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% "On the utility of Robinson-Amitsur ultrafilters"
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\title{On the utility of Robinson--Amitsur ultrafilter}
\author{Pasha Zusmanovich}
\date{September 13, 2010}
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\begin{frame}{A theorem from 1960s}
\begin{block}{Theorem (Amitsur, Robinson)}
If a prime associative ring $R$ embeds in the direct product of associative
division rings, then $R$ embeds in an associative division ring.
\end{block}
\uncover<2->{
\begin{block}{Reminder 1}
An (associative) ring $R$ is called \textit{prime} if one of the following
equivalent conditions holds:
\begin{enumerate}[(i)]
\item
$\forall \> I,J \triangleleft R \>\> I,J \ne 0 \Rightarrow IJ \ne 0$;
\item
$\forall a,b\in R, a,b\ne 0 \>\exists x\in R: axb \ne 0$.
\end{enumerate}
\end{block}
}
\uncover<3->{
\begin{block}{Reminder 2}
A ring $R$ is called \textit{division ring} if
$\>\forall a,b\in R \>\exists x,y \in R : ax = b \>\&\> ya = b$.
\end{block}
}
\end{frame}
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\begin{frame}{The proof}
The proof uses an ultrafilter constructed from the given embedding
$R \subseteq \prod_{i\in \mathbb I} A_i$.
\begin{overprint}
\onslide<1>
\onslide<2>
\begin{block}{Reminder}
An \textit{ultrafilter} $\mathcal U$ on a set $\mathbb I$ is a set
of subsets of $\mathbb I$ satisfying the following conditions:
\begin{enumerate}
\item $\varnothing \notin \mathcal U$
\item
$\mathbb X, \mathbb Y\in \mathcal U \Rightarrow \mathbb X \cap \mathbb Y \in \mathcal U$
\item
$\mathbb X\in \mathcal U, \mathbb X \subset \mathbb Y \Rightarrow \mathbb Y\in \mathcal U$
\item
$\forall \mathbb X \subset \mathbb I \quad (\mathbb X \in \mathcal U) \vee (\mathbb I\backslash \mathbb X\in \mathcal U)$
(maximality)
\end{enumerate}
\end{block}
\onslide<3>
\begin{block}{Reminder}
An \textit{\st{ultra}filter} $\mathcal U$ on a set $\mathbb I$ is a set
of subsets of $\mathbb I$ satisfying the following conditions:
\begin{enumerate}
\item $\varnothing \notin \mathcal U$
\item
$\mathbb X, \mathbb Y\in \mathcal U \Rightarrow \mathbb X \cap \mathbb Y \in \mathcal U$
\item
$\mathbb X\in \mathcal U, \mathbb X \subset \mathbb Y \Rightarrow \mathbb Y\in \mathcal U$
\item
\st{$\forall \mathbb X \subset \mathbb I \quad (\mathbb X \in \mathcal U) \vee (\mathbb I\backslash \mathbb X\in \mathcal U)$
(maximality)}
\end{enumerate}
\end{block}
\onslide<4>
\begin{block}{Reminder}
An \textit{ultrafilter} $\mathcal U$ on a set $\mathbb I$ is a set
of subsets of $\mathbb I$ satisfying the following conditions:
\begin{enumerate}
\item $\varnothing \notin \mathcal U$
\item
$\mathbb X, \mathbb Y\in \mathcal U \Rightarrow \mathbb X \cap \mathbb Y \in \mathcal U$
\item
$\mathbb X\in \mathcal U, \mathbb X \subset \mathbb Y \Rightarrow \mathbb Y\in \mathcal U$
\item
$\forall \mathbb X \subset \mathbb I \quad (\mathbb X \in \mathcal U) \vee (\mathbb I\backslash \mathbb X\in \mathcal U)$
(maximality)
\end{enumerate}
\end{block}
\begin{block}{Examples}
A \textit{principal ultrafilter}: $\set{\mathbb X \subset \mathbb I}{i\in \mathbb X}$ for some $i\in \mathbb I$.
A \textit{cofinite filter}: $\set{\mathbb X \subset \mathbb I}{\mathbb I\backslash \mathbb X \text{ is finite}}$.
