\chapter{A Theory of Solvability}\label{solvability}

A normal subgroup $N$ of a group $G$ is {\bf solvable} if
there is a finite chain of normal subgroups of~$G$,
\[
\{1\} = N_0 \subseteq N_1
\subseteq \cdots \subseteq N_k = N\,,
\]
such that each factor $N_{i+1}/N_i$ is abelian, or
equivalently such that $[N_{i+1}, N_{i+1}] \subseteq N_i$
for all $i$.  An interval $I[M,N]$ in the normal subgroup
lattice of~$G$ is a {\bf solvable interval} if $N/M$ is a
solvable normal subgroup of~$G/M$.  Normal subgroups $M$ and
$N$ of~$G$ are {\bf solvably related}, written $M\solv N$,
if they belong to a solvable interval.  We will call the
relation ${}\solv{}$ ``the solvability relation'', and refer
to the study of this solvability relation as the ``theory of
solvability'' for groups.  Among the important features of
the theory of solvability for groups~are:

\begin{itemize}
\item The solvability relation is a congruence on the normal
  subgroup lattice of any group.

\smallskip

\item The solvability relation is preserved by homomorphisms
  in the sense that if $M$, $N$ and $P$ are normal subgroups
  of~$G$ with $P\leq M\cap N$, then $M$ is solvably related
  to~$N$ if and only if $M/P$ is solvably related to~$N/P$
  in~$G/P$.

\smallskip

\item If $N$ is a minimal normal subgroup of~$G$, then $N$
  is solvable if and only if it is abelian.
\end{itemize}

The ring-theoretic concept that corresponds directly to that
of a solvable normal subgroup is that of a nilpotent
ideal. In the early days of ring theory the concept of
solvability was developed in analogy with the group
theoretical concept via the Wedderburn radical ( = the
largest nilpotent ideal, if it exists).  Other radicals soon
made their appearance, which permitted new approaches to
solvability not bound to {\em finite} chains of congruences
with abelian factors. Moreover, in the case of some
radicals, the resulting ``solvability theory'' was no longer
bound to the abelian property.  For example, if $\m r$ is a
subdirectly irreducible ring that is not simple, then the
Brown--McCoy radical ( = the intersection of maximal
two-sided ideals) contains the monolith {\em whether or not
  the monolith is abelian}.  Thus, it can happen that a
minimal nonabelian congruence is contained in the
Brown--McCoy radical, hence is ``solvable'' in the
Brown--McCoy sense.

A third approach to solvability, valid for arbitrary finite
algebras, is developed by D.~Hobby and R.~McKenzie in
Chapter~7 
of~\cite{hobby-mckenzie}.\index{Hobby, D.}\index{McKenzie, R.}  
This approach does not use chains
of congruences with abelian factors nor does it use radicals
to define the solvability concept.  Rather, one starts by
defining a pair $(0,1)\in A^2$ from an algebra $\m a$ to be
a {\bf $2$-snag} if there is a polynomial $p(x,y)\in
\Pol_2(\m a)$ such that ${p(0,0) = p(0,1) = p(1,0) = 0}$ and
$p(1,1) = 1$.  Hobby and McKenzie prove that if $\m a$ is
a finite algebra and $\alpha \prec \beta$ are congruences on
$\m a$, then $\C c(\beta,\beta;\alpha)$ holds if and only if
$\beta - \alpha$ contains no $2$-snag.  From this it is
clear that a chain of congruences:
\[
\alpha = 
\alpha_0\prec \alpha_1\prec\cdots\prec\alpha_k 
= \beta
\]
has abelian factors $\alpha_{i+1}/\alpha_i$ if and only if
$\beta - \alpha$ contains no $2$-snag. Starting from this
observation, it can be argued that the relation $\alpha\solv
\beta$, defined by the condition that $\alpha$ and $\beta$
contain the same $2$-snags, is a complete congruence on
$\Conb(\m a)$ for any locally finite algebra.

Our aim in this chapter is to synthesize ideas from the
solvability theories for groups, rings, and finite algebras
in order to develop a solvability theory for a wide class of
varieties.  The theory resembles the theory for groups in
that it is defined in terms of chains of congruences with
abelian factors.  However, as we will show in
Section~\ref{radical_approach}, the theory may viewed as an
extension of the theory of solvability for rings that is
based on the prime radical ( = the intersection of all prime
congruences), since for rings our theory defines an ideal to
be solvable precisely when it is contained in the prime
radical.  Finally, anyone familiar with the theory developed
by Hobby and McKenzie in~\cite{hobby-mckenzie} will
recognize that our ``solvability obstructions''
(Definition~\ref{XXXdefn}) play the role of their $2$-snags,
while Theorems~\ref{HMa}, \ref{3permute} and~\ref{weak_omit}
of this chapter are analogues of Theorem~7.9~(3),
Corollary~7.13 and Lemma~6.5 of~\cite{hobby-mckenzie}.

\section{Varieties With A Weak Difference Term}\index{weak~difference~term|(}

There is an obstacle to the development of a very general
solvability theory that we must deal with first.  Any
absolutely free algebra in any given signature is abelian in
the sense of Chapter~\ref{centrality}.  If there were a
solvability theory that applied to all such algebras, where
the solvability relation is a congruence preserved by
homomorphisms, then such a theory must identify every
congruence on every algebra as solvable.  Such a theory
would, of course, be trivial.  We cannot avoid this type of
problem even if we restrict our attention to varieties
satisfying a nontrivial idempotent Maltsev condition, since
there exist algebras (such as $\m a = \left\lb [0,1];
  \frac{x+y}{2}\right\rb$, the unit interval of the real
line under the averaging operation) that are abelian in the
sense of Chapter~\ref{centrality}, have a Taylor term
($f(x,y) = \frac{x+y}{2}$), but which have simple nonabelian
homomorphic images (in this case, $\m a/\theta$ where
$\theta = [0,1)\times [0,1)\cup \{(1,1)\}$).  This
phenomenon, that solvable algebras have nonsolvable
homomorphic images, prevents the development of a powerful
solvability theory.

As it happens, there is a weakest idempotent Maltsev
condition that guarantees that homomorphic images of
solvable algebras are solvable, and this Maltsev condition
is a strong enough hypothesis for the theorems that we have
in mind. We start this chapter by discussing the idempotent
Maltsev condition on which the solvability theory is based,
and then we discuss the theory.

\bigskip

\begin{df}\label{weak_difference}
  Let ${\vr V}$ be a variety.  A ternary term $d(x,y,z)$ is
  a {\bf weak difference
    term}\index{weak~difference~term!defined} for~${\vr V}$
  if whenever $\m a\in {\vr V}$,\, $\theta\in \Con(\m a)$
  and $(a,b)\in\theta$ it is the case that
  \[
  d(a,a,b) \equiv b \equiv d(b,a,a)
  \pmod{[\theta,\theta]}\,.
  \]
\end{df}

The following theorem is included to show that our
development of a solvability theory only for varieties with
a weak difference term is reasonable.

\begin{thm}\label{justify_weak}
  The class of varieties with a weak difference term is
  definable by a nontrivial idempotent Maltsev condition.
  This idempotent Maltsev condition is the weakest
  conjunction of idempotent Maltsev conditions that implies
  that homomorphic images of solvable algebras are solvable.
  It is also the weakest conjunction of idempotent Maltsev
  conditions that implies that abelian algebras are affine.
\end{thm}

\begin{proof}
  The first and third statements are proved in Theorem~4.8
  of~\cite{k-szendrei2}. We prove the second only.

  If $\vr V$ has a weak difference term, then it can be
  shown (by using Lemma~\ref{lift_delta} and following the
  same argument that works for groups) that homomorphic
  images of solvable algebras are solvable.

  Conversely assume that $\vr V$ satisfies some conjunction
  of idempotent Maltsev conditions that implies that
  homomorphic images of solvable algebras are solvable. At
  least one of these Maltsev conditions must be nontrivial,
  since there exist solvable algebras with nonsolvable
  homomorphic images (e.g. $\left\lb [0,1];
    \frac{x+y}{2}\right\rb$).  The idempotent
  reduct~$\Id({\vr V})$ of~$\vr V$ satisfies the same
  idempotent Maltsev conditions as $\vr V$, so homomorphic
  images of solvable algebras in~$\Id({\vr V})$ are again
  solvable.  Let $\vr A$~be the subvariety of~$\Id({\vr V})$
  that is generated by the abelian algebras in~$\Id({\vr
    V})$. Since the class of abelian algebras in any
  language is axiomatizable by the universal Horn sentences
  of the term condition, this class is closed under the
  formation of subalgebras and products. From our assumption
  that homomorphic images of solvable algebras are solvable
  we get that $\vr A$ consists of solvable algebras.  In
  particular, $\vr A$ contains no subvariety term equivalent
  to the variety of semilattices.  Since $\vr A$~also
  satisfies the nontrivial idempotent Maltsev conditions
  satisfied by~$\vr V$, it follows from Lemma~2.5
  of~\cite{idem} that $\vr A$ must satisfy an idempotent
  Maltsev condition that fails in the variety of
  semilattices. From Theorem~4.10 of~\cite{k-szendrei2}, it
  follows that the variety $\vr A$ is affine, and in
  particular congruence permutable.  If $d(x,y,z)$ is a
  Maltsev operation for~$\vr A$, then it will be a weak
  difference term for~$\vr V$, as is explained in the proof
  of Theorem~4.8 of~\cite{k-szendrei2}.
\end{proof}

This theorem has a corollary which will prove useful
in what follows.

\begin{cor}\label{justify_weak_cor}
  If a variety has a Hobby--McKenzie term, then it has a
  weak difference term.
\end{cor}

\begin{proof}
  Theorem~4.10 of~\cite{k-szendrei2} proves that if $\vr V$
  satisfies an idempotent Maltsev condition that fails in
  the variety of semilattices, then abelian algebras in~$\vr
  V$ are affine.  Theorem~\ref{rect_Malt_cond2} proves that
  the hypothesis of this statement is equivalent to the
  existence of a Hobby--McKenzie term, while
  Theorem~\ref{justify_weak} proves that the conclusion of
  the statement implies the existence of a weak difference
  term.
\end{proof}

The concept of a weak difference term entered mathematics
during the development of commutator theory.  Recall from
the introduction that A.~I.~Maltsev\index{Maltsev, A.~I.}
showed in~\cite{maltsev} that a variety $\vr V$ has
permuting congruences if and only if it has a term
$d(x,y,z)$ such that ${\vr V}\models d(x,x,y) \approx y
\approx d(y,x,x)$.  In~\cite{herrmann},
C.~Herr\-mann\index{Herrmann, C.}  proved that abelian
algebras in a congruence modular variety $\vr V$ are affine
by constructing a term $d(x,y,z)$ from the Day terms for
congruence modularity and showing that $d(x,y,z)$ is a
Maltsev term for any abelian algebra in~$\vr
V$.\footnote{This fact, together with the fact that the
  class of congruence modular varieties is definable by an
  idempotent Maltsev condition, already implies that
  $d(x,y,z)$ is a weak difference term for~$\vr V$, as the
  proof of Theorem~\ref{justify_weak} shows.}  Then
H.~P.~Gumm\index{Gumm, H.~P.}  observed in~\cite{gumm} that
the term constructed by Herrmann has the stronger properties
that the identity $d(x,x,y)\approx y$ holds in~$\vr V$, and
$d(b,a,a) \equiv b\pmod{[\theta,\theta]}$ for any $\m a\in
{\vr V}$, $\theta\in \Con(\m a)$, and $(a,b)\in\theta$.
Such a term is now called a {\bf difference term}.
Difference terms play a central role in most developments of
modular commutator theory and in some extensions of the
theory to nonmodular varieties
(see~\cite{diff,lipparini3,lipparini2}).  Weak difference
terms were first identified in~\cite{hobby-mckenzie}.  They
occur far more frequently than ordinary difference terms,
and are equally useful in most circumstances.

\begin{exmps}
  As noted in Theorem~\ref{justify_weak}, any variety with a
  weak difference term satisfies a nontrivial idempotent
  Maltsev condition, and therefore has a Taylor term.  The
  converse is not true in general, as we explained in the
  first paragraph of this section, but the converse does
  hold for certain classes of varieties.  We describe some
  of these classes now, thereby providing a number of
  examples of varieties with weak difference terms.

  Theorem~9.6 of~\cite{hobby-mckenzie} proves the converse
  for locally finite varieties: that is, any locally finite
  variety that has a Taylor term has a weak difference term.

  Any idempotent abelian (or solvable) variety that has a
  Taylor term has a weak difference term.  (The proof of
  this is contained in the proof of
  Theorem~\ref{justify_weak}.)

  A variety of semigroups that has a Taylor term has a weak
  difference term.  Indeed, suppose that $f(x_1,\ldots,x_n)
  = x_{i_0}\cdots x_{i_k}$ is a Taylor term for~$\vr V$.
  Then necessarily $k\geq 1$. The idempotence of~$f$ implies
  that ${\vr V}\models x^{k+1}\approx x$.  Any semigroup $\m
  s$ satisfying this equation has the property that $(s^k)^2
  = s^{k-1}s^{k+1} =s^{k-1}s=s^k$ for any $s\in S$, so
  $k$-th powers of elements of~$\m s$ are idempotent.  If
  $e, f\in S$ are idempotent, and $ef\neq fe$, then it can
  be shown that the subsemigroup of~$\m s$ generated by
  $\{e,f\}$ has a homomorphism onto a nontrivial left zero
  or right zero semigroup.  But a nontrivial left zero
  semigroup fails the $i_0$-th Taylor identity for
  $f(x_1,\ldots,x_n) = x_{i_0}\cdots x_{i_k}$, and a
  nontrivial right zero semigroup fails the $i_k$-th Taylor
  identity.  It follows that $\vr V$ has no nontrivial left
  or right zero semigroups, and hence that no $\m s\in {\vr
    V}$ has a pair of noncommuting idempotents.  This fact
  may be expressed by saying that the identity
  $x^ky^k\approx y^kx^k$ holds in~$\vr V$. Now we argue that
  the term $d(x,y,z) = xy^{2k-1}z$ is a weak difference term
  for the variety of semigroups axiomatized by
  $x^{k+1}\approx x$ and $x^ky^k\approx y^kx^k$.  To justify
  this, it suffices to show that if $\m a$ is a semigroup in
  $\vr V$ and $a,b\in A$ are elements for which $\theta:=\Cg
  a(a,b)$ is abelian, then $d(a,a,b) = b = d(b,a,a)$.  To
  show that $b = d(a,a,b) = a^{2k}b = a^kb$ and $b = ba^k$
  it is enough to prove that $a^k = b^k$.  Since $e = a^k$
  and $f = b^k$ are idempotent and $(e,f) = (a^k,b^k)\in
  \theta$, it suffices to prove that idempotents related by
  an abelian congruence are equal.  The $\theta,\theta$-term
  condition applied to~$ee\underline{f} = ef =
  ef\underline{f}$ allows us to replace the underlined $f$
  with $e$ and obtain~$eee = efe$. Since idempotents
  commute, this is equivalent to~$e = ef$.  The same
  argument shows that $f = ef$, so $e=f$.

