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%-------------------------------------------------
\begin{document}
%{\sl MSC }

% {\sl UDC 512.7}
\medskip


\begin{center}
{\Large\sc Amalgam for zero-dimensional schemes}\end{center}
\medskip
\begin{center}
Nadezda V. TIMOFEEVA

\smallskip

Yaroslavl' State University

Sovetskaya str. 14, 150000 Yaroslavl', Russia

e-mail: {\it ntimofeeva@list.ru}
\end{center}
\bigskip




\begin{quote} The construction for the amalgamated sum
for zero-dimensional Grothendieck' schemes is given as a dual for
universal (fibred) product in the category of commutative
associative Artinian algebras with unity over a field $k=\overline
k$.

Bibliography: 3 titles.

{\it Keywords:} Artinian commutative algebras, zero-dimensional
Grothendieck' schemes, universal product, amalgamated sum.
\end{quote}
\section{Introduction}
\subsection{General problem} We start with  general {\it quotient problem} in the
category of algebraic schemes. The problem includes  following
ingredients:
\begin{itemize}
\item{a scheme $X$;}
\item{an equivalence relation $R \subset X\times X$;}
\item{morphisms $p_i: R \subset X\times X \to X$ defined as
composites of the immersion with the projection on $i$th factor.}
\end{itemize}
The subscheme $R \subset X\times X$ is said to be an {\it
equivalence relation}; this means that it satisfies three
requirements as follows:
\begin{enumerate}
\item{$R$ is {\it reflexive}, i.e. $R \supset \diag(X)$ where
$\diag (X)$ is an image of the diagonal immersion $\diag: X
\hookrightarrow X\times X$;}
\item{$R$ is {\it symmetric}, i.e. the immersion
$R \subset X \times X$ is stable under
the involution intertwining first and second factors of the product
$X\times X$;}
\item{$R$ is {\it transitive}, i.e.
$\pr_{13}(R \times X \cap X\times R) \subset R$ where the
intersection is taken in $X\times X\times X$. Here $\pr_{13}:
X\times X\times X\to X\times X$ is the projection on the product
of the 1st and 3rd factors.}
\end{enumerate}

The question is \cite[ch. 1, 4.3]{Dan} to construct (if possible)
the {\it universal quotient} $X/R$. This is an object  fitting
into the commutative diagram
\begin{equation*}\xymatrix{R\ar[d]_{p_1} \ar[r]^{p_2}& X \ar[d]\\
X \ar[r]& X/R}
\end{equation*}
such that for any other commutative square
\begin{equation*}\xymatrix{R\ar[d]_{p_1} \ar[r]^{p_2}& X \ar[d]\\
X \ar[r]& T}
\end{equation*}
there is a unique morphism $\tau: X/R \to T$ (in the appropriate
category) fitting into the commutative diagram
\begin{equation*}
\xymatrix{R\ar[d]_{p_1} \ar[r]^{p_2}& X \ar[d] \ar[ddr]\\
X \ar[r] \ar[drr]& X/R \ar[rd]^\tau &\\
&& T}
\end{equation*}
The best result should be to know precisely when the universal
quotient exist in some  category (for example, in the category of
algebraic schemes)  and when it does not.

\subsection{Examples} The basic examples are the notion
of algebraic space \cite{Knu} which appears when morphisms $p_i,
i=1,2$ assumed to be \'{e}tale, and open covering of a scheme $Y$
when $X=\bigsqcup U_{\alpha}$ is a disjoint union of open
subschemes of $Y$, $R=\bigsqcup R_{\alpha \beta}$ and morphisms
$R_{\alpha \beta} \to U_{\alpha}$ and $R_{\alpha \beta} \to
U_{\beta}$ are local isomorphisms. In the first case the object
$X/R$ belongs to the category of algebraic spaces. In the second
case the object $X/R$ is the scheme $Y$ whose open cover was
considered.

Another well-known example arises when the scheme $X$ is acted
upon by an algebraic group $G$ and $R$ is a subscheme induced by
$G$-equivalence \cite{MumFo}. In this case if $X/G=X/R$ exists its
universality means that it is a categorical quotient of the scheme
$X$ by $G$.

\subsection{Reformulation: amalgam of schemes} Now let $X$ be
a disjoint union of schemes $X=X_1\bigsqcup X_2$. Then
$$
X\times X=X_1 \times X_1\bigsqcup X_1 \times X_2 \bigsqcup
X_2\times X_1 \bigsqcup X_2 \times X_2
$$
 and the equivalence
relation is broken into 4 disjoint components
$$
R=R_{11} \bigsqcup R_{12}\bigsqcup R_{21}\bigsqcup R_{22}.
$$
The problem reduces to the search of the universal completion of
the diagram
\begin{equation*} \xymatrix{R_{ij}\ar[d]_{p_i} \ar[r]^{p_j}& X_j\\
X_i}
\end{equation*}
 Such a universal completion is called an {\it
amalgam} (or an {\it amalgamated sum}) of schemes $X_i$ and $X_j$
with respect to $R_{ij}$ and is denoted as $X_i\coprod_{R_{ij}}
X_j$.

\subsection{Schemes}
In the present article we work with zero-dimensional algebraic
schemes of finite type over a field $k=\overline k$. This means
that they are prime spectra of Artinian algebras over $k$ and one
can consider the dual picture in the category of Artinian
$k$-algebras. In this circumstance we have to construct the {\it
universal product} of Artinian $k$-algebras. Its functorial
behavior must be analogous to one of fibred product in the
category of schemes. As usually the {\it length} of Artinian
$k$-algebra is its dimension as of $k$-vector space. It is denoted
as $\length A$. If $Z=\Spec A$ then $\length Z:=\length A$.

