General theory of algebras

95 Pages·2007·1.378 MB·English

by  Don Pigozzi

GENERAL THEORY OF ALGEBRAS
DON PIGOZZI
1. Lattices
Anotionof\order"playsanimportantroleinthetheoryofalgebraicstructures. Manyof
the key results of the theory relate important properties of algebraic structures and classes
of such strutures to questions of order, e.g., the ordering of substructures, congruence
relations, etc. Order also plays an important role in the computational part of the theory;
for example, recursion can conveniently be de(cid:12)ned as the least (cid:12)xed point of an interative
procedure. Themostimportantkindoforderinginthegeneraltheoryofalgebrasisalattice
ordering, which turns out to be de(cid:12)nable by identitiesin terms of of the least-upper-bound
(the join) and greatest-lower-bound (the meet) operations.
De(cid:12)nition 1.1. A lattice is a nonempty set A with two binary operations _:A(cid:2)A ! A
(join) and ^:A(cid:2)A !A (meet) satisfying the following identities.
(L1) x_y = y_x x^y = y^x (commutative laws)
(L2) (x_y)_z = x_(y_z) (x^y)^z = x^(y^z) (transitive laws)
(L3) x_x = x x^x = x (idempotent laws)
(L4) x_(x^y) =x x^(x_y)= x (absorption laws)
Examples. (1) (2-element) Boolean algebra: A = fT;Fg.
a b a_b a^b
T T T T
T F T F
F T T F
F F F F
(2) Natural numbers: A = ! = f0;1;2;:::g. a _ b = LCM(a;b), the least common
multiple of a and b; a^b = GCD(a;b), the greatest common divisor of a and b.
1.1. Some Set Theory. Sets will normally be represented by uppercase Roman letters:
A;B;C;:::,andelementsofsetsbylowercaseRomanlettersa;b;c;:::. Thesetofallsubsets
of a set A is denoted by P(A).
f:A ! B: a function with domain A and codomain B. f(A) = ff(a) : a 2 Ag (cid:18) B is
the range of f.
ha ;:::;a i: ordered n-tuple, for n 2 !. ha ;:::;a i= hb ;:::;b i i(cid:11) (if and only if), for
1 n 1 n 1 n
all i(cid:20) n, a = b .
i i
Date:(cid:12)rst week.
1
2 DONPIGOZZI
A (cid:2)(cid:1)(cid:1)(cid:1)(cid:2)A = fha ;:::;a i: for all i (cid:20) n, a 2 A g: Cartesian (or direct) product.
1 n 1 n i i
A (cid:2)(cid:1)(cid:1)(cid:1)(cid:2)A = An if A = A for all i(cid:20) n: n-th Cartesian power of A.
1 n i
An n-ary operation on A is a function f from the n-th Cartesian power of A to itself,
i.e., f:An ! A. We write f(a ;:::;a ) for f(ha ;:::;a i). f is binary if n = 2. If f is
1 n 1 n
binary we often write afb instead of f(a;b); this is in(cid:12)x notation.
An n-ary relation on A is a subset R of the n-th Cartesian power of A, i.e., R(cid:18) An. R
is binary if n = 2. In this case aRa0 means the same as ha;a0i2 R.
De(cid:12)nition 1.2. A partially ordered set (poset) is a nonempty set A with a binary relation
(cid:20) (cid:26) A(cid:2)A satisfying the following conditions
(P1) x (cid:20) x (reflexive law)
(P2) x (cid:20) y and y (cid:20) z implies x (cid:20) z (transitive law)
(P3) x (cid:20) y and y (cid:20) x implies x = y. (antisymmetric law)
A is a linearly ordered set or a chain if it satis(cid:12)es
x (cid:20) y or y (cid:20) x
a (cid:20) b (cid:20) c means a (cid:20) b and b (cid:20) c. a < b means a (cid:20) b and a 6= b. a (cid:30) b means a < b
and, for all c 2 A, a (cid:20) c (cid:20) b implies a = c or c = b; we say that b covers a in this case,
and (cid:30) is called the covering relation. Note that, if A is (cid:12)nite, then a (cid:30) b i(cid:11) there exist
c ;:::;c 2 A such that a = c (cid:30) c (cid:30) (cid:1)(cid:1)(cid:1) (cid:30) c = b. So every (cid:12)nite poset is completely
0 n 0 1 n
determined by its covering relation.
