I}\nsubseteq\ker(\tau).$ Hence, $\ker(\tau)$ contains no non-zero
left ideals of $\mathcal{R},$ proving that $\tau$ is an $\mathbb{F}$-valued
trace of $\mathcal{R}.$

The non-commutative $\mathbb{F}_{2}$-algebra
$\mathcal{U}=\left\\{A=\begin{bmatrix}a&b\\\
0&c\end{bmatrix}:a,b,c\in\mathbb{F}_{2}\right\\}$ does not admit any
$\mathbb{F}_{2}$-valued trace map.

Consider the non-commutative non-unital ring $E=\langle
a,b\,|\,2a=2b=0,a^{2}=a,b^{2}=b,ab=b,ba=a\rangle.$ Note that $E$ has no
$\mathbb{F}_{2}$-algebra structure and the underlying set of $E$ is
$\\{0,a,b,c=a+b\\}.$ Consider the following action of $\mathbb{F}_{2}$ on $E$:
$0e=e0=0$ and $1e=e1=e$ for all $e\in E$. Then every element of $E$ can be
expressed as $as+ct$ for $s,t\in\mathbb{F}_{2}.$ The only non-zero projections
of $E$ onto $\mathbb{F}_{2}$ are $\tau_{i}:E\to\mathbb{F}_{2},i=1,2,3,$ where

$\displaystyle\tau_{1}(as+ct)$ $\displaystyle=a$
$\displaystyle\tau_{2}(as+ct)$ $\displaystyle=c$
$\displaystyle\tau_{3}(as+ct)$ $\displaystyle=a+c.$

It is not difficult to verify that $\ker(\tau_{i})$ contains non-zero left
ideals of $E,$ for $i=1,2,3.$

## Declarations

### Conflict of Interest

Both authors declare that they have no conflict of interest.

## Acknowledgements

The work of the first author was supported by Council of Scientific and
Industrial Research (CSIR) India, under the grant no.
09/0086(13310)/2022-EMR-I.

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