\input{\FilePrefix Tensor.Algebra.Eq}


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\Section{Tensor Product of Algebras}
\else
\section{Tensor Product of Algebras}
\fi
\label{Section: Tensor Product of Algebras}

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\begin{definition}
\label{definition: tensor product of algebras}
Let $A_1$, ..., $A_n$ be
free algebras over commutative ring $D$.\footnote{I give definition
of tensor product of $D$\Hyph algebras
following to definition in \citeBib{Serge Lang}, p. 601 - 603.
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This subsection
is written on the base of the section
\xRef{8433-5163}{Section: Tensor Product of Algebras}.
\fi
}
Let us consider category $\mathcal A$ whose objects are
polylinear over commutative ring $D$ mappings
\ShowEq{polylinear mappings category}
where $S_1$, $S_2$ are modules over ring $D$,
We define morphism $f\rightarrow g$
to be linear over commutative ring $D$ mapping $h:S_1\rightarrow S_2$
making diagram
\ShowEq{polylinear mappings category, diagram}
commutative.
Universal object
\ShowEq{tensor product of algebras}
of category $\mathcal A$
is called \AddIndex{tensor product of algebras $A_1$, ..., $A_n$}
{tensor product of algebras}.
\qed
\end{definition}

\begin{definition}
\label{definition: tensor power of algebra}
Tensor product
\ShowEq{tensor power of algebra}
is called
\AddIndex{tensor power of algebra}{tensor power of algebra} $A$.
\qed
\end{definition}

\begin{theorem}
\label{theorem: tensor product of algebras is module}
There exists tensor product of algebras.
\end{theorem}
\begin{proof}
Let $M$ be module over ring $D$ generated
by product $\Times$
of $D$\Hyph algebras $A_1$, ..., $A_n$.
Injection
\ShowEq{map i, algebra, tensor product, 1}
is defined according to rule
\ShowEq{map i, algebra, tensor product}
Let $N\subset M$ be submodule generated by elements of the following type
\ShowEq{kernel, algebra, tensor product}
where $d_i\in A_i$, $c_i\in A_i$, $a\in D$.
Let
\[
j:M\rightarrow M/N
\]
be canonical mapping on factor module.
Consider commutative diagram
\ShowEq{diagram top, algebra, tensor product}
Since elements \EqRef{kernel 1, algebra, tensor product}
and \EqRef{kernel 2, algebra, tensor product}
belong to kernel of linear mapping $j$,
then, from equation
\EqRef{map i, algebra, tensor product},
it follows
\ShowEq{f, algebra, tensor product}
From equations \EqRef{f 1, algebra, tensor product}
and \EqRef{f 2, algebra, tensor product}
it follows that mapping $f$ is polylinear over ring $D$.
Since $M$ is module with basis
$\Times$, then, according to theorem \citeBib{Serge Lang}-4.1
on p. 135,
for any module $V$ and any
polylinear over $D$ map
\ShowEq{diagram bottom, 1, algebra, tensor product}
there exists a unique homomorphism $k:M\rightarrow V$,
for which following diagram is commutative
\ShowEq{diagram bottom, algebra, tensor product}
Since $g$ is polylinear over $D$, then
\ShowEq{ker k subseteq N}
According to statement 
on p. \citeBib{Serge Lang}-119,
map $j$ is universal in the category of homomorphisms of
vector space $M$
whose kernel contains $N$.
Therefore, we have homomorphism
\[
h:M/N\rightarrow V
\]
which makes the following diagram commutative
\begin{equation}
\EqLabel{diagram right, algebra, tensor product}
\xymatrix{
&M/N\ar[dd]^h
\\
M\ar[dr]^k\ar[ur]_j
\\
&V
}
\end{equation}
We join diagrams
\EqRef{diagram top, algebra, tensor product},
\EqRef{diagram bottom, algebra, tensor product},
\EqRef{diagram right, algebra, tensor product},
and get commutative diagram
\ShowEq{diagram, algebra, tensor product}
Since $\mathrm{Im}f$ generates $M/N$,
then map $h$ is uniquely determined.
\end{proof}