  It is easily seen that any congruence $n$-permutable
  variety satisfies a nontrivial idempotent Maltsev
  condition, hence has a Taylor term. Less obvious is the
  theorem of P.~Lipparini\index{Lipparini, P.}
  in~\cite{lipparini1} that any congruence $n$-permutable
  variety has a weak difference term.  (This fact can be
  derived from Corollary~\ref{justify_weak_cor}.)
\end{exmps}\index{weak~difference~term|)}


\section{$\infty$-Solvability}

We now begin the development of a solvability theory for
varieties with a weak difference term.  Any successor
ordinal, considered as a lattice under its usual ordering,
is a meet continuous lattice.  By a {\bf continuous sequence
  of congruences}\index{continuous sequence of congruences}
on an algebra $\m a$ we mean a meet continuous lattice
homomorphism $\sigma: {\kappa+1}\rightarrow \Conb(\m a)$. We
write such a sequence as
$(\sigma_{\lambda})_{\lambda<\kappa+1}$ where
$\sigma_{\lambda}:=\sigma(\lambda)$.  It is not hard to
verify that $(\sigma_{\lambda})_{\lambda<\kappa+1}$ is a
continuous sequence of congruences if and only if
$\sigma_{\mu}\leq \sigma_{\lambda}$ for~$\mu<\lambda$ and
$\sigma_{\lambda} = \bigcup_{\mu<\lambda} \sigma_{\mu}$ for
all limit ordinals $\lambda$.


\begin{df}\label{XXXdefn}
  Let $\m a$ be an algebra with congruences $\alpha$,
  $\beta$, $\theta$, $\theta_{\lambda}$, $\delta$
  and~$\delta'$.

  If $\delta\leq \theta$ and $\C c(\theta,\theta;\delta)$
  holds then we say that {\bf $\theta$ is abelian over
    $\delta$},\index{abelian!over} and write $\delta\lhd
  \theta$.

  If $\kappa$ is an ordinal, and
  $(\theta_{\lambda})_{\lambda<\kappa+1}$ is a continuous
  sequence of congruences such that $\theta_{\lambda}\lhd
  \theta_{\lambda+1}$ for all $\lambda < \kappa$, then we
  say that {\bf $\theta_{\kappa}$ is $\kappa$-step solvable
    over $\theta_0$}.\index{solvable!$\kappa$-step}

  We say that {\bf $\theta$ is $\XXX$-solvable
    over~$\delta$}\index{$\XXX$-solvable!over}\index{solvable!$\XXX$-}
  if $\theta$ is $\kappa$-step solvable over $\delta$ for
  some $\kappa$. We may denote this by~$\delta\lhdlhd
  \theta$.\index{$\lhdlhd$} If $0\lhdlhd \theta$, then we
  say that {\bf $\theta$~is
    $\XXX$-solvable}.\index{$\XXX$-solvable!congruence} If
  $0\lhdlhd 1$, then we say that {\bf $\m a$ is
    $\XXX$-solvable}.\index{$\XXX$-solvable!algebra}

  If $\alpha\wedge\beta\lhdlhd \alpha\vee\beta$, then we say
  that {\bf $\alpha$ is solvably related to
    $\beta$}\index{solvably related to} and denote this by
  $\alpha\solv \beta$.\index{$\solv$}

  If $\delta < \theta$, and $\delta\lhd \delta'\leq \theta$
  implies $\delta = \delta'$, then we call the congruence
  interval $I[\delta,\theta]$ a {\bf solvability
    obstruction}\index{solvability obstruction} and we
  denote this by~$\delta\blacktriangleleft
  \theta$.\index{$\blacktriangleleft$}
\end{df}

Note that for comparable congruences $\delta\leq\theta$ the
relations $\delta\lhdlhd \theta$ and $\delta\solv \theta$
are the same.

\begin{exmp}
  If $\vr V$ is a variety whose members have the property
  that their principal congruences are abelian, then every
  member of~$\vr V$ is $\XXX$-solvable.  For if $\m a\in
  {\vr V}$, $(a_{\lambda})_{\lambda<\kappa}$ is an
  enumeration of~$\m a$, and $\theta_{\lambda}$~is the
  congruence generated by~$\{(a_{0},a_{\mu})\mid
  \mu<\lambda\}$, then it is not hard to see that
  \begin{enumerate}
  \item[(i)] $\theta_0=0$ and $\theta_{\kappa} = 1$,
  \item[(ii)] $\theta_{\lambda}\lhd \theta_{\lambda+1}$ for
    all $\lambda < \kappa$, and
  \item[(iii)] $\theta_{\lambda} = \bigcup_{\mu<\lambda}
    \theta_{\mu}$ for all limit ordinals.
  \end{enumerate}
  Items (i) and (iii) follow immediately from the definition
  of~$\theta_{\lambda}$.  For~(ii), note that
  $\theta_{\lambda+1} = \Cg
  a(a_0,a_{\lambda})\vee\theta_{\lambda}$.  Thus
  $\theta_{\lambda+1}/\theta_{\lambda}$ is a principal
  (hence abelian) congruence on $\m
  a/\theta_{\lambda}$. This implies that
  $\theta_{\lambda}\lhd \theta_{\lambda+1}$ for all
  $\lambda<\kappa$.

  An example of a variety in which principal congruences are
  abelian is the variety $\vr V$ of nonunital rings
  satisfying $x^2 \approx 0$. This identity implies that
  $0\approx (x+y)^2 \approx x^2 + xy + yx + y^2\approx
  xy+yx$, so the anticommutative law $xy \approx -yx$ holds
  in~$\vr V$.  This law implies that, for any $\m r\in {\vr
    V}$, all principal ideals have the form $(a) = R\cdot
  a$. Hence $(a)^2 = RaRa = \sum \{ra\cdot sa = -rsa^2\mid
  r, s\in R\} = (0).$ Since congruences correspond to
  ideals, and the commutator operation corresponds to ideal
  multiplication, this is indeed a variety in which
  principal congruences are abelian.  Hence $\vr V$ consists
  entirely of~$\XXX$-solvable algebras.

  However, there is no cardinal bound on solvability degree
  in~$\vr V$.  In fact, if $\kappa$ is any infinite
  cardinal, then $\vr V$ contains rings of solvability
  degree exactly $\kappa$.  One such ring, $\m r_{\kappa}$,
  may be presented by generators and relations relative
  to~$\vr V$ as follows.  Let $S$ be an infinite set of
  cardinality~$\kappa$.  Take as generators for~$\m
  r_{\kappa}$ all symbols $x_U$ where $U$ is a nonempty
  subset of~$S$.  Impose the relations
  \begin{itemize}
  \item $2x_U = 0$ for all nonempty $U\subseteq S$,
  \item $x_U\cdot x_V = 0$ if $U\cap V \neq \emptyset$, and
  \item $x_U\cdot x_V = x_{U\cup V}$ if $U\cap V =
    \emptyset$.
  \end{itemize}
  These relations force each element of~$\m r_{\kappa}$ to
  be expressible as a finite sum of generators
  $x_{U_1}+\cdots+x_{U_n}$ where $U_i\neq U_j$ if $i\neq j$.
  We leave it as an exercise to prove that the ring $\m
  r_{\kappa}$ is $\kappa$-step solvable, but not
  $\lambda$-step solvable for any ordinal $\lambda<\kappa$.
\end{exmp}


\begin{lm}\label{tol=cong}
  Let\/ $\vr V$ be a variety with a weak difference term.
  If\/ $\m a\in {\vr V}$ has a tolerance $T$ and a
  congruence $\delta$ such that\/ $\C C(T,T;\delta)$ holds,
  then $T':=\delta\circ T\circ \delta$ is the smallest
  congruence on $\m a$ containing~$T\cup
  \delta$. Furthermore, $\C C(T',T';\delta)$ holds.
\end{lm}

\begin{proof} 
  We know from the basic properties of the centralizer
  relation (see Theorem~\ref{basic_centrality}) that if $\C
  C(T,T;\delta)$ holds and $T':=\delta\circ T\circ \delta$,
  then $\C C(T',T';\delta)$ holds.  Since $(T\cup
  \delta)\subseteq T' = \delta\circ T\circ \delta\subseteq
  \Cg a(T\cup\delta)$, the only thing left to show is that
  $T'$ is a congruence.  For this we only need to establish
  that $T'\circ T'\subseteq T'$.

  Let $\theta = \Cg a(T')$. Since $\C C(T',T';\delta)$
  holds, it follows from Theorems~\ref{abel_tol=abel_cong}
  and~\ref{justify_weak} that $\C C(\theta,\theta;\delta)$
  holds.  If $(a,c)\in T'\circ T'$, then there is an element
  $b$ such that $(a,b), (b,c)\in T'\subseteq \theta$.  Thus
  \[
  a\equiv_{[\theta,\theta]} d(a,b,b)\equiv_{T'} d(b,b,c)
  \equiv_{[\theta,\theta]} c
  \]
  and $[\theta,\theta]\leq\delta$ imply that $(a,c)\in
  \delta\circ T'\circ \delta = T'$.  Since $(a,c)\in T'\circ
  T'$ was chosen arbitrarily, $T'\circ T'\subseteq T'$.
\end{proof}

\begin{lm}\label{lift_delta}
  Let\/ $\vr V$ be a variety with a weak difference term.
  If\/ $\m a\in {\vr V}$ has congruences $\theta$ and
  $\delta$ such that\/ $[\theta,\theta]\leq \delta$, then\/
  $\C C(\theta,\theta;\delta)$ holds.
\end{lm}

\begin{proof} 
  Assume that $[\theta,\theta]\leq\delta$. If $\C
  C(\theta,\theta;\delta)$ does not hold, then there are a
  polynomial $p(\wec{x},\wec{y})$ and tuples $\wec{a}
  \wrel{\theta}\wec{b}$ and $\wec{u} \wrel{\theta}\wec{v}$
  such that
  \[
  p(\wec{a},\wec{u})\equiv_{\delta} p(\wec{a},\wec{v})
  \]
  and
  \begin{equation}\label{line_zero}
    p(\wec{b},\wec{u})\not\equiv_{\delta}
    p(\wec{b},\wec{v})\,.
  \end{equation}
  Let $p'(\wec{x},\wec{y}) =
  d\big(p(\wec{x},\wec{y}),p(\wec{x},\wec{u}),
  p(\wec{b},\wec{u})\big)$.  Using that $d$ is an idempotent
  term that is Maltsev on $\theta$-classes modulo
  $[\theta,\theta]$, we get that
  \[
  p'(\wec{a},\underline{\wec{u}}) =
  d\big(p(\wec{a},\wec{u}),
  p(\wec{a},\wec{u}),p(\wec{b},\wec{u})\big)
  \equiv_{[\theta,\theta]}
  p(\wec{b},\wec{u})=
  p'(\wec{b},\underline{\wec{u}})\,.
  \]
  Since $\C C(\theta,\theta;[\theta,\theta])$ holds, we can
  change the underlined $\wec{u}$ to~$\wec{v}$ without
  changing the fact that the left hand side is
  $[\theta,\theta]$-related to the right hand side:
  \begin{equation}\label{line_one}
    p'(\wec{a},\underline{\wec{v}}) 
    \equiv_{[\theta,\theta]}
    p'(\wec{b},\underline{\wec{v}})\,.
  \end{equation}
  Moreover we have that
  \begin{equation}\label{line_two}
    \begin{array}{r@{}l}
      p'(\wec{a},\wec{v}) &{} = 
      d\big(p(\wec{a},\wec{v}),p(\wec{a},\wec{u}),
      p(\wec{b},\wec{u})\big)\\
      &{} \equiv_{\delta}
      d\big(p(\wec{a},\wec{u}),p(\wec{a},\wec{u}),
      p(\wec{b},\wec{u})\big)
      \equiv_{[\theta,\theta]}
      p(\wec{b},\wec{u})\,.
    \end{array}
  \end{equation}
  Here
  \[
  d\big(p(\wec{a},\wec{v}),p(\wec{a},\wec{u}),
  p(\wec{b},\wec{u})\big)
  \equiv_{\delta}
  d\big(p(\wec{a},\wec{u}),p(\wec{a},\wec{u}),
  p(\wec{b},\wec{u})\big)
  \]
  is a consequence of the fact that
  $p(\wec{a},\wec{v})\equiv_{\delta} p(\wec{a},\wec{u})$,
  and
  \[
  d\big(p(\wec{a},\wec{u}),p(\wec{a},\wec{u}),
  p(\wec{b},\wec{u})\big)
  \equiv_{[\theta,\theta]} p(\wec{b},\wec{u})
  \]
  is a consequence of the facts that
  $\wec{a}\wrel{\theta}\wec{b}$ and that $d$ is Maltsev on
  $\theta$-classes modulo $[\theta,\theta]$.  We also have
  that
  \begin{equation}\label{line_three}
    p'(\wec{b},\wec{v}) = 
    d\big(p(\wec{b},\wec{v}),p(\wec{b},\wec{u}),
    p(\wec{b},\wec{u})\big)
    \equiv_{[\theta,\theta]}
    p(\wec{b},\wec{v})\,,
  \end{equation}
  since $\wec{u}\wrel{\theta}\wec{v}$ and $d$ is Maltsev on
  $\theta$-classes modulo $[\theta,\theta]$.  Combining
  lines (\ref{line_one}), (\ref{line_two}), and
  (\ref{line_three}) yields
  \[
  p(\wec{b},\wec{u}) \equiv_{\delta} p'(\wec{a},\wec{v})
  \equiv_{[\theta,\theta]} p'(\wec{b},\wec{v})
  \equiv_{[\theta,\theta]} p(\wec{b},\wec{v})\,,
  \]
  or simply
  $p(\wec{b},\wec{u})\equiv_{\delta}p(\wec{b},\wec{v})$
  (since $[\theta,\theta]\leq\delta$).  This contradicts
  line (\ref{line_zero}), thereby completing the proof.
\end{proof}

The following corollary will be used in Section~\ref{ACD}.

\begin{cor}\label{ACD_cor2}
  Let\/ $\vr V$ be a variety with a weak difference term.
  Assume that $T$ is an abelian tolerance on some $\m a\in
  {\vr V}$, and that $a$ and~$b$ are distinct $T$-related
  elements.  If $\rho$ is a congruence on $\m a$ that is
  maximal with respect to~$(a,b)\notin\rho$, then $T/\rho$
  is a nontrivial abelian congruence on the subdirectly
  irreducible algebra $\m a/\rho$.
\end{cor}

\begin{proof}
  It follows from Lemma~\ref{tol=cong} (using $\delta = 0$)
  that $T$ is an abelian congruence on $\m a$.  Then, using
  Lemma~\ref{lift_delta} in the case where $\theta = T$ and
  $\delta = \rho$, we obtain that $\C C(T,T;\rho)$ holds, so
  $T/\rho$ is a nontrivial abelian tolerance on the
  subdirectly irreducible algebra $\m a/\rho$. Using
  Lemma~\ref{tol=cong}, we find that $T/\rho$ is a
  congruence.
\end{proof}


\begin{lm}\label{meet+join}
  Let\/ $\vr V$ be a variety with a weak difference term.
  Assume that\/ $\m a\in {\vr V}$ has congruences $\alpha$,
  $\beta$, and $\gamma$.
  \begin{enumerate}
  \item[(1)] If\/ $\alpha\wrel{\lhd} \beta$, then
    $(\alpha\wedge\gamma)\wrel{\lhd} (\beta\wedge\gamma)$
    and\/ $(\alpha\vee\gamma)\wrel{\lhd} (\beta\vee\gamma)$.
  \item[(2)] If\/ $\alpha\lhdlhd \beta$, then
    $(\alpha\wedge\gamma)\lhdlhd (\beta\wedge\gamma)$ and\/
    $(\alpha\vee\gamma)\lhdlhd (\beta\vee\gamma)$.
  \end{enumerate}
\end{lm}

\begin{proof}
  To prove the part of item (1) that concerns the
  compatibility of~$\lhd$ with meet, recall that $\alpha\lhd
  \beta$ means that $\C C(\beta,\beta;\alpha)$ holds.  By
  item~(1) of Theorem~\ref{basic_centrality} we have that
  $\C C(\beta\wedge\gamma,\beta\wedge\gamma;\alpha)$ holds.
  By item~(8) of Theorem~\ref{basic_centrality} we have that
  $\C C(\beta\wedge\gamma,\beta\wedge\gamma;\gamma)$
  holds. Item~(6) of Theorem~\ref{basic_centrality} now
  yields that $\C
  C(\beta\wedge\gamma,\beta\wedge\gamma;\alpha\wedge\gamma)$
  holds, which means that $(\alpha\wedge\gamma)\lhd
  (\beta\wedge\gamma)$.