\end{block}
\end{overprint}
\end{frame}
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\begin{frame}{The proof (cont.)}
\begin{block}{Fact}
Any set $\mathcal S$ satisfying:
\begin{gather*}
\mathbb X, \mathbb Y\in \mathcal S \Rightarrow \mathbb X \cap \mathbb Y \ne \varnothing \\
\text{(finite intersection property)}
\end{gather*}
contained in some filter and hence (by Zorn's lemma) in some ultrafilter.
\end{block}
\uncover<2->{
\begin{block}{Reminder}
Let $\mathcal U$ be an ultrafilter on $\mathbb I$.
An \textit{ultraproduct} of a set of rings $\{A_i\}_{i\in \mathbb I}$
is the ring
$$
\prod_{\mathcal U} A_i =
\Big(\prod_{i\in \mathbb I} A_i\Big) \Big/ \set{f}{\set{i\in \mathbb I}{f_i = 0} \in \mathcal U} .
$$
If $A_i \simeq A$, we have an \textit{ultrapower} $A^{\mathcal U}$.
\end{block}
}
\end{frame}
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\begin{frame}{The proof (cont.)}
\begin{block}{{\L}o\'s' theorem}
For any first-order sentence $\varphi$,
$$
\prod_{\mathcal U} A_i \models \varphi
\quad\Leftrightarrow\quad
\set{i\in \mathbb I}{A_i \models \varphi} \in \mathcal U .
$$
In particular,
$$
Th \Big(\prod_{\mathcal U} A_i\Big) \supseteq \bigcap_{i\in \mathbb I} Th(A_i)
$$
and
$$
A^{\mathcal U} \equiv A .
$$
\end{block}
\end{frame}
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\begin{frame}{The proof (cont.)}
The proof uses an ultrafilter constructed from the given embedding
$R \subseteq \prod_{i\in \mathbb I} A_i$.
\medskip
Define $\mathcal S = \set{\set{i\in \mathbb I}{f_i \ne 0}}{f\in R, f\ne 0}$.
Which (ultra)filter properties it satisfies?
\begin{overprint}
\onslide<1>
\onslide<2>
\begin{enumerate}
\item $\varnothing \notin \mathcal S$
\end{enumerate}
\onslide<3>
\begin{enumerate}
\item $\varnothing \notin \mathcal S$
\item
$\mathbb X, \mathbb Y\in \mathcal S \overset{?}\Rightarrow \mathbb X \cap \mathbb Y \in \mathcal S$
\end{enumerate}
\onslide<4>
\begin{enumerate}
\item $\varnothing \notin \mathcal S$
\item
\st{$\mathbb X, \mathbb Y\in \mathcal S \Rightarrow \mathbb X \cap \mathbb Y \in \mathcal S$}
$\phantom{\overset{?}\Rightarrow}$
$\mathbb X, \mathbb Y\in \mathcal S \Rightarrow \exists \mathbb Z\in \mathcal S: \mathbb Z \subset \mathbb X \cap \mathbb Y$
\end{enumerate}
\onslide<5>
\begin{enumerate}
\item $\varnothing \notin \mathcal S$
\item
\st{$\mathbb X, \mathbb Y\in \mathcal S \Rightarrow \mathbb X \cap \mathbb Y \in \mathcal S$}
$\phantom{\overset{?}\Rightarrow}$
$\mathbb X, \mathbb Y\in \mathcal S \Rightarrow \exists \mathbb Z\in \mathcal S: \mathbb Z \subset \mathbb X \cap \mathbb Y$
$\mathbb X = \set{i\in \mathbb I}{f_i \ne 0}$, $\>0\ne f\in R$
$\mathbb Y = \set{i\in \mathbb I}{g_i \ne 0}$, $0\ne g\in R$
Since $R$ is prime, there is $x\in R$ such that $fxg \ne 0$,
and
\begin{align*}
\mathbb Z = \set{i\in \mathbb I}{(fxg)_i = f_ix_ig_i \ne 0} &\subset \set{i\in \mathbb I}{f_i \ne 0} \,= \mathbb X \\
&\subset \set{i\in \mathbb I}{g_i \ne 0} = \mathbb Y .
