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
: & - 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
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