  To prove the compatibility of~$\lhd$ with join, assume
  that $\C C(\beta,\beta;\alpha)$ holds.  Then we have
  $[\beta,\beta]\leq\alpha\leq\alpha\vee\gamma$, so by
  Lemma~\ref{lift_delta} we get that $\C
  C(\beta,\beta;\alpha\vee\gamma)$ holds.  Using
  Lemma~\ref{tol=cong} with $T = \beta$ and $\delta =
  \alpha\vee\gamma$ we deduce that
  $(\alpha\vee\gamma)\circ\beta\circ(\alpha\vee\gamma) =
  \beta\vee(\alpha\vee\gamma) = \beta\vee\gamma$ and that
  $\C C(\beta\vee\gamma,\beta\vee\gamma;\alpha\vee\gamma)$
  holds. This shows that $(\alpha\vee\gamma)\lhd
  (\beta\vee\gamma)$, so we proved item~(1).

  For (2), assume that
  $(\theta_{\lambda})_{\lambda<\kappa+1}$ is a continuous
  sequence of congruences for which $\alpha = \theta_0$,
  $\beta = \theta_{\kappa}$ and $\theta_{\lambda}\lhd
  \theta_{\lambda+1}$ for all $\lambda$.  The sequences
  $(\theta_{\lambda}\wedge \gamma)_{\lambda<\kappa+1}$ and
  $(\theta_{\lambda}\vee \gamma)_{\lambda<\kappa+1}$ are
  also continuous, since meeting or joining with the fixed
  element $\gamma$ preserves order and unions of chains, so
  item (1) of this lemma implies that
  $(\theta_{\lambda}\wedge \gamma)_{\lambda<\kappa+1}$ and
  $(\theta_{\lambda}\vee \gamma)_{\lambda<\kappa+1}$ are
  sequences witnessing $(\alpha\wedge\gamma)\lhdlhd
  (\beta\wedge\gamma)$ and $(\alpha\vee\gamma)\lhdlhd
  (\beta\vee\gamma)$, respectively.
\end{proof}


\begin{thm}\label{solv_is_join_comp}
  Let\/ $\vr V$ be a variety with a weak difference term.
  If\/ $\m a\in {\vr V}$, then $\solv$ is a congruence on\/
  $\Conb(\m a)$ that is compatible with the complete join
  operation.
\end{thm}

\begin{proof}
  Recall that $\alpha\solv \beta$ holds if and only if
  $(\alpha\wedge\beta)\lhdlhd (\alpha\vee\beta)$.  It is
  easy to see that this relation is reflexive and symmetric.
  To show that it is transitive, assume that $\alpha\solv
  \beta\solv \gamma$.  Then $(\alpha\wedge\beta)\lhdlhd
  (\alpha\vee\beta)$ and $(\beta\wedge\gamma)\lhdlhd
  (\beta\vee\gamma)$, so by the second statement of
  Lemma~\ref{meet+join} we get that
  \begin{align*}
    (\alpha\wedge\beta\wedge\gamma) =
    (\alpha\wedge\beta)\wedge(\beta\wedge\gamma) 
    &{} \lhdlhd
    (\alpha\wedge\beta)\wedge(\beta\vee\gamma) =
    (\alpha\wedge\beta)\\
    \noalign{\noindent\mbox{and}}
    (\alpha\vee\beta) =
    (\alpha\vee\beta)\vee(\beta\wedge\gamma) 
    &{} \lhdlhd
    (\alpha\vee\beta)\vee(\beta\vee\gamma) =
    (\alpha\vee\beta\vee\gamma)\,.
  \end{align*}
  Thus we have $(\alpha\wedge\beta\wedge\gamma) \lhdlhd
  (\alpha\wedge\beta) \lhdlhd (\alpha\vee\beta) \lhdlhd
  (\alpha\vee\beta\vee\gamma)$. The well-ordered chains of
  congruences that witness these three instances of
  $\XXX$-solvability may be concatenated to produce a
  single such chain from $\alpha\wedge\beta\wedge\gamma$ to
  $\alpha\vee\beta\vee\gamma$, thus proving that
  $(\alpha\wedge\beta\wedge\gamma)\lhdlhd
  (\alpha\vee\beta\vee\gamma)$.  Since
  $(\alpha\wedge\beta\wedge\gamma)\leq
  \alpha\wedge\gamma\leq\alpha\vee\gamma\leq
  (\alpha\vee\beta\vee\gamma)$, we can deduce that
  $(\alpha\wedge\gamma)\lhdlhd (\alpha\vee\gamma)$ if we
  show first that $\delta\leq \delta'\leq\theta'\leq\theta$
  and~$\delta\lhdlhd \theta$ imply~$\delta'\lhdlhd \theta'$.
  We show this by using Lemma~\ref{meet+join} again: joining
  $\delta\lhdlhd \theta$ with~$\delta'$ we obtain
  \[
  \delta' = (\delta' \vee \delta) \lhdlhd (\delta' \vee
  \theta) = \theta\,.
  \]
  Meeting with $\theta'$ we find that
  \[
  \delta' =
  (\delta' \wedge \theta') \lhdlhd
  (\theta\wedge\theta')=\theta'\,.
  \]
  This reasoning completes the argument that $\solv$ is
  transitive, hence an equivalence relation, and also shows
  that subintervals of~$\XXX$-solvable intervals are
  again~$\XXX$-solvable.  Hence $\solv$\,-classes are convex
  sublattices of~$\Conb(\m a)$.  This implies that
  $\alpha\solv \beta$ if and only if $\alpha$ and $\beta$
  belong to some $\XXX$-solvable interval.  This
  reformulation of the definition of the $\solv$\,-relation
  makes it easy to see that it is a congruence: $\alpha\solv
  \beta$ if and only if $\delta:= \alpha\wedge\beta\lhdlhd
  \alpha\vee\beta =:\theta$.  But $\delta\lhdlhd \theta$
  implies $\delta\wedge\gamma\lhdlhd \theta\wedge\gamma$ and
  $\delta\vee\gamma\lhdlhd \theta\vee\gamma$ by
  Lemma~\ref{meet+join}.  Since $\alpha\wedge\gamma,
  \beta\wedge\gamma\in
  I[\delta\wedge\gamma,\theta\wedge\gamma]$ and
  $\alpha\vee\gamma, \beta\vee\gamma\in
  I[\delta\vee\gamma,\theta\vee\gamma]$, we conclude that
  $\alpha\solv \beta$ implies that
  $(\alpha\wedge\gamma)\solv (\beta\wedge\gamma)$ and
  $(\alpha\vee\gamma)\solv (\beta\vee\gamma)$.

  It remains to show that $\solv$ is compatible with the
  complete join operation of~$\Conb(\m a)$.  Suppose that
  $\alpha_i\solv \beta_i$ for all $i\in I$. Let ${\delta_i =
    \alpha_i\wedge\beta_i}$ and $\theta_i =
  \alpha_i\vee\beta_i$.  Then $\delta_i\lhdlhd \theta_i$ for
  all $i$. If $\Delta:=\bigvee_{i\in I}\delta_i$ and
  $\Theta:=\bigvee_{i\in I}\theta_i$, then both
  $\bigvee_{i\in I}\alpha_i$ and $\bigvee_{i\in I}\beta_i$
  belong to the interval $I[\Delta,\Theta]$.  We will be
  done if we show that $\Theta$ is $\XXX$-solvable over
  $\Delta$.

  Consider the collection $\mathcal P$ of all sequences of
  the form $\sigma = (\sigma_{\lambda})_{\lambda<\kappa+1}$
  where $\sigma_{\lambda}\in\Con(\m a)$ and
  \begin{enumerate}
  \item[(0)] $\sigma_0 = \Delta$, $\sigma_{\kappa}\in
    I[\Delta,\Theta]$,
  \item[(1)] $\sigma_{\lambda}\lhd \sigma_{\lambda+1}$ for
    all $\lambda < \kappa$, and
  \item[(2)] $\sigma_{\lambda} = \bigcup_{\mu<\lambda}
    \sigma_{\mu}$ for limit ordinals $\lambda$.
  \end{enumerate}
  The elements of~$\mathcal P$ are the continuous sequences
  of congruences that witness the $\XXX$-solvability over
  $\Delta$ of some elements $\sigma_{\kappa}\in
  I[\Delta,\Theta]$.  The set $\mathcal P$ is nonempty,
  since it contains the one-term sequence $(\Delta)$.
  $\mathcal P$~can be partially ordered by the rule:
  $\sigma\leq \sigma'$ if $\sigma'$ is an extension
  of~$\sigma$.  That is, $\sigma\leq \sigma'$ if the domain
  of~$\sigma$ is contained in the domain of~$\sigma'$ and
  the functions agree on their common domain.  This is an
  inductive ordering of~$\mathcal P$, as we now explain.
  Assume that $(\sigma^{\mu})_{\mu<\nu}$ is an increasing
  sequence of sequences in~$\mathcal P$.  Define $\Sigma'$
  to be the union of the functions~$\sigma^{\mu}$.  Then
  $\Sigma'$~is a function from the ordinal $\rho:=
  \bigcup_{\mu<\nu} {\rm dom}(\sigma^{\mu})$ into~$\Conb(\m
  a)$ that preserves all existing joins.  $\Sigma'$ may not
  be in~$\mathcal P$ because elements of~$\mathcal P$ are
  functions from successor ordinals into~$\Conb(\m a)$ and
  $\rho$ may be a limit ordinal.  But if it is, then there
  is a unique meet continuous extension of~$\Sigma'$
  to~$\rho+1$: it is the sequence $\Sigma =
  (\Sigma_{\lambda})_{\lambda<\rho+1}$ where
  ${(\Sigma_{\lambda})_{\lambda<\rho} = \Sigma'}$ and
  $\Sigma_{\rho}$ is defined to be $\bigcup_{\lambda<\rho}
  \Sigma_{\lambda}$.  The facts that $\Sigma$ is a meet
  continuous homomorphism from a successor ordinal
  into~$\Conb(\m a)$, $\Sigma$~extends each~$\sigma^{\mu}$,
  and $\Sigma_{\lambda} = \sigma^{\mu}_{\lambda}\lhd
  \sigma^{\mu}_{\lambda+1} = \Sigma_{\lambda+1}$ for some
  $\mu$ for each $\lambda<\rho$ imply that $\Sigma$ is an
  upper bound in~$\mathcal P$ for all the
  sequences~$\sigma^{\mu}$.

  Since $\mathcal P$ is nonempty and inductively ordered,
  Zorn's lemma applies to show that $\mathcal P$ has a
  maximal element $(\psi_{\lambda})_{\lambda<\kappa+1}$.
  The final congruence in this sequence, $\psi_{\kappa}$, is
  $\XXX$-solvable over $\Delta$ and is a maximal congruence
  in the interval $I[\Delta,\Theta]$ among those that are
  $\XXX$-solvable over $\Delta$.  For each $i\in I$ we have
  $\delta_i\lhdlhd \theta_i$ and $\delta_i\leq
  \Delta\leq\psi_{\kappa}$, so by Lemma~\ref{meet+join},
  \[
  \psi_{\kappa} = (\psi_{\kappa}\vee\delta_i) \lhdlhd
  (\psi_{\kappa}\vee\theta_i)\leq \Theta\,.
  \]
  The maximality condition on $\psi_{\kappa}$ implies that
  $\psi_{\kappa} = \psi_{\kappa}\vee\theta_i$, or
  $\psi_{\kappa}\geq \theta_i$. Since this is true for all
  $i\in I$, we get that $\psi_{\kappa}\geq \bigvee_{i\in I}
  \theta_i = \Theta$.  This proves that $\Delta\lhdlhd
  \psi_{\kappa}=\Theta$, as desired.
\end{proof}

\begin{df}
  Assume that $\vr V$ is a variety with a weak difference
  term, and $\m a\in {\vr V}$. If $\alpha\in\Con(\m a)$,
  then
  \[
  \rad(\alpha) := \bigvee \{\beta\in\Con(\m a)\mid
  \alpha\solv \beta\}
  \]
  is called the {\bf $\XXX$-solvable radical
    of~$\alpha$}.\index{rad@$\rad(\alpha)$} The {\bf
    $\XXX$-solvable radical of~$\m
    a$}\index{$\XXX$-solvable!radical}\index{solvable radical!$\XXX$--} 
  is~$\rad(0)$. 
  A {\bf radical congruence}\index{radical congruence} 
  is a congruence~$\alpha$ of $\m a$ such that 
  ${\rad(\alpha) = \alpha}$.
\end{df}

\begin{thm}\label{rad_stuff}
  Let\/ $\vr V$ be a variety with a weak difference term.
  Assume that\/ $\m a\in{\vr V}$ and that\/
  $\alpha,\beta\in\Con(\m a)$. Then
  \begin{enumerate}
  \item[(1)] $\rad(\alpha)$ is the largest congruence
    $\solv$\,-related to~$\alpha$.
  \item[(2)] $\rad\big(\rad(\alpha)\big) = \rad(\alpha)$.
  \item[(3)] $\rad(\alpha\wedge\beta) =
    \rad(\alpha)\wedge\rad(\beta)$.
  \item[(4)] $\alpha\obst \beta$ if and only if $\alpha
    <\beta$ and $\alpha = \beta\wedge\rad(\alpha)$.
  \item[(5)] If\/ ${\rm Rad}(\m a)$\index{Rad@${\rm Rad}(\m
      a)$} denotes the meet subsemilattice of\/~$\Conb(\m
    a)$ consisting of the radical congruences, then the
    restriction of the natural homomorphism
    \[
    \nu:\Conb(\m a)\rightarrow~\Conb(\m a)/{\solv}:
    \alpha\mapsto \alpha/{\solv}
    \]
    to\/~${\rm Rad}(\m a)$ is an isomorphism onto~$\Conb(\m
    a)/{\solv}$\,.
  \item[(6)] The natural homomorphism $\nu$ from item (5) is
    meet continuous, and its image $\Conb(\m a)/{\solv}$
    is meet semidistributive.
  \end{enumerate}
\end{thm}

\begin{proof}
  For item (1), the fact that $\solv$ is compatible with the
  complete join operation implies that the join of all
  congruences $\solv$\,-related to~$\alpha$ is
  $\solv$\,-related to~$\alpha$, and it is necessarily the
  largest congruence $\solv$\,-related to~$\alpha$.