Let $A$ be an Artinian commutative associative algebra with unity;
then it can be represented as a quotient algebra of $n$-generated
polynomial algebra $\phi: k[x_1, \dots ,x_n] \twoheadrightarrow A$
where $n$ depends on the structure of $A$ but can be chosen $n\le
\length A$ if necessary. The representing homomorphism $\phi$
corresponds to the immersion of zero-dimensional scheme $\Spec A$
into $k$-affine space $\A^n=\Spec k[x_1, \dots ,x_n]$. By means of
this immersion we interpret the abstract zero-dimensional
$k$-scheme $Z$ as a closed subscheme $Z\subset \A^n$ of affine
space. We call the isomorphism $A \cong k[x_1, \dots , x_n]/\ker
\phi$ (resp., the immersion $Z \subset \A^n$) the {\it affine
representation} of the algebra $A$ (resp., scheme $Z$).

Since any Artinian algebra over a field can be split into a direct
sum of local Artinian algebras then it is enough to develop the
construction of universal product for local algebras only.

We say that the subscheme $Z \subset \A^n$ (resp. the  algebra
$k[x_1, \dots, x_n]/\ker \phi $) is {\it monomial subscheme}
(resp. {\it monomial algebra}) if the ideal $\ker \phi \subset
k[x_1, \dots x_n]$ is monomial ideal, i.e. if it admits a system
of monomial generators. Monomial algebras (and corresponding
monomial subschemes) can be depicted by means of $n$-dimensional
Young's diagrams. If such an algebra (or subscheme) has length $l$
then its Young's diagram consists of $l$ $n$-dimensional blocks.
\begin{remark} Our notion of monomiality is formulated with
respect to the representing
homomorphism $\phi$ chosen and depends strongly on it.
\end{remark}
\begin{remark} Monomiality of a subscheme $Z\subset \A^n$ implies
that $Z$ is supported at the origin $0\in \A^n$.
\end{remark}
\begin{example} A pair of monomial zero-dimensional subschemes
allows to describe the case when the amalgam is compute in obvious
way. Let $Z_i\subset \A^n$, $i=1,2$, be monomial subschemes and
$Y_i$ their Young's diagrams. Then the subscheme of the
intersection $Z_0:=Z_1 \cap Z_2$ is represented by the Young's
diagram $Y_0:=Y_1 \cap Y_2$ which is a maximal common part of
diagrams $Y_1$ and $Y_2$. It plays the role of the relation $R$ in
this example. The amalgam $Z=Z_1\coprod_{Z_0}Z_2$ is represented
by the union of diagrams $Y:=Y_1 \cup Y_2$. It is easily seen that
$\length Z=\length Z_1 +\length Z_2-\length Z_0.$ The universal
property of $Z$ as of an amalgam is verified immediately.
\end{example}

We prove the following result.
\begin{theorem} The category of commutative associative Artinian
algebras with unity over a field $k,$ $\overline k=k$, can be
supplied with universal products. Dually, the category of
0-dimensional schemes over $k,$ $\overline k=k$, is supplied with
amalgamated sums.
\end{theorem}
First we fix the terminology and describe the algorithm which
constructs algebra to be a universal product of algebras by means
of their affine representations. Then we prove that choice of
different affine representations leads to isomorphic algebras.
Finally we prove the universality and hence confirm that the
obtained algebra is indeed a product.

\section{Construction}
\subsection{Glossary}
We  work with polynomial rings of view $k[x_1, \dots, x_n],$ its
ideals and quotients.

The {\it basis} of the ideal $I$ is any system of generators of
$I$ as $k[x_1, \dots, x_n]$-module.

The basis is {\it minimal} if it does not span $I$ whenever any of
elements is excluded.

The {\it total degree of the monomial} $x_1^{\alpha_1} \dots
x_n^{\alpha_n}$ is the sum $\alpha_1+ \dots +\alpha_n$.

The {\it total degree of the polynomial} is the maximum of total
degrees of all its monomials with nonzero coefficients.

The {\it $k$-basis} of polynomial algebra (which is not obliged to
be with unity) is its basis if the algebra is considered as
$k$-vector space.

The {\it $k$-relation} between elements of the algebra $A$ (which
is not obliged to be with unity) is a nontrivial $k$-linear
function (nontrivial $k$-linear combination) in these elements
which equals 0 in $A$.

Fix the natural ordering of  variables $x_1, \dots, x_n$; then we
can use shorthand notation $x^{\alpha}$ for the monomial
$x_1^{\alpha_1} \dots x_n^{\alpha_n}$. The symbol $\alpha$ denotes
the row of degrees $\alpha_1, \dots , \alpha_n$.
\subsection{Algorithm}
\subsubsection*{Reduction to surjective morphisms} We are given three
algebras $A_0, A_1, A_2$ with morphisms
$A_1\stackrel{f_1}{\longrightarrow}
A_0\stackrel{f_2}{\longleftarrow } A_2$. Show that we can assume
that both $f_i$, $i=1,2$ are surjective. Since the morphisms $f_i$
must include into commutative squares of the form
\begin{equation}\label{cd1} \xymatrix{A_0 &\ar[l]_{f_2}
A_2\\
\ar[u]^{f_1} A_1& \ar[l]^{\chi_1} A_T
\ar[u]_{\chi_2}}\end{equation} then for any such a commutative
square the composite morphisms $f_1 \circ \chi_1$ and $f_2 \circ
\chi_2$ have coincident images in $A_0.$ By commutativity of the
square $\im f_1 \circ \chi_1 =\im f_2 \circ \chi_2 \subset \im
f_i$, $i=1,2$. This means that $\im f_i \circ \chi_i \subset \im
f_1 \cap \im f_2$ and the morphism $\chi_i$ factors through the
subalgebra $f_i^{-1}(\im f_1 \cap \im f_2) \subset A_i$ for
$i=1,2$. Then all algebras of interest $A_T$ complete the
square
\begin{equation*}
\xymatrix{\im f_1 \cap \im f_2& \ar@{->>}[l]_{f'_2} {f_2}^{-1}(\im
f_1 \cap
\im f_2)\\
\ar@{->>}[u]^{f'_1} f_1^{-1}(\im f_1 \cap \im f_2)& \ar[l] A_T
\ar[u]}
\end{equation*}
to commute.