TheHassediagram ofa(cid:12)niteposetisagraphicalrepresentationofofitscoveringrelation,
where a (cid:30) b if there is a edge that goes up from a to b. Here are the Hasse diagrams of
some posets. The set of natural numbers ! with the natural ordering, although in(cid:12)nite, is
also determined by its covering relation. But the set of real numbers R with the natural
ordering is not; the covering relation is empty.
Let A be a poset with partialordering(cid:20). Let X be a subset of elements of A. UB(X)=
fy 2 A : x(cid:20) y for every x 2Xg; the set of upper bounds of X. The least-upper-bound of
X, in symbols LUB(X), is the smallest element of UB(X), if it exists, i.e., LUB(X) is the
unique element a of UB(X) such that a (cid:20) y for all y 2 UB(X). The set of lower bounds,
LB(X) and the greatest-lower-bound, GLB(X) are de(cid:12)ned by interchanging (cid:20) and (cid:21).
First Week
GENERAL THEORY OF ALGEBRAS 3
De(cid:12)nition 1.3. A poset A is a lattice ordered set (a loset) if every pair of elements has a
least-upper-bound (LUB) and a greatest-lower-bound (GLB).
Among the posets displayed above only the third fails to be a loset.
The lattice hA;_;^i will be denoted by A; in general boldface letters will be used to
denotelatticesandposetsand thecorrespondinglowercaseletterwilldenotetheunderlying
set of the lattice or poset. The underlying set is also called the universe or carrier of the
lattice or poset.
Theorem 1.4. (i) If hA;(cid:20)i is a loset, then hA;LUB;GLBi is a lattice.
(ii) Conversely, if hA;_;^i is a lattice, then hA;(cid:20)i is a loset where a (cid:20) b if a_b = b
(equivalently, a^b = a).
Proof. (i). Theaxioms(L1){(L4)oflatticesmustbeveri(cid:12)ed. (L4)saysthatLUB(a;GLB(a;b))=
a. But GLB(a;b)(cid:20) a by de(cid:12)nition so the above equality is obvious.
(L2). We must show that
(1) LUB(a;LUB(b;c))= LUB(LUB(a;b);c):
Letdbetheleft-handsideofthisequation. d (cid:21) aandd(cid:21) LUB(b;c). Thesecondinequality
implies d (cid:21) b, d (cid:21) c. From d (cid:21) a and d (cid:21) b we get d (cid:21) LUB(a;b), which together with
d(cid:21) c gives d (cid:21) LUB(LUB(a;b);c)). This gives one of the two inclusions of (1). The proof
ofthe otherissimilar. The veri(cid:12)cationof theremaininglatticeaxiomsisleftasanexercise.
(ii). We note (cid:12)rst of all that, if a_b= b, then a^b= a^(a_b)= a by (L4). Similarly,
a^b= a implies a_b = (a^b)_b = b by (L1) and (L4). We verify (P1){(P4).
(P1). a (cid:20) a i(cid:11) a^a = a.
(P2). We must verify that a_b= b and b_c = c implies a_c= c. a_c =a_(b_c)=
(a_b)_c= b_c= c.
(P3). Suppose a_b= b and b_a = a. Then a= b by the commutativity of _. (cid:3)
For any set A, hP(A);(cid:18)i is clearly a loset with LUB(X;Y) =X [Y and GLB(X;Y) =
X \Y. Thus by the theorem hP;[;\i is a lattice.
