
The structure of free algebras 71
. . < Number of vectors: 20
. . Length of vectors: 8
. . V 0: ( 1, 1, 1, 1, 2, 2, 2, 2)
. . V 1: ( 1, 1, 2, 2, 1, 1, 2, 2)
. . V 2: ( 1, 2, 1, 2, 1, 2, 1, 2)
. . V 3: ( 0, 0, 0, 0, 0, 0, 0, 0)
. . V 4: ( 1, 1, 2, 2, 2, 2, 2, 2)
. . V 5: ( 1, 2, 1, 2, 2, 2, 2, 2)
. . V 6: ( 1, 2, 2, 2, 1, 2, 2, 2)
. . V 7: ( 1, 1, 1, 1, 1, 1, 2, 2)
. . V 8: ( 1, 1, 1, 1, 1, 2, 1, 2)
. . V 9: ( 1, 1, 1, 2, 1, 1, 1, 2)
.. V 10:(1,1,1,1,1,1,1,1)
.. V 11:(1,2,2,2,2,2,2,2)
.. V 12:(1,1,1,2,2,2,2,2)
.. V 13:(1,1,2,2,1,2,2,2)
.. V 14:(1,2,1,2,1,2,2,2)
.. V 15:(1,1,1,1,1,2,2,2)
.. V 16:(1,1,1,2,1,1,2,2)
.. V 17:(1,1,1,2,1,2,1,2)
.. V 18:(1,1,1,1,1,1,1,2)
.. V 19:(1,1,1,2,1,2,2,2)
Table 4
but we write it as an ordered sum in order to emphasize the origin of the two components of
the algebra as determined by those n
1
variables that are mapped to 1 and those n
2
that are
mapped to 0. We shall see that this decomposition is an example of a more general one of
the form F
V
0
(n
0
) ∪ F
V
1
(n
1
) for two varieties V
0
and V
1
with n
0
+ n
1
= n.
Next we consider how the different Ge(X, Z) fit together. We view F
S
(n)asGe(X, U)
for U the set of all valuations to S and B and we view F
B
(n)asGe(X,W)withW = B
X
.
Let Z
1
,...,Z
2
n
be a list of all the classes of ∼
B
.Letg be the canonical onto homomorphism
g : F
S
(n) → F
B
(n) that maps generators to generators. Since g is onto, for each ∼
B
class Z
there is a term m
Z
(x
1
,...,x
n
)forwhichv(g(m
Z
)) = 1 if v ∈ Z,andv(g(m
Z
)) = 0 otherwise.
Then for any
t ∈ Ge(X, Z)thetermp
Z,t
= t ∧ m
Z
is such that v(p
Z,t
)=v(t)ifv ∈ Z and
v(p
Z,t
)=0ifv ∈ Z.Let(t
1
,...,t
2
n
) ∈ Ge(X, Z
1
) ×···×Ge(X, Z
2
n
) be arbitrary. Form
the term
t(x
1
,...,x
n
)=
1≤k≤2
n
p
Z
k
,t
k
.
For every 1 ≤ k ≤ n and every v ∈ Z
k
we have v(t
k
)=v(t). Therefore we conclude that
F
S
(n)
∼
=
Ge(X, U)
∼
=
Ge(X, Z
1
) ×···×Ge(X, Z
2
n
).
By combining this with the characterization of Ge(X, Z) given in the previous paragraph