Chapter 4 New results on the commutator in varieties with
4.2 A new result on affine behavior in difference term varieties
In his 1995 paper,Varieties with a difference term, Keith Kearnes obtains, as Lemma 2.9, the following partial generalization of a result of Freese and McKenzie concerning the commutator in congruence modular varieties, given in (1987) as Theorem 5.7.
Theorem 4.8. For A ∈ V, a variety with difference term d and α ∈ ConA, the
following conditions are necessary and sufficient to exhibit [α, α] = 0A:
• For any fundamental operation s (and hence for any term operation) of arity,
say, n, and xiα yiα zi, i= 1, . . . , n,
d(s(x), s(y), s(z))) =s(d(x1, y1, z1), . . . , d(xn, yn, zn)).
and
• For any x α y,
y=d(y, x, x) =d(x, x, y).
Remark 4.9. It seems rather clear (though I haven’t fastidiously checked it) that
forA∈ V, a variety with a weak difference termw, a similar result holds with “weak difference term” replacing “difference term.”
Now, it turns out that a stronger version of this theorem is also true—in fact, Theorem 5.7 from Freese and McKenzie (1987) extends intact to varieties with a dif- ference term, as shown in the following. Following its proof, we shall apply this result to extend Freese and McKenzie’s characterization of the center of an algebra in a congruence modular varieties to algebras; it works also if one only assumes the pres- ence of a difference term. This latter finding enables us to similarly characterize the upward central series of any algebra in a difference term variety. This, in turn, we use to prove the equivalence of upward and downward nilpotence in a difference term va- riety, which is enough to implicitly demonstrate that the equational characterization
of nilpotence in congruence modular varieties, given by Freese and McKenzie (1987) as Theorem 14.2, works also in difference term varieties. We do give an explicit proof of this, however.
Theorem 4.10. For A ∈ V,a variety with difference termd andβ ≤α fromConA,
the following conditions are necessary and sufficient to exhibit [β, α] = 0A:
• For any fundamental operation s (and hence fors any term operation) of arity, say, n, and all xiβ yiα zi, i= 1, . . . , n,
d(s(x), s(y), s(z))) =s(d(x1, y1, z1), . . . , d(xn, yn, zn)).
and
• For any x β y, y=d(y, x, x) =d(x, x, y).
Proof. We begin with a brief
Lemma 4.11. Let A ∈ V, a variety with a difference term, and α≥β ∈ConA for
which [β, α] = 0A. Then for any x, y ∈A, if
(i) hx, zi∆βαhy, zi for some z
or (ii) hz, xi∆β
αhz, yi for some z
then x=y.
Proof. We prove that (ii) implies the conclusion, the other case having a similar
proof. So suppose (ii) holds for some x, y, and z. In light of Mal’cev’s description of congruence generation (Proposition A.9) we get thatx β y. Thus, using the definition
of ∆β
α, we find that
hz, xi∆βαhz, xi hz, xi∆βαhz, yi hx, xi∆βαhy, yi.
Applying d “vertically" (and using also that [β, β]≤[β, α] = 0A), we get that
hx, xi∆βαhy, xi.
Now, by Remark A.38, the claim holds.
Let A ∈ V and α, β ∈ ConA, be as described in the hypotheses. First, suppose that [β, α] = 0A. Then since d is a difference term and [β, β] ≤ [β, α] = 0A, we get
the second bullet immediately. Now, choose fundamental operation s of arity, say, n
andxiβ yiα zi,i= 1, . . . , n. Also, letx, y, z ∈Awithx β y α z. Sincedis a difference term and [β, β]≤[α, β] = 0A, we can apply d vertically to
hx, xi∆βαhy, yi hy, xi∆βαhy, xi hz, xi∆βαhz, xi
to obtain that hd(x, y, z), xi∆βαhz, yi. We now apply this observation in two ways. Since s(x)β s(y)α s(z), we have that
hd(s(x), s(y), s(z)), s(x)i∆βαhs(z), s(y)i.
Also, for each i= 1, . . . , n we have that
hd(xi, yi, zi), xii∆βαhzi, yii.
Applying sto the tuple given on the left and right sides of the line above then yields
After using transitivity of ∆β
α, we can use Lemma 4.11 (ii) to obtain the first bullet.
