Abstract Algebra Exam 3 Review Problems and Solutions (Basic Ring Theory and Field Theory)

#abstractalgebra #abstractalgebrareview #ringtheory

Links and resources
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(0:00) Types of problems
(0:35) Abelian groups of order 72 (isomorphism classes)
(3:20) Number of Abelian groups of order 2592 (use partitions of integer powers)
(7:02) Definition of a ring R
(11:33) Definition of a unit in a commutative ring with identity
(12:54) Definition of a zero divisor in a commutative ring
(14:56) Definition of a field F (could also define an integral domain)
(17:04) Definition of an ideal of a ring (two-sided ideal)
(19:04) Ideal Test
(20:28) Principal Ideal definition
(22:14) Principal Ideal Domain (PID) definition
(23:36) Prime Ideals, Maximal Ideals, and Factor Rings (Quotient Rings). Relationship to integral domains and fields.
(29:29) Irreducible element definition (in an integral domain)
(31:40) Z8 units and zero divisors, U(Z8) group of units
(34:09) Ring homomorphisms from Z12 to Z20
(38:38) Integral domains, fields, PIDs, UFDs, EDs (True/False)
(41:15) Z[x] is a UFD but not a PID (Z[x] is a Unique Factorization Domain but not a Principal Ideal Domain)
(42:29) Long division in Z3[x] (& synthetic division mod 3) (Division algorithm over a field)
(48:36) Reducibility test of degree 2 polynomial over field Z5
(50:43) Eisenstein's Criterion for irreducibility over the rationals Q
(52:53) Tricky factorization to prove reducibility over Q
(54:19) Mod p Irreducibility test for degree 3 polynomial over Q
(57:11) Prove fields have no nontrivial proper ideals
(1:01:21) Prove the intersection of ideals is an ideal (use the Ideal Test)
(1:05:32) Mod p Irreducibility test for degree 4 polynomial over Q
(1:12:33) Factor ring calculations in Z3[x]/A, where A is a maximal principal ideal generated by an irreducible polynomial over Z3
(1:25:12) Part of proof that Z[sqrt(-5)] is not a UFD (it's an Integral Domain that is not a Unique Factorization Domain). Need properties of a norm defined on Z[(-5)^(1/2)] and the definition of irreducible in an integral domain.

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[luiszapata]

Thank you so much professor, it really helped me to prepare for my exam, quality of quality content!!!

[eddiejennings]

Super closure and associated slurping are growing on me very rapidly as a teaching tool. This lecture is a gift to the match community as always.

[ahmedabdulateefibrahim]

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[cioewsl]

I think at 18:48 there's an error. An Ideal Is not Always a subring. If It Is a subring It contains the 1 and so every element of the ring R Is in the Ideal. Ideal=ring.