According to proof of theorem
\ref{theorem: tensor product of algebras is module}
\[A_1\otimes...\otimes A_n=M/N\]
If $d_i\in A_i$, we write
\ShowEq{map j, algebra, tensor product}

\begin{theorem}
\label{theorem: tensor product and polylinear mapping}
Let $A_1$, ..., $A_n$ be
algebras over commutative ring $D$.
Let
\ShowEq{map f, 1, algebra, tensor product}
be polylinear mapping defined by equation
\ShowEq{map f, algebra, tensor product}
Let
\ShowEq{map g, algebra, tensor product}
be polylinear mapping into $D$\Hyph module $V$.
There exists an $D$\Hyph linear mapping
\ShowEq{map h, algebra, tensor product}
such that the diagram
\ShowEq{map gh, algebra, tensor product}
is commutative.
\end{theorem}
\begin{proof}
Equation
\EqRef{map f, algebra, tensor product}
follows
from equations
\EqRef{map i, algebra, tensor product} and
\EqRef{map j, algebra, tensor product}.
An existence of the mapping $h$ follows from the definition
\ref{definition: tensor product of algebras}
and constructions made in the proof of the theorem
\ref{theorem: tensor product of algebras is module}.
\end{proof}

We can write equations \EqRef{f 1, algebra, tensor product}
and \EqRef{f 2, algebra, tensor product}
as
\ShowEq{tensors, algebra, tensor product}

\begin{theorem}
\label{theorem: tensor product and product in algebra}
Let $A$ be
algebra over commutative ring $D$.
There exists a linear mapping
\ShowEq{tensor product and product in algebra}
\end{theorem}
\begin{proof}
The theorem is corollary of the theorem
\ref{theorem: tensor product and polylinear mapping}
and the definition
\ref{definition: algebra over ring}.
\end{proof}
\else
\newpage
For $a_i\in A_i$, $i=1$, ..., $n$,
$a_1\otimes...\otimes a_n\in A_1\otimes...\otimes A_n$.
According to definition
\ShowEq{tensors, algebra, tensor product}

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\fi
\begin{remark}
\label{remark: representation of tensor, ring}
Representation of tensor as
\ShowEq{tensor representation, algebra}
is ambiguous.
Since $r\in D$, then
\[
(d_1 r)
\otimes
d_2
=d_1
\otimes
(rd_2 )
\]
We can increase or decrease number of summands
using algebraic operations.
Following transformation
\begin{align*}
&d_1
\otimes
d_2
+c_1
\otimes
c_2
\\
=&d_1
\otimes
(d_2-c_2+c_2)
+c_1
\otimes
c_2
\\
=&d_1
\otimes
(d_2-c_2)
+d_1
\otimes
c_2
+c_1
\otimes
c_2
\\
=&d_1
\otimes
(d_2-c_2)
+(d_1
+c_1)
\otimes
c_2
\end{align*}
is between possible transformations.
\qed
\end{remark}
\fi

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\newpage
\fi
\begin{theorem}
\label{theorem: standard component of tensor, algebra}
Tensor product $\Tensor A$ of free
finite dimensional algebras
$A_1$, ..., $A_n$ over the commutative ring $D$ is free
finite dimensional algebra.

Let
\ShowEq{tensor product of algebras, basis i}
be the basis of algebra $A_i$ over ring $D$.
We can represent any tensor $a\in\Tensor A$ in the following form
\ShowEq{tensor canonical representation, algebra}


Expression
$\ShowSymbol{standard component of tensor, algebra}$
is called
\AddIndex{standard component of tensor}
{standard component of tensor, algebra}.
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\qed
\end{theorem}
\newpage
\else
\end{theorem}
\begin{proof}
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See the proof of the theorem
\xRef{8433-5163}{theorem: standard component of tensor, algebra}.
\else
Algebras
$A_1$, ..., $A_n$
are modules over the ring $D$.
According to theorem
\ref{theorem: tensor product of algebras is module},
$\Tensor A$ is module. 

Vector $a_i\in A_i$ has expansion
\ShowEq{expansion i relative basis}
relative to basis $\Basis e_i$.
From equations
\EqRef{tensors 1, algebra, tensor product},
\EqRef{tensors 2, algebra, tensor product},
it follows
\ShowEq{expansion i relative basis, 1}
Since set of tensors $\Tensor a$
is the generating set of module $\Tensor A$,
then we can write tensor
$a\in\Tensor A$
in form
\ShowEq{tensor canonical representation 1, algebra}
where
\ShowEq{tensor canonical representation 2, algebra}
Let
\ShowEq{tensor canonical representation 3, algebra}
Then equation \EqRef{tensor canonical representation 1, algebra}
has form
\EqRef{tensor canonical representation, algebra}.