Denote $A'_1\stackrel{f'_1}{\twoheadrightarrow}A'_0
\stackrel{f'_2}{\twoheadleftarrow} A'_2$ where $A'_0=\im f_1 \cap
\im f_2,$ $A'_i=f_i^{-1}(A'_0)$ and $f'_i=f_i|_{A'_i},$ $i=1,2.$

Let there is a universal (fibred) product $A'_1\times _{A_0'}
A'_2$ for the special case
$A'_1\stackrel{f'_1}{\twoheadrightarrow}A'_0
\stackrel{f'_2}{\twoheadleftarrow} A'_2$. Confirm that $A'_1\times
_{A_0'} A'_2$ is also a universal product for the initial
arbitrary data \linebreak $A_1\stackrel{f_1}{\longrightarrow}A_0
\stackrel{f_2}{\longleftarrow} A_2$. By the construction of
algebras $A'_i$, $i=0,1,2,$ and by the definition of the algebra
$A'_1\times _{A_0'} A'_2$ it includes into the commutative diagram
\begin{equation*}\xymatrix{A_0&&&\ar[lll]_{f_2}A_2\\
&\ar@{^(->}[ul]A'_0&\ar[l]A'_2 \ar@{^(->}[ur]
\\
&A'_1 \ar[u] \ar@{^(->}[ld]&\ar[l]A'_1\times _{A_0'} A'_2\ar[u]
\\
A_1\ar[uuu]}
\end{equation*}
Let $A_T$ fit into the commutative diagram (\ref{cd1}). Then by
the universality of $A'_1\times _{A_0'} A'_2$ as a product there
is a unique homomorphism $\varphi: A_T\to A'_1\times _{A_0'} A'_2$
which fits in the commutative diagram
\begin{equation*}\xymatrix{A_0&&&\ar[lll]_{f_2}A_2\\
&\ar@{^(->}[ul]A'_0&\ar[l]_{f'_1}A'_2 \ar@{^(->}[ur]
\\
&A'_1 \ar[u]^{f'_1} \ar@{^(->}[ld]&\ar[l]A'_1\times _{A_0'}
A'_2\ar[u]
\\
A_1\ar[uuu]^{f_1}&&& \ar[lll] \ar[ull] \ar@{.>}[ul]_{\varphi}
\ar[uul] A_T \ar[uuu]}
\end{equation*}
This diagram provides universality of the algebra $A'_1\times
_{A_0'} A'_2$ as a product for initial data, i.e. $$A'_1\times
_{A_0'} A'_2=A_1\times _{A_0} A_2.$$

  Now we can replace everywhere our initial arbitrary data by
the more special case when morphisms $f_i$, $i=1,2$, are
surjective.


\subsubsection*{Easy case} Assume that
$A_i \cong k[x_1, \dots, x_n]/I_i $, $i=0,1,2,$ so that $A_0=A_1
\otimes_{k[x_1, \dots, x_n]} A_2$. This assumption says that the
subscheme $\Spec A_0 \cong Z_0\subset \A^n$ is an intersection of
subschemes $Z_1\subset \A^n$ and $Z_2 \subset \A^n$, $Z_i \cong
\Spec A_i$, $i=1,2$.

Choose minimal bases in ideals $I_i$, $i=1,2$. Let $d_i$ be the
maximum of total degrees of all elements in chosen minimal basis
of $I_i$. Now take $d:=\max \{d_1, d_2\}$ and form a fuzzy point
algebra $F=k[x_1, \dots, x_n]/{\mathfrak m}^d$ where ${\mathfrak
m}=(x_1, \dots, x_n)$ is maximal ideal of the origin. The algebra
$F$ includes into the commutative square
\begin{equation*}\xymatrix{A_0 &\ar@{->>}[l]_{f_2}
A_2\\
\ar@{->>}[u]^{f_1} A_1& \ar@{->>}[l] F
\ar@{->>}[u]}\end{equation*}

Consider $k$-basis elements in $F$. The list $M$ of monomial
$k$-generators of algebra $A$ is formed as follows. We include the
element $x^\alpha \in F$ into $M$ if and only if it is taken to a
nonzero element at least in one of $A_1$ or $A_2$. When all
$k$-basis elements in $F$ were considered the formation of the
list $M$ is completed. The $k$-span $<M>$ is a monomial algebra.
The product algebra of interest $A$ is quotient algebra of $<M>$.

It rests to form the list $L$ of $k$-relations on $M$. The
$k$-relation $\sum a_{\alpha} x^{\alpha}$ between elements of $F$
is include into $L$ iff it equals 0 in both of $A_1$, $A_2$.

Now set $A=<M>/<L>$.

Since $\overline k=k$ then $\length A =| M| -|L |= \length A_1
+\length A_2 -\length A_0.$

\subsubsection*{Hard case} How to build up affine representations
of algebras $A_0,A_1,A_2$ to validate the Easy case, i.e. such
that $A_0=A_1 \otimes_{k[x_1, \dots, x_n]} A_2$?

