
50 J. Berman
V has cardinality at most f (n). The number of valuations v from a set X of size n to a
particular algebra A is bounded above by |A|
n
,whichisatmostf(n)
n
.
Let V be a locally finite variety, X = {x
1
,...,x
n
},andU ⊆ val(X, V). We present a
concrete representation of Ge(X, U) as a rectangular array of elements from Alg(v)forv
ranging over U .Eachv ∈ U determines a column in this array and each element of Ge(X, U)
determines a row. The first n rows are indexed by
x
1
,...,x
n
.Ift(x
1
,...,x
n
)isanyterm,then
t denotes t
Ge(X,U)
(x
1
,...,x
n
). Since the x
i
generate Ge(X, U), every element of Ge(X, U)
is of the form
t for some term t.Forv ∈ U and t = t(x
1
,...,x
n
) ∈ Ge(X, U), the entry in
row
t and column v is v(t) ∈ Alg(v). Note that v(t)=t
Alg(v)
(v(x
1
),...,v(x
n
)). In column
v, every element of Alg(v) appears at least once since v is a valuation. Moreover, column v
codes the projection pr
v
of Ge(X, U) onto the algebra Alg(v).
In the special case that V is generated by a finite algebra A and U = A
X
,thenU is free
for V by Corollary 1.4, the number of columns of Ge(X, U)is|A|
n
, and the number of rows
of Ge(X, U)is|F
V
(n)|.
We next present three examples of how Ge(X, U) may be used to describe the structure
of free algebras.
1.8 Example Let S be the variety of semilattices. We show how to find a normal form for an
arbitrary term and how to use this normal form to determine the structure of free semilattices.
If t(x
1
,...,x
n
) is any semilattice term, then by associativity we may ignore the parentheses
and write t as a string of variables. By commutativity we may sort the variables in the string
by increasing subscripts. Idempotence allows us to conclude S|= t ≈ x
i
1
x
i
2
...x
i
k
where
1 ≤ i
1
<i
2
< ···<i
k
≤ n and var(t)={x
i
1
,x
i
2
,...,x
i
k
}. Thus a concrete representation of
F
S
(X) as a join semilattice is the set of nonvoid subsets of X with the operation of union.
In particular |F
S
(X)| =2
n
− 1.
The variety S is generated by the 2-element semilattice S
2
with universe {0, 1}. Without
loss of generality S
2
is a join semilattice. The set val(X, S
2
)has2
n
− 2 elements. Thus,
F
S
(X) is isomorphic to Ge(X, val(X, S
2
)). If we view Ge(X, val(X, S
2
)) as an array, then
it has 2
n
− 2 columns and 2
n
− 1rows. IfU ⊆ val(X, S
2
) consists of the valuations v
i
for
1 ≤ i ≤ n where v
i
(x
j
) = 1 if and only if i = j,thenU is free for S. To see this, it suffices to
show that the projection pr
U
is one-to-one. Suppose t = t(x
1
,...,x
n
)ands = s(x
1
,...,x
n
)
are distinct elements of Ge(X, val(X, S
2
)). Let v be any valuation into S
2
that separates s
and
t,withsayv(t) = 1. So there exists x
i
∈ var(t) − var(s). Then v
i
(x
i
)=1=v
i
(t) while
v
i
(s) = 0. So a valuation v
i
in U separates t and s. Hence U is free for S. The array for
Ge(X, U)hasn columns and 2
n
− 1 rows. A cardinality argument shows that no proper
subset of U is free for S. Although U is free for S, it is not independent since there is no
t
having v(t) = 0 for all v. However, if S were the variety of join semilattices with constant 0,
then U would be both free and independent for S with F
S
(n)
∼
=
S
n
2
for S
2
the 2-element join
semilattice with constant 0.
1.9 Example Let B be the variety of Boolean algebras. This variety is generated by the 2-
element Boolean algebra B
2
= {0, 1}, ∧, ∨,
, 0, 1, which is the only subdirectly irreducible
algebra in the variety. Note that val(X, B
2
)=B
X
2
.Forv ∈ val(X, B
2
)lett
v
be the term
x
v(x
1
)
1
∧···∧x
v(x
n
)
n
, where x
0
i
= x
i
and x
1
i
= x
i
.Thenv(t
v
) = 1 but w(t
v
) = 0 for every
valuation w = v. The array for Ge(X, val(X, B
2
)) has 2
n
columns. By considering joins of the
2
n
different t
v
we see that there are 2
2
n
rows. So val(X, B
2
)isfreeforB and is independent