\end{align*}
\end{enumerate}
\onslide<6>
\begin{enumerate}
\item $\varnothing \notin \mathcal S$
\item
\st{$\mathbb X, \mathbb Y\in \mathcal S \Rightarrow \mathbb X \cap \mathbb Y \in \mathcal S$}
$\phantom{\overset{?}\Rightarrow}$
$\mathbb X, \mathbb Y\in \mathcal S \Rightarrow \exists \mathbb Z\in \mathcal S: \mathbb Z \subset \mathbb X \cap \mathbb Y$
\item
$\mathbb X\in \mathcal U, \mathbb X \subset \mathbb Y \overset{?}\Rightarrow \mathbb Y\in \mathcal U$
\item
$\forall \mathbb X \subset \mathbb I \quad (\mathbb X \in \mathcal U) \vee (\mathbb I\backslash \mathbb X\in \mathcal U)$ ?
\end{enumerate}
\onslide<7>
\begin{enumerate}
\item $\varnothing \notin \mathcal S$
\item
\st{$\mathbb X, \mathbb Y\in \mathcal S \Rightarrow \mathbb X \cap \mathbb Y \in \mathcal S$}
$\phantom{\overset{?}\Rightarrow}$
$\mathbb X, \mathbb Y\in \mathcal S \Rightarrow \exists \mathbb Z\in \mathcal S: \mathbb Z \subset \mathbb X \cap \mathbb Y$
\item
\st{$\mathbb X\in \mathcal U, \mathbb X \subset \mathbb Y \Rightarrow \mathbb Y\in \mathcal U$}
$\phantom{\overset{?}\Rightarrow}$
\item
\st{$\forall \mathbb X \subset \mathbb I \quad (\mathbb X \in \mathcal U) \vee (\mathbb I\backslash \mathbb X\in \mathcal U)$}
\end{enumerate}
$\mathcal S$ contained in some ultrafilter $\mathcal U$.
$$
R = R \Big/
\Big(
R \cap \set{f \in \prod_{i\in \mathbb I} A_i}{\set{i\in \mathbb I}{f_i = 0} \in \mathcal U}
\Big)
\subseteq \prod_{\mathcal U} A_i .
$$
\end{overprint}
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\begin{frame}{A generalization}
\begin{block}{Theorem (``Robinson--Amitsur for algebras'')}
If a prime (nonassociative) algebra $R$ embeds in
the direct product $\prod_{i\in \mathbb I} A_i$, then $R$ embeds in an ultraproduct
$\prod_{\mathcal U} A_i$.
\end{block}
\uncover<2->{
\begin{block}{Definition}
A (nonassociative) algebra is called \textit{prime} if one of the following
equivalent conditions holds:
\begin{enumerate}[(i)]
\item
$\forall \> I,J \triangleleft R \>\> I,J \ne 0 \Rightarrow IJ \ne 0$;
\item
$\forall a,b\in R, a,b\ne 0$ there is a nonzero word in elements of $R$ containing $a,b$.
\end{enumerate}
\end{block}
}
\end{frame}
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\begin{frame}{A question}
Does the converse is true (at least in the associative case)?
Suppose that for an algebra $R$ the following holds:
for any set of algebras $\{A_i\}_{i\in \mathbb I}$, if $R$ embeds in the direct product
$\prod_{i\in \mathbb I} A_i$, then $R$ embeds in an ultraproduct
$\prod_{\mathcal U} A_i$. Does this imply that $R$ is prime? that $R$ satisfies
any other natural structural condition?
\end{frame}
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\begin{frame}{Birkhoff meets Robinson--Amitsur}
\begin{block}{Birkhoff's theorem}
A class of algebras is a variety iff it is closed under
subalgebras, quotients and direct products.
\end{block}
\uncover<2->{
\begin{block}{Corollary (Birkhoff + Robinson--Amitsur)}
A prime relatively free algebra $\mathscr F$ in a variety $Var(A)$ embeds in an
ultrapower $A^{\mathcal U}$.
