Advanced Modern Algebra, Part 1 by Joseph J. Rotman

By Joseph J. Rotman

This re-creation, now in components, has been considerably reorganized and lots of sections were rewritten. this primary half, designed for a primary yr of graduate algebra, includes classes: Galois thought and Module thought. issues coated within the first direction are classical formulation for strategies of cubic and quartic equations, classical quantity idea, commutative algebra, teams, and Galois idea. issues within the moment direction are Zorn's lemma, canonical types, internal product areas, different types and boundaries, tensor items, projective, injective, and flat modules, multilinear algebra, affine types, and Grobner bases.

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If A and Rare (not necessarily commutative) rings, a (ring) homomorphism is a function cp: A --+ R such that (i) cp(l) = 1, (ii) cp( a+ a') = cp( a) (iii) cp(aa') + cp( a') for all a, a' E A, = cp(a)cp(a') for all a, a' EA. A ring homomorphism that is also a bijection is called an isomorphism. Rings A and R are called isomorphic, denoted by A~R, if there is an isomorphism cp : A --+ R. We continue to focus on commutative rings. 23. (i) Let R be a domain and let F = Frac(R) denote its fraction field.

Ii) If b 2: a are integers, then gcd(a, b) = gcd(b - a, a). Proof. 27. • We will abbreviate gcd(a, b) to (a, b) in the next three paragraphs. 18 allows us to consider (b- a, a) instead; indeed, we can continue reducing the numbers, (b - 2a, a), (b - 3a, a), ... , (b - qa, a) as long as b - qa > 0. Since the natural numbers b - a, b - 2a, ... , b- qa are strictly decreasing, the Least Integer Axiom says that we must reach a smallest such integer: r = b - qa; that is, r < a. Now (b, a) = (r, a).

74 + 0 . 73 + 6 . 72 + 6 . 7 + 4 = 12005 + 0 + 294 + 42 + 4 = 12345. We have written 12345 in "base 7:" it is 50664. ~ This idea works for any integer b 2 2. 30. If b 2 2 is an integer, then every positive integer h has an expression in base b: there are unique integers di with 0 :::; di < b such that h = dkbk + dk-1bk-I +···+do. Proof. We first prove the existence of such an expression, by induction on h. By the Division Algorithm, h = qb + r, where 0 :::; r < b. Since b 2 2, we have h = qb + r 2 qb 2 2q.

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