By V. A. Krechmar

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**Extra info for A Problem Book in Algebra **

**Example text**

Our definition of a countable set differs from that of [Bishop 1967], [Bridges 1979] and [Bishop-Bridges 1985) in that we do not require that a countable set be nonempty (or even that we can decide whether or not it is emptyl; for discrete sets, it is equivalent to the definition in [Brouwer 1981]. We are unlikely to be able to construct a Brouwerian example of a subset of m that is not countable. However, any acceptable proof of the theorem T that every subset of m is countable could probab1y be converted into a proof that every subset of m is recursively enumerab1e, which is false.

Lattiee above has two maximal finite ehains, one of length 2 and one of length 3. For modular lattiees this ean't happen. linearly ordered sets C and D We say that two are pieeewise iSOIIIOrphie if there exist elements el" .. •. ,dn such that (i) {x (ii) (x {x C C x S cl) is isomorphie to c x :> e n ) is isomorphie to {x C TJ C (iii ) There is a permutation a of x <; cl l C D [1, ... ,n-I) isomorphie to [da; ,d ai +1] for eaeh i } x :> d n } so that [ci ,ci+l] is < Tl. We leave the proof that pieeewise isomorphism is a transitive relation as Exercise 4.

Commulatiue R. Then I i PROOF. then If E In ei ther ca se we can I \ UiESP i . ng = 0, #F 2 and for each i E S, ~ enlarge (xl"" ,xm ), without enlarging I, so that S by induction on #F. We proceed by If '''rn J such that a i E Pi \ u8 \{ i )p j . If #8 2, r \ UiE8p i . If #8 > 2, then Pi is prime for some E {xl"" j = 2iE~~i elements Otherwise choose 8 in F minimizing #8. j. •• ,n} such that {xl"" ,x m} induction I \ UiP f I 1 , •.. l or gener'ated R such of ideals that the a product P far some f.