Start with principal ideals in $A_0$: only those which are maximal
under inclusion are necessary. Take a generator of each such an
ideal. Choose any maximal $k$-linearly independent subset in the
set of generators chosen. Since $A_0$ is Artinian algebra of
finite type, this subset is obliged to be finite; let it consist
of $s$ elements $g_1, \dots, g_s$. Each element $g_j$  corresponds
to the variable $x_j$ in $k[x_1, \dots, x_s].$ The construction
done fixes a homomorphism of $k$-algebras $k[x_1, \dots,
x_s]\twoheadrightarrow A_0$. Let $K_0:=\ker h_0$. It is clear that
this is a finitely generated $k[x_1, \dots, x_s]$-module.


Let $g_j'$ be one of preimages of the element  $g_j\in A_0$ in
$A_1$, $g''_j$ be one of preimages of the same element in $A_2$.
We complete the set  $g'_1, \dots ,g'_s$ to form the set of
$k$-linearly independent generators of the algebra $A_1$, by
adding elements  $g'_{s+1}, \dots. g'_m \in A_1$. Analogously, the
set  $g''_1, \dots , g''_s$ is completed to the set of
$k$-linearly independent generators of the algebra $A_2$ by adding
elements $g''_{m+1}, \dots, g''_n$. We put the variables $x_{s+1},
\dots,x_m$ in the correspondence to the elements $g'_{s+1}, \dots
,g'_{m}$ and the variables $x_{m+1}, \dots, x_n$ to the elements
 $g''_{m+1}, \dots, g''_n$.




This leads to  homomorphisms $h_i:k[x_1,
\dots,x_n]\twoheadrightarrow A_i$, $i=0,1,2$, $n\ge s$ defined by
following correspondences
\begin{eqnarray*}&&h_0(x_j)=g_j,\;\; j=1,\dots,s,\\
&&h_0 (x_j)=0,\;\;j=s+1, \dots , n,\\
&&h_1(x_j)=g'_j,\;\; j=1,\dots,m,\\
&&h_1(x_j)=0,\;\; j=m+1,\dots,n,\\
&&h_2(x_j)=g''_j,\;\; j=1, \dots,s,m+1, \dots,n,\\
&&h_2(x_j)=0,\;\; j=s+1,\dots,m.
\end{eqnarray*}
In particular,  $h_0=f_i\circ h_i$.

Affine representations of algebras $A_0, A_1, A_2$ and their
homomorphisms lead to immersions of their spectra into affine
space according to the following commutative diagram
\begin{equation*}\xymatrix{\A^n&\ar@{_(->}[l]_{h_2^{\sharp}} \,\Spec A_2\\
\ar@{_(->}[u]^{h_1^{\sharp}} \Spec A_1&\ar@{_(->}[l]
\ar@{_(->}[ul]_{h_0^{\sharp}} \,\Spec A_0 \ar@{_(->}[u]}
\end{equation*}
By some modification of representing homomorphisms  $h_i$,
$i=0,1,2,$ we will easily achieve that in the appropriate affine
space
$$h_0^{\sharp}(\Spec A_0)=h_1^{\sharp}(\Spec A_1) \cap
h_2^{\sharp}(\Spec A_2).$$ For this purpose choose any system of
generators $\kappa_1,\dots,\kappa_r$ of the ideal $K_0$ (as
$k[x_1, \dots, x_n]$-module). Add to the set of variables $x_1,
\dots, x_n$ additional variables (whose number equals to the
number  $r$ of generators if $K_0$ chosen) $y_1, \dots, y_r$, and
consider in  $A_1$-algebra $A_1[y_1, \dots, y_r]$ an ideal $K'_0$
generated by all relations of view $y_l-\kappa_l(g'_1, \dots,
g'_m),$ $l=1,\dots, r$. Then we have a commutative diagram
\begin{equation*}\xymatrix{
k[x_1, \dots, x_n, y_1, \dots,y_r] \ar@{->>}[d]_{h'_1}
\ar@{->>}[rd]^>>>>>>>>{h'_0} \ar@{->>}[r]^>>>>>{h'_2}&A_2
\ar@{->>}[d]^{f_2}\\A_1[y_1, \dots y_r]/K'_0 \ar@{->>}[r] &A_0}
\end{equation*}
There $h'_i (x_j)=h_i(x_j)$ for $j=1,\dots ,n$ and for $i=0,1,2$,
but $h'_i(y_j)=0$ for $i=0,2$ and $j=1,\dots, r$.
\begin{lemma} There is an isomorphism of $k$-algebras $A_1[y_1, \dots y_r]/K'_0 \cong
A_1$.
\end{lemma}
\begin{proof} Let $K_1=\ker h_1$. Denote by $J$ the ideal
in the ring  $k[x_1, \dots, x_n, y_1, \dots, y_r]$ which is
generated by relations $y_l-\kappa_l(x_1, \dots, x_n),$
$l=1,\dots,r$. Obviously, $J \subset \ker h'_1$. The isomorphism
 of $k[x_1, \dots, x_n, y_1,\dots,y_r]$-modules $A_1[y_1, \dots
y_r]/K'_0 \cong A_1$ follows from the exact diagram
\begin{equation*}\xymatrix{&0 \ar[d]&0 \ar[d]\\
& J\ar[d]
\ar[r]^<<<<<<<<<{\sim}&(y_1, \dots, y_r)\ar[d]&\\
0\ar[r]&\ker h'_1 \ar[d] \ar[r]&k[x_1, \dots, x_n, y_1,\dots,y_r]
\ar[d] \ar[r]&A_1[y_1, \dots,y_r]/K'_0 \ar[d]^{\wr}
\ar[r]&0\\
0\ar[r]&K_1 \ar[d] \ar[r] &k[x_1, \dots ,x_n] \ar[d] \ar[r]&A_1
\ar[r]&0\\
&0&0}\end{equation*} where the isomorphism  $J \cong (y_1, \dots,
y_r)$ is an isomorphism of free modules of equal ranks.
 The $k$-algebra isomorphism  $A_1[y_1, \dots y_r]/K'_0 \cong A_1$
follows from the fact that the quotient algebra $A_1[y_1, \dots
y_r]/K'_0 $ admits the same system of generators  $g'_1, \dots,
g'_m$ as an algebra  $A_1$, with same relations (equal to
generators of the ideal $K_1$).
\end{proof}