  For (2), $\alpha\solv \rad(\alpha)$ and $\rad(\alpha)\solv
  \rad\big(\rad(\alpha)\big)$ hold by (1), so $\alpha\solv
  \rad\big(\rad(\alpha)\big)$.  Since $\rad(\alpha)$ is the
  largest congruence $\solv$\,-related to~$\alpha$, it
  follows that $\rad\big(\rad(\alpha)\big)\leq
  \rad(\alpha)$.  The reverse inclusion follows from the
  facts that $\rad(\alpha)\solv \rad(\alpha)$ and
  $\rad\big(\rad(\alpha)\big)$ is the largest congruence
  $\solv$\,-related to~$\rad(\alpha)$.

  For (3), $\alpha\solv \rad(\alpha)$ and $\beta\solv
  \rad(\beta)$, so $\alpha\wedge\beta\solv
  \rad(\alpha)\wedge\rad(\beta)$ since $\solv$ is a
  congruence.  Since $\rad(\alpha\wedge\beta)$ is the
  largest congruence $\solv$\,-related to~$\alpha\wedge\beta$
  we get that $\rad(\alpha)\wedge\rad(\beta)\leq
  \rad(\alpha\wedge\beta)$.  On the other hand,
  $\alpha\wedge\beta\solv \rad(\alpha\wedge\beta)$, so
  \[
  \alpha = \alpha\vee(\alpha\wedge\beta) \solv
  \alpha\vee\rad(\alpha\wedge\beta)\leq \rad(\alpha)
  \]
  from the definition of~$\rad(\alpha)$.  Thus,
  $\rad(\alpha\wedge\beta)\leq \rad(\alpha)$, and similarly
  $\rad(\alpha\wedge\beta)\leq \rad(\beta)$, so
  $\rad(\alpha\wedge\beta)\leq
  \rad(\alpha)\wedge\rad(\beta)$.  This completes the proof
  of (3).

  For (4), assume that $\alpha \obst \beta$.  Then
  $\alpha<\beta$ according to the definition of~$\obst$. The
  congruence $\alpha' := \beta\wedge\rad(\alpha)$ satisfies
  $\alpha\leq \alpha'\leq \rad(\alpha)$ and is
  in~$I[\alpha,\beta]$, so it is a congruence in
  $I[\alpha,\beta]$ that is $\XXX$-solvable over
  $\alpha$. If this fact is witnessed by a chain of
  congruences $(\theta_{\lambda})_{\lambda<\kappa+1}$
  starting at $\theta_0 = \alpha$ and ending at
  $\theta_{\kappa} = \alpha'$, and satisfying the conditions
  of Definition~\ref{XXXdefn}, then using the fact that
  $\alpha\obst \beta$ and induction it follows that
  $\theta_{\lambda}=\alpha$ for all $\lambda$, so $\alpha =
  \alpha'= \beta\wedge\rad(\alpha)$.  Conversely, if
  $\alpha\obst \beta$ fails to hold and $\alpha < \beta$,
  then there is a congruence $\alpha'\in I[\alpha,\beta]$
  such that $\alpha\lhd\alpha'$ and $\alpha<\alpha'$. Since
  $\alpha\solv \alpha'$ we get that $\alpha < \alpha'\leq
  \beta\wedge\rad(\alpha)$, so $\alpha =
  \beta\wedge\rad(\alpha)$ fails to hold.

  For (5), it is clear that $\nu:{\rm Rad}(\m
  a)\rightarrow~\Conb(\m a)/{\solv}$ is an order-preserving
  function between meet semilattices. To prove that it is an
  isomorphism it suffices to show that it is bijective, or
  equivalently that the elements of~${\rm Rad}(\m a)$ form a
  transversal for the congruence~$\solv$\,.  The fact that
  $\alpha\solv \rad(\alpha)$ (proved in (1)) together with
  the fact that $\rad(\alpha)\in {\rm Rad}(\m a)$ (an
  equivalent form of (2)) shows that every
  \hbox{$\solv$\,-class} contains a radical congruence.  But
  we proved in (1) that every radical element is the largest
  element of its $\solv$\,-class, so each $\solv$\,-class
  has at most one radical congruence.  This shows that the
  elements of~${\rm Rad}(\m a)$ form a transversal
  for~$\solv$\,.

  For (6), the fact that $\solv$ is compatible with the
  complete join operation implies that it is a meet
  continuous congruence, hence the quotient modulo this
  congruence is meet continuous and the natural homomorphism
  $\nu:\Conb(\m a)\rightarrow \Conb(\m a)/{\solv}$ is a meet
  continuous homomorphism.

  To prove that $\Conb(\m a)/{\solv}$ is meet
  semidistributive, suppose that
  $(\alpha/{\solv})\wedge(\beta/{\solv}) =
  (\alpha/{\solv})\wedge(\gamma/{\solv}) =
  (\delta/{\solv})$.  From item (5) we get that
  $\rad(\alpha)\wedge\rad(\beta) =
  \rad(\alpha)\wedge\rad(\gamma) = \rad(\delta)$ in~${\rm
    Rad}(\m a)$.  Since ${\rm Rad}(\m a)$ is a meet
  subsemilattice of~$\Conb(\m a)$, these equalities hold
  in~$\Conb(\m a)$.  Theorem~\ref{basic_centrality}~(8)
  implies that $\C C\big(\rad(\beta),\, \rad(\alpha);\,
  \rad(\delta)\big)$ and $\C C\big(\rad(\gamma),\,
  \rad(\alpha);\, \rad(\delta)\big)$ hold.
  Theorem~\ref{basic_centrality}~(5) now yields that $\C
  C\big(\rad(\beta)\vee\rad(\gamma),\, \rad(\alpha);\,
  \rad(\delta)\big)$ holds.  If
  \[
  \theta :=
  \rad(\alpha)\wedge
  \big(\rad(\beta)\vee\rad(\gamma)\big)\,,
  \]
  then $\theta\geq \rad(\delta)$ and $\C C\big(\theta,
  \theta;\rad(\delta)\big)$ holds by
  Theorem~\ref{basic_centrality}~(1).  This shows that
  $\rad(\delta)\lhd\theta$.  But $\rad(\delta)\obst 1$ or
  else $\rad(\delta)=1$, by items (2) and (4) of this
  theorem, and in either case we get
  \[
  \rad(\delta) = \theta =
  \rad(\alpha)\wedge
  \big(\rad(\beta)\vee\rad(\gamma)\big)\,.
  \]
  Using the isomorphism proved in (5) once again, we
  conclude that $(\alpha/{\solv})\wedge
  \big((\beta/{\solv})\vee (\gamma/{\solv})\big) =
  (\delta/{\solv})$.  This proves that $\Conb(\m a)/{\solv}$
  is meet semidistributive.
\end{proof}

\begin{thm}\label{solv=omit_obst}
  Let\/ $\vr V$ be a variety with a weak difference term.
  If\/ $\m a\in {\vr V}$ has congruences $\delta\leq\theta$,
  then $\theta$ is $\XXX$-solvable over $\delta$ if and only
  if the interval $I[\delta,\theta]$ contains no solvability
  obstruction.
\end{thm}

\begin{proof}
  Assume first that $\theta$ is $\XXX$-solvable over
  $\delta$, and that $I[\alpha,\beta]$ is an arbitrarily
  chosen but nontrivial subinterval of~$I[\delta,\theta]$.
  Then ${\delta = \delta\wedge\alpha\solv \theta\wedge\alpha
    = \alpha}$, so $\rad(\alpha) =
  \rad(\delta)\geq\theta\geq\beta$.  This shows that $\alpha
  < \beta = \beta\wedge\rad(\alpha)$.  By item (4) of
  Theorem~\ref{rad_stuff}, $I[\alpha,\beta]$ is not a
  solvability obstruction.

  Conversely, assume that $I[\delta,\theta]$ contains no
  solvability obstruction.  Let $\alpha =
  \theta\wedge\rad(\delta)$.  Then since
  $\delta\leq\alpha\leq\rad(\delta)$ we get that
  $\delta\solv \alpha$, and hence that $\rad(\alpha) =
  \rad(\delta)$.  This means that $\alpha =
  \theta\wedge\rad(\alpha)$.  According to item (4) of
  Theorem~\ref{rad_stuff}, either $\alpha\obst \theta$ or
  else $\alpha = \theta$.  The first possibility is ruled
  out by the assumption that $I[\delta,\theta]$ contains no
  solvability obstruction, so it must be that $\alpha =
  \theta$. Since $\delta\solv \alpha = \theta$, the proof is
  complete.
\end{proof}

\begin{lm}\label{weak_permute} 
  Let\/ $\vr V$ be a variety with a weak difference term.
  Assume that\/ $\m a\in {\vr V}$ has congruences $\gamma$,
  $\gamma'$, $\delta$ and $\delta'$ such that $\gamma'\lhd
  \gamma$ and $\delta'\lhd \delta$. Then
  $\gamma\circ\delta\subseteq
  \gamma'\circ\delta\circ\gamma\circ\delta'$.
\end{lm}

\begin{proof}
  Choose any $(a,c)\in \gamma\circ\delta$. There is a $b\in
  A$ such that $a\equiv_{\gamma} b\equiv_{\delta} c$, so
  \[
  a\equiv_{\gamma'} d(a,b,b) \equiv_{\delta} d(a,b,c)
  \equiv_{\gamma} d(b,b,c) \equiv_{\delta'} c
  \]
  shows that $(a,c)\in
  \gamma'\circ\delta\circ\gamma\circ\delta'$.
\end{proof}

\begin{thm}\label{sim_cong} 
  Let\/ $\vr V$ be a variety with a weak difference term.
  If\/ $\m a\in {\vr V}$, then $\solv$\,-classes
  of\/~$\Conb(\m a)$ consist of permuting congruences.  In
  particular, these classes are modular sublattices.
\end{thm}

\begin{proof} 
  We assume that the theorem is false and argue to a
  contradiction.  If some $\solv$\,-class of~$\Conb(\m a)$
  contains a pair of nonpermuting congruences, then there is
  a subinterval $I[\sigma,\tau]$ of that $\solv$\,-class
  that contains the two congruences.  (For example, we could
  take $\sigma$ to be the meet and $\tau$ to be the join of
  the two nonpermuting congruences.)  Let $A$~denote the
  ordered set of all congruences $\theta\in I[\sigma,\tau]$
  such that every congruence in~$I[\sigma,\theta]$ permutes
  with every congruence in~$I[\sigma,\tau]$:
  \[
  A = \big\{\theta\in I[\sigma,\tau]\mid
  \forall\theta'\in I[\sigma,\theta],\,
  \forall\psi\in I[\sigma,\tau]\;
  (\theta'\circ \psi = \psi\circ\theta')\big\}\,.
  \]
  $A$ is nonempty (since $\sigma\in A$) and it is closed
  under updirected unions, therefore $A$ has maximal
  elements.  Let $\alpha$ be a maximal element of~$A$.  It
  must be that $\alpha < \tau$ since otherwise all
  congruences in~$I[\sigma,\tau]$ permute, contrary to
  assumption.  Since $I[\sigma,\tau]$ is $\XXX$-solvable
  and $\sigma\leq \alpha < \tau$, there is a congruence
  $\widehat{\alpha}$ such that $\alpha < \widehat{\alpha}
  \leq \tau$ and $\alpha\lhd \widehat{\alpha}$.  The set
  \[
  B = \big\{\theta\in I[\sigma,\tau]\mid
  \forall\theta'\in I[\sigma,\theta],\,
  \forall\psi\in I[\sigma,\widehat{\alpha}]\;
  (\theta'\circ \psi = \psi\circ\theta')\big\}
  \]
  of all $\theta\in I[\sigma,\tau]$ such that every
  congruence in~$I[\sigma,\theta]$ permutes with every
  congruence in~$I[\sigma,\widehat{\alpha}]$ is also
  nonempty and closed under updirected unions, so $B$ has a
  maximal element $\beta$.  We cannot have $\beta = \tau$
  since this would force $\widehat{\alpha}\in A$,
  contradicting the maximality of~$\alpha$.  Since $\beta <
  \tau$, there is a congruence $\widehat{\beta}$ such that
  $\beta < \widehat{\beta}\leq \tau$ and $\beta\lhd
  \widehat{\beta}$.

  By the maximality of~$\beta$ and the fact that $\beta <
  \widehat{\beta}$, there is some $\delta\leq
  \widehat{\beta}$ which fails to permute with some
  $\gamma\leq \widehat{\alpha}$.  Every congruence
  below~$\beta$ permutes with every congruence below
  $\widehat{\alpha}$, so $\delta\not\leq \beta$. Every
  congruence below $\alpha$ permutes with every congruence
  in~$I[\sigma,\tau]$, so $\gamma\not\leq \alpha$.
  Therefore $\gamma' = \alpha\wedge\gamma$ is strictly less
  than $\gamma$ and $\delta' = \beta\wedge\delta$ is
  strictly less than~$\delta$.  The four congruences
  $\gamma$, $\gamma'$, $\delta$ and $\delta'$ have the
  following properties:
  \begin{enumerate}
  \item[(i)] $\gamma' < \gamma\leq \widehat{\alpha}$,\;
    $\gamma'\leq \alpha$;
  \item[(ii)] $\delta' < \delta\leq \widehat{\beta}$,\;
    $\delta'\leq \beta$;
  \item[(iii)] $\gamma'\lhd \gamma$, $\delta'\lhd \delta$;
    and
  \item[(iv)] any two congruences in the four element
    set~$\{\gamma,\gamma',\delta,\delta'\}$ permute, except
    $\gamma\circ\delta\not\subseteq \delta\circ\gamma$.
  \end{enumerate}
  (The first, second and fourth items follow from the way we
  chose the congruences $\alpha$, $\widehat{\alpha}$,
  $\beta$, $\widehat{\beta}$, $\gamma$, $\gamma'$, $\delta$,
  and $\delta'$.  The third item follows from
  Lemma~\ref{meet+join} since, for example, $\alpha\lhd
  \widehat{\alpha}$ so $\gamma' = \alpha\wedge \gamma\lhd
  \widehat{\alpha}\wedge \gamma = \gamma$.)  By item (iii)
  and Lemma~\ref{weak_permute} we have
  $\gamma\circ\delta\subseteq
  \gamma'\circ\delta\circ\gamma\circ\delta'$.  But by items
  (i), (ii) and (iv) we have
  \[
  \gamma'\circ\delta\circ\gamma\circ\delta' =
  \delta\circ\gamma'\circ\gamma\circ\delta' =
  \delta\circ\gamma\circ\delta' =
  \delta\circ\delta'\circ\gamma = \delta\circ\gamma\,.
  \]
  Thus $\gamma\circ\delta\subseteq \delta\circ\gamma$, in
  contradiction with (iv). This is a contradiction to the
  assumption that $I[\sigma,\tau]$ contains a pair of
  nonpermuting congruences.