If hA;(cid:20)i is a loset, then a (cid:20) b i(cid:11) LUB(a;b)= b (equivalently, GLB(a;b)= a). Thus, if
we start with a loset hA;(cid:20)i and form a lattice hA;LUB;GLBi by (i) and then a loset by
(ii) we get back the original loset. Conversely, the following lemma shows that if we start
with a lattice,form a loset by (ii)and then a lattice by (i), we get back the originallattice.
Lemma 1.5. Let hA;_;^i be a lattice, and de(cid:12)ne a (cid:20) b and as in part (ii) of the theorem.
Then, for all a;b2 A, LUB(a;b)= a_b and GLB(a;b)= a^b.
Proof. a^(a_b) = a. So a (cid:20) a_b. b^(a_b) = b^(b_c) = b. So b (cid:20) a_b. Suppose
a;b (cid:20) c. Then a _c = c and b_c = c. Thus (a_ b)_c = a_(b_ c) = a_c = c. So
a_b (cid:20) c. Hence LUB(a;b) = a_b. The proof that GLB(a;b) = a_b is obtain from the
above by interchanging \(cid:20)" and \(cid:21)" and interchanging \_" and \^". (cid:3)
So the mappings between lattices and losets given in Theorem 1.4 are inverses on one
another; the lattice hA;_;^iand the loset hA;(cid:20)i are essentiallythe same and we normally
will not distinguish between them in the sequel.
De(cid:12)nition 1.6. An isomorphism between lattices A = hA;_;^i and B = hB;_;^i is a
bijection (i.e., a one-one correspondence) h:A$B such that, for all a;a0 2 A, h(a_a0)=
First Week
4 DONPIGOZZI
h(a)_h(a0) and h(a)^a0)= h(a)^h(a0). A and B are isomorphic, in symbols A(cid:24)= B, if
(cid:24)
there is an isomorphism h between them. We write h:A =B.
De(cid:12)nition 1.7. An order-preserving map between posets A = hA;(cid:20)i and B = hB;(cid:20)i is
a function h:A ! B such that, for all a;a0 2 A, a (cid:20) a0 implies h(a)(cid:20) h(a0). A mapping h
is strictly order-preserving if a(cid:20) a0 i(cid:11) h(a) (cid:20) h(a0).
A mapping h is (strictly) order-preserving map between two lattices if it (strictly) pre-
serves that lattice orderings.
Theorem 1.8. Let A = hA;_;^i and B = hB;_;^i be lattices and Let h:A ! B. Then
h:A (cid:24)= B i(cid:11) h is a strictly order-preserving bijection, i.e., h is a bijection and h and h−1
are both order-preserving.
Proof. =): Let a;a0 2 A. We must show that h(LUB(a;a0)) = LUB(h(a);h(a0)) and
h(GLB(a;a0))=GLB(h(a);h(a0)). Let a00 = LUB(a;a0). a;a0 (cid:20) a00. So h(a);h(a0)(cid:20) h(a00).
Supposeh(a);h(a0) (cid:20) b2 B. Thena= h−1(h(a));b= h−1(h(a0))(cid:20) h−1(b). Soa00 (cid:20) h−1(b)
and h(a00) (cid:20) h−1(h−1(b))=b. The proof for GLB is similar.
(=: Exercise. (cid:3)
De(cid:12)nition 1.9. Let A = hA;_ ;^ i, B = hB;_ ;^ i be lattices. A is a sublattice of
A A B B
B if A (cid:18) B and a_ a0 = a_ a0 and a^ a0 = a^ a0 for all a;a0 2 A.
B A B A
1 1 1
c’ c a b a b c
B
3
c’ b’ a’
0
0
0
The lattice on the left is a sublattice of B (the three-atom Boolean algebra).
3
Let A = hA;(cid:20) i, B = hB;(cid:20) i be posets. B is a subposet of A if B (cid:18) A and, for all
A B
b;b02 B, b(cid:20) b0 i(cid:11) b(cid:20) b0.
B A
Suppose A and B are losets. In general it is not true that B is a sublattice of A if B is
a subposet of A. For example, the second lattice in the above (cid:12)gure is a subposet of B
3
but not a sublattice.
