Axioms for the integers by Brian Osserman

By Brian Osserman

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A polynomial P in some R,, is given. It is desired to produce ;I polynomial $ ( P ) such that P = $(P)modulo fl;;* and such that each niononiial of $ ( P ) has Property A . ( 1 ) Set Q equal to I-’. ( 3 ) Choose any term PM of Q , /3 # 0 (mod S), which does not Iiave Property A . If no such term can be found, go t o step 6. Otherwisc. go t o ctcp 3. ( 3 ) Determine (possibly e m p t y ) inonoriiids M , and M 2 such that the monomial ,ill chosen in step 2 has the form ill, . Y ~ . where Y , ~ Ii ~2.

This means that (3) is not fulfilled in r(m,a, K ) when k $ K . Thus, identity relation (3) is fulfilled in the group r(m,n, K ) that we constructed if and only if k E K . If we take as K the set of all prime numbers not equal to a given prime number I, then we find that relation ( 3 ) for k = 1 does not follow from the other identities in system ( 3 ) . The fact that the system of group identities ( 3 ) is irreducible implies immediately that a continuum exists of systems of group identities of the form ( 3 ) that are not pairwise equivalent.

Proposition 2. yfy the restrictions stated in the Collection Algorithm and P can be factored in the form P , P, P 3 , (respectively P, P3 or P , P , 1, where P , does not involve x,. Then each Pi is an element of some RMi,n and Pisatisfy the restrictions of the Collection Algorithm, and (respectively or Lemma 2. For each n 2 1 , H t is a proper subspace of R,, and x12 x 2 2 ... x n 2 together with H; span R,. S. , A non-solvable group of exponent 5 46 Proof. vl 2 . 2s , , 2 and p,,(P)= Y modulo H:.

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