  For the final statement of the theorem, recall that any
  lattice of permuting equivalence relations is modular.
\end{proof}

This result may be used to establish a new characterization
of varieties with a weak difference term.  In the following
theorem, a lattice~$\m l$ has an $\sdm/\Mod$ {\bf
  factorization}\index{factorization} in a category $\vr C$
of lattices if there is a $\vr C$-morphism of~$\m l$ onto a
meet semidistributive lattice whose kernel classes are
modular.

\begin{thm}\label{HMa}
  The following are equivalent for a variety\/ $\vr V$.
  \begin{enumerate}
  \item[(1)] $\vr V$ has a weak difference term.
  \item[(2)] For every\/ $\m a\in {\vr V}$,\; $\Conb(\m a)$
    has an $\sdm/\Mod$ factorization in the category of
    meet continuous lattices.
  \item[(3)] For every\/ $\m a\in {\vr V}$,\; $\Conb(\m a)$
    has an $\sdm/\Mod$ factorization in the category of
    ordinary lattices.
  \item[(4)] For every\/ $\m a\in {\vr V}$, each\/
    $\sdm$-failure in~$\Conb(\m a)$ is modular.
  \item[(5)] If\/ $\m l$ belongs to the meet continuous
    congruence variety of\/~$\vr V$, then each\/
    $\sdm$-failure in~$\m l$ is modular.
  \end{enumerate}
\end{thm}

\begin{proof}
  If $\vr V$ has a weak difference term, then the natural
  homomorphism $\nu\colon \Conb(\m a)\to \Conb(\m
  a)/{\solv}$ is a homomorphism of meet continuous lattices
  whose image is meet semidistributive according to
  Theorem~\ref{rad_stuff}~(6).  The kernel classes are
  modular according to Theorem~\ref{sim_cong}. This proves
  that (1)$\mimplies$(2).

  The implications (2)$\mimplies$(3)$\mimplies$(4) are
  straightforward; the first since any factorization in the
  category of meet continuous lattices is a factorization in
  the category of lattices, and the second because all
  \hbox{$\sdm$-failures} must be collapsed by any
  homomorphism into a meet semidistributive lattice.

  Assume that $\sdm$-failures of~$\Conb(\m a)$ are modular
  for any $\m a\in {\vr V}$.  As explained in the third
  paragraph of Remark~\ref{sdm_rem}, the class of meet
  continuous lattices whose $\sdm$-failures are modular is
  definable by a meet continuous identity.  If (4) holds,
  then the congruence lattices of algebras in~$\vr V$
  satisfy this identity, so the meet continuous congruence
  variety of~$\vr V$ also satisfies it, hence (5) holds.

  By Theorem~\ref{pixley-wille_extension} the class of
  varieties $\vr U$ that satisfy the meet continuous
  congruence identity expressing the fact that
  \hbox{$\sdm$-failures} are modular is definable by
  idempotent Maltsev conditions.  To show that
  (5)$\mimplies$(1) we must show that these Maltsev
  conditions are strong enough to force the existence of a
  weak difference term.  By Theorem~\ref{justify_weak}, it
  is enough to show that if $\vr V$ is a variety satisfying
  (5) then the abelian algebras in~$\vr V$ are affine.

  So let $\vr V$ be any variety satisfying condition (5),
  and let $\vr A$ be the subvariety of~$\vr V$ that is
  generated by the abelian algebras of~$\vr V$. Then $\vr
  A$~must also satisfy item (5), and in particular (since
  the variety of sets does not satisfy (5)) this implies
  that $\vr A$ has a Taylor term.  Let ${\m f = \m f_{\vr
      A}(4)}$ be the $4$-generated free algebra of this
  variety. The algebra $\m f$ is abelian since $\vr A$ is
  generated by abelian algebras, the abelian property is
  preserved by subalgebras and products, and free algebras
  may be constructed as subalgebras of products of
  generating algebras.  Since $\m f$ is abelian, there is a
  congruence $\Delta$ on $\m f^2$ that has the diagonal $D =
  \{(f,f)\mid f\in F\}$ as a class.  It follows from
  Lemma~4.4 of~\cite{k-szendrei2} (in the case
  $\gamma=\delta=1$ of that result) that since $\vr A$ has a
  Taylor term, the congruence $\Delta$ is a lattice
  complement of the coordinate projection kernels $\eta_1,
  \eta_2\in \Con(\m f^2)$.  Since $\Delta\wedge \eta_1 =
  \Delta\wedge \eta_2 = 0$, the interval
  $I[0,\Delta\wedge(\eta_1\vee\eta_2)] = I[0,\Delta]$ is an
  $\sdm$-failure. By item~(5), this interval is
  modular. Since the diagonal $D$ is a $\Delta$-class, it
  follows from Lemma~2.4 of~\cite{hobby-mckenzie} that
  restriction is a lattice homomorphism from the interval
  $I[0,\Delta]$ to~$\Conb(\m f^2|_D)$, where $\m f^2|_D$ is
  the algebra induced on~$D$ by~$\m f^2$, so $\Conb(\m
  f^2|_D)$ is modular.  The operations of~$\m f^2|_D$ are
  the polynomial operations of~$\m f^2$ under which $D$ is
  closed, each of which agrees with some polynomial
  operation of~$\m f$ acting coordinatewise on~$D$.  Thus,
  $\m f^2|_D$ is isomorphic as a nonindexed algebra to the
  algebra $\m f|_F$ that $\m f$ induces on itself. But the
  congruence lattices of~$\m f$ and~$\m f|_F$ are the same,
  so $\m f = \m f_{\vr A}(4)$ has a modular congruence
  lattice.  By Day's Theorem,~\cite{day1}, this implies that
  $\vr A$ is congruence modular. Since $\vr A$ is generated
  by abelian algebras, modular commutator theory implies
  that all algebras in~$\vr A$ are affine.
\end{proof}

\begin{exmp}
  There exists a purely lattice-theoretic proof that
  (2)$\mimplies$(3)$\mimplies$(4) in Theorem~\ref{HMa}, but
  not for the reverse implications. Those implications are
  false even for congruence lattices of individual algebras.

  For example, if\, $\m l$ is a lattice of height $\leq 3$,
  then any $\sdm$-failure has height $\leq 2$, hence is
  modular. Thus, any finite, nonmodular, simple lattice of
  height $\leq 3$ has the properties that it is algebraic
  (hence isomorphic to the congruence lattice of some
  algebra), has modular $\sdm$-failures, but has no
  $\sdm/\Mod$ factorization.  There are many graph lattices
  satisfying the required conditions (cf.\
  Section~\ref{4.3.5}).

  There also exist algebraic lattices with $\sdm/\Mod$
  factorizations in the category of ordinary lattices and no
  such factorization in the category of meet continuous
  lattices, such as the one in Figure~\ref{fig22}.

\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(45,45)
\put(0,5){\circle*{1.2}}
\put(5,0){\circle*{1.2}}
\put(5,5){\circle*{1.2}}
\put(5,10){\circle*{1.2}}
\put(10,5){\circle*{1.2}}
\put(5,0){\line(-1,1){5}}
\put(5,0){\line(0,1){5}}
\put(5,0){\line(1,1){5}}
\put(5,10){\line(-1,-1){5}}
\put(5,10){\line(0,-1){5}}
\put(5,10){\line(1,-1){5}}
%
\put(0,15){\circle*{1.2}}
\put(5,15){\circle*{1.2}}
\put(5,20){\circle*{1.2}}
\put(10,15){\circle*{1.2}}
\put(5,10){\line(-1,1){5}}
\put(5,10){\line(0,1){5}}
\put(5,10){\line(1,1){5}}
\put(5,20){\line(-1,-1){5}}
\put(5,20){\line(0,-1){5}}
\put(5,20){\line(1,-1){5}}
%
\put(0,25){\circle*{1.2}}
\put(5,25){\circle*{1.2}}
\put(5,30){\circle*{1.2}}
\put(10,25){\circle*{1.2}}
\put(5,20){\line(-1,1){5}}
\put(5,20){\line(0,1){5}}
\put(5,20){\line(1,1){5}}
\put(5,30){\line(-1,-1){5}}
\put(5,30){\line(0,-1){5}}
\put(5,30){\line(1,-1){5}}
%
%
\put(5,33){\circle*{.6}}
\put(5,35){\circle*{.6}}
\put(5,37){\circle*{.6}}
\put(5,40){\circle*{1.2}}
%
\put(45,20){\circle*{1.2}}
\put(45,20){\line(-2,1){40}}
\put(45,20){\line(-2,-1){40}}
\end{picture}
\end{center}
\caption{No $\sdm/\Mod$ factorization in~${\vr L}_{MC}$}\label{fig22}
\end{figure}
\end{exmp}

We next show that the relation $\solv$ behaves well with
respect to the class operators $\Ho, \Su$ and $\Pd_{fin}$.

\begin{thm}\label{hom+sub+prod}
  Let\/ $\vr V$ be a variety with a weak difference term.
  \begin{enumerate}
  \item[(1)] If\/ $\m a\in {\vr V}$ and\/ $\alpha, \beta\geq
    \gamma$ are congruences on\/~$\m a$, then $\alpha\solv
    \beta$ in\/~$\Conb(\m a)$ if and only if\/
    $\alpha/\gamma\solv \beta/\gamma$ in\/~$\Conb(\m
    a/\gamma)$.
  \item[(2)] If\/ $\m b$ is a subalgebra of\/~$\m a\in {\vr
      V}$, then $\alpha\solv \beta$ in\/~$\Conb(\m a)$
    implies that $\alpha|_B\solv \beta|_B$ in\/~$\Conb(\m
    b)$.
  \item[(3)] If\/ $\m a\in{\vr V}$ has congruences $\alpha$
    and\/ $\beta$ and\/ $\m b\in{\vr V}$ has congruences
    $\gamma$ and\/~$\delta$, then $\alpha\times\gamma\solv
    \beta\times\delta$ in\/~$\Conb(\m a\times \m b)$ if and
    only if\/ $\alpha\solv \beta$ in~$\Conb(\m a)$ and
    $\gamma\solv\delta$ in\/~$\Conb(\m b)$.
  \end{enumerate}
  In particular, the class of\/~$\XXX$-solvable algebras
  in~$\vr V$ is closed under the formation of homomorphic
  images, subalgebras and finite products.
\end{thm}

\begin{proof}
  It follows from Theorem~\ref{basic_centrality}~(10) that
  if $\gamma\leq\delta\leq\theta$, then $\delta\lhd\theta$
  if and only if $\delta/\gamma\lhd \theta/\gamma$.  It is
  also true that a sequence
  $(\theta_{\lambda})_{\lambda<\kappa}$ of congruences on
  $\m a$, each containing $\gamma$, is continuous if and
  only if $(\theta_{\lambda}/\gamma)_{\lambda<\kappa}$
  is. The property $\alpha\solv \beta$ in~$\Conb(\m a)$ is
  the property that there exists a continuous sequence of
  congruences witnessing $\XXX$-solvability from
  $\alpha\wedge\beta$ to~$\alpha\vee\beta$, which is
  therefore equivalent to the existence of a continuous
  sequence of congruences witnessing $\XXX$-solvability from
  $(\alpha\wedge\beta)/\gamma =
  (\alpha/\gamma)\wedge(\beta/\gamma)$ to
  $(\alpha\vee\beta)/\gamma =
  (\alpha/\gamma)\vee(\beta/\gamma)$.  Hence $\alpha\solv
  \beta$ is equivalent to~$\alpha/\gamma\solv \beta/\gamma$
  when $\gamma\leq \alpha\wedge\beta$. This proves (1).

  It follows from Theorem~\ref{basic_centrality}~(9) that
  restriction to~$\m b$ preserves the relation~$\lhd$
  (i.e., $\delta\lhd\theta$ in~$\Conb(\m a)$ implies
  $\delta|_{\m b}\lhd \theta|_{\m b}$).  Since restriction
  to~$\m b$ also preserves unions of chains, it must
  preserve~$\lhdlhd$.  Hence $\XXX$-solvable intervals
  in~$\Conb(\m a)$ restrict to~$\XXX$-solvable intervals
  in~$\Conb(\m b)$.  Since $\alpha\solv \beta$ if and only
  if $\alpha$ and $\beta$ lie in a single $\XXX$-solvable
  interval, $\solv$ is also preserved under restriction
  to~$\m b$. This proves (2).