Let G = hG;(cid:1);−1;ei be a group, and let Sub(G) = fH : H < Gg be the set of
(underlyingTsets of) all subgroups of G. hSub(G);(cid:18)i is a loset, where H^K = H\K and
H _ K = fL < G : H;K < Gg. hSub(G);(cid:18)i is a subposet of hP(G);(cid:18)i but is not a
sublattice. H _K = H [K i(cid:11) H (cid:18) K or K (cid:18) H.
First Week
GENERAL THEORY OF ALGEBRAS 5
De(cid:12)nition 1.10. A latticeA= hA;_;^iisdistributive each of joinand meet distributives
over the other, i.e.,
(D1) x^(y_z)=(x^y)_(x^z);
(D2) x_(y^z)=(x_y)^(x_z):
Theorem 1.11. Either one of the two distributive laws is su(cid:14)cient, i.e., in any lattice A,
(D1) implies (D2) and (D2) implies (D1).
Also, in every lattice A, the following inidenitities hold
(2) x^(y_z)(cid:21) (x^y)_(x^z);
(3) x_(y^z)(cid:20) (x_y)^(x_z):
Thus either of the two opposite inidentities is su(cid:14)cient for distributivity.
Proof. (D1) =) (D2).
(x_y)^(x_z) = ((x_y)^x)_((x_y)^z); (D1)
= x_((x^z)_(y^z)); (L1), (L4) and (D1)
= (x_(x^z))_(y^z); (L2)
= x_(y^z); (L4):
The proof of (D1) =) (D2) is obtained from the above by interchanging \_" and \^".
Proof of (2). x^y (cid:20) x and x^z (cid:20) x together imply
(4) (x^y)_(x^z) (cid:20) x:
x^y (cid:20) y (cid:20) y_z and x^z (cid:20) z (cid:20) y_z together imply
(5) (x^y)_(x^z) (cid:20) y_z:
(4) and (5) together together imply (2).
The proof of (3) is obtained by interchanging \_" and \^" and \(cid:20)" and \(cid:21)". (cid:3)
In every lattice, x (cid:20) y and z (cid:20) w together imply both x^z (cid:20) y^w and x_z (cid:20) y_w.
To see this we note that x^z (cid:20) x (cid:20) y and x^z (cid:20) z (cid:20) w together imply x^z (cid:20) y ^w.
Proof of the other implication is obtained by the usual interchanges.
As an special case, we get that x (cid:20) y implies each of x ^ z (cid:20) y ^ z, z ^ x (cid:20) z ^ y,
x_z (cid:20) y_z, and z_x(cid:20) z _y. Thus, for every lattice A and every a2 A, the mappings
x7!x^a;a^x;x_a;a_x are all order preserving.
Department of Mathematics, Iowa StateUniversity, Ames, IA 50011,USA
E-mail address: [email protected]
First Week
De(cid:12)nition 1.12. A poset A = hA;(cid:20)i is complete if, for every X (cid:18) A, LUB(X) and
GLB(X) both exist.
W V
We denote LUB(X) by X (if it exists) and GLB(X) by X.
If A is a lattice and X = fx ;:::;x g is (cid:12)nite, then LUB(X) and GLB(X) always exist
1 n
and equal x _(cid:1)(cid:1)(cid:1)_x and x ^(cid:1)(cid:1)(cid:1)^x , respectively. Thus every (cid:12)nite latticeis complete.
1 n 1 n
h!;(cid:20)i, hQ;(cid:20)i (Q the rational numbers), and hR;(cid:20)i (R the real numbers) are not com-
plete, but the natural numbers and the real numbers can be made complete by adjoining a
largest element 1 to the natural numbers and both a smallest element −1 and a largest
element +1 to the reals. The rationals cannot be completed so easily; in fact, for every
irrational number r, fq 2 Q : qW< rg fTails to haVve a leaSst upper bound. For any set A,
hP(A);(cid:18)i is a complete lattice. K = K and K = K for every K (cid:18) P(A).
