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4) The subset r0, 1s of the real line is a complete lattice, but it is not algebraic. In the next chapter we will encounter two situations where algebraic lattices arise, namely as lattices of subuniverses of algebras and as lattices of congruences on algebras. E XERCISES §4 1. Show that the binary relations on a set A form a lattice under ❸. 2. 6 is indeed an equivalence relation on A. 3. If I is a closed and bounded interval of the real line with the usual ordering, and P is a nonempty subset of I with the same ordering, show that P is a complete sublattice of I iff P is a closed subset of I.

6. If α : A Ñ B is a homomorphism and X is a subset of A then α Sg♣X q ✏ Sg♣αX q. P ROOF. From the definition of E (see §3) and the fact that α is a homomorphism we have αE ♣Y q ✏ E ♣αY q for all Y ❸ A. Thus, by induction on n, αE n ♣X q ✏ E n ♣αX q for n ➙ 1; hence α Sg♣X q ✏ α X ✟ ❨ E ♣X q ❨ E 2 ♣X q ❨ ☎ ☎ ☎ ✏ αX ❨ αE ♣X q ❨ αE 2♣X q ❨ ☎ ☎ ☎ ✏ αX ❨ E ♣αX q ❨ E 2♣αX q ❨ ☎ ☎ ☎ ✏ Sg♣αX q. 7. Let α : A Ñ B be a homomorphism. Then the kernel of α, written ker♣αq, and sometimes just ker α, is defined by ker♣αq ✏ t①a, b② € A2 : α♣aq ✏ α♣bq✉.

A sequence of consecutive numbers. P ROOF. Let B be an irredundant basis with ⑤B ⑤ irredundant bases A with ⑤A⑤ ↕ i. ✏ j. Let K be the set of The idea of the proof is simple. We will think of B as the center of S, and measure the distance from B using the “rings” Cnk 1 ♣B q ✁ Cnk ♣B q. We want to choose a basis A0 in K such that A0 is as close as possible to B, and such that the last ring which contains elements of A0 contains as few elements of A0 as possible. We choose one of the latter elements a0 and replace it by n or fewer closer elements b1 , .

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Algebraic surfaces and golomorphic vector bundles by Friedman.


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