  Applying item (1) to~$\m a\times \m b$ (with the
  coordinate projection kernels $\eta_1$ and~$\eta_2$
  playing the role of~$\gamma$ in (1)) we obtain that
  $\alpha\solv \beta$ in~$\Conb(\m a)$ if and only if
  $\alpha_1\solv \beta_1$ in~$\Conb(\m a\times \m b)$ and
  similarly $\gamma\solv \delta$ in~$\Conb(\m b)$ if and
  only if $\gamma_2\solv \delta_2$ in~$\Conb(\m a\times \m
  b)$.  Since $\XXX$-solvability is a congruence, the
  conjunction of~$\alpha\solv \beta$ and $\gamma\solv
  \delta$ implies that
  \[
  \alpha\times\gamma := \alpha_1\wedge \gamma_2 \solv
  \beta_1\wedge \delta_2 =: \beta\times\delta\,.
  \]
  Conversely, if $\alpha\times\gamma \solv
  \beta\times\delta$, then again because $\XXX$-solvability
  is a congruence we get
  \[
  \alpha_1 = \eta_1\vee(\alpha\times\gamma) \solv
  \eta_1\vee(\beta\times\delta) = \beta_1\,,
  \]
  and similarly $\gamma_2\solv \delta_2$.  As proved above,
  these instances of~$\solv$ hold if and only if
  $\alpha\solv \beta$ in~$\Conb(\m a)$ and $\gamma\solv
  \delta$ in~$\Conb(\m b)$.
\end{proof}

By a {\bf partial Maltsev 
operation}\index{Maltsev!partial --- operation}\index{partial Maltsev operation} 
on a set
$A$ we mean a function $m\colon D\to A$, for some
domain~$D\subseteq A^3$ containing all triples $(x,y,y)$ and
$(y,y,x)$, $x, y\in A$, which satisfies $m(x,y,y) = x =
m(y,y,x)$ for all $x, y\in A$.

\begin{thm}\label{tol=cong2_prep}
  Let\/ $\vr V$ be a variety with a weak difference term.
  Assume that\/ $\m a\in {\vr V}$,\, $\theta\in\Con(\m a)$
  is $\XXX$-solvable, and\/ $S\subseteq A^n[\theta]$ is a
  subset.  If\/ $S$ is compatible with the operations
  of\/~$\m a$, then $S$ is also compatible with some partial
  Maltsev operation on $A$.
\end{thm}

\begin{proof}
  Recall from the opening paragraph of
  Chapter~\ref{elementary} that $A^n[\theta]$ is defined to
  be $ \{\wec{x}\in A^n\mid x_i\equiv_{\theta} x_j
  \;\mbox{\rm for all $i, j$}\}$. Assume that ${S\subseteq
    A^n[\theta]}$ is compatible with the operations of~$\m
  a$. We will show that $S$ is compatible with the partial
  Maltsev operation on $A$ with smallest domain, which is
  \[
  D = \{(a,b,b)\mid a, b\in A\}\cup \{(a,a,b)\mid a, b\in
  A\}\,.
  \]
  We must show that if $\wec{a}, \wec{b}, \wec{c}\in S$ and
  the tuple
  \[
  m(\wec{a}, \wec{b}, \wec{c}) =
  \big(m(a_1,b_1,c_1),m(a_2,b_2,c_2),
  \ldots,m(a_n,b_n,c_n)\big) =: \wec{d}
  \]
  is defined in all coordinates, then $\wec{d}\in S$.  To be
  defined in all coordinates, it must be the case that $a_i
  = b_i$ or $b_i = c_i$ (or both) for all~$i$.  Assume that
  this is so.  Let $I \subseteq N:=\{1,\ldots,n\}$ be the
  set of coordinates where $a_i = b_i$.  If $I = \emptyset$,
  then $\wec{b}=\wec{c}$, so $\wec{d} = \wec{a}\in S$.  If
  $I = N$, then $\wec{a}=\wec{b}$, so $\wec{d} = \wec{c}\in
  S$. Thus the only nontrivial cases are those where
  $\emptyset\neq I\neq N$.  Restrict attention to those, and
  set $J = N - I$.

  Let $\m s$ be the subalgebra of $\m a^n$ that is supported
  by~$S=A^n[\theta]$.  Elements $\wec{u}, \wec{v}\in S$ are
  related by~$\theta_i$ (${} = \pi_i^{-1}(\theta)$) if and
  only if $u_i\equiv_\theta v_i$. Since $S\subseteq
  A^n[\theta]$ this implies that $u_j\equiv_{\theta}
  u_i\equiv_{\theta} v_i\equiv_{\theta} v_j$ for any~$j$,
  hence that $\theta_i\subseteq \theta_j$ for any $i$ and
  $j$.  Let $\widehat{\theta}$ denote the congruence
  $\theta_1 = \theta_2 = \cdots = \theta_n$ of~$\m s$.  It
  follows from Theorem~\ref{hom+sub+prod}~(1) and (2) that
  $\theta_i\solv \eta_i$ for each $i$, so $\widehat{\theta}
  = \bigwedge \theta_i\solv \bigwedge \eta_i = 0$, proving
  that $\widehat{\theta}$ is $\XXX$-solvable.

  If $i\in I$, then $\eta_I\leq \eta_i\leq \theta_i =
  \widehat{\theta}$, and a similar argument proves that
  $\eta_J\leq \widehat{\theta}$.  Since $\widehat{\theta}$
  is $\XXX$-solvable, $\eta_I\solv
  \widehat{\theta}\solv\eta_J$.  This and
  Theorem~\ref{sim_cong} prove that $\eta_I$ and $\eta_J$
  are permuting congruences of~$\m s$. According to the
  definition of~$I$ and $J$, we have $ \wec{a}\wrel{\eta_I}
  \wec{b}\wrel{\eta_J} \wec{c}, $ so there must exist a
  tuple $\wec{x}\in S$ such that $ \wec{a}\wrel{\eta_J}
  \wec{x}\wrel{\eta_I} \wec{c}.  $ The tuple $\wec{x}$
  satisfies the condition that $x_i=c_i$ if $i\in I$ and
  $x_j = a_j$ if $j\in N-I=J$. That is, ${x_i = c_i}$ if
  $m(a_i,b_i,c_i) = m(a_i,a_i,c_i) = c_i$, and $x_j = a_j$
  if $m(a_j,b_j,c_j) = m(a_j,c_j,c_j) = a_j$.  This proves
  that $\wec{x} = \wec{d}$, hence $\wec{d} = \wec{x} \in S$
  as desired.
\end{proof}

It is a general fact that if $\m a$ is an algebra, $m$ is an
operation on~$A$, and for every cardinal $\kappa$ any
relation $S\subseteq A^{\kappa}$ compatible with the
operations of~$\m a$ is also compatible with $m$, then $m$
is a term operation of~$\m a$. If instead one only has that
any {\it finitary} relation $S$ on $A$ that is compatible
with the operations of~$\m a$ is also compatible with $m$,
then $m$ is only guaranteed to be a {\bf local term
  operation}\index{local term operation} of~$\m a$. This
means that for any finite subset $F\subseteq A$ there is a
term operation $t_F$ that agrees with $m$ on $F$.
Theorem~\ref{tol=cong2_prep} suggests the possibility that
if $\vr V$ has a weak difference term and $\m a\in {\vr V}$
is $\XXX$-solvable, then $\m A$ has a local Maltsev term
operation.  Admittedly Theorem~\ref{tol=cong2_prep} refers
only to partial Maltsev operations, but that is exactly what
one would expect of a result like
Theorem~\ref{tol=cong2_prep} if the local Maltsev term
operation were not unique.  These observations lead to
speculations recorded in the following problem.

\begin{prb}\label{6.21}
  Assume that $\vr V$ has a weak difference term, and that
  $\m a\in {\vr V}$ is $\XXX$-solvable.  Does $\m a$ have a
  local Maltsev term operation?  Does $\m a$ have a Maltsev
  term operation?  If $\vr V$ is generated by
  $\XXX$-solvable algebras, then must $\vr V$ be congruence
  permutable?
\end{prb}

Although affirmative answers to these questions seem
unlikely, it is true that for each finite $n$ there is a
term that is Maltsev on classes of \hbox{$n$-step} solvable
congruences of algebras in~$\vr V$.  It would be interesting
if this property extended to congruences with higher degrees
of solvability.

\begin{cor}\label{tol=cong2}
  Let\/ $\vr V$ be a variety with a weak difference term.
  If\/ $\m a\in {\vr V}$,\, $\rho\solv 0$ in\/~$\Conb(\m a)$
  and\/ $S\subseteq \rho$ (${} = A^2[\rho]$) is a reflexive
  compatible binary relation on\/ $\m a$, then $S$ is a
  congruence.
\end{cor}

\begin{proof}
  This is proved in the same way that one proves that a
  reflexive compatible binary relation on a Maltsev algebra
  is a congruence, since that argument works equally well
  with a partial Maltsev operation.

  For the record, the details are as follows.  If
  $S\subseteq \rho$ is reflexive and compatible, then $S$ is
  compatible with a partial Maltsev operation according to
  Theorem~\ref{tol=cong2_prep}.  Thus, if $(a,b)\in S$, then
  since $S$ is reflexive $(a,a), (b,b)\in S$ also, so $(b,a)
  = m\big((a,a),(a,b),(b,b)\big)\in S$.  This proves that
  $S$ is symmetric.  Suppose that $(a,b), (b,c)\in S$, then
  ${(a,c) = m\big((a,b),(b,b),(b,c)\big)\in S}$.  This
  proves that $S$ is transitive, hence a congruence.
\end{proof}

\begin{thm}\label{3permute}
  Let\/ $\vr V$ be a variety with a weak difference term.
  \begin{enumerate}
  \item[(1)] If\/ $\m a\in {\vr V}$ has congruences
    $\alpha'\lhdlhd \alpha$ and\/ $\beta'\lhdlhd \beta$, then
    \[
    \alpha\vee\beta = (\alpha'\vee\beta')\circ\alpha\circ
    \beta\circ(\alpha'\vee\beta')\,.
    \]
  \item[(2)] If\/ $\m a\in {\vr V}$ has congruences $\alpha$
    and\/ $\beta$ with $\beta\solv \alpha\wedge\beta$, then
    \[
    \alpha\vee\beta = \alpha\circ\beta\circ\alpha\,.
    \]
  \end{enumerate}
\end{thm}

\begin{proof}
  For item (1), the relation $R =
  (\alpha'\vee\beta')\circ\alpha\circ
  \beta\circ(\alpha'\vee\beta')$ is an
  $(\alpha'\vee\beta')$-closed compatible reflexive relation
  on $\m a$ whose transitive closure is $\alpha\vee\beta$.
  To prove this part of the theorem it suffices to show that
  $R$ is a congruence.

  Since $\alpha'\lhdlhd \alpha$ and $\beta'\lhdlhd \beta$,
  it follows from Lemma~\ref{meet+join} and the transitivity
  of~$\lhdlhd$ that $(\alpha'\vee\beta')\lhdlhd
  (\alpha\vee\beta)$.  In the quotient algebra $\m
  a/(\alpha'\vee\beta')$ we have that
  \[
  \rho:=(\alpha\vee\beta)/(\alpha'\vee\beta') \solv
  (\alpha'\vee\beta')/(\alpha'\vee\beta') = 0
  \]
  by Theorem~\ref{hom+sub+prod}. The relation $S :=
  R/(\alpha'\vee\beta')$ is a compatible reflexive relation
  of~$\m a/(\alpha'\vee\beta')$, and $S\subseteq \rho$,
  hence $S$ is a congruence on ${\m a/(\alpha'\vee\beta')}$
  according to Corollary~\ref{tol=cong2}.  But $R$ is
  $(\alpha'\vee\beta')$-closed, so for the natural
  homomorphism $\nu: \m a\rightarrow\m
  a/(\alpha'\vee\beta'): a\mapsto a/(\alpha'\vee\beta')$ we
  have $R = \nu^{-1}\big(\nu(R)\big)$. We know that $\nu(R)
  = R/(\alpha'\vee\beta') = S$ is a congruence, and the
  inverse image of a congruence is a congruence. Hence $R$
  is a congruence.

  Item (2) is the special case of (1) where $\alpha =
  \alpha'\geq \beta'=\alpha\wedge\beta$.
\end{proof}

Theorem~\ref{sim_cong} follows from
Theorem~\ref{3permute}~(1) when
$\alpha'=\beta'=\alpha\wedge\beta$.

\begin{thm}\label{pre-N5}
  Let\/ $\vr V$ be a variety with a weak difference term.
  If\/ $\m a\in {\vr V}$ has congruences $\alpha$ and\/
  $\beta$ with $\beta\solv \alpha\wedge\beta$ and
  $\alpha\leq\gamma\leq\alpha\vee\beta$, then $\gamma =
  \alpha\vee(\beta\wedge\gamma)$.  Hence the map $J:
  I[\alpha\wedge\beta,\beta] \rightarrow
  I[\alpha,\alpha\vee\beta]: x\mapsto x\vee \alpha$ is
  surjective.
\end{thm}

\begin{proof}
  The hypotheses of Theorem~\ref{3permute}~(2) are
  satisfied, therefore $\alpha\vee\beta =
  \alpha\circ\beta\circ\alpha$.  Arbitrarily choose
  $(a,b)\in\gamma$.  Since $\gamma\leq \alpha\vee\beta =
  \alpha\circ\beta\circ\alpha$, there are elements $u, v\in
  A$ such that
  \begin{equation}\label{a_u_v_b}
    a\equiv_{\alpha}u
    \equiv_{\beta}v
    \equiv_{\alpha}b\,.
  \end{equation}
  This means that $u\equiv_{\alpha}a \equiv_{\gamma}b
  \equiv_{\alpha}v$, so
  $(u,v)\in\beta\wedge(\alpha\vee\gamma)=
  \beta\wedge\gamma$.  This allows us to strengthen the
  $\beta$-relation in line (\ref{a_u_v_b}) to a
  $(\beta\wedge\gamma)$-relation, and in the strengthened
  form that line indicates that
  $(a,b)\in\alpha\vee(\beta\wedge\gamma)$.  Since $(a,b)$
  was chosen arbitrarily, $\gamma\subseteq
  \alpha\vee(\beta\wedge\gamma)$.  The reverse inclusion
  follows from the fact that $\alpha\leq\gamma$.

  For the final statement, note that if ${\gamma\in
    I[\alpha,\alpha\vee\beta]}$, then $\beta\wedge\gamma$ is
  in the interval $I[\alpha\wedge\beta,\beta]$ and we have
  proved above that $\gamma = J(\beta\wedge\gamma)$.
\end{proof}

\begin{thm}\label{weak_omit}
  Let\/ $\vr V$ be a variety with a weak difference term.
  Assume that\/ $\m a\in {\vr V}$.
  \begin{enumerate}
  \item[(1)] If\/ $\m n_5$, with the labeling indicated in
    Figure~\ref{fig23}, appears as a sublattice
    of\/~$\Conb(\m a)$, then
    $\beta\nsolv\beta\wedge\alpha$.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(20,35)
\put(0,20){\circle*{1.2}}
\put(10,5){\circle*{1.2}}
\put(10,35){\circle*{1.2}}
\put(20,15){\circle*{1.2}}
\put(20,25){\circle*{1.2}}
%
\put(10,5){\line(-2,3){10}}
\put(10,5){\line(1,1){10}}
\put(10,35){\line(-2,-3){10}}
\put(10,35){\line(1,-1){10}}
\put(20,25){\line(0,-1){10}}
%
\put(-4.5,18){$\beta$}
\put(22,14){$\alpha$}
\put(22,24){$\gamma$}
\put(5.5,0){$\beta\wedge \alpha$}
\end{picture}
\end{center}
\caption{$\beta\nsolv\beta\wedge\alpha$}\label{fig23}
\end{figure}
\item[(2)] If\/ $\m d_2$ appears as a sublattice
  of\/~$\Conb(\m a)$, then no two distinct congruences in
  the sublattice are $\XXX$-solvably related.
\end{enumerate}
\end{thm}

\begin{proof}
  In item (1), we have
  $\alpha\leq\gamma\leq\alpha\vee\beta$.  Therefore, if we
  had $\beta\solv \beta\wedge\alpha$, then all of the
  hypotheses of Theorem~\ref{pre-N5} would be met.  This
  would force $\gamma = \alpha\vee(\beta\wedge\gamma)$.  But
  $\alpha\vee(\beta\wedge\gamma) = \alpha\neq\gamma$, so we
  cannot have $\beta\solv \beta\wedge\alpha$.