W
TheoremV1.13. Let A = hA;(cid:20)i be a poset. For every X (cid:18) A, X exists i(cid:11), for every
X (cid:18) A, X exists. Thus a poset and in particular a lattice is complete i(cid:11) every subset
has a LUB, equivalently, i(cid:11) every subset has a GLB.
Proof. =). Assume LUB(X) exists for every X (cid:18) A. It su(cid:14)ces to show that, for every
X (cid:18) A, GLB(X)= LUB(LB(X)); recall that LB(X)= fy 2 A : for every x 2 X, y (cid:20) xg,
the set of lower bounds of X. Let a = LUB(LB(X)). For every x 2 X, x 2 UB(LB(X)).
Thus, for every x 2 X, x (cid:21) LUB(LB(X)) = a. So a 2 LB(X), and clearly, for every
y 2 LB(X), a (cid:21) y.
(=. The proof is the dual of the one above, i.e., it is obtained by interchanging \(cid:20)"
and \(cid:21)", \LUB" and \GLB", and \UB" and \LB". (cid:3)
WFor aTny groupVG, SubT(G)= hSub(G);(cid:18)Si is a complete lattice. For every K (cid:18) Sub(G),
K = K and K = fH 2 Sub(G) : K (cid:18) Hg.
Why doesn’t h!;(cid:20)i contradict the above theorem? Isn’t the greatest lower bound of
any set of natural numbers the smallest natural number in the set? This is true for any
nonempty set, but ; has no GLB in !. For any poset A, GLB(;) (if it exists) is the largest
element of A, and LUB(;) is the smallest element of A.
Let A = h! [fa;b;1g; (cid:20)i, where (cid:20) is the natural order on !, for every n 2 !, n <
a;b;1, and a;b < 1. Every (cid:12)nite subset of A has a LUB (including ;), but GLB(a;b)
does not exist. So the requirement that X ranges over all subsets of A in the theorem is
critical. But if A is a (cid:12)nite poset, and LUB(a;b)exists for every pair of elements of A and
A has a smallest element, then A is a complete lWattice.V
Some notatioVn: if AWis a complete lattice, 1= A= ; will denote the largestelement
of A, and 0 = A = ; will denote the smallest element.
De(cid:12)nition 1.14. Let A = hA;_;^i and B = hB;_;^i be complete lattices. B is a
W W V V
complete sublattice of A if, for every X (cid:18) A, X = X and X = X.
B A B A
hf−2g[(−1;+1)[f+2g;(cid:20)i isa complete latticeand a sublatticeof the completelattice
hf−1g[R[f+1g; (cid:20)i but not a complete sublattice.
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7
De(cid:12)nition 1.15. Let A be a set. E (cid:18) A2 is an equivalence relation on A if
(E1) xRx;
(E2) xEy and yEz imply xEz,
(E3) xEy implies yEx. (symmetric law)
Eq(A) will denote the set of all equivalence relations on A. Let K (cid:18) Eq(A).
\
(6) K 2 Eq(A):
Check (E1){(E3). T
(E2). Assume ha;bi;hb;ci 2T K. For every E 2 K, ha;bi;hb;ci 2 E. Thus, for every
E 2 K, ha;ci2 E. So ha;ci2 K. (E1) anWd (E3)Tare veri(cid:12)ed similarly.
So hEq(A);(cid:18)i is a complete lattice with K = K and
^ \ [
(7) K = fE 2 Eq(A): K (cid:26) Eg:
The smallest equivalence relation on A is the identity or diagonal relation, (cid:1) = fha;ai:
A
a 2 Ag, read \delta A". The largest equivalence relation is the universal relation, r =
A
A(cid:2)A, read \nabla A".
The description of the join operation in (7) is what can be called a \coinductive" or
\fromabove"characterization;itisveryusefulfortheoreticalpurposes, forexampleproving
general propositions about the join, but it does not give much information about what the
elements of the join of K look likein terms of the elements of the subgroups or the ordered
pairs of the equivalence relations of K. For this we need an \inductive" or \from below"
characterization.