\textbf{Proof.}
By Birkhoff's theorem, $\mathscr F = B/I$ for some $B \subseteq A^{\mathbb I}$
and $I \triangleleft B$. By the universal property of $\mathscr F$, it embeds in
$B$ and hence in $A^{\mathbb I}$. Apply Robinson--Amitsur for algebras.
\end{block}
}
\end{frame}
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\begin{frame}{Criterion for absence of non-trivial identities of algebras}
\begin{block}{Theorem}
For an algebra $R$ over a field $K$ belonging to one of the following variety of algebras:
all algebras, associative algebras, or Lie algebras, the following is equivalent:
\begin{enumerate}[(i)]
\item $R$ does not satisfy a nontrivial identity;
\item A free algebra embeds in an ultrapower of $R$;
\item A free algebra embeds in an algebra elementary equivalent to $R$.
\end{enumerate}
\end{block}
\uncover<2->{
(ii) $\Rightarrow$ (iii) $\Rightarrow$ (i) are trivial.
(i) $\Rightarrow$ (ii) ``almost'' follows from ``Birkhoff + Robinson--Amitsur''.
Two issues:
\begin{itemize}
\item Whether a free algebra is prime?
\item ``Birkhoff + Robinson--Amitsur'' gives embedding on the level of $K$-algebras,
not of $K^{\mathcal U}$-algebras.
\end{itemize}
}
\end{frame}
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\begin{frame}{In which varieties free algebras are prime?}
\begin{tabular}{|p{1.8cm}|c|l|p{2.6cm}|}
\hline
\centering \textbf{name} & \textbf{identities} & & \centering \textbf{why?}
\tabularnewline \hline
all algebras & & yes & no zero divisors \\ \cline{1-3}
associative & $(xy)z = x(yz)$ & yes & \\ \hline
Lie & $xy = - yx$ & yes & Shirshov--Witt \\
& $(xy)z + (zx)y + (yz)x = 0$ & & theorem \\ \hline
alternative & $(xx)y = x(xy)$ & no & explicit example \\
& $(xy)y = x(yy)$ & & \\ \hline
Jordan & $xy = yx$ & no & explicit example \\
& $(xy)(xx) = x(y(xx))$ & & \\ \hline
\end{tabular}
\end{frame}
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\begin{frame}{$K$-algebras vs. $K^{\mathcal U}$-algebras}
``Birkhoff + Robinson--Amitsur'' establishes embedding (under appropriate
conditions) of $K$-algebras:
$$
\mathscr F \subseteq A^{\mathcal U} .
$$
To be able to apply {\L}o\'s' theorem, one needs embedding of
$K^{\mathcal U}$-algebras:
$$
K^{\mathcal U}\mathscr F \subseteq A^{\mathcal U} .
$$
By the universal property of the tensor product, we have a surjection
$$
\mathscr F \otimes_K K^{\mathcal U} \to K^{\mathcal U}\mathscr F ,
$$
but, generally, this is far from being an isomorphism. This is so, however,
if $\mathscr F$ does not have commutative subalgebras of dimension $>1$,
in particular, for absolutely free, free associative and free Lie algebras.
\end{frame}
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\begin{frame}{``Robinson--Amitsur for groups''}
\begin{block}{Theorem}
If a group $G$, all whose abelian subgroups are cyclic either of prime order, or of
infinite order, embeds in the direct product $\prod_{i\in \mathbb I} F_i$ of groups,
then $G$ embeds in an ultraproduct $\prod_{\mathcal U} F_i$.
\end{block}
\uncover<2->{
\begin{block}{Question}
Does the converse is true?
Suppose that for a group $G$ the following holds:
for any set of groups $\{F_i\}_{i\in \mathbb I}$, if $G$ embeds in the direct product
$\prod_{i\in \mathbb I} F_i$, then $G$ embeds in an ultraproduct
$\prod_{\mathcal U} F_i$. Could $G$ be characterized in terms of some structural
properties?