\begin{proposition} The homomorphisms $h'_i$ lead to the
expression $$A_0=A_1 \otimes_{k[x_1, \dots, x_n,y_1,\dots ,y_r]}
A_2.$$
\end{proposition}
\begin{proof} To prove the proposition one needs to confirm that
$A_0$ is universal as a coproduct. Let Q be a $k$-algebra supplied
with two homomorphisms $A_1 \stackrel{\varphi_1}{\longrightarrow}
Q \stackrel{\varphi_2}{\longleftarrow} A_2$ such that the
following diagram commutes:
\begin{equation*}
\xymatrix{k[x_1, \dots, x_n,y_1,\dots, y_r] \ar@{->>}[d]_{h'_1}
\ar@{->>}[r]^>>>>>>{h'_2}
& A_2 \ar@{->>}[d]_{f_2} \ar[rdd]^{\varphi_2}\\
A_1 \ar@{->>}[r]^{f_1} \ar[rrd]_{\varphi_1}& A_0&\\
&&Q}
\end{equation*} Commutativity of the ambient contour
guarantees that $\varphi_1(A_1)=\varphi_2(A_2)$ and hence we can
replace Q by $\varphi_1(A_1)=\varphi_2(A_2)$. However we preserve
the notation $Q$ but assume that homomorphisms $\varphi_i$ are
surjective. By the construction of homomorphisms $h'_i,$ the
algebra $A_0$ is generated by the images of first $s$ variables
$x_1, \dots, x_s$.

Then by the commutativity of the ambient contour
$\varphi_1h'_1(x_j)=\varphi_2h'_2(x_j)$ for $j=1,\dots,n$. In
particular, by the construction of homomorphisms  $h_i$ (and of
homomorphisms $h'_i$ built up by means of them),
$\varphi_1h'_1(x_j)=\varphi_2h'_2(x_j)=0$ for $j=s+1, \dots, n$.
For  $j=1,\dots s$ we have $0=\varphi_1\circ h'_1(y_l-
\kappa_l(x_j))=\varphi_2\circ h'_2(y_l-\kappa_l(x_j))=
-\varphi_2h'_2\kappa_l(x_j)$ for $l=1,\dots r$ since
$h'_2(y_l)=0$. This implies that homomorphisms  $\varphi_i$,
$i=1,2,$ factor through the algebra $A_0.$


The homomorphism $\varphi_0: A_0 \to Q$ is uniquely defined on
generators $h'_0(x_j)$ of $A_0$ by the correspondence $h_0(x_j)
\mapsto \varphi_i \circ h'_i (x_j),$ $i=1,2,$ $j=1, \dots, s.$
\end{proof}

\begin{remark} Proposition 1 means that the constructed affine representations
are such that the intersection of images of schemes $\Spec A_1$
and $\Spec A_2$ under the immersion into  $\Spec k[x_1, \dots,
x_n, y_1,\dots,y_r]=\A^{n+r}$ is the image of $\Spec A_0.$
\end{remark}
\begin{example} Let we are given algebras $A_0=k[x]/(x^2),$
$A_1=A_2=k[x]/(x^3).$ \linebreak The construction of affine
representations according to the Hard case leads to
$A_0=k[x,y]/(x^2,y),$ $A_1=k[x,y]/(x^3, y-x^2),$
$A_2=k[x,y]/(x^3,y).$ Now we can apply the Easy case:
$F=k[x,y]/{\mathfrak m}^3=k[x,y]/(x^3, x^2y,xy^2,y^3)$,
$M=\{1,x,y,x^2\},$ $L=\emptyset$. This implies
$<M>/<L>=k[x,y]/(x^3,xy,y^2).$
\end{example}
\begin{remark} In the further considerations we use the notation
of the view  $$A_1 \stackrel{h_1}{\longleftarrow} k[x_1, \dots,
x_n] \stackrel{h_2}{\longrightarrow} A_2$$ assuming that
homomorphisms  $h_i$. $i=1,2,$ are surjective and that they admit
application of the Easy case, i.e. Proposition 1 holds for them.
\end{remark}

\subsection{Different affine representations chosen}
Prove that different affine representations of algebras $A_0, A_1,
A_2$ lead to isomorphic product algebras $A$.