  For item (2), the relation $\solv$ restricts to a
  congruence on $\m d_2$, which is a subdirectly irreducible
  lattice.  If the copy of~$\m d_2$ under consideration is
  labeled as in Figure~\ref{fig24}, then the critical
  intervals are $I[\mu,\nu]$,\, $I[\alpha,\tau]$, and
  $I[\delta,\theta]$. Therefore, if any two distinct
  elements of this sublattice are $\solv$\,-related then
  $\mu\solv \nu$, $\alpha\solv \tau$, and $\delta\solv
  \theta$.
\begin{figure}[ht]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(30,30)
\put(0,10){\circle*{1.2}}
\put(0,20){\circle*{1.2}}
\put(15,0){\circle*{1.2}}
\put(15,20){\circle*{1.2}}
\put(15,30){\circle*{1.2}}
\put(30,10){\circle*{1.2}}
\put(30,20){\circle*{1.2}}
%
\put(0,10){\line(3,-2){15}}
\put(0,10){\line(3,2){15}}
\put(0,10){\line(0,1){10}}
\put(15,30){\line(-3,-2){15}}
\put(15,30){\line(0,-1){10}}
\put(15,30){\line(3,-2){15}}
\put(30,10){\line(-3,-2){15}}
\put(30,10){\line(-3,2){15}}
\put(30,10){\line(0,1){10}}
%
\put(14,15){$\alpha$}
\put(14,32){$\tau$}
\put(-4.5,9){$\mu$}
\put(-4.5,19){$\nu$}
\put(32,9){$\delta$}
\put(32,19){$\theta$}
\end{picture}
\end{center}
\caption{ \ }\label{fig24}
\end{figure}
Assume that this is so and choose $\alpha'$ so that $\alpha
< \alpha'\leq\tau$ and $\alpha\lhd\alpha'$. By
Theorem~\ref{pre-N5} the maps $I[\mu,\nu]\to
I[\alpha,\tau]\colon x\mapsto x\vee\alpha$ and
$I[\delta,\theta]\to I[\alpha,\tau]\colon x\mapsto
x\vee\alpha$ are surjective, so there exist $\mu'$ and
$\delta'$ such that $\mu < \mu'\leq\nu$,\, $\delta < \delta'
\leq \theta$ and $\mu'\vee\alpha = \alpha' =
\delta'\vee\alpha$.  Since $I[\mu,\mu']$ and
$I[\delta,\delta']$ are perspective with $I[\alpha,\alpha']$
we get that $\mu\lhd \mu'$ and $\delta\lhd\delta'$ from
Theorem~\ref{perspective_abelian_taylor}
(or Lemma~\ref{meet+join}).

\begin{clm}
  The congruences $\mu'$, $\alpha$, and\/ $\delta'$ and
  generate a sublattice isomorphic to\/~$\m d_2$.
\end{clm}

Our argument uses Lemma~\ref{presentation} with $x =
\mu'$,\, $y = \alpha$ and $z = \delta'$, as well as the fact
that $\nu$, $\alpha$ and $\theta$ generate a copy of~$\m
d_2$.

For item (i) of Lemma~\ref{presentation},
\[
x\wedge z = \mu'\wedge\delta'\leq \nu\wedge\theta\leq \alpha
= y\,.
\]

For item (ii),
\[
z\vee(y\wedge x) = \delta'\vee(\alpha\wedge \mu') =
\delta'\vee \mu = \alpha' = \alpha\vee \mu'\ge \mu' = x\,.
\]
Here we have used the facts that $I[\mu,\mu']$ and
$I[\delta,\delta']$ are perspective up to
$I[\alpha,\alpha']$, and that $\delta'\vee\mu =
\delta'\vee(\delta\vee\mu) = \delta'\vee\alpha = \alpha'$.

Item (iii) is similar to (ii). For (iv),
\[
(x\wedge y)\vee(y\wedge z) = (\mu'\wedge
\alpha)\vee(\alpha\wedge\delta') = \mu\vee\delta = \alpha\,.
\]

Since $x = \mu'\not\leq \alpha = y$, the sublattice
generated by~$\{\mu', \alpha, \delta'\}$ is isomorphic to
$\m d_2$. This proves the claim.

\medskip

From the assumption that $\Conb(\m a)$ has a sublattice
isomorphic to~$\m d_2$ with $\XXX$-solvable critical
intervals we have shown that $\Conb(\m a)$ has a sublattice
isomorphic to~$\m d_2$ with abelian critical intervals.
This contradicts Theorem~\ref{taylor_forbidden}~(2).
\end{proof}

The rest of the results in this section provide some
information about failures of join semidistributivity in the
congruence lattice of an algebra with a weak difference
term.

\begin{lm}\label{sdj_failure_cent_lm}
  Assume that\/ $\vr V$ has a weak difference term.  If\/
  $\m a\in {\vr V}$ has congruences $\alpha, \beta$ and
  $\gamma$ satisfying
  \begin{enumerate}
  \item[(1)] $\alpha\vee\beta\solv \alpha \solv
    \alpha\vee\gamma$, and
  \item[(2)] $\beta\wedge\gamma\leq \alpha$,
  \end{enumerate}
  then $\big(\beta\vee(\alpha\wedge\gamma)\big)\wedge
  \big(\gamma\vee(\alpha\wedge\beta)\big) =
  (\alpha\wedge\beta)\vee (\alpha\wedge\gamma)$.
\end{lm}

\begin{proof}
  Since $\beta\wedge\gamma$ lies below all congruences of
  interest, we may factor by it and assume that
  $\beta\wedge\gamma = 0$.

  In the next claim we label some of the congruences in the
  sublattice of~$\Conb(\m a)$ generated by~$\{\alpha, \beta,
  \gamma\}$, and prove some of the more obvious
  relationships between them.

\begin{clm}\label{get_stuff_right}
  Let\/ $\delta =
  (\alpha\wedge\beta)\vee(\alpha\wedge\gamma)$,\,
  $\widehat{\beta} = \beta\vee\delta$,\,
  $\widehat{\gamma} = \gamma\vee\delta$,\,
  $\theta = \widehat{\beta}\wedge\widehat{\gamma}$,\,
  $\beta' = \beta\wedge\theta$, and\/
  $\gamma' = \gamma\wedge\theta$.
  Then
  \begin{enumerate}
  \item[(i)] $\delta\leq \alpha$,
  \item[(ii)] $\beta\wedge\delta = \beta\wedge\alpha$ and\/
    $\gamma\wedge\delta = \gamma\wedge\alpha$,
  \item[(iii)] $\beta\solv \beta\wedge\delta$ and\/
    $\gamma\solv \gamma\wedge\delta$,
  \item[(iv)] $\delta\leq\theta\leq\widehat{\beta}$ and\/
    $\delta\leq\theta\leq\widehat{\gamma}$,
  \item[(v)] $\widehat{\beta}\solv \delta\solv
    \widehat{\gamma}$,
  \item[(vi)] $\widehat{\beta}\vee\alpha = \beta\vee\alpha$
    and\/ $\widehat{\gamma}\vee\alpha = \gamma\vee\alpha$,
  \item[(vii)] $\beta'\vee\delta = \theta =
    \gamma'\vee\delta$.
  \end{enumerate}
\end{clm}

Some of these congruences and relationships are indicated in
Figure~\ref{fig25} (on the next page).

\begin{figure}[t]
\begin{center}
\setlength{\unitlength}{1mm}
\begin{picture}(30,30)
\put(0,15){\circle*{1.2}}
\put(10,5){\circle*{1.2}}
\put(10,25){\circle*{1.2}}
\put(15,0){\circle*{1.2}}
\put(15,10){\circle*{1.2}}
\put(15,20){\circle*{1.2}}
\put(20,5){\circle*{1.2}}
\put(20,25){\circle*{1.2}}
\put(30,15){\circle*{1.2}}
%
\put(15,0){\line(-1,1){15}}
\put(15,0){\line(1,1){15}}
\put(15,10){\line(-1,-1){5}}
\put(15,10){\line(1,-1){5}}
\put(15,10){\line(0,1){10}}
\put(15,20){\line(-1,1){5}}
\put(15,20){\line(1,1){5}}
\put(0,15){\line(1,1){10}}
\put(30,15){\line(-1,1){10}}
%
\put(-5,14){$\beta$}
\put(32,14){$\gamma$}
\put(-4,3){$\alpha\wedge\beta$}
\put(23,3){$\alpha\wedge\gamma$}
\put(-21,25){$\widehat{\beta} = \beta\vee(\alpha\wedge\gamma)$}
\put(23,25){$\gamma \vee (\alpha\wedge\beta) = \widehat{\gamma}$}
\put(10.5,18.5){$\theta$}
\put(10.5,8.5){$\delta$}
\end{picture}
\end{center}
\caption{ \ }\label{fig25}
\end{figure}

We omit the proofs of the parts of the claim that concern
$\gamma$ if those parts are symmetric to the arguments we
give for~$\beta$.

For item (i) of Claim~\ref{get_stuff_right}, note that
$\delta = (\alpha\wedge\beta)\vee (\alpha\wedge\gamma)$ is a
join of two elements below $\alpha$, so $\delta\leq \alpha$.

For item (ii), since $\delta\leq \alpha$ we get that
$\beta\wedge\delta \leq \beta\wedge\alpha$. On the other
hand, both $\beta$ and $\delta$ are above
$\beta\wedge\alpha$, so $\beta\wedge\delta\geq
\beta\wedge\alpha$.

For item (iii), start with $\alpha\solv \alpha\vee\beta$,
which holds by assumption (1) of the lemma.  Using the fact
that $\solv$ is a congruence, meet both sides of
$\alpha\solv \alpha\vee\beta$ with $\beta$ to
obtain~$\beta\wedge\alpha\solv \beta$.  Applying the result
from (ii) of this claim yields $\beta\wedge\delta\solv
\beta$.

For item (iv), the fact that $\theta =
\widehat{\beta}\wedge\widehat{\gamma} =
(\beta\vee\delta)\wedge(\gamma\vee\delta)$ is a meet of
elements above $\delta$ is enough to prove that $\delta\leq
\theta$, while $\theta\leq\widehat{\beta}$ is immediate from
the definition $\theta =
\widehat{\beta}\wedge\widehat{\gamma}$.

For item (v), join the relation $\beta\wedge\delta\solv
\beta$ from (iii) with $\delta$ to obtain $\delta\solv
\beta\vee\delta = \widehat{\beta}$.

For item (vi), $\widehat{\beta}\vee\alpha =
(\beta\vee\delta)\vee\alpha = \beta\vee (\delta\vee\alpha) =
\beta\vee\alpha$, where the last equality follows from
item~(i).

For item (vii), use the fact established in (iii) that
$\beta\solv \beta\wedge\delta$, the fact established in (iv)
that $\delta\leq\theta\leq\widehat{\beta} =\beta\vee\delta$,
and Theorem~\ref{pre-N5} to deduce that $\theta =
\delta\vee(\beta\wedge\theta)$.  Since $\beta' =
\beta\wedge\theta$, this yields $\theta = \beta'\vee\delta$.
This completes the proof of Claim~\ref{get_stuff_right}.

\medskip

By examining the definitions of~$\delta$ and $\theta$, the
reader will verify that the statement of the next claim is
precisely what is asserted in the statement of this lemma.

\begin{clm}\label{get_meet_right}
  $\delta = \theta$.
\end{clm}

Assume instead that $\delta \neq \theta$. We show that the
sublattice of~$\Conb(\m a)$ generated by~$\{\beta', \gamma',
\delta\}$ is isomorphic to~$\m d_2$.  In order to show this,
we verify that the congruences $x = \beta'$,\, $y = \delta$,
and $z = \gamma'$ satisfy the relations from the
presentation of~$\m d_2$ given in
Lemma~\ref{presentation}. Throughout the argument we will
make free use of Claim~\ref{get_stuff_right}.

To establish relation (i) of Lemma~\ref{presentation}, we
have
\[
x\wedge z = \beta'\wedge\gamma' \leq \beta\wedge\gamma =
0\leq \delta = y\,.
\]
For relation (ii),
\[
\begin{array}{r@{}l}
z\vee (y\wedge x) 
&{} = \gamma'\vee(\delta\wedge\beta')\\
&{} = (\gamma\wedge\theta)\vee
\big(\delta\wedge(\beta\wedge\theta)\big)\\
&{} = \big(\gamma\wedge\theta)\vee(\beta\wedge\delta\big)\\
&{} = (\gamma\wedge\theta)\vee(\gamma\wedge\delta)\vee
(\beta\wedge\delta)\\
&{} = \gamma'\vee\big((\gamma\wedge\alpha)\vee
(\beta\wedge\alpha)\big)\\
&{} = \gamma'\vee \delta\\
&{} = \theta\geq \beta\wedge\theta = \beta' = x\,.
\end{array}
\]
The proof that $x\vee (y\wedge z) \geq z$ is similar to the
one just given.  For relation (iv),
\[
\begin{array}{r@{}l}
(x\wedge y)\vee(y\wedge z)
&{} = 
(\beta'\wedge\delta)\vee (\delta\wedge\gamma')\\
&{} =\big((\beta\wedge\theta)\wedge\delta\big)\vee
\big(\delta\wedge(\gamma\wedge\theta)\big)\\
&{} =(\beta\wedge\delta)\vee(\gamma\wedge\delta)\\
&{} =(\beta\wedge\alpha)\vee(\gamma\wedge\alpha)\\
&{} =\delta = y\,.
\end{array}
\]
Finally, $x = \beta'\not\leq \delta = y$ since
$\beta'\vee\delta = \theta$ by
Claim~\ref{get_stuff_right}~(vii), and we are assuming that
$\delta\neq\theta$.