Theorem 1.16. Let H;K be subgroups of G = hG;(cid:1);−1;ei.
[
H _K = HK[HKHK[HKHKHK[(cid:1)(cid:1)(cid:1)= (HK)n;
1(cid:20)n2!
where (HK)n = fh (cid:1)k (cid:1)(cid:1)(cid:1)h (cid:1)k :h ;:::;h 2 H;k ;:::;k 2Kg.
1 1 n n 1 n 1 n
S
Proof. Let L= (HK)n. We must show three things.
1(cid:20)n2!
(8) L2 Sub(G);
(9) H;K (cid:18) L;
(10) for all M 2 Sub(G), H;K (cid:18) M implies L(cid:18) M.
Proof of (8). Clearly L 6= ;. Let a;b 2 L. We must show ab 2 L (following convention
we often omit the \(cid:1)" when writing the product of elements of groups) and a−1 2 L.
a 2 (HK)n and b 2 (HK)m for some n;m 2 !. So ab 2 (HK)n+m (cid:18) L, and a−1 2
K−1H−1(cid:1)(cid:1)(cid:1)K−1H−1 = (KH)n = feg(KH)nfeg = H(KH)nK = (HK)n+1 (cid:18) L.
(9). H = Hfeg(cid:18) HK (cid:18) L and K (cid:18) fegK (cid:18) HK (cid:18) L.
(10). Suppose H;K (cid:18) L 2 Sub(G). We prove by induction on n that (HK)n (cid:18) M.
HK (cid:18) MM (cid:18) M. Assume (HK)n (cid:18) M. Then (HK)n+1 (cid:18) (HK)nHK (cid:18) MM (cid:18) M. (cid:3)
week 2
8
A similar argument gives a inductive characterization of the join of an arbitrary set of
subgroups fH :i2 Ig. We leave as an exercise the proof that
i
_ [
H = H (cid:1)(cid:1)(cid:1)H ;
i i1 in
i2I hh1;:::;hni2I(cid:3)
(cid:3)
where I is the set of all (cid:12)nite sequences of elements of I.
Wenext obtainasimilar\inductive"characterizationofthe joinofequivalence relations.
For this purpose we need to explain some basic results in the \calculus of relations". Let
A;B;C be arbitrary sets and R (cid:18) A(cid:2)B and S (cid:18) B(cid:2)C. By the relative product of R and
S, in symbols R(cid:14)S, we mean the relation
fha;ci2 A(cid:2)C :there exists a b2 B such that aRbScg:
The relative product is a binary operation on the set of all binary relations on any set A
such that (cid:1) acts as an identity(i.e.,(cid:1) (cid:14)R= R(cid:14)(cid:1) =R). Also r acts like an in(cid:12)nity
A A A A
element on reflexive relations, i.e., (cid:1) (cid:18) A implies r (cid:14) R = R (cid:14) r = r . We also
A A A A
have a unary converse operation that has some of the properties of the inverse of a group
(but is not a group inverse). R(cid:20) = fha;bi: hb;ai2 Rg, i.e., aR(cid:20)b i(cid:11) bR(cid:20)a. a(R(cid:14)S)^a0 i(cid:11)
a0(R(cid:14)S)a i(cid:11) there exists a b such that a0RbSa i(cid:11) there exists a b such that aS(cid:20)bR(cid:20)a0 i(cid:11)
aS(cid:20)(cid:14)R(cid:20)a0. So (R(cid:14)S)^ = S(cid:20)(cid:14)R(cid:20).
Wenotethatthenotationofthecalculusofrelationscanbeusedtoformulatethede(cid:12)ning
conditions of an equivalence relation in very simple terms. The reflexive law: (cid:1) (cid:18) E; the
A
transitive law: E(cid:14)E (cid:18) E; the symmetric law: E(cid:20) (cid:18) E.
Theorem 1.17. Let E;F 2 Eq(A).