Let us have two pairs of different affine representations defined
by pairs of homomorphisms $$A_1 \stackrel{h_1}{\twoheadleftarrow}
k[x_1, \dots, x_n] \stackrel{h_2}{\twoheadrightarrow} A_2 \quad
\mbox{ \rm and}\quad A_1 \stackrel{h'_1}{\twoheadleftarrow}
k[x_1', \dots, x_m'] \stackrel{h'_2}{\twoheadrightarrow} A_2$$
such that $A_0=A_1 \otimes_{k[x_1', \dots, x_m']} A_2=A_1
\otimes_{k[x_1, \dots, x_n]} A_2$. They correspond to two pairs of
immersions of prime spectra $$ \Spec A_1 \stackrel
{h_1^{\sharp}}{\hookrightarrow} \A^n
\stackrel{h_2^{\sharp}}{\hookleftarrow} \Spec A_2 \quad \mbox{\rm
and} \quad \Spec A_1 \stackrel {{h_1'}^{\sharp}}{\hookrightarrow}
\A^m \stackrel{{h_2'}^{\sharp}}{\hookleftarrow} \Spec A_2
$$
such that $h_1^{\sharp}(\Spec A_1) \cap h_2^{\sharp}(\Spec
A_2)=h_0^{\sharp}(\Spec A_0)$ and ${h'_1}^{\sharp}(\Spec A_1) \cap
{h'_2}^{\sharp}(\Spec A_2)={h'_0}^{\sharp}(\Spec A_0)$. This leads
to immersions $\overline {h_i^{\sharp}}: \Spec A_i \hookrightarrow
\A^n \times \A^m=\A^{n+m}$ as composites with the diagonal
immersion $\diag$:
$$\overline {h_i^{\sharp}}:\Spec A_i
\stackrel{\diag}{\hookrightarrow} \Spec A_i \times \Spec A_i
\stackrel{(h_i^{\sharp},
{h_i'}^{\sharp})}{-\!\!-\!\!\!\longrightarrow} \A^n \times \A^m.$$
Now confirm that in this case also $\overline{h_1^{\sharp}}(\Spec
A_1) \cap \overline{h_2^{\sharp}}(\Spec
A_2)=\overline{h_0^{\sharp}}\Spec A_0$.

The homomorphisms $\overline h_i$ are defined as composites
\begin{eqnarray}
k[x_1, \dots, x_n]\otimes _k k[x_1', \dots, x_m']\stackrel{(h_i,
h_i')}{-\!\!\!\longrightarrow}& A_i \otimes_k A_i
&\stackrel{\delta_i}{\to} A_i, \nonumber
\\
g_1 \otimes g_2 \;\quad \mapsto\, & h_i(g_1) \otimes h_i'(g_2)
&\mapsto h_i(g_1) \cdot h_i'(g_2). \nonumber
\end{eqnarray}

Let $Q$ be an algebra together with two homomorphisms $\varphi_i:
A_i \to Q$ such that $ \varphi_1\circ \overline h_1 = \varphi_2
\circ \overline h_2$. To confirm that $A_0=A_1 \otimes _{k[x_1,
\dots, x_n]\otimes _k k[x_1', \dots, x_m']}A_2$ it is necessary to
construct a unique homomorphism $\varphi_0: A_0 \to A_T$ such that
$\varphi_i=\varphi_0 \circ f_i$. Consider commutative diagrams
\begin{equation*} \xymatrix{A_i\otimes_k A_i \ar[d]_{\delta_i} \ar[r]^{(f_i,
f_i)}& A_0 \otimes _k A_0 \ar[d]^{\delta_0} \ar[ddr]^{\varphi'_0}\\
A_i \ar[r]^{f_i} \ar[drr]_{\varphi_i}&A_0 \ar@{..>}[dr]_<<<{\varphi_0}\\
&& Q}
\end{equation*}
for $i=1,2$. Let for any reducible tensor $x\otimes x' \in A_i
\otimes_k A_i$ $$\begin{array}{ll}(f_i, f_i)(x\otimes
x')=\overline x \otimes \overline x' \in A_0 \times_k A_0;&
\delta_0 (\overline x \otimes \overline x')=\overline x\cdot
\overline
x' \in A_0;\\
f_i(x\cdot x')=\overline x \otimes \overline x' \in A_0;&\delta_i
(x\otimes x')=x\cdot x'\in A_i.
 \end{array}$$
Since by  universality of $$A_0 \otimes_k A_0= (A_1 \otimes_k
A_1)\otimes_{k[x_1,\dots, x_n]\otimes_k k[x'_1, \dots, x'_m]} (A_2
\otimes_k A_2)$$ the homomorphism $\varphi'_0$ is uniquely defined
as
$$\varphi'_0(\overline x \otimes \overline
x')=\varphi_i(x)\cdot \varphi_i(x') \in Q$$ then $\varphi_0:
A_0\to Q$ is also uniquely defined as $\varphi_0(\overline
y)=\varphi_i(y)$ for $\overline y=f_i(y)$, $y\in A_i$.

Now it rests to form product algebras for three different affine
representations \begin{equation*}\xymatrix{k[x_1, \dots, x_n]
\ar@{->>}[d]_{h_i}&\ar@{->>}[l] k[x_1, \dots, x_n] \otimes _k
k[x_1', \dots, x_m'] \ar@{->>}[d]_{\overline h_i}\ar@{->>}[r]&
k[x_1', \dots, x_m']
\ar@{->>}[d]^{h'_i}\\
A_i\ar@{=}[r]& A_i& \ar@{=}[l] A_i}
\end{equation*}
Horizontal homomorphisms are defined by correspondences
$x'_l\mapsto 0$, $l=1,\dots, m$ � $x_j \mapsto 0$, $j=1,\dots, n$
respectively.