We now assume that the bulleted conditions hold and prove that [β, α] = 0A
(using, of course, too, that β ≤ α). We claim that under our assumptions, we get that
hx, yi∆αβhu, vi if and only if xβ y α u and v =d(y, x, u).
Let ∆0 be the binary relation on βcharacterized by the second part in the line above. Notice first that for hx, yi and hu, vi as described on the right in the displayed line above, v = d(y, x, u)β d(y, y, u) = u, and hence ∆0 is indeed a subset of β ×β. It is important to note, too, that, since β ≤ α, we have that x α y α u α v, for any
hhx, yi,hu, vii ∈∆0.
We first show that ∆0 is a congruence on β. To see that ∆0 is reflexive, take any
hx, yi ∈β. Then, sinceβ ≤α, we get x β y α x and, by the second bullet,
y=d(y, x, x),
which puts hx, yi∆0hx, yi. Now assume that hx, yi∆0hu, vi. Using that β ≤ α, we have that v = d(y, x, u)α d(x, x, x) = x. Thus, u β v α x. Also, by the first bullet, using d0 :=d in the place ofs, we discover that
d(v, u, x) =d0(d(y, x, u), d(x, x, u), d(x, x, x)) =d(d0(y, x, x), d0(x, x, x), d0(u, u, x)) =d(d0(y, x, x), x, x) =y,
where we have used the second bullet twice to obtain the last equality. Thus, we find that ∆0 is symmetric.
To see that it is transitive, suppose that
hx, yi∆0hu, vi∆0hz, wi.
For ease of reference, notice that this entails that x β y α u β v α z, v =d(y, x, u), and
too, thatu α v α z. Using the equations just above and the bulleted conditions (again, with d0 :=d in the place ofs), we also have that
d(y, x, z) =d0(d(y, x, x), d(x, x, x), d(u, u, z)) =d(d0(y, x, u), d0(x, x, u), d0(x, x, z)) =d(v, u, z) =w.
It follows that ∆0 is transitive.
That ∆0 respects the operations ofβfollows immediately from the first bullet, the fact that β and α respect the operations of A, that operations on β are computed coordinate-wise, and thatβ ≤α. We observe this computation now. Suppose that f
is a fundamental operation of arityn , andhxi, yii∆0hui, viifor i= 1, . . . , n. Writing
out what this means, we get that for i = 1, . . . , n, xiβ yiα ui and vi = d(yi, xi, ui).
Since α and β respect f, we immediately get that
f(x)β f(y)α f(u).
Using the first bullet, withf in the place ofs, we also get that
f(v) =f(d(y1, x1, u1), . . . , d(yn, xn, un)) =d(f(y), f(x), f(u)).
Finally, we observe that in β, we compute that, for any n-tuplesa and b fromA,
fβ(ha1, b1i, . . . ,han, bni) = hfA(a), fA(b)i.
Thus, ∆0 is a congruence onβ.
Note that for any x α y, reflexivity of β and the equation y = d(x, x, y) puts
hx, xi∆0hy, yi; thus we have that the generators of ∆α
β are contained in congruence
∆0 and hence ∆α β ⊆∆
Now take hx, yi∆0hu, vi. Then x β y α u and v = d(y, x, u). In particular, x α u
and so hx, xi∆α
βhu, ui. Thus, using also the second bullet,
hx, yi=hd(x, x, x), d(y, x, x)i
=d(hx, yi,hx, xi,hx, xi) ∆αβd(hx, yi,hx, xi,hu, ui)
=hd(x, x, u), d(y, x, u)i
=hu, vi
We thus conclude that ∆0 = ∆α
β, as claimed.
We will use this observation in concert with another claim, which we add now. Under the present assumptions, we hold that
[α, β]⊆ {hx, yi | hx, xi∆αβhx, yi}.
Let
θ :={hx, yi ∈[α, β]| hx, xi∆αβhx, yi}.
We wish to show that θ comprises the whole of [α, β]. Using our characterization of ∆α
β just given we will then quickly finish the proof.
