10

PETER H. KRAUSS AND DAVID M. CLARK

Lemma 1.12 Suppose (I, 3"(1ft)) is compact whereas (I, S'(ff)) is

not. If J is the set of prime filters on J* then

J = h(l) U {*}.

Proof: Since I C %, by Lemma 1.10, DR,€ J. Suppose l^€ J, where

IT^ X and let IL = V*n W, . By Lemma 1 .10, 7- 6 is a prime filter on fe, .

Since (I,?(R)) is compact, D U. # 0. Let i € 0 K. . By Lemma 1.10,

IT= h(i). •

Our second topic for review are redundancies in subdirect products. We

shall be concerned with redundancies which play a central role in the investi-

gation of quasi primal algebras (See Krauss [27]). Again let

Jg-cn&. I i € I be subdirect. For i, j € I we define i ~ j(#) if

there exists an isomorphism f from (A. onto Oi. such that for all

x € B, f(x(i)) = x(j). J c 1 is called a jr-transversal if for every

i, j € J, i~j(d5-) if and only if i = j. J is called ^-complete if for

every i € I there exists j € J such that i ~ j(^6-). &- is called

irredundant if I is a tr -transversal.

Lemma 1.13 If £r is an algebra then for every i, j € I, i~j(4?)

if and only if 6. = 9* . I

Lemma 1.1U If J is a complete &-transversal then

TTT

: & - U01 I j € J

is a subdirect embedding and T\ (&) is irredundant. •

Lemma 1 -1U tells us that we can always "make a subdirect product irredun-

dant" by "removing redundancies". We shall see that many special properties

of subdirect representations we are interested in are preserved by this

process.