Therefore, set of tensors $\TensorBasis i$
is the generating set of module $\Tensor A$.
Since the dimension of module $A_i$, $i=1$, ..., $n$, is finite,
then the set of tensors $\TensorBasis i$ is finite.
Therefore, the set of tensors $\TensorBasis i$
contains a basis of module $\Tensor A$, and the module $\Tensor A$
is free module over the ring $D$.

We define the product of tensors like $\Tensor a$ componentwise
\ShowEq{product of tensors, algebra, tensor product}
In particular, if for any $i$, $i=1$, ..., $n$,
$a_i\in A_i$ has inverse,
then tensor
\ShowEq{inverse tensor, algebra, 2}
is inverse tensor to tensor
\ShowEq{inverse tensor, algebra, 1}

The definition of the product
\EqRef{product of tensors, algebra, tensor product}
agreed with equation
\EqRef{tensors 2, algebra, tensor product}
because
\ShowEq{product of tensors, algebra, tensor product, 1}
The distributive property of multiplication over
addition
\ShowEq{product of tensors, algebra, tensor product, 2}
follows from the equation
\EqRef{tensors 1, algebra, tensor product}.
The equation
\EqRef{product of tensors, algebra, tensor product, 2}
allows us to define the product for any tensors $a$, $b$.
\fi
\end{proof}

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\begin{remark}
\label{remark: tensor product of algebras}
According to the remark
\ref{remark: opposite algebra},
we can define different structures of algebra
in the tensor product of algebras.
For instance, algebras $A_1\otimes A_2$,
$A_1\otimes A_2^*$, $A_1^*\otimes A_2$
are defined in the same module.
\qed
\end{remark}
\fi
\fi

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\begin{theorem}
Let $\Basis e_i$ be the basis of the algebra $A_i$ over the ring $D$.
Let $B_{i\cdot}{}^{\gi j}_{\gi{kl}}$ be
structural constants of the algebra $A_i$ relative the basis $\Basis e_i$.
Structural constants of the tensor product $\Tensor A$
relative to the basis $\TensorBasis i$ have form
\ShowEq{structural constants tensor algebra}
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\qed
\end{theorem}
\newpage
\else
\end{theorem}
\begin{proof}
Direct multiplication of tensors
$\TensorBasis i$ has form
\ShowEq{structural constants tensor algebra, proof, 1}
According to the definition of structural constants
\ShowEq{structural constants tensor algebra, proof, 2}
The equation
\EqRef{structural constants tensor algebra}
follows from comparison
\EqRef{structural constants tensor algebra, proof, 1},
\EqRef{structural constants tensor algebra, proof, 2}.

From the chain of equations
\ShowEq{structural constants tensor algebra, proof, 3}
it follows that definition of product
\EqRef{structural constants tensor algebra, proof, 2}
with structural constants
\EqRef{structural constants tensor algebra}
agreed with the definition of product
\EqRef{product of tensors, algebra, tensor product}.
\end{proof}

\begin{theorem}
For tensors $a$, $b\in\Tensor A$,
standard components of product satisfy to equation
\ShowEq{standard components of product, tensor product}
\end{theorem}
\begin{proof}
According to the definition
\ShowEq{standard components of product, tensor product, 1}
At the same time
\ShowEq{standard components of product, tensor product, 2}
The equation
\EqRef{standard components of product, tensor product}
follows from equations
\EqRef{standard components of product, tensor product, 1},
\EqRef{standard components of product, tensor product, 2}.
\end{proof}

\begin{theorem}
If the algebra $A_i$, $i=1$, ..., $n$, is associative,
then the tensor product $\Tensor A$ is associative algebra.
\end{theorem}
\begin{proof}
Since
\ShowEq{tensor product, associative algebra, 1}
then
\ShowEq{tensor product, associative algebra, 2}
\end{proof}
\fi
\fi
{"config": "arxiv", "file": "1006.2597/Tensor.Algebra.English.tex"}