Congruence modular varieties 315
a variety has permuting congruences if and only if its free algebra on three generators has
permuting congruences. Since V is locally finite, it therefore suffices to show that the finite
algebras in V have permuting congruences. Thus suppose that A ∈Vis finite and {α, β}⊆
Con A.LetB be the algebra A
2
(α) ≤ A × A consisting of all pairs x, y with xαy.Let
θ (on B) be the congruence β × β|
B
,sothatx, yθu, v if and only if xβu and yβv.By
assumption, the congruences β, α ∩ β,θ are uniform. Let b, c, e be the respective block-sizes
for these congruences. Choose any a ∈ A.Thena, a/θ consists of all x, y∈A × A such
that x ∈ a/β and y ∈ x/(α ∩ β); thus e = bc.
Now, to conclude this proof, we assume that there are elements u, v, w in A with uαvβw
and u, w ∈ β ◦ α, and we derive a contradiction. This assumption implies that there is no
x ∈ A with x, w∈u, v/θ. Note that for any z ∈ A,ifx, z∈u, v/θ for some x = x
0
then x, z∈u, v/θ precisely for the elements x ∈ x
0
/(α ∩β). Thus |u, v/θ| = b
c where b
is the number of z ∈ v/β such that such u, z∈β ◦α. Wehavethatb
<bsince w ∈ v/β and
u, w ∈ β ◦α.Thus|u, v/θ| = b
c<bc= |u, v/θ|, which is the promised contradiction. 2
13.3 Lemma In a directly representable variety, every finite subdirectly irreducible algebra
is Abelian or simple.
Proof Assume that V is a directly representable variety. By Theorem 13.2, V is congruence-
modular; in fact, it is a Maltsev variety. Since subdirectly irreducible algebras are directly
indecomposable, V has a finite bound on the size of its finite subdirectly irreducible algebras.
As we noted in our proof of Theorem 12.2, this implies that the locally finite variety V has
resb(V) <ω,andV|=(C1). NowletA be a finite subdirectly irreducible algebra in V.
To get a contradiction, we suppose that A is neither Abelian nor simple. Where β is the
monolith of A, this supposition means that 0
A
≺ β<1
A
and β ≤ [1
A
, 1
A
]. Applying (C1),
we find that β =[β, 1
A
].
Choose any positive integer n. Once again, we consider the algebra A
n
(β)ofβ-constant n-
tuples. Our goal this time is to prove that A
n
(β) is directly indecomposable. Since |A
n
(β)| >
2
n
, this will contradict the ground assumption that V is directly representable. (This is the
fourth time that we have used this fruitful construction after Section 11.)
Suppose, for sake of contradiction, that A
n
(β) is not directly indecomposable. Then it
possesses a pair of congruences δ
0
,δ
1
such that 0 <δ
ε
< 1, δ
0
∨ δ
1
=1,δ
0
∧ δ
1
=0. We
write η
i
for the kernel of the projection homomorphism of A
n
(β)toA at the i coordinate
(as usual) and put η
i
=
j=i
η
j
.Wewriteβ
i
for the kernel of the homomorphism of A
n
(β)
to A/β through the i coordinate, so that β
i
= β
j
for alll {i, j}⊆{0,...,n−1},andwewrite
simply β for this congruence. The fact that A |= β =[β, 1] gives A
n
(β) |= β = η
i
∨ [β,1]
for each i<n. (Here we have used Theorem 8.3(6), again.) For i<nwe have η
i
≺ β,
β = η
i
∨ η
i
,andη
i
∧ η
i
=0. Thenbymodularity,0≺ η
i
.
We can now show that for each i<nand ε ∈{0, 1},ifδ
ε
≤ η
i
then η
i
≤ δ
ε
. Indeed, if
δ
ε
≤ η
i
,thenβ ≤ η
i
∨ δ
ε
, giving β = η
i
∨ (β ∧ δ
ε
). Then
[η
i
,δ
1−ε
] ≤ [β,δ
1−ε
]=[η
i
∨ (β ∧ δ
ε
),δ
1−ε
] ≤ η
i
∨ [δ
ε
,δ
1−ε
]=η
i
. (13.5)
Thus [η
i
,δ
1−ε
] ≤ η
i
∧ η
i
=0. Thensince[β,1] = [η
i
∨ η
i
, 1] ≤ η
i
we have
0 =[η
i
, 1] = [η
i
,δ
0
∨ δ
1
]=[η
i
,δ
0
] ∨ [η
i
,δ
1
]=[η
i
,δ
ε
]. (13.6)