\end{block}
}
\end{frame}
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\begin{frame}{Criterion for absence of non-trivial identities of groups}
\begin{block}{Theorem}
For a group $G$ the following is equivalent:
\begin{enumerate}[(i)]
\item $G$ does not satisfy a nontrivial identity;
\item A nonabelian free group embeds in an ultrapower of $G$;
\item A nonabelian free group embeds in a group elementary equivalent to $G$.
\end{enumerate}
\end{block}
\uncover<2->{
\begin{block}{Question}
Semigroups?
\end{block}
}
\end{frame}
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\begin{frame}{Application: PI}
\begin{block}{Theorem (Regev, 1972)}
If $A, B$ are PI, then $A\otimes B$ is PI.
\end{block}
\uncover<2->{
\begin{block}{Theorem (Procesi--Small, 1968)}
If $A = M_n(K)$, and $B$ is PI, then $A\otimes B$ is PI.
\end{block}
}
\uncover<3->{
\begin{block}{Theorem}
If $A$ is finite-dimensional, $B$ is PI, then $A\otimes B$ is PI.
\end{block}
\textbf{Proof}. Suppose the contrary. Then a free associative algebra of exponential
growth embeds in
$$
(A \otimes_K B)^{\mathcal U} \simeq
(A \otimes_K K^{\mathcal U}) \otimes_{K^{\mathcal U}} B^{\mathcal U}
$$
of polynomial growth, a contradiction.
}
\end{frame}
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\begin{frame}{Application: algebras with same identities}
\begin{block}{Theorem (Razmyslov, 1982)}
If $\mathfrak g_1, \mathfrak g_2$ are finite-dimensional simple Lie algebras
over an algebraically closed field of characteristic $0$, then
$Var(\mathfrak g_1) = Var(\mathfrak g_2)$ iff $\mathfrak g_1 \simeq \mathfrak g_2$.
\end{block}
\begin{block}{Theorem (Drensky--Racine, 1992)}
Ditto for finite-dimensional simple Jordan algebras.
\end{block}
\uncover<2->{
\textbf{Joint proof}.
Using the fact that in these classes of algebras primeness is equivalent to simplicity,
and by the embedding machinery above, reduced to the case where
$\mathfrak g_1 \subseteq \mathfrak g_2$, which follows easily
from the known structural results.
}
\end{frame}
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\begin{frame}{Application: Tarski's monsters}
\uncover<2->{
Rips (197?), Olshanskii (1979)
}
\uncover<3->{
\begin{block}{Girth of groups}
Schleimer (2000), Akhmedov (2003)
\end{block}
}
\uncover<4->{
\begin{block}{Growth sequence}
Number of generators of $G \times \dots \times G$ ($n$ times).
Wise (2002): groups with growth sequence equal to $2$.
\end{block}
}
\uncover<5->{
\begin{block}{Theorem}
\begin{enumerate}[(i)]
\item
A Tarski's monster of type $p$ does not satisfy any nontrivial identity
except $x^p = 1$ and its consequences, iff it has infinite relative girth.
\item
A Tarski's monster of type $\infty$ does not satisfy any nontrivial identity
iff it has infinite girth.
\item
Growth sequence of Tarski's monsters satisfying conditions (i) or (ii), is $2$.
\end{enumerate}
\end{block}
}
\end{frame}
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\begin{frame}{Speculations}
\begin{block}{Lie-algebraic analogs of Tarski's monsters}
Do there exist infinite-dimensional Lie algebras all whose proper
subalgebras are $1$-dimensional?
\end{block}
\uncover<2->{
\begin{block}{Tits' alternative}
Tits (1972): a linear group contains either a solvable subgroup
of finite index, or a nonabelian free subgroup.
Platonov (1967): a linear group which satisfies a nontrivial identity,
contains a solvable subgroup of finite index.
\end{block}
}
\uncover<3->{
\begin{block}{Jacobson's problem}
In a Lie $p$-algebra, does $x^{p^{n(x)}} = x$ imply abelianity?
Weaker question: does it imply a nontrivial identity?
\end{block}
}
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{\Huge That's all. Thank you.}
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Based on \textsf{arXiv:0911.5414} .
Slides at \texttt{http://justpasha.org/math/iceland-2010.pdf} .
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