Denoting by $\overline A$ the product algebra for $\overline h_i:
k[x_1, \dots, x_n] \otimes _k k[x_1', \dots,
x_m']\twoheadrightarrow A_i$, $i=0,1,2,$ we come to the
commutative diagram
\begin{equation}\label{cd2}\xymatrix{k[x_1, \dots, x_n] \ar@{->>}[d]_{h}&\ar@{->>}[l]
k[x_1, \dots, x_n] \otimes _k k[x_1', \dots, x_m']
\ar@{->>}[d]_{\overline h}\ar@{->>}[r]& k[x_1', \dots, x_m']
\ar@{->>}[d]^{h'}\\
A&\ar@{->>}[l] \overline A\ar@{->>}[r]&  A'}
\end{equation}
associated to the immersions
\begin{eqnarray*}&&\xymatrix{&\A^n\;\; \ar@{^(->}[rr]&&\;\;\;
\A^n \times \A^m \;\;&&\ar@{_(->}[ll]
\;\; \A^m,&}\\
&&\xymatrix{\;\;(x_1,\dots,x_n)\ar@{|->}[r]&(x_1, \dots, x_n,
0,\dots,
0),}\\
&&\xymatrix{&&&\!\!(0,\dots, 0,x'_1,\dots,x'_m)&\ar@{|->}[l]
(x'_1, \dots, x'_m).}\end{eqnarray*} Vertical homomorphisms in
(\ref{cd2}) define affine representations for product algebras
built up by means of the method  described in the Easy case. Lower
horizontal morphisms are surjective by the construction of the
Easy case: a nonzero element of the algebra $A$ (resp., of the
algebra $A'$) has a nonzero preimage in the polynomial ring
$k[x_1, \dots,x_n]$ (resp., in $k[x_1', \dots, x_m']$). This
preimage has a nonzero preimage in the product $k[x_1, \dots,x_n]
\otimes_k k[x_1', \dots, x_m']$. The nonzero element obtained in
this way has a nonzero image as least in one of algebras $A_1,
A_2$ and hence in  $\overline A$. Since three algebras in the
lower row are Artinian and
 $\length A
=\length A' =\length \overline A$, then  $A=A'=\overline A$.

\subsection{Universality}
It rests to prove that the construction of product algebra $A$ is
indeed universal, i.e. $A$ is true fibred product.

Let $A_T$ be an Artinian  $k$-algebra together with two
homomorphisms $\chi_i: A_T \to A_i$, $i=1,2$ such that $\chi_1
\circ f_1=\chi_2\circ f_2$. We construct a homomorphism $\varphi$
to complete the diagram \begin{equation*}
\xymatrix{&\ar[ld]_{\chi_1}A_T \ar@{..>}[d]^{\varphi} \ar[rd]^{\chi_2}\\
A_1& \ar[l]_{\overline f_1} A \ar[r]^{\overline f_2}& A_2}
\end{equation*}
Choose  appropriate affine representations of algebras $A_T, A,
A_i$, $i=1,2$. Take a fuzzy point algebra (of an appropriate
degree) $F\twoheadleftarrow k[x_1, \dots, x_n]$. Affine
representations chosen supply algebras $A_T, A, A_i$ with
epimorphisms $F\twoheadrightarrow A_T,$ $F \twoheadrightarrow A,$
$F\twoheadrightarrow A_i$.

Choose an arbitrary element $\alpha \in A_T$; it is taken to
$\chi_1(\alpha) \in A_1$ and to $\chi_2(\alpha)\in A_2$ so that
$f_1 \chi_1(\alpha) =f_2 \chi_2(\alpha).$ Any $\gamma$ of
preimages of $\alpha $ in $F$ is also taken to $f_1
\chi_1(\alpha)=f_2 \chi_2(\alpha) \in A_0$ and to some element
$\overline \alpha \in A$. We claim that $\overline \alpha$ does
not depend on the choice of $\gamma$.

Choose another $\gamma' \in F$ taken to $\alpha$, then $\gamma
-\gamma'$ maps to zero in $A_T$ and hence it is mapped to zero in
both $A_i,$ $i=1,2$. Then by the construction of $A$ (cf. Easy
case) $\gamma -\gamma'$ is taken to zero in $A$. This shows that
there is a homomorphism $\varphi:A_T \to A$ such that
$\chi_i=\varphi \circ \overline f_i$, $i=1,2,$ as required.

It rests to confirm ourselves that the homomorphism $\varphi: A_T
\to A$ does not depend on the choice of affine representations of
algebras  $A_T$, $A_i$, $i=0,1,2.$ For this sake assume that there
is 2 different collections of affine representations include into
commutative diagrams
\begin{equation}\label{tworep}\xymatrix{&\ar@{->>}[ldd]_{h_1} k[x_1, \dots, x_n] \ar@{->>}[d]^{h_T} \ar@{->>}[rdd]^{h_2}\\
&\ar[ld] A_T \ar[rd]&\\
A_1\ar@{->>}[dr]&\ar[l]A\ar[r]&\ar@{->>}[dl]A_2\\
&A_0&}\quad \xymatrix{&\ar@{->>}[ldd]_{h'_1} k[x'_1, \dots, x'_m] \ar@{->>}[d]^{h'_T} \ar@{->>}[rdd]^{h'_2}\\
&\ar[ld] A_T \ar[rd]&\\
A_1\ar@{->>}[dr]&\ar[l]A\ar[r]&\ar@{->>}[dl]A_2\\
&A_0&}
\end{equation}