[
E_F = E(cid:14)F [E(cid:14)F (cid:14)E(cid:14)F [(cid:1)(cid:1)(cid:1)= (E(cid:14)F)n:
1(cid:20)n2!
S
Proof. Let G = (E(cid:14)F)n. We show that G2 Eq(A). (cid:1) = (cid:1) (cid:14)(cid:1) (cid:18) E(cid:14)F (cid:18) G.
1(cid:20)n2! A A A
Assume ha;bi;hb;ci 2 G, i.e., that there exist n;m 2 ! such that a(E(cid:14)F)nb(E(cid:14)F)mc.
T(cid:0)hen ha;ci 2 (E (cid:14) F)n (cid:14) (E (cid:14) F)m = (E (cid:14) F)n+m (cid:18) G. We also have that hb;ai 2
(E(cid:14)F)n)^ = (F(cid:20)(cid:14)E(cid:20))n = (F (cid:14)E)n (cid:18) (E(cid:14)F)n+1 (cid:18) G. The proof that G = E _F is left
at an exercise. (cid:3)
We also leave as an exercise the following inductive characterization of the join of an
arbitrary set fE :i2 Ig of equivalence relations on a set A.
i
_ [
E = E (cid:14)(cid:1)(cid:1)(cid:1)(cid:14)E :
i i1 in
i2I hh1;:::;hni2I(cid:3) S
Exercise. Let R (cid:18) A2 be an arbitrary binary relation on A. Prove that Rn,
1(cid:20)n2!
where Rn = R(cid:14)R(cid:14)(cid:1)(cid:1)(cid:1)(cid:14)R with n repetitions of R, is the smallest transitive relation that
includes R. It is called the transitive closure of R.
Every equivalence relation is uniquely determined by its corresponding partition.
P is a partition of a set A if
(cid:15) PS(cid:18) P(AS)nf;g,
(cid:15) P (= fX :X 2 Pg) = A,
(cid:15) for all X;Y 2 P, X 6= Y implies X \Y = ;.
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Let E 2 Eq(A). For each a 2 A, let [a] = fx2 A :xEag, called the equivalence class of
E
a (over E); [a] is also denoted by a=E.
E
f[a] : a 2 Ag is a partition of A (exercise). Conversely, if P is a partition of A,
E
de(cid:12)ne a (cid:17)P by the condition that there is an X 2 P such that a;b 2 X. Then (cid:17)P is an
equivalence relationwhosepartitionisP (exercise). Moreover,foreach equivalence relation
E, (cid:17)f[a] :a2Ag= E.
E
Part(A) denotes the set of partitions of A, and Part(A)= hPart(A);(cid:20)i, where (cid:20) is the
partial ordering on Part(A) de(cid:12)ned as follows: P (cid:20) Q if, for each X 2 P, there exists a
Y 2 Q such that X (cid:18) Y, equivalently, each equivalence class of Q is a union of equivalence
classes of P. The mapping P 7! (cid:17)P is bijection between the posets Eq(A) and Part(A)
that is strictly order-preserving (exercise). Thus Part(A) is a complete lattice and the
above mapping is a lattice isomorphism.
Itisusuallyeasiertopicturethepartitionofaspeci(cid:12)cequivalencerelationratherthanthe
relation itself. The following characterizations of the join and meet operations in Part(A)
are left as exercises. In the lattice Part(A),
P ^Q =fX \Y :X 2 P;Y 2 Q;X \Y 6= ;g:
A (cid:12)nite sequence X ;Y ;:::;X ;Y , where X ;:::;X 2 P and Y ;:::;Y 2 P is called
1 1 n n 1 n 1 n
connected if X \Y 6= ; for all i (cid:20) n and Y \X 6= ; for all i<n. Exercise: show that,
i i i i+1
for every a 2 A, b 2 [a]P[Q i(cid:11) there exists a connected sequence X1;Y1;:::;Xn;Yn such
that a 2 X and b2 Y .