We first show that θ is a congruence. Reflexivity is immediate from the reflex- ivity of [α, β] and ∆αβ. Now, suppose that hx, yi ∈ [α, β] and hx, xi∆αβhx, yi. Using reflexivity of ∆α
β, we get that
hx, xi∆αβhx, xi,
hx, xi∆αβhx, yi,
hy, yi∆αβhy, yi.
Of course, by definition of ∆αβ, we have x β y, and, using the second bullet, we can apply d “vertically” to the left and right sides above to obtain
We now need only notice symmetry of [α, β] to conclude thatθ is symmetric.
Similarly, to find that θ is transitive, we suppose that x θ y θ z. Immediately, we get hx, zi ∈[α, β], by the transitivity of that relation. We apply d in the same way as above, this time to the left and right sides of
hx, xi∆αβhx, yi hy, yi∆αβhy, yi hy, yi∆αβhy, zi
to conclude that hx, xi∆α βhx, zi.
That θ respects the operations of A is an immediate consequence of the facts that [α, β] and ∆α
β both respect the operations of their respective algebras and that
operations are computed coordinate-wise onβ. So, θis a congruence on Acontained in [α, β].
Now, by Theorem A.37, we have that [α, β] is the least congruenceγ onAso that
γ∩β is the union of ∆α
β-classes. We contend that θ is also the union of ∆αβ-classes.
To see this, suppose thathx, yi ∈θ and thathx, yi∆α
βhu, vi. Sinceθ ⊆[α, β] and the
latter is a union of ∆α
β-classes, we get thathu, vi ∈[α, β]. But alsohx, xi∆αβhx, yiand
so, by transitivity of ∆α
β,hx, xi∆αβhu, vi. By Mal’cev’s characterization of congruence
generation A.9, we evidently have that x α u. Thus, hu, ui∆α
βhx, xi∆αβhu, vi, which
puts hu, vi ∈θ.Thus, the contention is good, and so θ is all of [α, β].
Now, take an arbitrary x[β, α]y. By Theorem A.40, x[α, β]y. Then, as we have just seen, hx, xi∆α
βhx, yi. But since ∆αβ = ∆0, we get that y=d(x, x, x) =x. Thus,
[β, α] = 0A, as claimed. This finishes the proof.
The following was noted by Kearnes.
Corollary 4.12. In a varietyV with a difference term, every abelian algebra is affine
For an explanation of what is meant by “affine” see Theorem 3.5.
Theorem 4.10 enables us to construct useful generators for [β, α], for any con- gruences β and α of A such that β ≤ α, for a given algebra A in a difference term variety.
Theorem 4.13. Let V a be variety with a difference term d and equipped with fun-
damental operation symbols F. Let A ∈ V, and let β, α ∈ ConA such that β ≤ α.
Then [β, α] = CgG, where
G:={hdA(fA(x), fA(y), fA(z)), fA(dA(x1, y1, z1), . . . , dA(xn, yn, zn))i |
f ∈F and xiβ yiα zi} ∪ {hy, dA(y, x, x)i |x β y}
Proof. Writed fordA. Let θ:= CgG, where Gis as above. Using Theorem 4.10 and a basic commutator fact that holds in any algebra—namely that
[β/[β, α], α/[β, α]] = 0A/[β,α]
—one soon finds that θ ⊆[β, α]: Take any f ∈ F of arity, say, n. Let us writef for
fA, as well. Then, since [β, α]⊆α∩β, we get that givenxiβ yiα zi(i= 1, . . . , n), xi/[β, α]β/[β, α]yi/[β, α]β/[β, α]zi/[β, α],
and so, by Theorem 4.10,
d(f(x), f(y), f(z)) [β, α]f(d(x1, y1, z1), . . . , d(xn, yn, zn)).
Similarly, if x β y, then using the theorem again, we get that
y[β, α]d(y, x, x).
It follows that θ⊆[β, α], and, in particular, θ ⊆α∩β.
To get the reverse inclusion, we need only check that C(β, α;θ). Now, evidently, by Theorem 4.10 we have that [β/θ, α/θ] = 0A/θ. Thus, C(β/θ, α/θ; 0A/θ). By Proposition A.28 we then get that C(β, α;θ).
4.3 The equivalence of upward and downward nilpotence in varieties