Then consider the algebra $k[x_1,\dots,x_n]\otimes_k k[x'_1,
\dots, x'_m]$ and its homomorphisms on quotient algebras
$k[x_1,\dots,x_n]\stackrel{\lambda}{\longleftarrow}k[x_1,\dots,x_n]\otimes_k
k[x'_1, \dots, x'_m]\stackrel{\rho }{\longrightarrow}k[x'_1,
\dots, x'_m]$ which are defined by their kernels  $\ker
\lambda=(x'_1, \dots, x'_m)$, $\ker \rho=(x_1, \dots, x_n)$ and
correspond to immersions of subspaces $\A^n
\stackrel{\lambda^{\sharp}}{\hookrightarrow} \A^n \times \A^m
\stackrel{\rho^{\sharp}}{\hookleftarrow}\A^m$ as linear
subvarieties. Representations  (\ref{tworep}) define induced
affine representations for algebras $A_T$, $A_i$, $i=0,1,2,$
according to the following rule: $\overline h_T(x_j\otimes
x'_l)=h_T(x_j)\cdot h'_T(x'_l)$, $\overline h_i(x_j\otimes
x'_l)=h_i(x_j)\cdot h'_i(x'_l)$, $i=0,1,2$, $j=1,\dots,n,$
$l=1,\dots,m.$ This rule corresponds to the composites of
immersions of prime spectra $\overline h_T^{\sharp}:\Spec A_T
\stackrel{\diag}{\hookrightarrow} \Spec A_T \times \Spec A_T
\stackrel{(h^{\sharp}_T,{h'_T}^{\sharp})}{\hookrightarrow} \A^n
\times \A^m$, $\overline h_i^{\sharp}:\Spec A_i
\stackrel{\diag}{\hookrightarrow} \Spec A_i \times \Spec A_i
\stackrel{(h^{\sharp}_i,{h'_i}^{\sharp})}{\hookrightarrow}\A^n
\times \A^m$.

Let $d$ be the maximal of indices of nilpotent elements in
algebras $A_T,$ $A_i$, $i=0,1,2.$ Take maximal ideals ${\mathfrak
m}=(x_1, \dots, x_n) \subset k[x_1, \dots, x_n],$ ${\mathfrak
m}'=(x'_1, \dots, x'_m)\subset k[x'_1, \dots,x'_m]$ �
$\overline{\mathfrak m}=(x_1,\dots,x_n,x'_1,\dots,x'_m)\subset$ $
k[x_1, \dots,x_n] \otimes_k k[x'_1, \dots ,x'_m]$ and form
algebras $F=k[x_1, \dots, x_n]/{\mathfrak m}^d,$ $F'=k[x'_1,
\dots, x'_m]/{{\mathfrak m}'}^d,$ $\overline F=k[x_1, \dots,x_n]
\otimes_k k[x'_1, \dots ,x'_m]/\overline{\mathfrak m}^d$. They
include into the commutative diagram
\begin{equation*}\xymatrix{k[x_1,\dots,x_n]\ar@{->>}[d]
&\ar[l]_<<<<{\lambda}k[x_1,\dots,x_n]\otimes_k k[x'_1, \dots,
x'_m]\ar@{->>}[d] \ar[r]^>>>>{\rho }&k[x'_1, \dots,
x'_m]\ar@{->>}[d]\\
F&\ar@{->>}[l]\overline F \ar@{->>}[r]&F'}\end{equation*} Now we
perform the construction of homomorphisms $\varphi, \varphi'$ and
$\overline \varphi$ using algebras $F, F'$ and $\overline F$
respectively, as described in the beginning of this subsection.
This lead to the commutative diagram
\begin{equation*}\xymatrix{F\ar@{->>}[dd] \ar@{->>}[dr]&&\ar@{->>}[ll]\overline F \ar@{->>}[dd] \ar@{->>}[dr]\ar@{->>}[rr]&&F'\ar@{->>}[dd]
\ar@{->>}[dr]\\
&\ar@{->>}[ld]_{\varphi}A_T\ar@{=}[rr]&&\ar@{->>}[ld]_{\overline \varphi}A_T\ar@{=}[rr]&&\ar@{->>}[ld]_{\varphi'}A_T\\
A\ar@{=}[rr]&&A\ar@{=}[rr]&&A}\end{equation*} which implies that
 $\varphi=\varphi'=\overline \varphi.$


\begin{thebibliography}{99}
\bibitem{Dan} Danilov V.I. Algebraic manifolds and schemes.
(Russian) // Sovremennye Problemy Matematiki. Fundamental'nye
napravleniya (Contemporary Problems of Math. Fundamental
Directions) Vol. 23 (Itogi nauki i tehn. VINITI RAN USSR), 1988,
172 -- 302.
\bibitem{Knu} {\it Knutson D.}  Algebraic Spaces.
Lecture Notes in Mathematics, 203. Springer, 1971.
\bibitem{MumFo}  {\it Mumford D., Fogarty J.} Geometric
Invariant Theory. Second enlarged Ed. Berlin -- Heidelberg -- New
York, Springer-Verlag, 1982.

\end{thebibliography}


\end{document}
The homomorphism $\chi: T \to A$ is defined as follows. Take a
principal ideal in $T$. We can choose principal ideals which are
maximal under inclusion.  Take a generating element $t\in T$ in
such an ideal.

1307.2192.extracted_bib
\begin{thebibliography}{99}
\bibitem{Dan} Danilov V.I. Algebraic manifolds and schemes.
(Russian) // Sovremennye Problemy Matematiki. Fundamental'nye
napravleniya (Contemporary Problems of Math. Fundamental
Directions) Vol. 23 (Itogi nauki i tehn. VINITI RAN USSR), 1988,
172 -- 302.
\bibitem{Knu} {\it Knutson D.}  Algebraic Spaces.
Lecture Notes in Mathematics, 203. Springer, 1971.
\bibitem{MumFo}  {\it Mumford D., Fogarty J.} Geometric
Invariant Theory. Second enlarged Ed. Berlin -- Heidelberg -- New
York, Springer-Verlag, 1982.

\end{thebibliography}