1 n
Exercise. For each n 2 !nf0g, de(cid:12)ne (cid:17) (mod n) 2 Z2 by a (cid:17) b (mod n) if a = b+kn
for some k 2 Z, i.e., nja − b. Show (cid:17) (mod n) 2 Eq(Z). Describe the partition of (cid:17)
(mod n). Describe (cid:17) (mod n)^(cid:17) (mod m) and (cid:17) (mod n)_(cid:17) (mod m).
The latticesof subgroups and equivalence relationshave special properties. In the sequel
we write X0 (cid:18) X to mean that X0 is a (cid:12)nite subset of X.
!
De(cid:12)nitionW1.18. Let A be a lattice. An element c of A is compact if, for every X (cid:18) A
such that X exists,
_ _
c(cid:20) X implies there exists a X0 (cid:18) X such that c (cid:20) X0.
!
The set of compact elements of A is denoted by Comp(A).
A is compactly generated if every element of A is the join of the compact elements less
than or equal to it, i.e., for every a2 A,
_
a = fc2 CompA: c(cid:20) ag:
A lattice is algebraic if it is complete and compactly generated.
We note that an element is compactly generated i(cid:11) it is the join of some set of compact
elements, since in this case it must be the join of all compact elements less than or equal
to it.
Examples.
(cid:15) Every (cid:12)nite lattice is algebraic.
(cid:15) The lWattice h! [ f1g;(cid:20)i is algebraic. 1 is the only noncompact element and
1= !.
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(cid:15) h[0;1];(cid:20)i, where [0;1] = fx 2 R : 0 (cid:20) x (cid:20) 1g, is complete but not algebraic; 0 is
the only compact element and hence the only element that is the join of compact
elements.
(cid:15) hSub(G);(cid:18)i, for every group G, and hEq(A);(cid:18)i, for every nonempty set A, are
algebraic lattices, but this will be shown until later.
1.2. Closedset systemsand closureoperators. AfamilyK ofsetsissaidtobeupward
directed by inclusion, or upward directed for short, or even shorter, simply directed, if each
pair of sets in K is included in a third member of K, i.e.,
for all X;Y 2 K there exists a Z 2 K such that X _Y (cid:18) Z.
De(cid:12)nition 1.19. A closed set system consists of a nonempty set A and a C (cid:18) P(A) such
that C is closed under intersections of arbitrary subsets, i.e.,
\
for every K (cid:18) C, K 2 C.
AclosedsetsystemhA;Ciisalgebraic ifC isclosedunder unionsofupwarddirectedsubsets,
i.e., [
for every directed K (cid:18) C, K 2 C.
T
Note that by de(cid:12)nition C always contains A since A = ;. Since Sub(G) and Eq(A)
are closed under intersections of arbitrary subsets, to show they form algebraic closed-set
systems it su(cid:14)ces to show they are closed under unions of directed subsets.
The union of any, not necessarily directed, K (cid:18) Sub(G)Scontains the identity and is
closed under inverse. Assume that K is directed. Let a;b 2 K, and let H;L 2SK such
tShat a 2 H and b 2 L. Choose M 2 K such that K [L (cid:18) M. Then ab 2 M (cid:18) K. So
K 2 Sub(G).
The union of any, not necessarily directed, K (cid:18) Eq(A)Sincludes (cid:1)A and is closed under
converse. Assume that K is directed. Let ha;bi;hb;ci2 K, and let H;L 2 K suchSthat
ha;bSi2 H and hb;ci2 L. Choose M 2 K such that K [L (cid:18) M. Then ha;ci2 M (cid:18) K.
So K 2 Eq(A).
Eachofthede(cid:12)ningconditionsofagroupinvolvesonlya(cid:12)nitenumberofgroupelements,
and, similarly,each of the conditions that de(cid:12)ne equivalence relationsinvolves only a (cid:12)nite
number of ordered pairs. This is the common property that guarantees subgroups and
equivalence relations form an algebraic lattice. This vague observation will be made more
precise in the next chapter.
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