Thus, our assumption that $\delta \neq \theta$ and
Lemma~\ref{presentation} together imply that the sublattice
generated by~$\{\beta', \gamma', \delta\}$ is isomorphic to
$\m d_2$.  It follows from Claim~\ref{get_stuff_right}~(iv)
and (v) that $I[\delta, \theta]$ is an $\XXX$-solvable
interval in this sublattice.  This is forbidden by
Theorem~\ref{weak_omit}, hence Claim~\ref{get_meet_right}
and the lemma are proved.
\end{proof}

\begin{thm}\label{sdj_failure_cent}
  Assume that a variety $\vr V$ has a weak difference term.
  If\/ $\m a\in {\vr V}$ has congruences $\alpha$, $\beta$
  and $\gamma$ satisfying
  \begin{enumerate}
  \item[(1)] $\alpha\vee\beta\solv \alpha \solv
    \alpha\vee\gamma$, and
  \item[(2)] $\beta\wedge\gamma\leq \alpha$,
  \end{enumerate}
  then $\C C(\alpha\vee\beta, \alpha\vee\gamma;\alpha)$
  holds.
\end{thm}

\begin{proof}
  As in the proof of Lemma~\ref{sdj_failure_cent_lm}, we may
  factor by~$\beta\wedge\gamma$, and thus assume that
  $\beta\wedge\gamma=0$.  As in that proof, let $\delta =
  (\alpha\wedge\beta)\vee(\alpha\wedge\gamma)$,
  $\widehat{\beta} = \beta\vee\delta$, and $\widehat{\gamma}
  = \gamma\vee\delta$.

\begin{clm}\label{get-meet-again}
  $\alpha\wedge \widehat{\beta} =
  \alpha\wedge\widehat{\gamma}= \widehat{\beta}\wedge
  \widehat{\gamma}$.
\end{clm}

We proved in Lemma ~\ref{sdj_failure_cent_lm} that $\delta =
\widehat{\beta}\wedge\widehat{\gamma}$, so it suffices (by
symmetry) to prove only that $\alpha\wedge\widehat{\beta} =
\delta$.  Both $\alpha$ and $\widehat{\beta}$ are above
$\delta$, according to the definitions of~$\delta$ and
$\widehat{\beta}$, so we certainly have $\alpha\wedge
\widehat{\beta}\geq \delta$.  This forces $\alpha\wedge
\widehat{\beta}\in I[\delta,\widehat{\beta}] =
I[\delta,\beta\vee\delta]$.  The perspective interval
$I[\beta\wedge\delta, \beta] = I[\beta\wedge\alpha,\beta]$
is $\XXX$-solvable by assumption, so Theorem~\ref{pre-N5}
guarantees that
\[
\alpha\wedge\widehat{\beta} =
\delta\vee\big(\beta\wedge
(\alpha\wedge\widehat{\beta})\big)\,.
\]
But we can simplify the expression on the right as follows:
\[
\delta\vee\big(\beta\wedge
(\alpha\wedge\widehat{\beta})\big) = 
\delta\vee\big(\alpha\wedge
(\beta\wedge\widehat{\beta})\big) = 
\delta\vee(\alpha\wedge\beta) = \delta\,.
\]
This proves the claim.

\begin{clm}\label{cent1}
  $\C C(\widehat{\beta}, \widehat{\gamma};\alpha)$ holds.
\end{clm}

It follows from Claim~\ref{get-meet-again} that
$\widehat{\gamma}\wedge
\big(\widehat{\beta}\vee(\widehat{\gamma}\wedge\alpha)\big)
\leq \alpha$, so Claim~\ref{cent1} is a direct application
of Theorem~\ref{basic_centrality}~(8).

\medskip

Now we complete the proof that $\C C(\alpha\vee\beta,
\alpha\vee\gamma; \alpha)$ holds.  It follows from
Theorem~\ref{basic_centrality}~(7) or (8) that $\C C(\alpha,
\widehat{\gamma}; \alpha)$ holds.  From this,
Claim~\ref{cent1}, and Theorem~\ref{basic_centrality}~(5) we
get that $\C C(\alpha\vee\widehat{\beta},
\widehat{\gamma};\alpha)$ holds.  We established that
$\alpha\vee\widehat{\beta} = \alpha\vee\beta$ in
Claim~\ref{get_stuff_right}~(vi) of Lemma
~\ref{sdj_failure_cent_lm}, so $\C C(\alpha\vee\beta,
\widehat{\gamma};\alpha)$ holds.  Using
Theorem~\ref{basic_centrality}~(3) we finally get that $\C
C(\alpha\vee\beta,
\alpha\circ\widehat{\gamma}\circ\alpha;\alpha)$ holds.  This
claim, and this theorem, will be established if we show that
$\alpha\circ\widehat{\gamma}\circ\alpha =
\alpha\vee\widehat{\gamma}$ (${} = \alpha\vee\gamma$).  For
this, meet both sides of the relation $\alpha\vee\gamma\solv
\alpha$, which is part of assumption (1) of the theorem,
with the congruence $\widehat{\gamma}$.  The result is that
$\widehat{\gamma}\solv \alpha\wedge \widehat{\gamma}$.  From
Theorem~\ref{3permute}~(2) we conclude that
$\alpha\circ\widehat{\gamma}\circ\alpha =
\alpha\vee\widehat{\gamma}$, as desired.
\end{proof}

There is a special case of the previous theorem worth
recording.

\begin{cor}\label{sdj_failure=abel}
  Assume that\/ $\vr V$ has a weak difference term, that\/
  $\m a\in {\vr V}$, and that $I$ is an\/ $\sdj$-failure
  in\/~$\Conb(\m a)$.  If\/ $I$ is $\XXX$-solvable, then $I$
  is abelian.
\end{cor}

\begin{proof}
  An $\sdj$-failure is a nontrivial interval
  $I=I[\alpha,\alpha\vee\beta]$ determined by congruences
  $\alpha, \beta$ and $\gamma$ satisfying
  $\alpha\vee\beta=\alpha\vee\gamma$ and
  $\beta\wedge\gamma\leq \alpha$.  The assumption that $I$
  is $\XXX$-solvable means that $\alpha \solv
  \alpha\vee\beta$ (${} = \alpha\vee\gamma$).  This
  establishes hypothesis (1) of
  Theorem~\ref{sdj_failure_cent}.  Hypothesis (2) holds
  simply because $I$ is an $\sdj$-failure.  Hence we have
  the conclusion of Theorem~\ref{sdj_failure_cent}, that $\C
  C(\alpha\vee\beta, \alpha\vee\gamma; \alpha)$
  (equivalently $\C C(\alpha\vee\beta, \alpha\vee\beta;
  \alpha)$) holds.  This is precisely what it means for~$I$
  to be abelian.
\end{proof}


\section{An Alternative Development}\label{radical_approach}

In the preceding section we developed the theory of
solvability in analogy with the theory of solvability for
groups, using chains of congruences with abelian factors. In
this short section we show that it is possible to develop
the theory of solvability in analogy with the theory of
solvability for rings. Namely, we show that the basic
notions of the preceding section, $\rad$, $\solv$ and
$\lhdlhd$, can be defined in terms of prime congruences.

\begin{df}\label{YYYdefn}
  Let $\m a$ be an algebra.  A congruence $\theta$ on $\m a$
  is {\bf semiprime}\index{semiprime} if $\theta
  \blacktriangleleft 1$; it is {\bf prime}\index{prime} if
  it is semiprime and meet irreducible in~$\Conb(\m a)$.

  An {\bf $m$-sequence}\index{m-sequence@$m$-sequence} is a
  sequence $\sigma = (\theta_{\lambda})_{\lambda<\omega}$ of
  principal congruences where
  $[\theta_{\lambda},\theta_{\lambda}]\geq
  \theta_{\lambda+1}$ holds for all $\lambda$.  The
  $m$-sequence $\sigma$ {\bf avoids} a congruence $\theta$
  if $\theta_{\lambda}\not\leq\theta$ for all $\lambda$,
  otherwise it {\bf intersects} $\theta$.
\end{df}

Our definitions for ``prime'' and ``semiprime'' agree with
those from ring theory.  The concept of an $m$-sequence
generalizes the concept of a multiplicatively closed subset
of a commutative ring, or an $m$-system in a noncommutative
ring.  All of these notions are definable in terms of the
centralizer relation.

Observe that if $(\theta_{\lambda})_{\lambda<\omega}$ is an
$m$-sequence, then it is descending, since
$\theta_{\lambda}\geq [\theta_{\lambda},\theta_{\lambda}]
\geq \theta_{{\lambda}+1}$ for all ${\lambda}$.

\begin{thm}\label{ring1} 
  Let\/ $\vr V$ be a variety with a weak difference term.
  Assume that\/ $\m a\in {\vr V}$, and\/ $\theta\in \Con(\m
  a)$ is a proper congruence.  The following conditions
  concerning $\theta$ are equivalent.
  \begin{enumerate}
  \item[(1)] $\theta$ is semiprime.
  \item[(2)] $\theta$ is an intersection of prime
    congruences.
  \item[(3)] For every principal congruence
    $\theta_0\not\leq\theta$ there is an $m$-sequence
    $\sigma = (\theta_0,\theta_1,\ldots)$ starting at\/
    $\theta_0$ that avoids $\theta$.
  \item[(4)] $\theta$ is a radical congruence.
  \end{enumerate}
\end{thm}

\begin{proof}
  It follows immediately from the definitions that (1)
  and~(4) are equivalent, so we prove only that
  (1)$\mimplies$(3)$\mimplies$(2)$\mimplies$(1).

  Assume that $\theta$ is semiprime and that
  $\theta_0\not\leq\theta$.  If
  $[\theta_0,\theta_0]\leq\theta$, then since
  $[\theta_0,\theta_0]\lhd\theta_0$ we get from
  Lemma~\ref{meet+join} that
  \[
  \theta = \theta\vee[\theta_0,\theta_0]\lhd
  \theta\vee\theta_0\,.
  \]
  But since $\theta < \theta\vee\theta_0$, this contradicts
  the assumption that $\theta\blacktriangleleft 1$ (i.e.,
  that $\theta$ is semiprime). Thus it must be that
  $[\theta_0,\theta_0]\not\leq \theta$, and this means that
  there is a principal congruence $\theta_1\leq
  [\theta_0,\theta_0]$ such that $\theta_1\not\leq \theta$.
  The same argument applied to~$\theta_1$ produces a
  principal congruence $\theta_2$ such that $\theta_2\leq
  [\theta_1,\theta_1]$ and $\theta_2\not\leq
  \theta$. Repeating the argument indefinitely produces an
  $m$-sequence $(\theta_0,\theta_1,\ldots)$ starting at
  $\theta_0$ that avoids $\theta$.

  Now suppose that Condition (3) holds for~$\theta$. Let
  $\theta'$ be the (possibly empty) intersection of the
  prime congruences containing $\theta$.  If $\theta <
  \theta'$, then there is a principal congruence
  $\theta_0\leq \theta'$ such that $\theta_0\not\leq
  \theta$. By Condition (3), there is an $m$-sequence
  $\sigma = (\theta_0,\theta_1,\ldots)$ that avoids
  $\theta$. Extend $\theta$ to a congruence $\pi$ that is
  maximal for the property that $\sigma$~avoids~$\pi$. (The
  fact that such a congruence exists uses Zorn's Lemma, and
  relies on the fact that each $\theta_i$ in~$\sigma$ is
  compact.) We claim that $\pi$ is a prime congruence. To
  see that it is meet irreducible, assume instead that $\pi
  = \alpha\wedge\beta$ where $\pi < \alpha,\beta$.  Then by
  the maximality of~$\pi$ it must be that $\sigma$
  intersects both $\alpha$ and $\beta$, so $\theta_i\leq
  \alpha$ and $\theta_j\leq \beta$ for some $i$ and~$j$. But
  then for~$k = \max(i,j)$ we have $\theta_k\leq
  \alpha\wedge\beta = \pi$, since $\sigma$ is a descending
  chain of congruences, and this contradicts the fact that
  $\sigma$~avoids~$\pi$. To show that $\pi$ is semiprime,
  assume instead that there is a congruence $\delta$ such
  that $\pi \lhd \delta$ and $\pi < \delta$.  Then the
  maximality condition on $\pi$ forces $\sigma$ to intersect
  $\delta$, so for some $i$ we have
  $\theta_i\leq\delta$. But now the monotonicity of the
  commutator operation implies that
  \[
  \theta_{i+1}\leq [\theta_i,\theta_i]\leq
  [\delta,\delta]\leq \pi\,.
  \]
  This contradicts the fact that $\sigma$ avoids $\pi$.
  This concludes the proof that $\pi$ is prime.  Since $\pi$
  extends $\theta$, and $\theta'$ is the intersection of
  prime congruences containing $\theta$, it follows that
  $\theta'\leq \pi$. But now we have the contradiction that
  $\theta_0\leq\theta'\leq \pi$ while $\sigma =
  (\theta_{\lambda})_{\lambda<\omega}$ avoids $\pi$.  This
  final contradiction proves that $\theta'=\theta$, and
  hence that Condition (2) holds.

  Finally, assume that Condition (2) holds.  Then $\theta =
  \bigcap_{i\in I} \pi_i$ where each $\pi_i$ is a prime
  congruence.  If $\theta\lhd \delta$, then (since
  $\theta\leq\pi_i$) we get $\pi_i\lhd \pi_i\vee\delta$ for
  each $i$.  Since each $\pi_i$ is semiprime we deduce that
  $\pi_i = \pi_i\vee\delta$, or that $\delta\leq\pi_i$ for
  all $i$.  Hence $\delta\leq \bigcap_{i\in I} \pi_i =
  \theta$.  This shows that $\theta\lhd \delta$ implies that
  $\theta = \delta$, so $\theta\blacktriangleleft 1$.  Hence
  Condition (1) holds.
\end{proof}

The following corollary shows that the concepts introduced
earlier in the chapter can be introduced via prime
congruences.

\begin{cor}
  Let\/ $\vr V$ be a variety with a weak difference term.
  Assume that\/ $\m a\in {\vr V}$ has congruences $\alpha$
  and $\beta$.
  \begin{enumerate}
  \item[(1)] $\rad(\alpha)$ equals the intersection of the
    prime congruences containing $\alpha$.
  \item[(2)] $\alpha\solv \beta$ if and only if $\alpha$ and
    $\beta$ are contained in the same prime congruences.
  \item[(3)] $\alpha\lhdlhd\beta$ if and only if
    $\alpha\leq\beta$ and every $m$-sequence that intersects
    $\beta$ also intersects $\alpha$.
  \end{enumerate}
\end{cor}

\begin{proof}
  Item (1) follows from the fact, proved in
  Theorem~\ref{ring1}, that the radical congruences are
  exactly the intersections of prime congruences. Item (2)
  follows from (1) and the fact that $\alpha\solv \beta$ if
  and only if $\rad(\alpha) = \rad(\beta)$.

  For item (3), assume that $\alpha\lhdlhd\beta$ and that
  the $m$-sequence $\sigma$ intersects $\beta$. If $\sigma$
  avoids $\alpha$, then $\alpha$ can be extended to a prime
  $\pi$ that $\sigma$~avoids, as in the proof of
  Theorem~\ref{ring1}~(3)$\mimplies$(2).  The prime~$\pi$
  cannot contain~$\beta$ since $\sigma$ avoids $\pi$ but
  intersects $\beta$. Therefore $\pi$ contains $\alpha$ and
  not $\beta$.  Since $\alpha\lhdlhd\beta$, this contradicts
  item (2) of this corollary.  For the other direction,
  assume that $\alpha\nlhdlhd\beta$.  Then, since
  $\alpha\leq \beta$, item~(2) guarantees that there is a
  prime $\pi$ containing $\alpha$ and not containing
  $\beta$. If $\theta_0$ is a principal congruence
  satisfying $\theta_0\leq \beta$ and $\theta_0\not\leq\pi$,
  then Theorem~\ref{ring1} guarantees the existence of an
  $m$-sequence starting at $\theta_0$ and avoiding $\pi$
  since the prime $\pi$ is semiprime.  This $m$-sequence
  also avoids $\alpha$ since $\alpha\leq\pi$, but it does
  not avoid $\beta$ since $\theta_0\leq\beta$.  Thus (3)
  holds.
\end{proof}

