By Joseph J. Rotman

ISBN-10: 1470415542

ISBN-13: 9781470415549

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, contains classes: Galois conception and Module conception. themes coated within the first path are classical formulation for suggestions of cubic and quartic equations, classical quantity thought, commutative algebra, teams, and Galois conception. issues within the moment path are Zorn's lemma, canonical varieties, internal product areas, different types and bounds, tensor items, projective, injective, and flat modules, multilinear algebra, affine kinds, and Grobner bases.

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**Additional info for Advanced Modern Algebra, Part 1**

**Example text**

Definition. If O' = (so, s 1, ... , sn, 0, 0, ... ) is a nonzero polynomial, then there is n ;:::: 0 with Sn ;/; 0 and Si = 0 for all i > n. We call Sn the leading coefficient of 17, we call n the degree of 17, and we denote the degree by n If the leading coefficient Sn = deg(<7). = 1, then 17 is called monic. The zero polynomial 0 does not have a degree because it has no nonzero coefficients. 10 Notation. If Risa commutative ring, then R[[x]] denotes the set of all formal power series over R, and R[x] ~ R[[x]] denotes the set of all polynomials over R.

Prove or disprove and salvage if possible. ") = b, gcd(a2 ,b2 ) = (gcd(a,b)) 2 , gcd(a, b) = gcd(a, b + ka) (k E Z), (i) gcd(O, b) (ii) (iii) (iv) gcd(a, a) =a, 15 Divisibility (v) gcd(a, b) = gcd(b, a), = 1, gcd(a,b) = -gcd(-a,b). 2. If x is a real number, let LxJ denote the largest integer n with n :::; x. ) Show that the quotient q in the Division Algorithm is Lb/aJ. 3. Let p1,p2,p3, ... be the list of the primes in ascending order: p1 = 2, p2 = 3, p3 = 5, . . Define f k = p1p2 · · ·Pk + 1 for k 2: 1.

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. It follows that q < h; otherwise, q 2 h, giving the contradiction h 2 2q 2 2h. By the inductive hypothesis, h = qb + r = (d~bk + · · · + d~)b + r = d~bk+I + · · · + d~b + r. We prove uniqueness by induction on h. Suppose that h = dkbk + · · · + dib +do= embm + · · · + eib + eo, where 0 S ej < b for all j; that is, h = (dkbk-I + · · · + d 1)b + do and h = (embm-I + · · · + e 1)b + eo.

### Advanced Modern Algebra, Part 1 by Joseph J. Rotman

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